id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-4900 | Existence of Set with Singleton Intersections with Disjoint Collection | Let $\CC$ be a set of sets all of which are pairwise disjoint.
Then:
:there exists a set $A$ such that $\forall S \in \CC: A \cap S$ is a singleton
{{iff}}
:the axiom of choice holds. | {{proof wanted}}
{{AoC}} | Let $\CC$ be a [[Definition:Set of Sets|set of sets]] all of which are [[Definition:Pairwise Disjoint|pairwise disjoint]].
Then:
:there exists a [[Definition:Set|set]] $A$ such that $\forall S \in \CC: A \cap S$ is a [[Definition:Singleton|singleton]]
{{iff}}
:the [[Axiom:Axiom of Choice|axiom of choice]] holds. | {{proof wanted}}
{{AoC}} | Existence of Set with Singleton Intersections with Disjoint Collection | https://proofwiki.org/wiki/Existence_of_Set_with_Singleton_Intersections_with_Disjoint_Collection | https://proofwiki.org/wiki/Existence_of_Set_with_Singleton_Intersections_with_Disjoint_Collection | [
"Set Intersection",
"Singletons",
"Disjoint Sets"
] | [
"Definition:Set of Sets",
"Definition:Pairwise Disjoint",
"Definition:Set",
"Definition:Singleton",
"Axiom:Axiom of Choice"
] | [] |
proofwiki-4901 | Complementary Idempotent is Idempotent | Let $\HH$ be a Hilbert space.
Let $I$ be an identity operator on $\HH$.
Let $A$ be an idempotent operator.
Then the complementary idempotent $I - A$ is also idempotent. | {{begin-eqn}}
{{eqn|l = \paren {I - A}^2
|r = I^2 - I A - A I + A^2
}}
{{eqn|r = I^2 - 2 A + A^2
|c = {{Defof|Identity Operator}}
}}
{{eqn|r = I - A
|c = {{Defof|Idempotent Operator}}
}}
{{end-eqn}}
That is, $I - A$ is idempotent.
{{qed}} | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $I$ be an [[Definition:Identity Operator|identity operator]] on $\HH$.
Let $A$ be an [[Definition:Idempotent Operator|idempotent operator]].
Then the [[Definition:Complementary Idempotent|complementary idempotent]] $I - A$ is also [[Definition:Idempoten... | {{begin-eqn}}
{{eqn|l = \paren {I - A}^2
|r = I^2 - I A - A I + A^2
}}
{{eqn|r = I^2 - 2 A + A^2
|c = {{Defof|Identity Operator}}
}}
{{eqn|r = I - A
|c = {{Defof|Idempotent Operator}}
}}
{{end-eqn}}
That is, $I - A$ is [[Definition:Idempotent Operator|idempotent]].
{{qed}} | Complementary Idempotent is Idempotent | https://proofwiki.org/wiki/Complementary_Idempotent_is_Idempotent | https://proofwiki.org/wiki/Complementary_Idempotent_is_Idempotent | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Identity Mapping",
"Definition:Idempotent Operator",
"Definition:Complementary Idempotent",
"Definition:Idempotent Operator"
] | [
"Definition:Idempotent Operator"
] |
proofwiki-4902 | Range of Idempotent is Kernel of Complementary Idempotent | Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $A$ be an idempotent operator.
Then:
:$\Rng A = \map \ker {I - A}$ | If $h \in \map \ker {I - A}$, we have:
:$\map {\paren {I - A} } h = {\mathbf 0}_\HH$
That is:
:$h - A h = {\mathbf 0}_\HH$
so that:
:$A h = h$
and so:
:$h \in \Rng A$
So we have:
:$\map \ker {I - A} \subseteq \Rng A$
Now let $h \in \Rng A$.
Then there exists $k \in \HH$ such that $h = A k$.
Then we have $h - A h =... | Let $\struct {\HH, \innerprod \cdot \cdot}$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $A$ be an [[Definition:Idempotent Operator|idempotent operator]].
Then:
:$\Rng A = \map \ker {I - A}$ | If $h \in \map \ker {I - A}$, we have:
:$\map {\paren {I - A} } h = {\mathbf 0}_\HH$
That is:
:$h - A h = {\mathbf 0}_\HH$
so that:
:$A h = h$
and so:
:$h \in \Rng A$
So we have:
:$\map \ker {I - A} \subseteq \Rng A$
Now let $h \in \Rng A$.
Then there exists $k \in \HH$ such that $h = A k$.
Then we have $h - ... | Range of Idempotent is Kernel of Complementary Idempotent | https://proofwiki.org/wiki/Range_of_Idempotent_is_Kernel_of_Complementary_Idempotent | https://proofwiki.org/wiki/Range_of_Idempotent_is_Kernel_of_Complementary_Idempotent | [
"Range of Idempotent is Kernel of Complementary Idempotent",
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Idempotent Operator"
] | [
"Definition:Idempotent Operator"
] |
proofwiki-4903 | Real Natural Logarithm Function is Continuous | The real natural logarithm function is continuous. | We have that the Natural Logarithm Function is Differentiable.
The result follows from Differentiable Function is Continuous.
{{qed}} | The [[Definition:Real Natural Logarithm|real natural logarithm function]] is [[Definition:Continuous Real Function|continuous]]. | We have that the [[Natural Logarithm Function is Differentiable]].
The result follows from [[Differentiable Function is Continuous]].
{{qed}} | Real Natural Logarithm Function is Continuous/Proof 1 | https://proofwiki.org/wiki/Real_Natural_Logarithm_Function_is_Continuous | https://proofwiki.org/wiki/Real_Natural_Logarithm_Function_is_Continuous/Proof_1 | [
"Real Natural Logarithm Function is Continuous",
"Continuous Real Functions",
"Natural Logarithms"
] | [
"Definition:Natural Logarithm/Positive Real",
"Definition:Continuous Real Function"
] | [
"Natural Logarithm Function is Differentiable",
"Differentiable Function is Continuous"
] |
proofwiki-4904 | Real Natural Logarithm Function is Continuous | The real natural logarithm function is continuous. | From Bounds of Natural Logarithm:
:$\dfrac 1 2 < \map \ln 2 < 1$
Fix $x \in \R$.
Consider $\dfrac x {\map \ln 2}$.
From Rationals are Everywhere Dense in Topological Space of Reals:
:$\forall \epsilon \in \R_{>0} \exists r \in \Q : \size {r - \dfrac x {\map \ln 2} } < \epsilon$
Thus:
{{begin-eqn}}
{{eqn | l = \size {r ... | The [[Definition:Real Natural Logarithm|real natural logarithm function]] is [[Definition:Continuous Real Function|continuous]]. | From [[Bounds of Natural Logarithm]]:
:$\dfrac 1 2 < \map \ln 2 < 1$
Fix $x \in \R$.
Consider $\dfrac x {\map \ln 2}$.
From [[Rationals are Everywhere Dense in Topological Space of Reals]]:
:$\forall \epsilon \in \R_{>0} \exists r \in \Q : \size {r - \dfrac x {\map \ln 2} } < \epsilon$
Thus:
{{begin-eqn}}
{{eqn | l... | Real Natural Logarithm Function is Continuous/Proof 2 | https://proofwiki.org/wiki/Real_Natural_Logarithm_Function_is_Continuous | https://proofwiki.org/wiki/Real_Natural_Logarithm_Function_is_Continuous/Proof_2 | [
"Real Natural Logarithm Function is Continuous",
"Continuous Real Functions",
"Natural Logarithms"
] | [
"Definition:Natural Logarithm/Positive Real",
"Definition:Continuous Real Function"
] | [
"Bounds of Natural Logarithm",
"Rational Numbers are Everywhere Dense in Set of Real Numbers/Topology",
"Logarithm of Power/Natural Logarithm/Rational Power",
"Real Number Ordering is Compatible with Multiplication",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Everywhere Dense",
"Monoto... |
proofwiki-4905 | Characterization of Projections | Let $\HH$ be a Hilbert space.
Let $A \in \map B \HH$ be an idempotent operator.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is a projection in the Hilbert space sense}}
{{item|(2):|$A$ is the orthogonal projection onto $\Img A$}}
{{item|(3):|$\norm A {{=}} 1$, where $\norm {\, \cdot \,}$ is the norm on bounded linear op... | === $(1)$ implies $(2)$ ===
Let $K = \Img A$.
From {{Corollary|Range of Idempotent is Kernel of Complementary Idempotent|2}}, $K$ is a closed linear subspace of $\HH$.
Hence we can define the orthogonal projection $P_K$.
We just need to show that:
:$A = P_K$
From Unique Point of Minimal Distance to Closed Linear Subs... | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $A \in \map B \HH$ be an [[Definition:Idempotent Operator|idempotent operator]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is a [[Definition:Projection (Hilbert Spaces)|projection in the Hilbert space sense]]}}
{{item|(2):|$A$ is the [[Definition:Orthog... | === $(1)$ implies $(2)$ ===
Let $K = \Img A$.
From {{Corollary|Range of Idempotent is Kernel of Complementary Idempotent|2}}, $K$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $\HH$.
Hence we can define the [[Definition:Orthogonal Projection|orthogonal projection]] $P_K$.
We just need to sh... | Characterization of Projections | https://proofwiki.org/wiki/Characterization_of_Projections | https://proofwiki.org/wiki/Characterization_of_Projections | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Idempotent Operator",
"Definition:Projection (Hilbert Spaces)",
"Definition:Orthogonal Projection",
"Definition:Norm/Bounded Linear Transformation",
"Definition:Hermitian Operator",
"Definition:Normal Operator"
] | [
"Definition:Closed Linear Subspace",
"Definition:Orthogonal Projection",
"Unique Point of Minimal Distance to Closed Linear Subspace of Hilbert Space iff Orthogonal",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Projection (Hilbert Spaces)",
"Definition:Projection (Hilbert S... |
proofwiki-4906 | Set of Linear Subspaces is Closed under Intersection | Let $\struct {V, +, \circ}_K$ be a $K$-vector space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed family of subspaces of $V$.
Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a subspace of $V$. | It needs to be demonstrated that $M$ is:
:$(1): \quad$ a closed algebraic structure under $+$
:$(2): \quad$ closed for scalar product $\circ$.
So let $a, b \in M$.
By definition of intersection, $a, b \in M_i$ for all $i \in I$.
As the $M_i$ are subspaces of $V$, $a + b \in M_i$ for all $i \in I$.
That is, by definitio... | Let $\struct {V, +, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
Let $\family {M_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Vector Subspace|subspaces]] of $V$.
Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a [[Definition:Vector Subspace|subspa... | It needs to be demonstrated that $M$ is:
:$(1): \quad$ a [[Definition:Closed Algebraic Structure|closed algebraic structure]] under $+$
:$(2): \quad$ [[Definition:Closed for Scalar Product|closed for scalar product]] $\circ$.
So let $a, b \in M$.
By definition of [[Definition:Set Intersection|intersection]], $a, b \... | Set of Linear Subspaces is Closed under Intersection | https://proofwiki.org/wiki/Set_of_Linear_Subspaces_is_Closed_under_Intersection | https://proofwiki.org/wiki/Set_of_Linear_Subspaces_is_Closed_under_Intersection | [
"Vector Subspaces"
] | [
"Definition:Vector Space",
"Definition:Indexing Set/Family",
"Definition:Vector Subspace",
"Definition:Vector Subspace"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Scalar Product",
"Definition:Set Intersection",
"Definition:Vector Subspace",
"Definition:Set Intersection",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Set Intersection",
"... |
proofwiki-4907 | Closed Linear Subspaces Closed under Intersection | Let $V$ be a topological vector space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed family of closed linear subspaces of $V$.
Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a closed linear subspace of $V$. | By Set of Linear Subspaces is Closed under Intersection, $M$ is a linear subspace of $V$.
By Topology Defined by Closed Sets, the intersection of closed sets is again closed.
As the $M_i$ are all closed, it follows that $M$ is closed.
Hence $M$ is a closed linear subspace of $V$.
{{qed}}
Category:Vector Subspaces
4qr0u... | Let $V$ be a [[Definition:Topological Vector Space|topological vector space]].
Let $\family {M_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Closed Linear Subspace|closed linear subspaces]] of $V$.
Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a [[Definition:Clo... | By [[Set of Linear Subspaces is Closed under Intersection]], $M$ is a [[Definition:Vector Subspace|linear subspace]] of $V$.
By [[Topology Defined by Closed Sets]], the [[Definition:Set Intersection|intersection]] of [[Definition:Closed Set (Topology)|closed sets]] is again [[Definition:Closed Set (Topology)|closed]]... | Closed Linear Subspaces Closed under Intersection | https://proofwiki.org/wiki/Closed_Linear_Subspaces_Closed_under_Intersection | https://proofwiki.org/wiki/Closed_Linear_Subspaces_Closed_under_Intersection | [
"Vector Subspaces"
] | [
"Definition:Topological Vector Space",
"Definition:Indexing Set/Family",
"Definition:Closed Linear Subspace",
"Definition:Closed Linear Subspace"
] | [
"Set of Linear Subspaces is Closed under Intersection",
"Definition:Vector Subspace",
"Topology Defined by Closed Sets",
"Definition:Set Intersection",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:... |
proofwiki-4908 | Orthogonal Difference is Closed Linear Subspace | Let $H$ be a Hilbert space.
Let $M, N$ be closed linear subspaces of $H$.
Then the orthogonal difference $M \ominus N$ is also a closed linear subspace of $H$. | By definition, $M \ominus N = M \cap N^\perp$.
By Orthocomplement is Closed Linear Subspace, $N^\perp$ is a closed linear subspace of $H$.
Hence the result, by Closed Linear Subspaces Closed under Intersection.
{{qed}}
Category:Hilbert Spaces
2lj3gg359w9cwzc1kxa4wbfnzjwhsak | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $M, N$ be [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$.
Then the [[Definition:Orthogonal Difference|orthogonal difference]] $M \ominus N$ is also a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$. | By definition, $M \ominus N = M \cap N^\perp$.
By [[Orthocomplement is Closed Linear Subspace]], $N^\perp$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
Hence the result, by [[Closed Linear Subspaces Closed under Intersection]].
{{qed}}
[[Category:Hilbert Spaces]]
2lj3gg359w9cwzc1kxa4wbf... | Orthogonal Difference is Closed Linear Subspace | https://proofwiki.org/wiki/Orthogonal_Difference_is_Closed_Linear_Subspace | https://proofwiki.org/wiki/Orthogonal_Difference_is_Closed_Linear_Subspace | [
"Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Closed Linear Subspace",
"Definition:Orthogonal Difference",
"Definition:Closed Linear Subspace"
] | [
"Orthocomplement is Closed Linear Subspace",
"Definition:Closed Linear Subspace",
"Closed Linear Subspaces Closed under Intersection",
"Category:Hilbert Spaces"
] |
proofwiki-4909 | Closed Linear Subspaces Closed under Setwise Addition | Let $H$ be a Hilbert space.
Let $M, N$ be closed linear subspaces of $H$.
Then $M + N$ is also a closed linear subspace of $H$, where $+$ denotes setwise addition. | By Linear Subspaces Closed under Setwise Addition, $M + N$ is a linear subspace of $H$.
Now to show that it is closed.
Let $P: H \to H$ denote the orthogonal projection on $M$.
Denote by $I - P$ the complementary projection of $P$.
Define $N' := \set {n - P n: n \in N}$.
$N'$ is a closed linear subspace of $H$.
{{expla... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $M, N$ be [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$.
Then $M + N$ is also a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$, where $+$ denotes [[Definition:Setwise Addition|setwise addition]]. | By [[Linear Subspaces Closed under Setwise Addition]], $M + N$ is a [[Definition:Vector Subspace|linear subspace]] of $H$.
Now to show that it is [[Definition:Closed Set (Topology)|closed]].
Let $P: H \to H$ denote the [[Definition:Orthogonal Projection|orthogonal projection]] on $M$.
Denote by $I - P$ the [[Defin... | Closed Linear Subspaces Closed under Setwise Addition | https://proofwiki.org/wiki/Closed_Linear_Subspaces_Closed_under_Setwise_Addition | https://proofwiki.org/wiki/Closed_Linear_Subspaces_Closed_under_Setwise_Addition | [
"Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Closed Linear Subspace",
"Definition:Closed Linear Subspace",
"Definition:Subset Product"
] | [
"Linear Subspaces Closed under Setwise Addition",
"Definition:Vector Subspace",
"Definition:Closed Set/Topology",
"Definition:Orthogonal Projection",
"Definition:Complementary Projection",
"Definition:Closed Linear Subspace",
"Range of Idempotent is Kernel of Complementary Idempotent",
"Properties of ... |
proofwiki-4910 | Linear Subspaces Closed under Setwise Addition | Let $V$ be a $K$-vector space.
Let $M, N$ be linear subspaces of $V$.
Then $L := M + N$ is also a linear subspace of $V$, where $+$ denotes setwise addition. | It needs to be demonstrated that $L$ is closed under $+$ and $\circ$.
So let $m_1 + n_1, m_2 + n_2 \in L$.
Then $\paren {m_1 + n_1} + \paren {m_2 + n_2} = \paren {m_1 + m_2} + \paren {n_1 + n_2} \in L$.
It follows that $L$ is closed under $+$.
Now let $\lambda \in K, m + n \in L$.
Then $\lambda \circ \paren {m + n} = \... | Let $V$ be a [[Definition:Vector Space|$K$-vector space]].
Let $M, N$ be [[Definition:Vector Subspace|linear subspaces]] of $V$.
Then $L := M + N$ is also a [[Definition:Vector Subspace|linear subspace]] of $V$, where $+$ denotes [[Definition:Setwise Addition|setwise addition]]. | It needs to be demonstrated that $L$ is [[Definition:Closed Algebraic Structure|closed]] under $+$ and $\circ$.
So let $m_1 + n_1, m_2 + n_2 \in L$.
Then $\paren {m_1 + n_1} + \paren {m_2 + n_2} = \paren {m_1 + m_2} + \paren {n_1 + n_2} \in L$.
It follows that $L$ is [[Definition:Closed Algebraic Structure|closed]]... | Linear Subspaces Closed under Setwise Addition | https://proofwiki.org/wiki/Linear_Subspaces_Closed_under_Setwise_Addition | https://proofwiki.org/wiki/Linear_Subspaces_Closed_under_Setwise_Addition | [
"Vector Subspaces"
] | [
"Definition:Vector Space",
"Definition:Vector Subspace",
"Definition:Vector Subspace",
"Definition:Subset Product"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closed for Scalar Product ",
"Definition:Vector Subspace",
"Category:Vector Subspaces"
] |
proofwiki-4911 | Complementary Projection is Projection | Let $\HH$ be a Hilbert space.
Let $A$ be a projection.
Then the complementary projection $I - A$ is also a projection. | By Characterization of Projections, $A$ is Hermitian.
Then $\paren {I - A}^* = I^* - A^* = I - A$ from Adjoint is Conjugate Linear.
So $I - A$ is also Hermitian.
From Complementary Idempotent is Idempotent, $I - A$ is idempotent.
Hence, applying Characterization of Projections, $I - A$ is a projection.
{{qed}} | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $A$ be a [[Definition:Projection (Hilbert Spaces)|projection]].
Then the [[Definition:Complementary Projection|complementary projection]] $I - A$ is also a [[Definition:Projection (Hilbert Spaces)|projection]]. | By [[Characterization of Projections]], $A$ is [[Definition:Hermitian Operator|Hermitian]].
Then $\paren {I - A}^* = I^* - A^* = I - A$ from [[Adjoint is Conjugate Linear]].
So $I - A$ is also [[Definition:Hermitian Operator|Hermitian]].
From [[Complementary Idempotent is Idempotent]], $I - A$ is [[Definition:Idempo... | Complementary Projection is Projection | https://proofwiki.org/wiki/Complementary_Projection_is_Projection | https://proofwiki.org/wiki/Complementary_Projection_is_Projection | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Projection (Hilbert Spaces)",
"Definition:Complementary Projection",
"Definition:Projection (Hilbert Spaces)"
] | [
"Characterization of Projections",
"Definition:Hermitian Operator",
"Adjoint is Conjugate Linear",
"Definition:Hermitian Operator",
"Complementary Idempotent is Idempotent",
"Definition:Idempotent Operator",
"Characterization of Projections",
"Definition:Projection (Hilbert Spaces)"
] |
proofwiki-4912 | Direct Sum of Subspace and Orthocomplement | Let $H$ be a Hilbert space.
Let $M$ be a closed linear subspace of $H$.
Denote by $M^\perp$ its orthocomplement.
Then the direct sum $M \oplus M^\perp$ is isomorphic to $H$. | Assert that $U: M \oplus M^\perp \to H: \left({m, m^\perp}\right) \mapsto m + m^\perp$ is an isomorphism.
According to the definition of isomorphism, it is sufficient to check that $U$ is surjective and that:
:$\left\langle{ U \left({m, m^\perp}\right), U \left({n, n^\perp}\right) }\right\rangle_H = \left\langle{ \left... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $M$ be a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
Denote by $M^\perp$ its [[Definition:Orthocomplement|orthocomplement]].
Then the [[Definition:Hilbert Space Direct Sum|direct sum]] $M \oplus M^\perp$ is [[Definition:Isomorphis... | Assert that $U: M \oplus M^\perp \to H: \left({m, m^\perp}\right) \mapsto m + m^\perp$ is an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]].
According to the definition of [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]], it is sufficient to check that $U$ is [[Definition:Surjection|surjective]] and th... | Direct Sum of Subspace and Orthocomplement | https://proofwiki.org/wiki/Direct_Sum_of_Subspace_and_Orthocomplement | https://proofwiki.org/wiki/Direct_Sum_of_Subspace_and_Orthocomplement | [
"Hilbert Spaces",
"Orthocomplements"
] | [
"Definition:Hilbert Space",
"Definition:Closed Linear Subspace",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Hilbert Space Direct Sum",
"Definition:Isomorphism (Hilbert Spaces)"
] | [
"Definition:Isomorphism (Hilbert Spaces)",
"Definition:Isomorphism (Hilbert Spaces)",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Orthogonal Projection",
"Definition:Orthogonal Projection",
"Orthogonal Projection onto Orthocomplement",
"Definition:Surjection",
"Definition:Inner Pro... |
proofwiki-4913 | Subset of Toset is Toset | Let $\left({S, \preceq}\right)$ be a totally ordered set.
Let $T \subseteq S$.
Then $\left({T, \preceq \restriction_T}\right)$ is also a totally ordered set.
In the above, $\preceq \restriction_T$ denotes the restriction of $\preceq$ to $T$. | As $\left({S, \preceq}\right)$ is a totally ordered set, the relation $\preceq$ is a total ordering, and is by definition:
* reflexive
* antisymmetric
* transitive
* connected
From Properties of Restriction of Relation, a restriction of a relation which has all those properties inherits them all.
Thus $\preceq \restric... | Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $T \subseteq S$.
Then $\left({T, \preceq \restriction_T}\right)$ is also a [[Definition:Totally Ordered Set|totally ordered set]].
In the above, $\preceq \restriction_T$ denotes the [[Definition:Restriction of Relation... | As $\left({S, \preceq}\right)$ is a [[Definition:Totally Ordered Set|totally ordered set]], the relation $\preceq$ is a [[Definition:Total Ordering|total ordering]], and is by definition:
* [[Definition:Reflexive Relation|reflexive]]
* [[Definition:Antisymmetric Relation|antisymmetric]]
* [[Definition:Transitive Relat... | Subset of Toset is Toset | https://proofwiki.org/wiki/Subset_of_Toset_is_Toset | https://proofwiki.org/wiki/Subset_of_Toset_is_Toset | [
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Totally Ordered Set",
"Definition:Restriction/Relation"
] | [
"Definition:Totally Ordered Set",
"Definition:Total Ordering",
"Definition:Reflexive Relation",
"Definition:Antisymmetric Relation",
"Definition:Transitive Relation",
"Definition:Connected Relation",
"Properties of Restriction of Relation",
"Definition:Restriction/Relation",
"Definition:Relation",
... |
proofwiki-4914 | Bretschneider's Formula | Let $ABCD$ be a general quadrilateral.
Then the area $\AA$ of $ABCD$ is given by:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2} }$
where:
:$a, b, c, d$ are the lengths of the sides of the quadrilateral
:$s = \dfrac {a + b + c + d} 2$ is t... | :400px
Let the area of $\triangle DAB$ and $\triangle BCD$ be $\AA_1$ and $\AA_2$.
From Area of Triangle in Terms of Two Sides and Angle:
:$\AA_1 = \dfrac {a b \sin \alpha} 2$ and $\AA_2 = \dfrac {c d \sin \gamma} 2$
From the second axiom of area:
:$\AA = \AA_1 + \AA_2$
so:
{{begin-eqn}}
{{eqn | l = \AA^2
| r = \... | Let $ABCD$ be a general [[Definition:Quadrilateral|quadrilateral]].
Then the area $\AA$ of $ABCD$ is given by:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2} }$
where:
:$a, b, c, d$ are the [[Definition:Length (Linear Measure)|lengths]]... | :[[File:Bretschneider's Formula.png|400px]]
Let the area of $\triangle DAB$ and $\triangle BCD$ be $\AA_1$ and $\AA_2$.
From [[Area of Triangle in Terms of Two Sides and Angle]]:
:$\AA_1 = \dfrac {a b \sin \alpha} 2$ and $\AA_2 = \dfrac {c d \sin \gamma} 2$
From the [[Axiom:Area Axioms|second axiom of area]]:
:$\AA ... | Bretschneider's Formula | https://proofwiki.org/wiki/Bretschneider's_Formula | https://proofwiki.org/wiki/Bretschneider's_Formula | [
"Areas of Quadrilaterals"
] | [
"Definition:Quadrilateral",
"Definition:Linear Measure/Length",
"Definition:Polygon/Side",
"Definition:Quadrilateral",
"Definition:Semiperimeter",
"Definition:Polygon/Opposite"
] | [
"File:Bretschneider's Formula.png",
"Area of Triangle in Terms of Two Sides and Angle",
"Axiom:Area Axioms",
"Law of Cosines",
"Equality is Transitive",
"Sum of Squares of Sine and Cosine",
"Cosine of Sum",
"Definition:Fraction/Numerator",
"Half Angle Formulas/Cosine"
] |
proofwiki-4915 | Transfinite Induction/Principle 1 | Let $\On$ denote the class of all ordinals.
Let $A$ denote a class.
Suppose that:
:For all elements $x$ of $\On$, if $x$ is a subset of $A$, then $x$ is an element of $A$.
Then $\On \subseteq A$. | {{NotZFC}}
{{AimForCont}} that $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.
By Epsilon Relation is Strongly Well-Founded on Ordinal Class, $\On \setminus A$ must have a strictly minimal element $y$ under $\in$.
By Elemen... | Let $\On$ denote the [[Definition:Class (Class Theory)|class]] of all [[Definition:Ordinal|ordinals]].
Let $A$ denote a [[Definition:Class (Class Theory)|class]].
Suppose that:
:For all elements $x$ of $\On$, if $x$ is a [[Definition:Subset|subset]] of $A$, then $x$ is an [[Definition:Element|element]] of $A$.
The... | {{NotZFC}}
{{AimForCont}} that $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From [[Set Difference is Subset]], $\On \setminus A$ is a [[Definition:Subclass|subclass]] of the [[Definition:Ordinal|ordinals]].
By [[Epsilon Relation is Strongly Well-Founded on Ordinal Class]], $\On \setminus A$ must... | Transfinite Induction/Principle 1 | https://proofwiki.org/wiki/Transfinite_Induction/Principle_1 | https://proofwiki.org/wiki/Transfinite_Induction/Principle_1 | [
"Transfinite Induction"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordinal",
"Definition:Class (Class Theory)",
"Definition:Subset",
"Definition:Element"
] | [
"Set Difference is Subset",
"Definition:Subclass",
"Definition:Ordinal",
"Epsilon Relation is Strongly Well-Founded on Ordinal Class",
"Definition:Strictly Minimal Element",
"Element of Ordinal is Ordinal",
"Definition:Subset",
"Definition:Class (Class Theory)",
"Definition:Ordinal",
"Definition:S... |
proofwiki-4916 | Transfinite Induction/Principle 1 | Let $\On$ denote the class of all ordinals.
Let $A$ denote a class.
Suppose that:
:For all elements $x$ of $\On$, if $x$ is a subset of $A$, then $x$ is an element of $A$.
Then $\On \subseteq A$. | <onlyinclude>
{{AimForCont}} $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.
By Class of All Ordinals is Well-Ordered by Subset Relation, $\On \setminus A$ must have a smallest element $y$.
Then every strict predecessor of ... | Let $\On$ denote the [[Definition:Class (Class Theory)|class]] of all [[Definition:Ordinal|ordinals]].
Let $A$ denote a [[Definition:Class (Class Theory)|class]].
Suppose that:
:For all elements $x$ of $\On$, if $x$ is a [[Definition:Subset|subset]] of $A$, then $x$ is an [[Definition:Element|element]] of $A$.
The... | <onlyinclude>
{{AimForCont}} $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From [[Set Difference is Subset]], $\On \setminus A$ is a [[Definition:Subclass|subclass]] of the [[Definition:Ordinal|ordinals]].
By [[Class of All Ordinals is Well-Ordered by Subset Relation]], $\On \setminus A$ must have... | Transfinite Induction/Principle 1/Proof 2 | https://proofwiki.org/wiki/Transfinite_Induction/Principle_1 | https://proofwiki.org/wiki/Transfinite_Induction/Principle_1/Proof_2 | [
"Transfinite Induction"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordinal",
"Definition:Class (Class Theory)",
"Definition:Subset",
"Definition:Element"
] | [
"Set Difference is Subset",
"Definition:Subclass",
"Definition:Ordinal",
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Definition:Smallest Element",
"Definition:Strictly Precede",
"Definition:Element/Class",
"Definition:Element/Class",
"Category:Transfinite Induction"
] |
proofwiki-4917 | Limit of Functions that Agree | Let $f$ and $g$ be real functions.
Let $f$ and $g$ agree for all $x$ in a deleted neighborhood of $c$.
Let the limit:
:$\ds \lim_{x \mathop \to c} \map f x$
exist.
Then the limit:
:$\ds \lim_{x \mathop \to c} \map g x$
also exists, and:
:$\ds \lim_{x \mathop \to c} \map f x = \lim_{x \mathop \to c} \map g x$ | By hypothesis:
:$\map f x = \map g x$
for all $x$ such that:
:$x \in \R: 0 < \size {\alpha - x} < \epsilon$
Putting $c$ for $\alpha$ and $\delta$ for $\epsilon$, this is equivalent to:
:$x \in \R: 0 < \size {x - c} < \delta$
By definition, that the limit of $\map f x$ exists is to say:
:$\exists L: \forall \epsilon \in... | Let $f$ and $g$ be [[Definition:Real Function|real functions]].
Let $f$ and $g$ [[Definition:Agreement of Mappings|agree]] for all $x$ in a [[Definition:Deleted Neighborhood (Real Analysis)|deleted neighborhood]] of $c$.
Let the [[Definition:Limit of Real Function|limit]]:
:$\ds \lim_{x \mathop \to c} \map f x$
exi... | [[Definition:By Hypothesis|By hypothesis]]:
:$\map f x = \map g x$
for all $x$ such that:
:$x \in \R: 0 < \size {\alpha - x} < \epsilon$
Putting $c$ for $\alpha$ and $\delta$ for $\epsilon$, this is equivalent to:
:$x \in \R: 0 < \size {x - c} < \delta$
By definition, that the [[Definition:Limit of Real Function... | Limit of Functions that Agree | https://proofwiki.org/wiki/Limit_of_Functions_that_Agree | https://proofwiki.org/wiki/Limit_of_Functions_that_Agree | [
"Limits of Real Functions"
] | [
"Definition:Real Function",
"Definition:Agreement/Mappings",
"Definition:Deleted Neighborhood/Real Analysis",
"Definition:Limit of Real Function",
"Definition:Limit of Real Function"
] | [
"Definition:By Hypothesis",
"Definition:Limit of Real Function",
"Definition:Limit of Real Function"
] |
proofwiki-4918 | Ordering of Reciprocals | Let $x, y \in \R$ be real numbers such that $x, y \in \openint 0 \to$ or $x, y \in \openint \gets 0$
Then:
:$x \le y \iff \dfrac 1 y \le \dfrac 1 x$ | By Reciprocal Function is Strictly Decreasing, the reciprocal function is strictly decreasing.
By Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing, the reciprocal function is a dual order embedding.
That is:
:$x \le y \iff \dfrac 1 y \le \dfrac 1 x$
{{qed}} | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]] such that $x, y \in \openint 0 \to$ or $x, y \in \openint \gets 0$
Then:
:$x \le y \iff \dfrac 1 y \le \dfrac 1 x$ | By [[Reciprocal Function is Strictly Decreasing]], the reciprocal function is [[Definition:Strictly Decreasing Mapping|strictly decreasing]].
By [[Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing]], the reciprocal function is a [[Definition:Dual Order Embedding|dual order embedding]].
... | Ordering of Reciprocals/Proof 1 | https://proofwiki.org/wiki/Ordering_of_Reciprocals | https://proofwiki.org/wiki/Ordering_of_Reciprocals/Proof_1 | [
"Real Numbers",
"Reciprocals",
"Inequalities",
"Ordering of Reciprocals"
] | [
"Definition:Real Number"
] | [
"Reciprocal Function is Strictly Decreasing",
"Definition:Strictly Decreasing/Mapping",
"Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing",
"Definition:Dual Order Embedding"
] |
proofwiki-4919 | Ordering of Reciprocals | Let $x, y \in \R$ be real numbers such that $x, y \in \openint 0 \to$ or $x, y \in \openint \gets 0$
Then:
:$x \le y \iff \dfrac 1 y \le \dfrac 1 x$ | By Reciprocal Function is Strictly Decreasing, the reciprocal function is strictly decreasing.
Thus
:$x \le y \implies \dfrac 1 y \le \dfrac 1 x$
Suppose then that $\dfrac 1 y \le \dfrac 1 x$.
If $x, y > 0$, then from Reciprocal of Strictly Positive Real Number is Strictly Positive:
$\dfrac 1 y, \dfrac 1 x > 0$
Simila... | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]] such that $x, y \in \openint 0 \to$ or $x, y \in \openint \gets 0$
Then:
:$x \le y \iff \dfrac 1 y \le \dfrac 1 x$ | By [[Reciprocal Function is Strictly Decreasing]], the reciprocal function is [[Definition:Strictly Decreasing Mapping|strictly decreasing]].
Thus
:$x \le y \implies \dfrac 1 y \le \dfrac 1 x$
Suppose then that $\dfrac 1 y \le \dfrac 1 x$.
If $x, y > 0$, then from [[Reciprocal of Strictly Positive Real Number is S... | Ordering of Reciprocals/Proof 2 | https://proofwiki.org/wiki/Ordering_of_Reciprocals | https://proofwiki.org/wiki/Ordering_of_Reciprocals/Proof_2 | [
"Real Numbers",
"Reciprocals",
"Inequalities",
"Ordering of Reciprocals"
] | [
"Definition:Real Number"
] | [
"Reciprocal Function is Strictly Decreasing",
"Definition:Strictly Decreasing/Mapping",
"Reciprocal of Strictly Positive Real Number is Strictly Positive",
"Reciprocal of Strictly Negative Real Number is Strictly Negative",
"Inverse of Multiplicative Inverse"
] |
proofwiki-4920 | Comparison Test for Divergence | Let $\ds \sum_{n \mathop = 1}^\infty b_n$ be a divergent series of positive real numbers.
Let $\sequence {a_n}$ be a sequence in $\R$.
Let:
:$\forall n \in \N_{>0}: b_n \le a_n$
Then the series $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges. | This is the contrapositive of the Comparison Test.
Hence the result, from the Rule of Transposition.
{{qed}} | Let $\ds \sum_{n \mathop = 1}^\infty b_n$ be a [[Definition:Divergent Series|divergent series]] of [[Definition:Positive Real Number|positive real numbers]].
Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence in $\R$]].
Let:
:$\forall n \in \N_{>0}: b_n \le a_n$
Then the [[Definition:Series|series]] $\ds \s... | This is the [[Definition:Contrapositive Statement|contrapositive]] of the [[Comparison Test]].
Hence the result, from the [[Rule of Transposition]].
{{qed}} | Comparison Test for Divergence | https://proofwiki.org/wiki/Comparison_Test_for_Divergence | https://proofwiki.org/wiki/Comparison_Test_for_Divergence | [
"Comparison Test"
] | [
"Definition:Divergent Series",
"Definition:Positive/Real Number",
"Definition:Sequence",
"Definition:Series",
"Definition:Divergent Series"
] | [
"Definition:Contrapositive Statement",
"Comparison Test",
"Rule of Transposition"
] |
proofwiki-4921 | Axiom of Choice implies Zorn's Lemma | Let the {{Axiom-link|Choice}} be accepted.
Then Zorn's Lemma holds. | For each $x \in X$, consider the lower closure $x^\preceq$:
:$x^\preceq = \set {y \in X: y \preceq x}$
Let $\mathbb S \subseteq \powerset X$ be the image of $\cdot^\preceq$ considered as a mapping from $X$ to $\powerset X$, where $\powerset X$ is the power set of $X$.
From Ordering is Equivalent to Subset Relation:
:$\... | Let the {{Axiom-link|Choice}} be accepted.
Then [[Zorn's Lemma]] holds. | For each $x \in X$, consider the [[Definition:Lower Closure of Element|lower closure]] $x^\preceq$:
:$x^\preceq = \set {y \in X: y \preceq x}$
Let $\mathbb S \subseteq \powerset X$ be the [[Definition:Image of Mapping|image]] of $\cdot^\preceq$ considered as a [[Definition:Mapping|mapping]] from $X$ to $\powerset X$, ... | Axiom of Choice implies Zorn's Lemma/Proof 1 | https://proofwiki.org/wiki/Axiom_of_Choice_implies_Zorn's_Lemma | https://proofwiki.org/wiki/Axiom_of_Choice_implies_Zorn's_Lemma/Proof_1 | [
"Axiom of Choice",
"Zorn's Lemma",
"Axiom of Choice implies Zorn's Lemma"
] | [
"Zorn's Lemma"
] | [
"Definition:Lower Closure/Element",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Power Set",
"Ordering is Equivalent to Subset Relation",
"Definition:Maximal/Element",
"Definition:Logical Equivalence",
"Definition:Maximal/Set",
"Definition:Chain (Order Theory)",... |
proofwiki-4922 | Axiom of Choice implies Zorn's Lemma | Let the {{Axiom-link|Choice}} be accepted.
Then Zorn's Lemma holds. | {{AimForCont}} that for each $x \in X$ there is a $y \in X$ such that $x \prec y$.
By the Axiom of Choice, there is a mapping $f: X \to X$ such that:
:$\forall x \in X: x \prec \map f x$
Let $\CC$ be the set of all chains in $X$.
By the premise, each element of $\CC$ has an upper bound in $X$.
Thus by the Axiom of Choi... | Let the {{Axiom-link|Choice}} be accepted.
Then [[Zorn's Lemma]] holds. | {{AimForCont}} that for each $x \in X$ there is a $y \in X$ such that $x \prec y$.
By the [[Axiom:Axiom of Choice|Axiom of Choice]], there is a [[Definition:Mapping|mapping]] $f: X \to X$ such that:
:$\forall x \in X: x \prec \map f x$
Let $\CC$ be the [[Definition:Set|set]] of all [[Definition:Chain (Order Theory)|c... | Axiom of Choice implies Zorn's Lemma/Proof 2 | https://proofwiki.org/wiki/Axiom_of_Choice_implies_Zorn's_Lemma | https://proofwiki.org/wiki/Axiom_of_Choice_implies_Zorn's_Lemma/Proof_2 | [
"Axiom of Choice",
"Zorn's Lemma",
"Axiom of Choice implies Zorn's Lemma"
] | [
"Zorn's Lemma"
] | [
"Axiom:Axiom of Choice",
"Definition:Mapping",
"Definition:Set",
"Definition:Chain (Order Theory)",
"Definition:Upper Bound of Set",
"Axiom:Axiom of Choice",
"Definition:Mapping",
"Definition:Upper Bound of Set",
"Definition:Mapping",
"Transfinite Recursion Theorem",
"Definition:Limit Ordinal",
... |
proofwiki-4923 | Axiom of Choice implies Zorn's Lemma | Let the {{Axiom-link|Choice}} be accepted.
Then Zorn's Lemma holds. | We have that Axiom of Choice implies Tukey's Lemma.
Then we have Tukey's Lemma implies Zorn's Lemma.
{{qed}} | Let the {{Axiom-link|Choice}} be accepted.
Then [[Zorn's Lemma]] holds. | We have that [[Axiom of Choice implies Tukey's Lemma]].
Then we have [[Tukey's Lemma implies Zorn's Lemma]].
{{qed}} | Axiom of Choice implies Zorn's Lemma/Proof 3 | https://proofwiki.org/wiki/Axiom_of_Choice_implies_Zorn's_Lemma | https://proofwiki.org/wiki/Axiom_of_Choice_implies_Zorn's_Lemma/Proof_3 | [
"Axiom of Choice",
"Zorn's Lemma",
"Axiom of Choice implies Zorn's Lemma"
] | [
"Zorn's Lemma"
] | [
"Axiom of Choice implies Tukey's Lemma",
"Tukey's Lemma implies Zorn's Lemma"
] |
proofwiki-4924 | Limit at Infinity of Real Identity Function | Let $I_\R: \R \to \R$ be the identity function on $\R$.
Then:
:$(1): \quad \ds \lim_{x \mathop \to +\infty} \map {I_\R} x = +\infty$
:$(2): \quad \ds \lim_{x \mathop \to -\infty} \map {I_\R} x = -\infty$ | We have that the Derivative of Identity Function is $1$.
Hence, by Derivative of Monotone Function, $x$ is strictly increasing.
Now, by the definition of infinite limit at infinity, the first assertion is:
:$\forall M \in \R_{>0}: \exists N \in \R_{>0}: x > N \implies \map f x > M$
For every $M$, choose $N = M$.
The se... | Let $I_\R: \R \to \R$ be the [[Definition:Identity Mapping|identity function]] on $\R$.
Then:
:$(1): \quad \ds \lim_{x \mathop \to +\infty} \map {I_\R} x = +\infty$
:$(2): \quad \ds \lim_{x \mathop \to -\infty} \map {I_\R} x = -\infty$ | We have that the [[Derivative of Identity Function]] is $1$.
Hence, by [[Derivative of Monotone Function]], $x$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
Now, by the definition of [[Definition:Infinite Limit at Infinity|infinite limit at infinity]], the first assertion is:
:$\forall M ... | Limit at Infinity of Real Identity Function | https://proofwiki.org/wiki/Limit_at_Infinity_of_Real_Identity_Function | https://proofwiki.org/wiki/Limit_at_Infinity_of_Real_Identity_Function | [
"Unbounded Mappings",
"Identity Mappings"
] | [
"Definition:Identity Mapping"
] | [
"Derivative of Identity Function",
"Derivative of Monotone Function",
"Definition:Strictly Increasing/Real Function",
"Definition:Limit of Real Function/Limit at Infinity/Positive/Increasing Without Bound",
"Category:Unbounded Mappings",
"Category:Identity Mappings"
] |
proofwiki-4925 | Push Theorem | Let $f$ be a real function which is continuous on the open interval $\openint a \to$, $a \in \R$, such that:
:$\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$
Let $g$ be a real function defined on some open interval $\openint b \to$ such that, for sufficiently large $x$:
:$\map g x > \map f x$
Then:
:$\ds \lim_{x ... | Let $\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$.
By the definition of infinite limit at infinity, this means:
:$\forall M_1 \in \R_{>0}: \exists N_1 \in \R_{>0}: x > N_1 \implies \map f x > M_1$
Now, the assertion that $\map g x \to +\infty$ is:
:$\forall M_1 \in \R_{>0}: \exists N_2 \in \R_{>0}: x > N_2 \imp... | Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function|continuous]] on the [[Definition:Open Real Interval|open interval]] $\openint a \to$, $a \in \R$, such that:
:$\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$
Let $g$ be a [[Definition:Real Function|real functi... | Let $\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$.
By the definition of [[Definition:Infinite Limit at Infinity|infinite limit at infinity]], this means:
:$\forall M_1 \in \R_{>0}: \exists N_1 \in \R_{>0}: x > N_1 \implies \map f x > M_1$
Now, the assertion that $\map g x \to +\infty$ is:
:$\forall M_1 \in... | Push Theorem | https://proofwiki.org/wiki/Push_Theorem | https://proofwiki.org/wiki/Push_Theorem | [
"Unbounded Mappings"
] | [
"Definition:Real Function",
"Definition:Continuous Real Function",
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Real Interval/Open",
"Definition:Sufficiently Large"
] | [
"Definition:Limit of Real Function/Limit at Infinity/Positive/Increasing Without Bound",
"Definition:Sufficiently Large"
] |
proofwiki-4926 | Intersection of Orthocomplements is Orthocomplement of Closed Linear Span | Let $H$ be a Hilbert space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed family of closed linear subspaces of $H$.
Then:
:$\ds \bigcap_{i \mathop \in I} M_i^\perp = \paren {\vee \set {M_i : i \in I} }^\perp$
where:
:$\perp$ denotes orthocomplementation
:$\vee$ denotes closed linear span. | By definition of set equality, it suffices to prove the following two inclusions:
:$\ds \bigcap_{i \mathop \in I} M_i^\perp \subseteq \paren {\vee \set {M_i : i \in I} }^\perp$
:$\paren {\vee \set {M_i : i \in I} }^\perp \subseteq \ds \bigcap_{i \mathop \in I} M_i^\perp$ | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $\family {M_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$.
Then:
:$\ds \bigcap_{i \mathop \in I} M_i^\perp = \paren {\vee \set {M_i : i \in I} }^\perp$
... | By definition of [[Definition:Set Equality/Definition 2|set equality]], it suffices to prove the following two [[Definition:Subset|inclusions]]:
:$\ds \bigcap_{i \mathop \in I} M_i^\perp \subseteq \paren {\vee \set {M_i : i \in I} }^\perp$
:$\paren {\vee \set {M_i : i \in I} }^\perp \subseteq \ds \bigcap_{i \mathop \i... | Intersection of Orthocomplements is Orthocomplement of Closed Linear Span | https://proofwiki.org/wiki/Intersection_of_Orthocomplements_is_Orthocomplement_of_Closed_Linear_Span | https://proofwiki.org/wiki/Intersection_of_Orthocomplements_is_Orthocomplement_of_Closed_Linear_Span | [
"Hilbert Spaces",
"Intersection of Orthocomplements is Orthocomplement of Closed Linear Span"
] | [
"Definition:Hilbert Space",
"Definition:Indexing Set/Family",
"Definition:Closed Linear Subspace",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Closed Linear Span"
] | [
"Definition:Set Equality/Definition 2",
"Definition:Subset"
] |
proofwiki-4927 | Zorn's Lemma implies Axiom of Choice | Let Zorn's Lemma be accepted.
Then the Axiom of Choice holds. | Let $X$ be a set.
Let $\FF$ be the set of partial choice functions defined as:
:<nowiki>$f \in \FF \iff \begin{cases}
\Dom f \subseteq \powerset X & \ \\
\Img f \subseteq X & \ \\
\forall A \in \Dom f: \map f A \in A & \ \end{cases}$</nowiki>
Let $\preceq$ be the relation defined on $\FF$ as:
:$\forall f_1, f_2 \in \FF... | Let [[Zorn's Lemma]] be accepted.
Then the [[Axiom:Axiom of Choice|Axiom of Choice]] holds. | Let $X$ be a [[Definition:Set|set]].
Let $\FF$ be the [[Definition:Set|set]] of partial [[Definition:Choice Function|choice functions]] defined as:
:<nowiki>$f \in \FF \iff \begin{cases}
\Dom f \subseteq \powerset X & \ \\
\Img f \subseteq X & \ \\
\forall A \in \Dom f: \map f A \in A & \ \end{cases}$</nowiki>
Let $\... | Zorn's Lemma implies Axiom of Choice | https://proofwiki.org/wiki/Zorn's_Lemma_implies_Axiom_of_Choice | https://proofwiki.org/wiki/Zorn's_Lemma_implies_Axiom_of_Choice | [
"Zorn's Lemma",
"Axiom of Choice"
] | [
"Zorn's Lemma",
"Axiom:Axiom of Choice"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Choice Function",
"Definition:Relation",
"Definition:Extension of Mapping",
"Definition:Partial Ordering",
"Definition:Empty Mapping",
"Definition:Element",
"Definition:Non-Empty Set",
"Definition:Chain (Order Theory)",
"Definition:Set Union",
"D... |
proofwiki-4928 | Orthocomplement Reverses Subset | Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space space.
Let $A, B$ be subsets of $V$ with $A \subseteq B$.
Then:
:$B^\perp \subseteq A^\perp$
where $\perp$ denotes orthocomplementation. | Let:
:$h \in B^\perp$
Then, from the definition of orthocomplement, we have:
:$h \perp b$ for each $b \in B$.
Since $A \subseteq B$, we in particular have:
:$h \perp a$ for each $a \in A$.
So, from the definition of orthocomplement:
:$h \in A^\perp$
So:
:$h \in B^\perp$ implies $h \in A^\perp$.
So, from the definitio... | Let $\struct {V, \innerprod \cdot \cdot}$ be an [[Definition:Inner Product Space|inner product space space]].
Let $A, B$ be [[Definition:Subset|subsets]] of $V$ with $A \subseteq B$.
Then:
:$B^\perp \subseteq A^\perp$
where $\perp$ denotes [[Definition:Orthocomplement|orthocomplementation]]. | Let:
:$h \in B^\perp$
Then, from the definition of [[Definition:Orthocomplement|orthocomplement]], we have:
:$h \perp b$ for each $b \in B$.
Since $A \subseteq B$, we in particular have:
:$h \perp a$ for each $a \in A$.
So, from the definition of [[Definition:Orthocomplement|orthocomplement]]:
:$h \in A^\perp$... | Orthocomplement Reverses Subset | https://proofwiki.org/wiki/Orthocomplement_Reverses_Subset | https://proofwiki.org/wiki/Orthocomplement_Reverses_Subset | [
"Inner Product Spaces",
"Orthocomplements"
] | [
"Definition:Inner Product Space",
"Definition:Subset",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement"
] | [
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Subset",
"Category:Inner Product Spaces",
"Category:Orthocomplements"
] |
proofwiki-4929 | Limit at Infinity of x^n | Let $x \mapsto x^n$, $n \in \R$ be a real function which is continuous on the open interval $\openint 1 {+\infty}$.
Let $n > 0$.
Then $x^n \to +\infty$ as $x \to +\infty$. | From Upper Bound of Natural Logarithm:
:$\forall n > 0: n \ln x < x^n$
which, by Multiple Rule for Continuous Real Functions, implies:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to +\infty} n \ln x
| r = n \lim_{x \mathop \to +\infty} \ln x
}}
{{end-eqn}}
From Logarithm Tends to Infinity:
:$n \ln x \to +\infty$ a... | Let $x \mapsto x^n$, $n \in \R$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Open Interval|continuous]] on the [[Definition:Open Real Interval|open interval]] $\openint 1 {+\infty}$.
Let $n > 0$.
Then $x^n \to +\infty$ as $x \to +\infty$. | From [[Upper Bound of Natural Logarithm]]:
:$\forall n > 0: n \ln x < x^n$
which, by [[Multiple Rule for Continuous Real Functions]], implies:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to +\infty} n \ln x
| r = n \lim_{x \mathop \to +\infty} \ln x
}}
{{end-eqn}}
From [[Logarithm Tends to Infinity]]:
:$n \ln... | Limit at Infinity of x^n | https://proofwiki.org/wiki/Limit_at_Infinity_of_x^n | https://proofwiki.org/wiki/Limit_at_Infinity_of_x^n | [
"Examples of Limits of Real Functions"
] | [
"Definition:Real Function",
"Definition:Continuous Real Function/Open Interval",
"Definition:Real Interval/Open"
] | [
"Upper Bound of Natural Logarithm",
"Combination Theorem for Continuous Functions/Real/Multiple Rule",
"Logarithm Tends to Infinity",
"Push Theorem"
] |
proofwiki-4930 | Monotonicity of Real Sequences | Let $\mathbb D$ be a subset of $\N$.
Let $\sequence {a_n}: \mathbb D \to \R$ be a real sequence.
Let $\mathbb X$ be a real interval such that $\mathbb D \subseteq \mathbb X$.
Let $f: \mathbb X \to \R, x \mapsto \map f x$ be a differentiable real function.
Suppose that for every $n \in \mathbb D$:
:$\map f n = a_n$
Then... | Consider the case where $D_x \map f x \ge 0$
Let $n \in \N$ be in the domain of $\sequence {a_n}$.
From Derivative of Monotone Function, the sign of $D_x f$ is indicative of the monotonicity of $f$.
Because Differentiable Function is Continuous and Continuous Real Function is Darboux Integrable, $D_x f$ is integrable.
... | Let $\mathbb D$ be a [[Definition:Subset|subset]] of $\N$.
Let $\sequence {a_n}: \mathbb D \to \R$ be a [[Definition:Real Sequence|real sequence]].
Let $\mathbb X$ be a [[Definition:Real Interval|real interval]] such that $\mathbb D \subseteq \mathbb X$.
Let $f: \mathbb X \to \R, x \mapsto \map f x$ be a [[Definitio... | Consider the case where $D_x \map f x \ge 0$
Let $n \in \N$ be in the [[Definition:Domain of Mapping|domain]] of $\sequence {a_n}$.
From [[Derivative of Monotone Function]], the sign of $D_x f$ is indicative of the [[Definition:Monotonicity|monotonicity]] of $f$.
Because [[Differentiable Function is Continuous]] and... | Monotonicity of Real Sequences | https://proofwiki.org/wiki/Monotonicity_of_Real_Sequences | https://proofwiki.org/wiki/Monotonicity_of_Real_Sequences | [
"Real Sequences"
] | [
"Definition:Subset",
"Definition:Real Sequence",
"Definition:Real Interval",
"Definition:Differentiable Mapping/Real Function/Interval",
"Definition:Real Function",
"Definition:Increasing/Sequence",
"Definition:Strictly Increasing/Sequence",
"Definition:Decreasing/Sequence",
"Definition:Strictly Dec... | [
"Definition:Domain (Set Theory)/Mapping",
"Derivative of Monotone Function",
"Definition:Monotonicity",
"Differentiable Function is Continuous",
"Continuous Real Function is Darboux Integrable",
"Definition:Darboux Integrable Function",
"Fundamental Theorem of Calculus",
"Relative Sizes of Definite In... |
proofwiki-4931 | Hausdorff's Maximal Principle is equivalent to Axiom of Choice | Every ordered set has a maximal chain {{iff}} the axiom of choice holds. | * Axiom of Choice implies Hausdorff's Maximal Principle
* Hausdorff's Maximal Principle implies Zermelo's Well-Ordering Theorem
* Zermelo's Well-Ordering Theorem is Equivalent to Axiom of Choice
{{qed}} | Every [[Definition:Ordered Set|ordered set]] has a [[Definition:Maximal Chain|maximal chain]] {{iff}} the [[Axiom:Axiom of Choice|axiom of choice]] holds. | * [[Axiom of Choice implies Hausdorff's Maximal Principle]]
* [[Hausdorff's Maximal Principle implies Zermelo's Well-Ordering Theorem]]
* [[Zermelo's Well-Ordering Theorem is Equivalent to Axiom of Choice]]
{{qed}} | Hausdorff's Maximal Principle is equivalent to Axiom of Choice | https://proofwiki.org/wiki/Hausdorff's_Maximal_Principle_is_equivalent_to_Axiom_of_Choice | https://proofwiki.org/wiki/Hausdorff's_Maximal_Principle_is_equivalent_to_Axiom_of_Choice | [
"Axiom of Choice",
"Hausdorff's Maximal Principle"
] | [
"Definition:Ordered Set",
"Definition:Maximal Chain",
"Axiom:Axiom of Choice"
] | [
"Axiom of Choice implies Hausdorff's Maximal Principle",
"Hausdorff's Maximal Principle implies Zermelo's Well-Ordering Theorem",
"Zermelo's Well-Ordering Theorem is Equivalent to Axiom of Choice"
] |
proofwiki-4932 | Young's Inequality for Products | Let $p, q \in \R_{> 0}$ be strictly positive real numbers such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then:
:$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs {{iff}}:
:$b = a^{p - 1}$ | :400px
In the above diagram:
:the {{Color|blue}} region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$
:the {{Color|red}} region corresponds to $\ds \int_0^\beta u^{q - 1} \rd u$.
From Positive Real Numbers whose Reciprocals Sum to 1, it is necessary for both $p > 1$ and $q > 1$.
{{begin-eqn}}
{{eqn | l = \frac 1 p... | Let $p, q \in \R_{> 0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]] such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then:
:$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs {{iff}}:
:$b = a^{p - 1}$ | :[[File:Holder's Ineq.jpg|400px]]
In the above diagram:
:the {{Color|blue}} region corresponds to $\ds \int_0^\alpha t^{p - 1} \rd t$
:the {{Color|red}} region corresponds to $\ds \int_0^\beta u^{q - 1} \rd u$.
From [[Positive Real Numbers whose Reciprocals Sum to 1]], it is necessary for both $p > 1$ and $q > 1$.
{... | Young's Inequality for Products/Geometric Proof | https://proofwiki.org/wiki/Young's_Inequality_for_Products | https://proofwiki.org/wiki/Young's_Inequality_for_Products/Geometric_Proof | [
"Young's Inequality for Products",
"Real Analysis",
"Inequalities"
] | [
"Definition:Strictly Positive/Real Number"
] | [
"File:Holder's Ineq.jpg",
"Positive Real Numbers whose Reciprocals Sum to 1",
"Definition:Positive/Real Number",
"Area of Parallelogram/Rectangle",
"Definition:Graph of Mapping",
"Definition:Intersection (Geometry)",
"Definition:Quadrilateral/Rectangle",
"Definition:Inequality",
"Definition:Inequali... |
proofwiki-4933 | Young's Inequality for Products | Let $p, q \in \R_{> 0}$ be strictly positive real numbers such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then:
:$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs {{iff}}:
:$b = a^{p - 1}$ | {{WLOG}} assume that $a^p \ge b^q$, otherwise swap $a$ with $b$ and $p$ with $q$.
Define $f : \hointr 0 \infty \to \R$ by:
:$\ds \map f t = \frac {t^p} p + \frac 1 q - t$
We have, from Derivative of Power and Sum Rule for Derivatives:
:$\ds \map {f'} t = t^{p - 1} - 1$
So for $t \ge 1$ we have:
:$\ds \map {f'} t \ge... | Let $p, q \in \R_{> 0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]] such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then:
:$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs {{iff}}:
:$b = a^{p - 1}$ | {{WLOG}} assume that $a^p \ge b^q$, otherwise swap $a$ with $b$ and $p$ with $q$.
Define $f : \hointr 0 \infty \to \R$ by:
:$\ds \map f t = \frac {t^p} p + \frac 1 q - t$
We have, from [[Derivative of Power]] and [[Sum Rule for Derivatives]]:
:$\ds \map {f'} t = t^{p - 1} - 1$
So for $t \ge 1$ we have:
:$\ds \... | Young's Inequality for Products/Proof by Calculus | https://proofwiki.org/wiki/Young's_Inequality_for_Products | https://proofwiki.org/wiki/Young's_Inequality_for_Products/Proof_by_Calculus | [
"Young's Inequality for Products",
"Real Analysis",
"Inequalities"
] | [
"Definition:Strictly Positive/Real Number"
] | [
"Power Rule for Derivatives",
"Sum Rule for Derivatives",
"Real Function with Positive Derivative is Increasing",
"Definition:Increasing/Real Function",
"Real Function with Strictly Positive Derivative is Strictly Increasing",
"Definition:Strictly Increasing/Real Function"
] |
proofwiki-4934 | Young's Inequality for Products | Let $p, q \in \R_{> 0}$ be strictly positive real numbers such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then:
:$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs {{iff}}:
:$b = a^{p - 1}$ | The result follows directly if $a = 0$ or $b = 0$.
{{WLOG}}, assume that $a > 0$ and $b > 0$.
Recall Exponential is Strictly Convex.
Consider:
:$x := \map \ln {a^p}$
:$y := \map \ln {b^q}$
:$\alpha := p^{-1}$
:$\beta := q^{-1}$
Note that {{hypothesis}}:
:$\alpha + \beta = 1$
Thus by definition of strictly convex real ... | Let $p, q \in \R_{> 0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]] such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then:
:$\forall a, b \in \R_{\ge 0}: a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs {{iff}}:
:$b = a^{p - 1}$ | The result follows directly if $a = 0$ or $b = 0$.
{{WLOG}}, assume that $a > 0$ and $b > 0$.
Recall [[Exponential is Strictly Convex]].
Consider:
:$x := \map \ln {a^p}$
:$y := \map \ln {b^q}$
:$\alpha := p^{-1}$
:$\beta := q^{-1}$
Note that {{hypothesis}}:
:$\alpha + \beta = 1$
Thus by definition of [[Definitio... | Young's Inequality for Products/Proof by Convexity | https://proofwiki.org/wiki/Young's_Inequality_for_Products | https://proofwiki.org/wiki/Young's_Inequality_for_Products/Proof_by_Convexity | [
"Young's Inequality for Products",
"Real Analysis",
"Inequalities"
] | [
"Definition:Strictly Positive/Real Number"
] | [
"Exponential is Strictly Convex",
"Definition:Convex Real Function/Definition 1/Strictly",
"Exponential of Natural Logarithm",
"Sum of Logarithms",
"Logarithm of Power",
"Exponential of Natural Logarithm"
] |
proofwiki-4935 | Restriction of Monotone Function is Monotone | The restriction of a monotone mapping is monotone. | A restriction does not introduce any new arguments.
Hence the result follows trivially from the definition of monotone mapping.
{{qed}}
{{handwaving}} | The [[Definition:Restriction of Mapping|restriction]] of a [[Definition:Monotone Mapping|monotone mapping]] is [[Definition:Monotone|monotone]]. | A [[Definition:Restriction of Mapping|restriction]] does not introduce any new [[Definition:Mapping#Domain, Codomain, Image, Preimage|arguments]].
Hence the result follows trivially from the definition of [[Definition:Monotone Mapping|monotone mapping]].
{{qed}}
{{handwaving}} | Restriction of Monotone Function is Monotone | https://proofwiki.org/wiki/Restriction_of_Monotone_Function_is_Monotone | https://proofwiki.org/wiki/Restriction_of_Monotone_Function_is_Monotone | [
"Analysis",
"Order Theory"
] | [
"Definition:Restriction/Mapping",
"Definition:Monotone (Order Theory)/Mapping",
"Definition:Monotone"
] | [
"Definition:Restriction/Mapping",
"Definition:Mapping",
"Definition:Monotone (Order Theory)/Mapping"
] |
proofwiki-4936 | Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering | Let $S$ be a set which is well-ordered by $\preccurlyeq$.
Let $\preccurlyeq_l$ be the lexicographic order on the set $T_n$ of all ordered $n$-tuples of $S$:
:$\tuple {x_1, x_2, \ldots, x_n} \prec \tuple {y_1, y_2, \ldots, y_n}$ {{iff}}:
::$\exists k: 1 \le k \le n$ such that $\forall 1 \le j < k: x_j = y_j$ but $x_k \p... | Consider $T_n$ where $n \in \N_{>0}$.
It is clear that $\struct {T_1, \preccurlyeq_l}$ is order isomorphic to $\struct {S, \preccurlyeq}$.
Thus as $\preccurlyeq$ is a well-ordering on $S$, $\preccurlyeq_l$ is a well-ordering on $T_1$.
Now, let us assume that $\preccurlyeq_l$ is a well-ordering on $T_k$ for some $k \in... | Let $S$ be a [[Definition:Set|set]] which is [[Definition:Well-Ordered Set|well-ordered]] by $\preccurlyeq$.
Let $\preccurlyeq_l$ be the [[Definition:Lexicographic Order|lexicographic order]] on the set $T_n$ of all [[Definition:Ordered Tuple|ordered $n$-tuples]] of $S$:
:$\tuple {x_1, x_2, \ldots, x_n} \prec \tuple {... | Consider $T_n$ where $n \in \N_{>0}$.
It is clear that $\struct {T_1, \preccurlyeq_l}$ is [[Definition:Order Isomorphism|order isomorphic]] to $\struct {S, \preccurlyeq}$.
Thus as $\preccurlyeq$ is a [[Definition:Well-Ordering|well-ordering]] on $S$, $\preccurlyeq_l$ is a [[Definition:Well-Ordering|well-ordering]] o... | Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering | https://proofwiki.org/wiki/Finite_Lexicographic_Order_on_Well-Ordered_Sets_is_Well-Ordering | https://proofwiki.org/wiki/Finite_Lexicographic_Order_on_Well-Ordered_Sets_is_Well-Ordering | [
"Lexicographic Order",
"Well-Orderings"
] | [
"Definition:Set",
"Definition:Well-Ordered Set",
"Definition:Lexicographic Order",
"Definition:Ordered Tuple",
"Definition:Well-Ordering"
] | [
"Definition:Order Isomorphism",
"Definition:Well-Ordering",
"Definition:Well-Ordering",
"Definition:Well-Ordering",
"Definition:Empty Set",
"Definition:Subset",
"Definition:Ordered Tuple",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Well-Ordered Set",
"Definition:Minimal/Element... |
proofwiki-4937 | Infinite Lexicographic Order on Well-Ordered Sets is not Well-Ordering | Let $\struct {S, \preccurlyeq}$ be a well-ordered set.
Let $\preccurlyeq_l$ be the lexicographic order on the set $T$ of all ordered tuples of $S$:
:$\tuple {x_1, x_2, \ldots, x_m} \prec \tuple {y_1, y_2, \ldots, y_n}$ {{iff}}:
::$\exists k: 1 \le k \le \map \min {m, n}$ such that $\forall 1 \le j < k: x_j = y_j$ but $... | Consider a set $S = \set {a, b}$ such that $a \prec b$.
Then the set $\set {\tuple b, \tuple {a, b}, \tuple {a, a, b}, \tuple {a, a, a, b}, \ldots}$ has no minimal element by $\preccurlyeq_l$.
{{handwaving}}
Thus $T$ is not well-ordered by $\preccurlyeq_l$.
{{qed}} | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Let $\preccurlyeq_l$ be the [[Definition:Lexicographic Order|lexicographic order]] on the set $T$ of all [[Definition:Ordered Tuple|ordered tuples]] of $S$:
:$\tuple {x_1, x_2, \ldots, x_m} \prec \tuple {y_1, y_2, \ldots, y_n}$ {{i... | Consider a set $S = \set {a, b}$ such that $a \prec b$.
Then the set $\set {\tuple b, \tuple {a, b}, \tuple {a, a, b}, \tuple {a, a, a, b}, \ldots}$ has no [[Definition:Minimal Element|minimal element]] by $\preccurlyeq_l$.
{{handwaving}}
Thus $T$ is not [[Definition:Well-Ordering|well-ordered]] by $\preccurlyeq_l$.
{... | Infinite Lexicographic Order on Well-Ordered Sets is not Well-Ordering | https://proofwiki.org/wiki/Infinite_Lexicographic_Order_on_Well-Ordered_Sets_is_not_Well-Ordering | https://proofwiki.org/wiki/Infinite_Lexicographic_Order_on_Well-Ordered_Sets_is_not_Well-Ordering | [
"Lexicographic Order",
"Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Lexicographic Order",
"Definition:Ordered Tuple",
"Definition:Well-Ordering"
] | [
"Definition:Minimal/Element",
"Definition:Well-Ordering"
] |
proofwiki-4938 | Transfinite Recursion Theorem/Theorem 1 | Let $G$ be a (class) mapping from $\On^{\On}$ to $\On$.
Let $K$ be a class of mappings $f$ that satisfy:
:the domain of $f$ is some ordinal $y$
:$\forall x \in y: \map f x = \map G {f {\restriction_x} }$
where $f {\restriction_x}$ denotes the restriction of $f$ to $x$.
Let $F = \bigcup K$ be the union of $K$.
Then:
:$... | First a lemma: | Let $G$ be a [[Definition:Class Mapping|(class) mapping]] from $\On^{\On}$ to $\On$.
Let $K$ be a [[Definition:Class (Class Theory)|class]] of [[Definition:Mapping|mappings]] $f$ that satisfy:
:the [[Definition:Domain of Mapping|domain]] of $f$ is some [[Definition:Ordinal|ordinal]] $y$
:$\forall x \in y: \map f x = ... | First a [[Definition:Lemma|lemma]]: | Transfinite Recursion Theorem/Theorem 1 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Theorem_1 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Theorem_1 | [
"Transfinite Recursion Theorem"
] | [
"Definition:Mapping/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Ordinal",
"Definition:Restriction/Mapping",
"Definition:Set Union/Set of Sets",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definit... | [
"Definition:Lemma"
] |
proofwiki-4939 | Squeeze Theorem for Absolutely Convergent Series | Let $\ds \sum \size {a_n}$ be an absolutely convergent series in $\R$.
Suppose that:
:$\ds -\sum \size {a_n} = \sum \size {a_n}$
Then $\ds \sum a_n$ equals the above two series. | From Absolutely Convergent Real Series is Convergent, the convergence of:
:$\ds \sum_{n \mathop = 1}^\infty \size {a_n}$
implies that of:
:$\ds \sum_{n \mathop = 1}^\infty a_n$
By Negative of Absolute Value:
:$\ds -\size {\sum_{n \mathop = 1}^j a_n} \le \sum_{n \mathop = 1}^j a_n \le \size {\sum_{n \mathop = 1}^j a_n}$... | Let $\ds \sum \size {a_n}$ be an [[Definition:Absolutely Convergent Series|absolutely convergent series]] in $\R$.
Suppose that:
:$\ds -\sum \size {a_n} = \sum \size {a_n}$
Then $\ds \sum a_n$ equals the above two series. | From [[Absolutely Convergent Real Series is Convergent]], the [[Definition:Convergent Series|convergence]] of:
:$\ds \sum_{n \mathop = 1}^\infty \size {a_n}$
implies that of:
:$\ds \sum_{n \mathop = 1}^\infty a_n$
By [[Negative of Absolute Value]]:
:$\ds -\size {\sum_{n \mathop = 1}^j a_n} \le \sum_{n \mathop = 1... | Squeeze Theorem for Absolutely Convergent Series | https://proofwiki.org/wiki/Squeeze_Theorem_for_Absolutely_Convergent_Series | https://proofwiki.org/wiki/Squeeze_Theorem_for_Absolutely_Convergent_Series | [
"Series"
] | [
"Definition:Absolutely Convergent Series"
] | [
"Absolutely Convergent Series is Convergent/Real Numbers",
"Definition:Convergent Series",
"Negative of Absolute Value",
"Triangle Inequality",
"Squeeze Theorem/Sequences/Real Numbers",
"Category:Series"
] |
proofwiki-4940 | N Choose k is not greater than n^k | :$\forall n \in \Z, k \in \Z: 1 \le k \le n: \dbinom n k < n^k$
where $\dbinom n k$ is a binomial coefficient.
Equality holds when $k = 0$ and $k = 1$. | First let $k > 1$.
{{begin-eqn}}
{{eqn | l = \binom n k
| r = \frac {n!} {k! \, \paren {n - k}!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | o = <
| r = \frac {n!} {\paren {n - k}!}
| c = as $k! > 0$ by {{Defof|Factorial}}
}}
{{eqn | r = \underbrace {n \paren {n - 1} \cdots (n - k + 1)}_{k \... | :$\forall n \in \Z, k \in \Z: 1 \le k \le n: \dbinom n k < n^k$
where $\dbinom n k$ is a [[Definition:Binomial Coefficient|binomial coefficient]].
Equality holds when $k = 0$ and $k = 1$. | First let $k > 1$.
{{begin-eqn}}
{{eqn | l = \binom n k
| r = \frac {n!} {k! \, \paren {n - k}!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | o = <
| r = \frac {n!} {\paren {n - k}!}
| c = as $k! > 0$ by {{Defof|Factorial}}
}}
{{eqn | r = \underbrace {n \paren {n - 1} \cdots (n - k + 1)}_{k ... | N Choose k is not greater than n^k/Proof 1 | https://proofwiki.org/wiki/N_Choose_k_is_not_greater_than_n^k | https://proofwiki.org/wiki/N_Choose_k_is_not_greater_than_n^k/Proof_1 | [
"Binomial Coefficients",
"N Choose k is not greater than n^k"
] | [
"Definition:Binomial Coefficient"
] | [
"Definition:Factorial",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Binomial Coefficient",
"Binomial Coefficient with Zero",
"Binomial Coefficient with One"
] |
proofwiki-4941 | N Choose k is not greater than n^k | :$\forall n \in \Z, k \in \Z: 1 \le k \le n: \dbinom n k < n^k$
where $\dbinom n k$ is a binomial coefficient.
Equality holds when $k = 0$ and $k = 1$. | We dismiss the cases where $k < 0$ by observing that in such cases $\dbinom n k = 0$ while $n^k > 0$.
Similarly we dismiss $k = 0$: we have $\dbinom n 0 = 1 = n^0$.
Let:
: $N = \set {1, \ldots, n}$
: $K = \set {1, \ldots, k}$
From Cardinality of Set of Strictly Increasing Mappings, $\dbinom n k$ is the number of strict... | :$\forall n \in \Z, k \in \Z: 1 \le k \le n: \dbinom n k < n^k$
where $\dbinom n k$ is a [[Definition:Binomial Coefficient|binomial coefficient]].
Equality holds when $k = 0$ and $k = 1$. | We dismiss the cases where $k < 0$ by observing that in such cases $\dbinom n k = 0$ while $n^k > 0$.
Similarly we dismiss $k = 0$: we have $\dbinom n 0 = 1 = n^0$.
Let:
: $N = \set {1, \ldots, n}$
: $K = \set {1, \ldots, k}$
From [[Cardinality of Set of Strictly Increasing Mappings]], $\dbinom n k$ is the number o... | N Choose k is not greater than n^k/Proof 2 | https://proofwiki.org/wiki/N_Choose_k_is_not_greater_than_n^k | https://proofwiki.org/wiki/N_Choose_k_is_not_greater_than_n^k/Proof_2 | [
"Binomial Coefficients",
"N Choose k is not greater than n^k"
] | [
"Definition:Binomial Coefficient"
] | [
"Cardinality of Set of Strictly Increasing Mappings",
"Definition:Strictly Increasing/Mapping",
"Cardinality of Set of All Mappings",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Strictly Increasing/Mapping",
"Definition:Element",
"Definition:Mapping",
"Definition:Strictly Increasing/Mapp... |
proofwiki-4942 | Set is Element of its Power Set | A set is an element of its power set:
:$S \in \powerset S$ | {{begin-eqn}}
{{eqn | q = \forall S
| l = S
| o = \subseteq
| r = S
| c = Set is Subset of Itself
}}
{{eqn | ll= \leadsto
| q = \forall S
| l = S
| o = \in
| r = \powerset S
| c = {{Defof|Power Set}}
}}
{{end-eqn}}
{{qed}} | A [[Definition:Set|set]] is an [[Definition:Element|element]] of its [[Definition:Power Set|power set]]:
:$S \in \powerset S$ | {{begin-eqn}}
{{eqn | q = \forall S
| l = S
| o = \subseteq
| r = S
| c = [[Set is Subset of Itself]]
}}
{{eqn | ll= \leadsto
| q = \forall S
| l = S
| o = \in
| r = \powerset S
| c = {{Defof|Power Set}}
}}
{{end-eqn}}
{{qed}} | Set is Element of its Power Set | https://proofwiki.org/wiki/Set_is_Element_of_its_Power_Set | https://proofwiki.org/wiki/Set_is_Element_of_its_Power_Set | [
"Power Set"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Power Set"
] | [
"Set is Subset of Itself"
] |
proofwiki-4943 | Power Set of Empty Set | The power set of the empty set $\O$ is the set $\set \O$. | From Empty Set is Element of Power Set and Set is Element of its Power Set:
:$\O \in \powerset \O$
From Empty Set is Subset of All Sets:
:$S \subseteq \O \implies S = \O$
That is:
:$S \in \powerset \O \implies S = \O$
Hence the only element of $\powerset \O$ is $\O$.
{{qed}} | The [[Definition:Power Set|power set]] of the [[Definition:Empty Set|empty set]] $\O$ is the set $\set \O$. | From [[Empty Set is Element of Power Set]] and [[Set is Element of its Power Set]]:
:$\O \in \powerset \O$
From [[Empty Set is Subset of All Sets]]:
:$S \subseteq \O \implies S = \O$
That is:
:$S \in \powerset \O \implies S = \O$
Hence the only element of $\powerset \O$ is $\O$.
{{qed}} | Power Set of Empty Set | https://proofwiki.org/wiki/Power_Set_of_Empty_Set | https://proofwiki.org/wiki/Power_Set_of_Empty_Set | [
"Power Set",
"Empty Set"
] | [
"Definition:Power Set",
"Definition:Empty Set"
] | [
"Empty Set is Element of Power Set",
"Set is Element of its Power Set",
"Empty Set is Subset of All Sets"
] |
proofwiki-4944 | Symmetric Difference is Subset of Union | The symmetric difference of two sets is a subset of their union:
:$S \symdif T \subseteq S \cup T$ | {{begin-eqn}}
{{eqn | l = S \symdif T
| r = \paren {S \cup T} \setminus \paren {S \cap T}
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | o = \subseteq
| r = \paren {S \cup T}
| c = Set Difference is Subset
}}
{{end-eqn}}
{{qed}} | The [[Definition:Symmetric Difference|symmetric difference]] of two [[Definition:Set|sets]] is a [[Definition:Subset|subset]] of their [[Definition:Set Union|union]]:
:$S \symdif T \subseteq S \cup T$ | {{begin-eqn}}
{{eqn | l = S \symdif T
| r = \paren {S \cup T} \setminus \paren {S \cap T}
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | o = \subseteq
| r = \paren {S \cup T}
| c = [[Set Difference is Subset]]
}}
{{end-eqn}}
{{qed}} | Symmetric Difference is Subset of Union | https://proofwiki.org/wiki/Symmetric_Difference_is_Subset_of_Union | https://proofwiki.org/wiki/Symmetric_Difference_is_Subset_of_Union | [
"Symmetric Difference",
"Set Union"
] | [
"Definition:Symmetric Difference",
"Definition:Set",
"Definition:Subset",
"Definition:Set Union"
] | [
"Set Difference is Subset"
] |
proofwiki-4945 | Sum of Projections/Binary Case | Let $H$ be a Hilbert space.
Let $P, Q$ be projections.
Then $P + Q$ is a projection {{iff}} $\Rng P \perp \Rng Q$. | === Necessary Condition ===
Suppose $P + Q$ is a projection. Then:
{{begin-eqn}}
{{eqn | l = P + Q
| r = \paren {P + Q}^2
| c = $P + Q$ is an idempotent
}}
{{eqn | r = P^2 + P Q + Q P + Q^2
}}
{{eqn | r = P + P Q + Q P + Q
| c = $P$ and $Q$ are idempotents
}}
{{eqn | ll= \leadstoandfrom
| l = PQ... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $P, Q$ be [[Definition:Projection (Hilbert Spaces)|projections]].
Then $P + Q$ is a [[Definition:Projection (Hilbert Spaces)|projection]] {{iff}} $\Rng P \perp \Rng Q$. | === Necessary Condition ===
Suppose $P + Q$ is a [[Definition:Projection (Hilbert Spaces)|projection]]. Then:
{{begin-eqn}}
{{eqn | l = P + Q
| r = \paren {P + Q}^2
| c = $P + Q$ is an [[Definition:Idempotent Operator|idempotent]]
}}
{{eqn | r = P^2 + P Q + Q P + Q^2
}}
{{eqn | r = P + P Q + Q P + Q
... | Sum of Projections/Binary Case | https://proofwiki.org/wiki/Sum_of_Projections/Binary_Case | https://proofwiki.org/wiki/Sum_of_Projections/Binary_Case | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Projection (Hilbert Spaces)",
"Definition:Projection (Hilbert Spaces)"
] | [
"Definition:Projection (Hilbert Spaces)",
"Definition:Idempotent Operator",
"Definition:Idempotent Operator",
"Definition:Idempotent Operator",
"Characterization of Projections",
"Definition:Positive",
"Definition:Negative",
"Definition:Inner Product",
"Definition:Projection (Hilbert Spaces)",
"De... |
proofwiki-4946 | Product of Projections | Let $H$ be a Hilbert space.
Let $P, Q$ be projections.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$P Q$ is a projection}}
{{item|(2):|$P Q {{=}} Q P$}}
{{item|(3):|$P + Q - P Q$ is a projection}}
{{end-itemize}}
{{MissingLinks|Provide proper linking to the def of addition and multiplication of operators}} | The proof proceeds by first showing that $(1)$ is equivalent to $(2)$.
Then, these are combined and shown equivalent to $(3)$. | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $P, Q$ be [[Definition:Projection (Hilbert Spaces)|projections]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$P Q$ is a [[Definition:Projection (Hilbert Spaces)|projection]]}}
{{item|(2):|$P Q {{=}} Q P$}}
{{item|(3):|$P + Q - P Q$ is a [[Definition:Projection... | The proof proceeds by first showing that $(1)$ is equivalent to $(2)$.
Then, these are combined and shown equivalent to $(3)$. | Product of Projections | https://proofwiki.org/wiki/Product_of_Projections | https://proofwiki.org/wiki/Product_of_Projections | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Projection (Hilbert Spaces)",
"Definition:Projection (Hilbert Spaces)",
"Definition:Projection (Hilbert Spaces)"
] | [] |
proofwiki-4947 | Difference of Projections | Let $H$ be a Hilbert space.
Let $P, Q$ be projections.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$P - Q$ is a projection}}
{{item|(2):|$P Q {{=}} Q$}}
{{item|(3):|$Q P {{=}} Q$}}
{{end-itemize}}
{{MissingLinks|Provide proper linking to the def of addition of operators}} | First it is shown that $(2)$ is equivalent to $(3)$.
Then, equivalence to $(1)$ is shown. | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $P, Q$ be [[Definition:Projection (Hilbert Spaces)|projections]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$P - Q$ is a [[Definition:Projection (Hilbert Spaces)|projection]]}}
{{item|(2):|$P Q {{=}} Q$}}
{{item|(3):|$Q P {{=}} Q$}}
{{end-itemize}}
{{Missing... | First it is shown that $(2)$ is equivalent to $(3)$.
Then, equivalence to $(1)$ is shown. | Difference of Projections | https://proofwiki.org/wiki/Difference_of_Projections | https://proofwiki.org/wiki/Difference_of_Projections | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Projection (Hilbert Spaces)",
"Definition:Projection (Hilbert Spaces)"
] | [] |
proofwiki-4948 | Union of Countable Sets of Sets | Let $\AA$ and $\BB$ be countable sets of sets.
Then:
:$\set {A \cup B: A \in \AA, B \in \BB}$
is also countable. | Since $\AA$ is countable, its contents can be arranged in a sequence:
:$\AA = \set {A_1, A_2, \ldots}$
Let $B \in \BB$.
Consider the sequence of sets:
:$\sequence {A_1 \cup B, A_2 \cup B, \ldots}$
We may leave out any possible repetitions, and obtain a countable set:
:$\set {A \cup B: A \in \AA}$
for every $B \in \BB$.... | Let $\AA$ and $\BB$ be [[Definition:Countable Set|countable]] [[Definition:Set of Sets|sets of sets]].
Then:
:$\set {A \cup B: A \in \AA, B \in \BB}$
is also [[Definition:Countable Set|countable]]. | Since $\AA$ is [[Definition:Countable|countable]], its contents can be arranged in a [[Definition:Sequence|sequence]]:
:$\AA = \set {A_1, A_2, \ldots}$
Let $B \in \BB$.
Consider the [[Definition:Sequence|sequence]] of sets:
:$\sequence {A_1 \cup B, A_2 \cup B, \ldots}$
We may leave out any possible repetitions, and ... | Union of Countable Sets of Sets/Proof 1 | https://proofwiki.org/wiki/Union_of_Countable_Sets_of_Sets | https://proofwiki.org/wiki/Union_of_Countable_Sets_of_Sets/Proof_1 | [
"Set Union",
"Countable Sets",
"Union of Countable Sets of Sets"
] | [
"Definition:Countable Set",
"Definition:Set of Sets",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Sequence",
"Definition:Sequence",
"Definition:Countable Set",
"Definition:Countable Set",
"Definition:Indexing Set/Family of Sets",
"Countable Union of Countable Sets is Countable",
"Definition:Set Union",
"Definition:Countable Set"
] |
proofwiki-4949 | Union of Countable Sets of Sets | Let $\AA$ and $\BB$ be countable sets of sets.
Then:
:$\set {A \cup B: A \in \AA, B \in \BB}$
is also countable. | Both $\AA$ and $\BB$ are countable.
Enumerate their elements by the sequences $\sequence {A_n}_{n \mathop \in \N}$ and $\sequence {B_n}_{n \mathop \in \N}$ respectively.
When one of $\AA$ and $\BB$ would be finite, achieve such sequences by allowing elements to occur multiple times.
Define $\CC := \set {A \cup B: A \in... | Let $\AA$ and $\BB$ be [[Definition:Countable Set|countable]] [[Definition:Set of Sets|sets of sets]].
Then:
:$\set {A \cup B: A \in \AA, B \in \BB}$
is also [[Definition:Countable Set|countable]]. | Both $\AA$ and $\BB$ are [[Definition:Countable Set|countable]].
Enumerate their elements by the [[Definition:Sequence|sequences]] $\sequence {A_n}_{n \mathop \in \N}$ and $\sequence {B_n}_{n \mathop \in \N}$ respectively.
When one of $\AA$ and $\BB$ would be [[Definition:Finite Set|finite]], achieve such [[Definitio... | Union of Countable Sets of Sets/Proof 2 | https://proofwiki.org/wiki/Union_of_Countable_Sets_of_Sets | https://proofwiki.org/wiki/Union_of_Countable_Sets_of_Sets/Proof_2 | [
"Set Union",
"Countable Sets",
"Union of Countable Sets of Sets"
] | [
"Definition:Countable Set",
"Definition:Set of Sets",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Sequence",
"Definition:Finite Set",
"Definition:Sequence",
"Definition:Lexicographic Order",
"Well-Ordering Principle",
"Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering",
"Definition:Well-Ordering",
"Definition:Non-Empty Set",
"Definition:Sm... |
proofwiki-4950 | Union of Countable Sets of Sets | Let $\AA$ and $\BB$ be countable sets of sets.
Then:
:$\set {A \cup B: A \in \AA, B \in \BB}$
is also countable. | Since $\AA$ and $\BB$ are countable, $\AA \times \BB$ is countable by Cartesian Product of Countable Sets is Countable.
Thus by Surjection from Natural Numbers iff Countable there is a surjection $f: \N \to \AA \times \BB$.
Let $\CC = \set {A \cup B: A \in \AA, B \in \BB}$.
Let $g: \AA \times \BB \to \CC$ be defined by... | Let $\AA$ and $\BB$ be [[Definition:Countable Set|countable]] [[Definition:Set of Sets|sets of sets]].
Then:
:$\set {A \cup B: A \in \AA, B \in \BB}$
is also [[Definition:Countable Set|countable]]. | Since $\AA$ and $\BB$ are [[Definition:Countable|countable]], $\AA \times \BB$ is countable by [[Cartesian Product of Countable Sets is Countable]].
Thus by [[Surjection from Natural Numbers iff Countable]] there is a [[Definition:Surjection|surjection]] $f: \N \to \AA \times \BB$.
Let $\CC = \set {A \cup B: A \in \A... | Union of Countable Sets of Sets/Proof 3 | https://proofwiki.org/wiki/Union_of_Countable_Sets_of_Sets | https://proofwiki.org/wiki/Union_of_Countable_Sets_of_Sets/Proof_3 | [
"Set Union",
"Countable Sets",
"Union of Countable Sets of Sets"
] | [
"Definition:Countable Set",
"Definition:Set of Sets",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Cartesian Product of Countable Sets is Countable",
"Surjection from Natural Numbers iff Countable",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Cartesian Product",
"Definition:Surjection",
"Composite of Surjections is Surjection",
"Definition:Countable ... |
proofwiki-4951 | Set of Finite Subsets of Countable Set is Countable | Let $A$ be a countable set.
Then the set of finite subsets of $A$ is countable. | By the definition of a countable set, there exists an injection $g: A \to \N$.
Let $\FF$ denote the set of all finite subsets of $A$.
Let $f: \FF \to \N$ be the mapping defined by:
:$\ds \map f F = \prod_{k \mathop \in \map g F} p_{k + 1}$
where $p_n$ denotes the $n$th prime number.
We define $\map f \O = 1$, the vacuo... | Let $A$ be a [[Definition:Countable Set|countable set]].
Then the [[Definition:Set of Sets|set]] of [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$ is [[Definition:Countable Set|countable]]. | By the definition of a [[Definition:Countable Set|countable set]], there exists an [[Definition:Injection|injection]] $g: A \to \N$.
Let $\FF$ denote the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$.
Let $f: \FF \to \N$ be the [[Definition:Mapping|mappin... | Set of Finite Subsets of Countable Set is Countable/Proof 1 | https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable | https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable/Proof_1 | [
"Set of Finite Subsets of Countable Set is Countable",
"Subsets",
"Countable Sets"
] | [
"Definition:Countable Set",
"Definition:Set of Sets",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Injection",
"Definition:Set of Sets",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Mapping",
"Definition:Prime Number",
"Definition:Continued Product/Vacuous Product",
"Expression for Integer as Product of Primes is Unique",
"Subset equals Preim... |
proofwiki-4952 | Set of Finite Subsets of Countable Set is Countable | Let $A$ be a countable set.
Then the set of finite subsets of $A$ is countable. | By definition of a countable set, there is an injection $f: A \to \N$.
Let $\le_f$ be the ordering induced by $f$ on $A$.
By Injection Induces Well-Ordering, $\le_f$ is a well-ordering.
Let $A^{\text {fin} }$ be the set of finite subsets of $A$.
For $n \in \N$, denote with $A^{\paren n}$ the set of subsets of $A$ that ... | Let $A$ be a [[Definition:Countable Set|countable set]].
Then the [[Definition:Set of Sets|set]] of [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$ is [[Definition:Countable Set|countable]]. | By definition of a [[Definition:Countable Set|countable set]], there is an [[Definition:Injection|injection]] $f: A \to \N$.
Let $\le_f$ be the [[Definition:Ordering Induced by Injection|ordering induced by $f$ on $A$]].
By [[Injection Induces Well-Ordering]], $\le_f$ is a [[Definition:Well-Ordering|well-ordering]].
... | Set of Finite Subsets of Countable Set is Countable/Proof 2 | https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable | https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable/Proof_2 | [
"Set of Finite Subsets of Countable Set is Countable",
"Subsets",
"Countable Sets"
] | [
"Definition:Countable Set",
"Definition:Set of Sets",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Countable Set"
] | [
"Definition:Countable Set",
"Definition:Injection",
"Definition:Ordering Induced by Injection",
"Injection Induces Well-Ordering",
"Definition:Well-Ordering",
"Definition:Set of Sets",
"Definition:Finite Subset",
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Element",
"Definition:Pr... |
proofwiki-4953 | Set of Finite Subsets of Countable Set is Countable | Let $A$ be a countable set.
Then the set of finite subsets of $A$ is countable. | Let $A^{\paren n}$ be the set of subsets of $A$ with no more than $n$ elements.
Thus:
:$A^{\paren 0} = \set \O$
:$A^{\paren 1} = A^{\paren 0} \cup \set {\set a: a \in A}$
and $\forall n \ge 0$:
:$A^{\paren {n + 1} } = \set {a^{\paren n} \cup a^{\paren 1}: a^{\paren n} \in A^{\paren n} \land a^{\paren 1} \in A^{\paren 1... | Let $A$ be a [[Definition:Countable Set|countable set]].
Then the [[Definition:Set of Sets|set]] of [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$ is [[Definition:Countable Set|countable]]. | Let $A^{\paren n}$ be the set of [[Definition:Subset|subsets]] of $A$ with no more than $n$ [[Definition:Element|elements]].
Thus:
:$A^{\paren 0} = \set \O$
:$A^{\paren 1} = A^{\paren 0} \cup \set {\set a: a \in A}$
and $\forall n \ge 0$:
:$A^{\paren {n + 1} } = \set {a^{\paren n} \cup a^{\paren 1}: a^{\paren n} \in ... | Set of Finite Subsets of Countable Set is Countable/Proof 3 | https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable | https://proofwiki.org/wiki/Set_of_Finite_Subsets_of_Countable_Set_is_Countable/Proof_3 | [
"Set of Finite Subsets of Countable Set is Countable",
"Subsets",
"Countable Sets"
] | [
"Definition:Countable Set",
"Definition:Set of Sets",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Countable Set"
] | [
"Definition:Subset",
"Definition:Element",
"Principle of Mathematical Induction",
"Definition:Countable Set",
"Definition:Countable Set",
"Definition:Countable Set",
"Definition:Cardinality",
"Definition:Countable Set",
"Union of Countable Sets of Sets",
"Definition:Countable Set",
"Principle of... |
proofwiki-4954 | Cantor's Theorem (Strong Version) | Let $S$ be a set.
Let $\map {\PP^n} S$ be defined recursively by:
:<nowiki>$\map {\PP^n} S = \begin{cases}
S & : n = 0 \\
\powerset {\map {\PP^{n - 1} } S} & : n > 0
\end{cases}$</nowiki>
where $\powerset S$ denotes the power set of $S$.
Then $S$ is not equivalent to $\map {\PP^n} S$ for any $n > 0$. | We temporarily introduce the notation:
:<nowiki>$a^n = \begin{cases}
a : & n = 0 \\
\set {a^{n - 1} } : & n > 0
\end{cases}$</nowiki>
where $a \in S$.
Thus $a^n$ consists of a single element $a^{n - 1} \in \map {\PP^{n - 1} } S$.
Let there be a bijection $f: S \to \QQ^n$ where $\QQ^n \subseteq \map {\PP^n} S$.
Then def... | Let $S$ be a [[Definition:Set|set]].
Let $\map {\PP^n} S$ be defined [[Principle of Recursive Definition|recursively]] by:
:<nowiki>$\map {\PP^n} S = \begin{cases}
S & : n = 0 \\
\powerset {\map {\PP^{n - 1} } S} & : n > 0
\end{cases}$</nowiki>
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.... | We temporarily introduce the notation:
:<nowiki>$a^n = \begin{cases}
a : & n = 0 \\
\set {a^{n - 1} } : & n > 0
\end{cases}$</nowiki>
where $a \in S$.
Thus $a^n$ consists of a single element $a^{n - 1} \in \map {\PP^{n - 1} } S$.
Let there be a [[Definition:Bijection|bijection]] $f: S \to \QQ^n$ where $\QQ^n \subsete... | Cantor's Theorem (Strong Version)/Proof 1 | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version) | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version)/Proof_1 | [
"Power Set",
"Cantor's Theorem"
] | [
"Definition:Set",
"Principle of Recursive Definition",
"Definition:Power Set",
"Definition:Set Equivalence"
] | [
"Definition:Bijection",
"Definition:Image (Set Theory)/Mapping/Element",
"Definition:Element",
"Definition:Proper Subset"
] |
proofwiki-4955 | Cantor's Theorem (Strong Version) | Let $S$ be a set.
Let $\map {\PP^n} S$ be defined recursively by:
:<nowiki>$\map {\PP^n} S = \begin{cases}
S & : n = 0 \\
\powerset {\map {\PP^{n - 1} } S} & : n > 0
\end{cases}$</nowiki>
where $\powerset S$ denotes the power set of $S$.
Then $S$ is not equivalent to $\map {\PP^n} S$ for any $n > 0$. | The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:There is no surjection from $S$ onto $\map {\PP^n} S$.
=== Basis for the Induction ===
$\map P 1$ is Cantor's Theorem.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ ... | Let $S$ be a [[Definition:Set|set]].
Let $\map {\PP^n} S$ be defined [[Principle of Recursive Definition|recursively]] by:
:<nowiki>$\map {\PP^n} S = \begin{cases}
S & : n = 0 \\
\powerset {\map {\PP^{n - 1} } S} & : n > 0
\end{cases}$</nowiki>
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:There is no [[Definition:Surjection|surjection]] from $S$ onto $\map {\PP^n} S$.
=== Basis for the Induction ===
$\map P 1$ is [[Cantor's Theorem]].
This... | Cantor's Theorem (Strong Version)/Proof 2 | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version) | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version)/Proof_2 | [
"Power Set",
"Cantor's Theorem"
] | [
"Definition:Set",
"Principle of Recursive Definition",
"Definition:Power Set",
"Definition:Set Equivalence"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Surjection",
"Cantor's Theorem",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Surjection",
"Definition:Surjection",
"Cantor's Theorem (Strong Version)/Proof 2/Induction Step"
] |
proofwiki-4956 | Cantor's Theorem (Strong Version) | Let $S$ be a set.
Let $\map {\PP^n} S$ be defined recursively by:
:<nowiki>$\map {\PP^n} S = \begin{cases}
S & : n = 0 \\
\powerset {\map {\PP^{n - 1} } S} & : n > 0
\end{cases}$</nowiki>
where $\powerset S$ denotes the power set of $S$.
Then $S$ is not equivalent to $\map {\PP^n} S$ for any $n > 0$. | By Power Set is Nonempty, $\powerset {\map {\PP^k} S}$ is non-empty.
By definition:
:$\map {\PP^{k + 1} } S = \powerset {\map {\PP^k} S}$
Then:
:$\map {\PP^{k + 1} } S \ne \O$
By Law of Excluded Middle, there are two choices:
:$S = \O$
or:
:$S \ne \O$
Suppose that $S = \O$.
By {{Corollary|Image of Empty Set is Empty S... | Let $S$ be a [[Definition:Set|set]].
Let $\map {\PP^n} S$ be defined [[Principle of Recursive Definition|recursively]] by:
:<nowiki>$\map {\PP^n} S = \begin{cases}
S & : n = 0 \\
\powerset {\map {\PP^{n - 1} } S} & : n > 0
\end{cases}$</nowiki>
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.... | By [[Power Set is Nonempty]], $\powerset {\map {\PP^k} S}$ is [[Definition:Non-Empty Set|non-empty]].
By definition:
:$\map {\PP^{k + 1} } S = \powerset {\map {\PP^k} S}$
Then:
:$\map {\PP^{k + 1} } S \ne \O$
By [[Law of Excluded Middle]], there are two choices:
:$S = \O$
or:
:$S \ne \O$
Suppose that $S = \O$.
... | Cantor's Theorem (Strong Version)/Proof 2/Induction Step/Proof 2 | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version) | https://proofwiki.org/wiki/Cantor's_Theorem_(Strong_Version)/Proof_2/Induction_Step/Proof_2 | [
"Power Set",
"Cantor's Theorem"
] | [
"Definition:Set",
"Principle of Recursive Definition",
"Definition:Power Set",
"Definition:Set Equivalence"
] | [
"Power Set is Nonempty",
"Definition:Non-Empty Set",
"Law of Excluded Middle",
"Definition:Surjection",
"Cantor's Theorem (Strong Version)/Proof 2",
"Definition:Surjection",
"Injection from Set to Power Set",
"Definition:Injection",
"Injection has Surjective Left Inverse Mapping",
"Definition:Surj... |
proofwiki-4957 | Ordering Induced by Injection is Ordering | Let $\left({T, \le}\right)$ be an ordered set, and let $S$ be a set.
Let $f: S \to T$ be an injection.
Then $\le_f$, the ordering induced by $f$, is an ordering. | To establish that $\le_f$ is an ordering on $S$, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering: | Let $\left({T, \le}\right)$ be an [[Definition:Ordered Set|ordered set]], and let $S$ be a [[Definition:Set|set]].
Let $f: S \to T$ be an [[Definition:Injection|injection]].
Then $\le_f$, the [[Definition:Ordering Induced by Injection|ordering induced by $f$]], is an [[Definition:Ordering|ordering]]. | To establish that $\le_f$ is an [[Definition:Ordering|ordering]] on $S$, we need to show that it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Antisymmetric Relation|antisymmetric]] and [[Definition:Transitive Relation|transitive]].
So, checking in turn each of the criteria for an [[Definition:Ordering|... | Ordering Induced by Injection is Ordering | https://proofwiki.org/wiki/Ordering_Induced_by_Injection_is_Ordering | https://proofwiki.org/wiki/Ordering_Induced_by_Injection_is_Ordering | [
"Order Theory",
"Injections"
] | [
"Definition:Ordered Set",
"Definition:Set",
"Definition:Injection",
"Definition:Ordering Induced by Injection",
"Definition:Ordering"
] | [
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Antisymmetric Relation",
"Definition:Transitive Relation",
"Definition:Ordering",
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Ordering",
"Defini... |
proofwiki-4958 | Injection Induces Total Ordering | Let $\left({T, \le}\right)$ be a totally ordered set, and let $S$ be a set.
Let $f: S \to T$ be an injection.
Then $\le_f$, the ordering induced by $f$, is a total ordering. | By Ordering Induced by Injection is Ordering, $\le_f$ is an ordering.
Let $s_1, s_2 \in S$.
Then as $\le$ is a total ordering, at least one of the following holds:
:$(1): \qquad f \left({s_1}\right) \le f \left({s_2}\right)$
:$(2): \qquad f \left({s_2}\right) \le f \left({s_1}\right)$
If $(1)$ holds, the definition of ... | Let $\left({T, \le}\right)$ be a [[Definition:Toset|totally ordered set]], and let $S$ be a [[Definition:Set|set]].
Let $f: S \to T$ be an [[Definition:Injection|injection]].
Then $\le_f$, the [[Definition:Ordering Induced by Injection|ordering induced by $f$]], is a [[Definition:Total Ordering|total ordering]]. | By [[Ordering Induced by Injection is Ordering]], $\le_f$ is an [[Definition:Ordering|ordering]].
Let $s_1, s_2 \in S$.
Then as $\le$ is a [[Definition:Total Ordering|total ordering]], at least one of the following holds:
:$(1): \qquad f \left({s_1}\right) \le f \left({s_2}\right)$
:$(2): \qquad f \left({s_2}\right)... | Injection Induces Total Ordering | https://proofwiki.org/wiki/Injection_Induces_Total_Ordering | https://proofwiki.org/wiki/Injection_Induces_Total_Ordering | [
"Total Orderings",
"Injections"
] | [
"Symbols:Abbreviations/T/Toset",
"Definition:Set",
"Definition:Injection",
"Definition:Ordering Induced by Injection",
"Definition:Total Ordering"
] | [
"Ordering Induced by Injection is Ordering",
"Definition:Ordering",
"Definition:Total Ordering",
"Definition:Ordering Induced by Injection",
"Definition:Total Ordering",
"Category:Total Orderings",
"Category:Injections"
] |
proofwiki-4959 | Injection Induces Well-Ordering | Let $\left({T, \le}\right)$ be a well-ordered set.
Let $S$ be a set.
Let $f: S \to T$ be an injection.
Then $\le_f$, the ordering induced by $f$, is a well-ordering. | By Ordering Induced by Injection is Ordering, $\le_f$ is an ordering.
Let $S' \subseteq S$ be non-empty.
Then as $\le$ is a well-ordering, the set:
:$f \left({S'}\right) = \left\{{f \left({s}\right): s \in S'}\right\}$
has a minimal element, say $f \left({s_0}\right)$.
That is, for all $s \in S', f \left({s_0}\right) \... | Let $\left({T, \le}\right)$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Let $S$ be a [[Definition:Set|set]].
Let $f: S \to T$ be an [[Definition:Injection|injection]].
Then $\le_f$, the [[Definition:Ordering Induced by Injection|ordering induced by $f$]], is a [[Definition:Well-Ordering|well-ordering]]. | By [[Ordering Induced by Injection is Ordering]], $\le_f$ is an [[Definition:Ordering|ordering]].
Let $S' \subseteq S$ be [[Definition:Non-Empty Set|non-empty]].
Then as $\le$ is a [[Definition:Well-Ordering|well-ordering]], the set:
:$f \left({S'}\right) = \left\{{f \left({s}\right): s \in S'}\right\}$
has a [[Def... | Injection Induces Well-Ordering | https://proofwiki.org/wiki/Injection_Induces_Well-Ordering | https://proofwiki.org/wiki/Injection_Induces_Well-Ordering | [
"Well-Orderings",
"Injections"
] | [
"Definition:Well-Ordered Set",
"Definition:Set",
"Definition:Injection",
"Definition:Ordering Induced by Injection",
"Definition:Well-Ordering"
] | [
"Ordering Induced by Injection is Ordering",
"Definition:Ordering",
"Definition:Non-Empty Set",
"Definition:Well-Ordering",
"Definition:Minimal/Element",
"Definition:Ordering Induced by Injection",
"Definition:Minimal/Element",
"Definition:Well-Ordering",
"Category:Well-Orderings",
"Category:Injec... |
proofwiki-4960 | K-Connectivity Implies Lesser Connectivity | Let $G = \struct {V, E}$ be a $k$-connected graph.
Then $G$ is $l$-connected for all $l \in \Z : 0 < l < k$. | Suppose that $G$ is $k$-connected.
Then:
:$\card V > k > l$
:$G$ is connected
:If $W$ is a vertex cut of $G$, then $\card W \ge k > l$ so $\card W \ge l$.
{{qed}}
Category:Connectedness (Graph Theory)
syax1man5j7s7nb575ekb7dgjj4g2gn | Let $G = \struct {V, E}$ be a [[Definition:K-Connected|$k$-connected]] [[Definition:Graph (Graph Theory)|graph]].
Then $G$ is [[Definition:K-Connected|$l$-connected]] for all $l \in \Z : 0 < l < k$. | Suppose that $G$ is [[Definition:K-Connected|$k$-connected]].
Then:
:$\card V > k > l$
:$G$ is [[Definition:Connected Graph|connected]]
:If $W$ is a [[Definition:Vertex Cut|vertex cut]] of $G$, then $\card W \ge k > l$ so $\card W \ge l$.
{{qed}}
[[Category:Connectedness (Graph Theory)]]
syax1man5j7s7nb575ekb7dgjj4g... | K-Connectivity Implies Lesser Connectivity | https://proofwiki.org/wiki/K-Connectivity_Implies_Lesser_Connectivity | https://proofwiki.org/wiki/K-Connectivity_Implies_Lesser_Connectivity | [
"Connectedness (Graph Theory)"
] | [
"Definition:K-Connected",
"Definition:Graph (Graph Theory)",
"Definition:K-Connected"
] | [
"Definition:K-Connected",
"Definition:Connected (Graph Theory)/Graph",
"Definition:Vertex Cut",
"Category:Connectedness (Graph Theory)"
] |
proofwiki-4961 | Hartogs' Lemma (Set Theory) | Let $S$ be a set.
Then there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$. | Define $\alpha = \set {\beta: \text{$\beta$ is an ordinal and there is an injection $\beta \to S$}}$.
First of all, it is to be shown that $\alpha$ is a set.
To this end, define the set $W$ by:
:$W = \set {\paren {S', \preceq}: \text{$S' \subseteq S$ and $\preceq$ well-orders $S'$}}$
By the Counting Theorem, each $w \i... | Let $S$ be a [[Definition:Set|set]].
Then there exists an [[Definition:Ordinal Number|ordinal]] $\alpha$ such that there is no [[Definition:Injection|injection]] from $\alpha$ to $S$. | Define $\alpha = \set {\beta: \text{$\beta$ is an ordinal and there is an injection $\beta \to S$}}$.
First of all, it is to be shown that $\alpha$ is a [[Definition:Set|set]].
To this end, define the [[Definition:Set|set]] $W$ by:
:$W = \set {\paren {S', \preceq}: \text{$S' \subseteq S$ and $\preceq$ well-orders $... | Hartogs' Lemma (Set Theory)/Proof 1 | https://proofwiki.org/wiki/Hartogs'_Lemma_(Set_Theory) | https://proofwiki.org/wiki/Hartogs'_Lemma_(Set_Theory)/Proof_1 | [
"Hartogs' Lemma (Set Theory)",
"Set Theory",
"Ordinals"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Injection"
] | [
"Definition:Set",
"Definition:Set",
"Counting Theorem",
"Definition:Ordinal",
"Axiom:Axiom of Replacement",
"Definition:Set",
"Injection Induces Well-Ordering",
"Definition:Set",
"Definition:Set",
"Definition:Ordinal",
"Definition:Injection",
"Definition:Inclusion Mapping",
"Composite of Inj... |
proofwiki-4962 | Hartogs' Lemma (Set Theory) | Let $S$ be a set.
Then there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$. | Let $W$ be the set of all well-orderings on subsets of $S$.
By the Counting Theorem, there exists a mapping $F: W \to \On$ defined by letting $\map F s$ be the ordinal which is isomorphic to $s$.
{{explain|"isomorphism" between ordinals}}
By Mapping from Set to Class of All Ordinals is Bounded Above, $F \sqbrk W$ has a... | Let $S$ be a [[Definition:Set|set]].
Then there exists an [[Definition:Ordinal Number|ordinal]] $\alpha$ such that there is no [[Definition:Injection|injection]] from $\alpha$ to $S$. | Let $W$ be the [[Definition:Set|set]] of all [[Definition:Well-Ordering|well-orderings]] on [[Definition:Subset|subsets]] of $S$.
By the [[Counting Theorem]], there exists a [[Definition:Mapping|mapping]] $F: W \to \On$ defined by letting $\map F s$ be the [[Definition:Ordinal|ordinal]] which is isomorphic to $s$.
{{... | Hartogs' Lemma (Set Theory)/Proof 2 | https://proofwiki.org/wiki/Hartogs'_Lemma_(Set_Theory) | https://proofwiki.org/wiki/Hartogs'_Lemma_(Set_Theory)/Proof_2 | [
"Hartogs' Lemma (Set Theory)",
"Set Theory",
"Ordinals"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Injection"
] | [
"Definition:Set",
"Definition:Well-Ordering",
"Definition:Subset",
"Counting Theorem",
"Definition:Mapping",
"Definition:Ordinal",
"Mapping from Set to Class of All Ordinals is Bounded Above",
"Definition:Upper Bound of Set",
"Definition:Injection",
"Injection to Image is Bijection",
"Definition... |
proofwiki-4963 | Bourbaki-Witt Fixed Point Theorem | Let $\struct {X, \le}$ be a non-empty chain complete ordered set (that is, an ordered set in which every chain has a supremum).
Let $f: X \to X$ be an inflationary mapping, that is, so that $\map f x \ge x$.
Then for every $x \in X$ there exists $y \in X$ where $y \ge x$ such that $\map f y = y$. | Let $\gamma$ be the Hartogs number of $X$.
Suppose that $x \in X$.
Define $g : \gamma \to X$ by transfinite recursion as follows:
:$\map g 0 = x$
:$\map g {\alpha + 1} = \map f {\map g \alpha}$
:$\map g \alpha = \sup \set {\map g \beta: \beta < \alpha}$ when $\alpha$ is a limit ordinal.
That $f$ is inflationary guaran... | Let $\struct {X, \le}$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Chain Complete Set|chain complete]] [[Definition:Ordered Set|ordered set]] (that is, an [[Definition:Ordered Set|ordered set]] in which every [[Definition:Chain (Order Theory)|chain]] has a [[Definition:Supremum of Set|supremum]]).
Let $f:... | Let $\gamma$ be the [[Definition:Hartogs Number|Hartogs number]] of $X$.
Suppose that $x \in X$.
Define $g : \gamma \to X$ by [[Principle of Transfinite Recursion|transfinite recursion]] as follows:
:$\map g 0 = x$
:$\map g {\alpha + 1} = \map f {\map g \alpha}$
:$\map g \alpha = \sup \set {\map g \beta: \beta < \... | Bourbaki-Witt Fixed Point Theorem | https://proofwiki.org/wiki/Bourbaki-Witt_Fixed_Point_Theorem | https://proofwiki.org/wiki/Bourbaki-Witt_Fixed_Point_Theorem | [
"Set Theory",
"Fixed Point Theorems"
] | [
"Definition:Non-Empty Set",
"Definition:Inductive Ordered Set",
"Definition:Ordered Set",
"Definition:Ordered Set",
"Definition:Chain (Order Theory)",
"Definition:Supremum of Set",
"Definition:Inflationary Mapping"
] | [
"Definition:Hartogs Number",
"Transfinite Recursion Theorem",
"Definition:Limit Ordinal",
"Definition:Inflationary Mapping",
"Definition:Chain (Order Theory)",
"Definition:Increasing/Mapping",
"Definition:Strictly Increasing/Mapping",
"Definition:Strictly Monotone/Mapping",
"Strictly Monotone Mappin... |
proofwiki-4964 | Convergent Sequence with Finite Number of Terms Deleted is Convergent | Let $\struct {X, d}$ be a metric space.
Let $\sequence {x_k}$ be a sequence in $X$.
Let $\sequence {x_k}$ be convergent.
Let a finite number of terms be deleted from $\sequence {x_k}$.
Then the resulting subsequence is convergent. | Suppose the sequence $\sequence {x_k}$ converges to $x \in X$.
That is for every $\epsilon > 0$ there is some index $N$ such that $\map d {x_n, x} < \epsilon$ for all $n \ge N$.
The same $N$ will work for the new sequence with finitely many terms removed, so the new sequence converges to the same point $x$ as the origi... | Let $\struct {X, d}$ be a [[Definition:Metric Space|metric space]].
Let $\sequence {x_k}$ be a [[Definition:Sequence|sequence in $X$]].
Let $\sequence {x_k}$ be [[Definition:Convergent Sequence (Metric Space)|convergent]].
Let a [[Definition:Finite Set|finite]] number of [[Definition:Term of Sequence|terms]] be del... | Suppose the sequence $\sequence {x_k}$ converges to $x \in X$.
That is for every $\epsilon > 0$ there is some index $N$ such that $\map d {x_n, x} < \epsilon$ for all $n \ge N$.
The same $N$ will work for the new sequence with finitely many terms removed, so the new sequence converges to the same point $x$ as the ori... | Convergent Sequence with Finite Number of Terms Deleted is Convergent | https://proofwiki.org/wiki/Convergent_Sequence_with_Finite_Number_of_Terms_Deleted_is_Convergent | https://proofwiki.org/wiki/Convergent_Sequence_with_Finite_Number_of_Terms_Deleted_is_Convergent | [
"Series",
"Sequences",
"Convergence Tests"
] | [
"Definition:Metric Space",
"Definition:Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Finite Set",
"Definition:Term of Sequence",
"Definition:Subsequence",
"Definition:Convergent Sequence/Metric Space"
] | [] |
proofwiki-4965 | Cauchy Condensation Test | Let $\sequence {a_n}: n \mapsto \map a n$ be a decreasing sequence of strictly positive terms in $\R$ which converges with a limit of zero.
That is, for every $n$ in the domain of $\sequence {a_n}$: $a_n > 0$, $a_n \ge a_{n + 1}$, and $a_n \to 0$ as $n \to +\infty$.
Then the series:
:$\ds \sum_{n \mathop = 1}^\infty a_... | === Necessary Condition ===
We will first show that:
:if the condensed series $\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges
then:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ converges as well.
Assume $\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges.
Consider the graph of $\sequence {a_n}$ and the ... | Let $\sequence {a_n}: n \mapsto \map a n$ be a [[Definition:Decreasing Sequence|decreasing sequence]] of [[Definition:Strictly Positive|strictly positive]] terms in $\R$ which [[Definition:Convergent Sequence|converges]] with a [[Definition:Limit of Sequence (Number Field)|limit]] of zero.
That is, for every $n$ in th... | === Necessary Condition ===
We will first show that:
:if the [[Definition:Condensed Series|condensed series]] $\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ [[Definition:Convergent Series|converges]]
then:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Convergent Series|converges]] as well.
Assume $\ds \sum... | Cauchy Condensation Test | https://proofwiki.org/wiki/Cauchy_Condensation_Test | https://proofwiki.org/wiki/Cauchy_Condensation_Test | [
"Series",
"Convergence Tests",
"Analytic Geometry"
] | [
"Definition:Decreasing/Sequence",
"Definition:Strictly Positive",
"Definition:Convergent Sequence",
"Definition:Limit of Sequence (Number Field)",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Series",
"Definition:Convergent Series",
"Definition:Condensed Series",
"Definition:Convergent Seri... | [
"Definition:Condensed Series",
"Definition:Convergent Series",
"Definition:Convergent Series",
"Definition:Convergent Series",
"Definition:Graph of Mapping",
"Definition:Series/Sequence of Partial Sums",
"File:Cauchycondensation1.png",
"Definition:Sequence",
"Definition:Quadrilateral/Rectangle",
"... |
proofwiki-4966 | Between Two Sets Exists Injection or Surjection | Let $S$ and $T$ be sets.
Then either or both of the following cases hold:
:$(1):$ There exists a mapping $f: S \to T$ such that $f$ is an injection
:$(2):$ There exists a mapping $f: S \to T$ such that $f$ is a surjection. | From Zermelo's Theorem, exactly one of the following cases holds:
:$(1): \quad S \le T$, that is, $T$ dominates $S$
:$(2): \quad S \equiv T$, that is, $T$ is equivalent to $S$
:$(3): \quad S \ge T$, that is, $S$ dominates $T$.
Suppose $(1)$ holds, that is, $T$ dominates $S$.
By definition, there exists an injection fro... | Let $S$ and $T$ be [[Definition:Set|sets]].
Then either or both of the following cases hold:
:$(1):$ There exists a [[Definition:Mapping|mapping]] $f: S \to T$ such that $f$ is an [[Definition:Injection|injection]]
:$(2):$ There exists a [[Definition:Mapping|mapping]] $f: S \to T$ such that $f$ is a [[Definition:Surj... | From [[Zermelo's Theorem (Set Theory)|Zermelo's Theorem]], exactly one of the following cases holds:
:$(1): \quad S \le T$, that is, $T$ [[Definition:Dominate (Set Theory)|dominates]] $S$
:$(2): \quad S \equiv T$, that is, $T$ is [[Definition:Set Equivalence|equivalent]] to $S$
:$(3): \quad S \ge T$, that is, $S$ [[Def... | Between Two Sets Exists Injection or Surjection | https://proofwiki.org/wiki/Between_Two_Sets_Exists_Injection_or_Surjection | https://proofwiki.org/wiki/Between_Two_Sets_Exists_Injection_or_Surjection | [
"Injections",
"Surjections"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Injection",
"Definition:Mapping",
"Definition:Surjection"
] | [
"Zermelo's Theorem (Set Theory)",
"Definition:Dominate (Set Theory)",
"Definition:Set Equivalence",
"Definition:Dominate (Set Theory)",
"Definition:Dominate (Set Theory)",
"Definition:Injection",
"Definition:Set Equivalence",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Mapping",
... |
proofwiki-4967 | Injection from Set to Power Set | For every set $S$, there exists an injection from $S$ to its power set $\powerset S$. | If $S = \O$, the empty mapping suffices, as it is vacuously an injection.
Let $f: S \to \powerset S$ be the mapping defined as:
:$\forall s \in S: \map f s = \set s$
Let $s, t \in S$ such that $\map f s = \map f t$.
Then $\set s = \set t$.
By definition of set equality it follows directly that $s = t$.
Hence $f$ is the... | For every set $S$, there exists an [[Definition:Injection|injection]] from $S$ to its [[Definition:Power Set|power set]] $\powerset S$. | If $S = \O$, the [[Definition:Empty Mapping|empty mapping]] suffices, as it is [[Definition:Vacuous Truth|vacuously]] an [[Definition:Injection|injection]].
Let $f: S \to \powerset S$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall s \in S: \map f s = \set s$
Let $s, t \in S$ such that $\map f s = \map f... | Injection from Set to Power Set | https://proofwiki.org/wiki/Injection_from_Set_to_Power_Set | https://proofwiki.org/wiki/Injection_from_Set_to_Power_Set | [
"Injections",
"Power Set"
] | [
"Definition:Injection",
"Definition:Power Set"
] | [
"Definition:Empty Mapping",
"Definition:Vacuous Truth",
"Definition:Injection",
"Definition:Mapping",
"Definition:Set Equality",
"Definition:Injection",
"Category:Injections",
"Category:Power Set"
] |
proofwiki-4968 | Minimum Degree is at Least Connectivity | Let $G = \struct {V, E}$ be a simple graph.
Then:
:$\map \delta G \ge \map \kappa G$
That is, the minimum degree of $G$ is at least its connectivity. | Pick a vertex $v \in G$ with $\map {\deg_G} v = \map \delta G$, that is, a vertex with minimum degree.
Recall that $\map {\Gamma_G} v$ is the neighborhood of $v$ in $G$.
Suppose that $V = \map {\Gamma_G} v \cup \set v$.
That is, that $v$ is adjacent to all other vertices of $G$.
Then:
:$\card V = \card {\map {\Gamma_G}... | Let $G = \struct {V, E}$ be a [[Definition:Simple Graph|simple graph]].
Then:
:$\map \delta G \ge \map \kappa G$
That is, the [[Definition:Minimum Degree|minimum degree]] of $G$ is at least its [[Definition:Connectivity|connectivity]]. | Pick a [[Definition:Vertex of Graph|vertex]] $v \in G$ with $\map {\deg_G} v = \map \delta G$, that is, a [[Definition:Vertex of Graph|vertex]] with minimum degree.
Recall that $\map {\Gamma_G} v$ is the [[Definition:Neighborhood (Graph Theory)|neighborhood]] of $v$ in $G$.
Suppose that $V = \map {\Gamma_G} v \cup \... | Minimum Degree is at Least Connectivity | https://proofwiki.org/wiki/Minimum_Degree_is_at_Least_Connectivity | https://proofwiki.org/wiki/Minimum_Degree_is_at_Least_Connectivity | [
"Connectedness (Graph Theory)"
] | [
"Definition:Simple Graph",
"Definition:Minimum Degree",
"Definition:Connectivity"
] | [
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Neighborhood (Graph Theory)",
"Definition:Vertex Deletion",
"Definition:Isolated Vertex",
"Definition:Vertex Cut",
"Definition:Vertex Cut"
] |
proofwiki-4969 | Homeomorphism Relation is Equivalence | Let $T_1$ and $T_2$ be topological spaces.
Let $T_1 \sim T_2$ denote that $T_1$ and $T_2$ are homeomorphic.
The relation $\sim$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $T_1 \sim T_2$ denote that $T_1$ and $T_2$ are [[Definition:Homeomorphism (Topological Spaces)|homeomorphic]].
The relation $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]]. | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Homeomorphism Relation is Equivalence | https://proofwiki.org/wiki/Homeomorphism_Relation_is_Equivalence | https://proofwiki.org/wiki/Homeomorphism_Relation_is_Equivalence | [
"Homeomorphisms (Topological Spaces)",
"Examples of Equivalence Relations"
] | [
"Definition:Topological Space",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-4970 | Identity Mapping is Homeomorphism | Let $T$ be a topological space.
The identity mapping $I_T: T \to T$ defined as:
:$\forall x \in T: \map {I_T} x = x$
is a homeomorphism. | We have Identity Mapping is Bijection.
We also have Identity Mapping is Continuous.
Hence, by definition, $I_T$ is a homeomorphism.
{{qed}}
Category:Homeomorphisms (Topological Spaces)
Category:Identity Mappings
onel9o918gu70i9ncm8rbow9peprcdu | Let $T$ be a [[Definition:Topological Space|topological space]].
The [[Definition:Identity Mapping|identity mapping]] $I_T: T \to T$ defined as:
:$\forall x \in T: \map {I_T} x = x$
is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]]. | We have [[Identity Mapping is Bijection]].
We also have [[Identity Mapping is Continuous]].
Hence, by definition, $I_T$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
{{qed}}
[[Category:Homeomorphisms (Topological Spaces)]]
[[Category:Identity Mappings]]
onel9o918gu70i9ncm8rbow9peprcdu | Identity Mapping is Homeomorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Homeomorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Homeomorphism | [
"Homeomorphisms (Topological Spaces)",
"Identity Mappings"
] | [
"Definition:Topological Space",
"Definition:Identity Mapping",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Identity Mapping is Bijection",
"Identity Mapping is Continuous",
"Definition:Homeomorphism/Topological Spaces",
"Category:Homeomorphisms (Topological Spaces)",
"Category:Identity Mappings"
] |
proofwiki-4971 | Composite of Homeomorphisms is Homeomorphism | Let $T_1, T_2, T_3$ be topological spaces.
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be homeomorphisms.
Then $g \circ f: T_1 \to T_3$ is also a homeomorphism. | By definition of homeomorphism, $f$ and $g$ are both bijections.
From Composite of Bijections is Bijection it follows that $g \circ f$ is also a bijection.
By definition of homeomorphism, $f$ and $g$ are both continuous mappings.
From Composite of Continuous Mappings is Continuous it follows that $g \circ f$ is also a ... | Let $T_1, T_2, T_3$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be [[Definition:Homeomorphism (Topological Spaces)|homeomorphisms]].
Then $g \circ f: T_1 \to T_3$ is also a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]]. | By definition of [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], $f$ and $g$ are both [[Definition:Bijection|bijections]].
From [[Composite of Bijections is Bijection]] it follows that $g \circ f$ is also a [[Definition:Bijection|bijection]].
By definition of [[Definition:Homeomorphism (Topological ... | Composite of Homeomorphisms is Homeomorphism | https://proofwiki.org/wiki/Composite_of_Homeomorphisms_is_Homeomorphism | https://proofwiki.org/wiki/Composite_of_Homeomorphisms_is_Homeomorphism | [
"Homeomorphisms (Topological Spaces)",
"Composite Mappings"
] | [
"Definition:Topological Space",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Definition:Homeomorphism/Topological Spaces",
"Definition:Bijection",
"Composite of Bijections is Bijection",
"Definition:Bijection",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Composite of Continuous Mappings is Continuous",
"Definition:Con... |
proofwiki-4972 | Identity Mapping is Continuous | Let $T = \struct {S, \tau}$ be a topological space.
The identity mapping $I_S: S \to S$ defined as:
:$\forall x \in S: \map {I_S} x = x$
is a continuous mapping. | Let $U \in \tau$.
We have Identity Mapping is Bijection.
So $I_S^{-1}$ is well-defined and:
:$\forall x \in U: \map {I_S^{-1} } x = x$
Thus $I_S^{-1} \sqbrk U = U \in \tau$.
Hence, by definition of continuous mapping, $I_S$ is continuous.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
The [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ defined as:
:$\forall x \in S: \map {I_S} x = x$
is a [[Definition:Continuous Mapping (Topology)|continuous mapping]]. | Let $U \in \tau$.
We have [[Identity Mapping is Bijection]].
So $I_S^{-1}$ is [[Definition:Well-Defined Mapping|well-defined]] and:
:$\forall x \in U: \map {I_S^{-1} } x = x$
Thus $I_S^{-1} \sqbrk U = U \in \tau$.
Hence, by definition of [[Definition:Continuous Mapping (Topology)|continuous mapping]], $I_S$ is [[De... | Identity Mapping is Continuous | https://proofwiki.org/wiki/Identity_Mapping_is_Continuous | https://proofwiki.org/wiki/Identity_Mapping_is_Continuous | [
"Continuous Mappings",
"Identity Mappings",
"Identity Mapping is Continuous"
] | [
"Definition:Topological Space",
"Definition:Identity Mapping",
"Definition:Continuous Mapping (Topology)"
] | [
"Identity Mapping is Bijection",
"Definition:Well-Defined/Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-4973 | Clopen Sets in Finite Complement Topology | Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then the only clopen sets of $T$ are $S$ and $\O$. | Let $U \in \tau$ be open in $T$.
Then by definition of finite complement topology, $S \setminus U$ is finite.
By definition of open set, $S \setminus U$ is closed.
As $S$ is infinite, it follows that $U$ must also be infinite.
Thus unless $U = S$, $S \setminus U$ can not be open.
Hence the result.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
Then the only [[Definition:Clopen Set|clopen]] sets of $T$ are $S$ and $\O$. | Let $U \in \tau$ be [[Definition:Open Set (Topology)|open]] in $T$.
Then by definition of [[Definition:Finite Complement Topology|finite complement topology]], $S \setminus U$ is [[Definition:Finite Set|finite]].
By definition of [[Definition:Open Set (Topology)|open set]], $S \setminus U$ is [[Definition:Closed Set ... | Clopen Sets in Finite Complement Topology | https://proofwiki.org/wiki/Clopen_Sets_in_Finite_Complement_Topology | https://proofwiki.org/wiki/Clopen_Sets_in_Finite_Complement_Topology | [
"Finite Complement Topologies",
"Examples of Clopen Sets"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Clopen Set"
] | [
"Definition:Open Set/Topology",
"Definition:Finite Complement Topology",
"Definition:Finite Set",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Infinite Set",
"Definition:Infinite Set",
"Definition:Open Set/Topology"
] |
proofwiki-4974 | Included Set Topology is Topology | Let $T = \struct {S, \tau_H}$ be an included set space.
Then $\tau_H$ is a topology on $S$, and $T$ is a topological space. | === {{Open-set-axiom|1|nolink}} ===
Let $U_1, U_2 \in \tau_H$.
By definition of included set topology:
:$H \subseteq U_1$ and $H \subseteq U_2$
From Intersection is Largest Subset:
:$H \subseteq U_1 \cap U_2$
By definition of included set topology:
:$U_1 \cap U_2 \in \tau_H$
{{qed|lemma}} | Let $T = \struct {S, \tau_H}$ be an [[Definition:Included Set Space|included set space]].
Then $\tau_H$ is a [[Definition:Topology|topology]] on $S$, and $T$ is a [[Definition:Topological Space|topological space]]. | === {{Open-set-axiom|1|nolink}} ===
Let $U_1, U_2 \in \tau_H$.
By definition of [[Definition:Included Set Topology|included set topology]]:
:$H \subseteq U_1$ and $H \subseteq U_2$
From [[Intersection is Largest Subset]]:
:$H \subseteq U_1 \cap U_2$
By definition of [[Definition:Included Set Topology|included set t... | Included Set Topology is Topology | https://proofwiki.org/wiki/Included_Set_Topology_is_Topology | https://proofwiki.org/wiki/Included_Set_Topology_is_Topology | [
"Included Set Topology"
] | [
"Definition:Included Set Topology",
"Definition:Topology",
"Definition:Topological Space"
] | [
"Definition:Included Set Topology",
"Intersection is Largest Subset",
"Definition:Included Set Topology",
"Definition:Included Set Topology",
"Definition:Included Set Topology"
] |
proofwiki-4975 | Space of Compact Linear Transformations is Banach Space | Let $\GF \in \set {\R,\C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\GF$.
Let $\struct {Y, \norm {\, \cdot \,}_Y}$ be a Banach space oer $\GF$.
Let $\map \KK {X, Y}$ be the space of compact linear transformations from $H$ to $K$.
Now $\map \KK {X, Y} \subseteq Y^X$, where $Y^X$ is the se... | First, from Restriction of Norm on Vector Space to Subspace is Norm, $\norm {\, \cdot \,}_{\map \KK {X, Y} }$ is a norm on $\map \KK {X, Y}$.
Now, we prove that $\map \KK {X, Y}$ is a closed vector subspace of $\map \BB {X, Y}$.
First, from Zero Linear Transformation is Compact, $\map \KK {X, Y} \ne \O$.
To then prove... | Let $\GF \in \set {\R,\C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a [[Definition:Normed Vector Space|normed vector space]] over $\GF$.
Let $\struct {Y, \norm {\, \cdot \,}_Y}$ be a [[Definition:Banach Space|Banach space]] oer $\GF$.
Let $\map \KK {X, Y}$ be the [[Definition:Space of Compact Linear Transformati... | First, from [[Restriction of Norm on Vector Space to Subspace is Norm]], $\norm {\, \cdot \,}_{\map \KK {X, Y} }$ is a [[Definition:Norm on Vector Space|norm]] on $\map \KK {X, Y}$.
Now, we prove that $\map \KK {X, Y}$ is a [[Definition:Closed Set|closed]] [[Definition:Vector Subspace|vector subspace]] of $\map \BB {X... | Space of Compact Linear Transformations is Banach Space | https://proofwiki.org/wiki/Space_of_Compact_Linear_Transformations_is_Banach_Space | https://proofwiki.org/wiki/Space_of_Compact_Linear_Transformations_is_Banach_Space | [
"Banach Spaces",
"Linear Transformations on Hilbert Spaces",
"Compact Linear Transformations"
] | [
"Definition:Normed Vector Space",
"Definition:Banach Space",
"Definition:Space of Compact Linear Transformations",
"Definition:Set of All Mappings",
"Definition:Pointwise Addition of Mappings",
"Definition:Pointwise Scalar Multiplication of Mappings",
"Definition:Norm/Bounded Linear Transformation",
"... | [
"Restriction of Norm on Vector Space to Subspace is Norm",
"Definition:Norm/Vector Space",
"Definition:Closed Set",
"Definition:Vector Subspace",
"Zero Linear Transformation is Compact",
"Definition:Vector Subspace",
"One-Step Vector Subspace Test",
"Linear Combination of Compact Linear Transformation... |
proofwiki-4976 | Compact Linear Transformation is Bounded | Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $\map {B_0} {\HH, \KK}$ be the space of compact linear transformations $\HH \to \KK$.
Let $\map B {\HH, \KK}$ be the space of bounded linear transformations $\HH \to \KK$.
Let $T \in \map {B_0} {\HH,... | Let $T : \HH \to \KK$ be a linear transformation.
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$.
Let $\norm \cdot_\KK$ be the inner product norm on $\KK$.
We show that:
:if $T$ is not bounded, then it is not compact.
That is:
:if $T \not \in \map B {\HH, \KK}$, then $T \not \in \map {B_0} {\HH, \KK}$.
Then... | Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $\map {B_0} {\HH, \KK}$ be the [[Definition:Space of Compact Linear Transformations|space of compact linear transformations]] $\HH \to \KK$.
Let $\map B {\HH, \KK}$ be t... | Let $T : \HH \to \KK$ be a [[Definition:Linear Transformation|linear transformation]].
Let $\norm \cdot_\HH$ be the [[Definition:Inner Product Norm|inner product norm]] on $\HH$.
Let $\norm \cdot_\KK$ be the [[Definition:Inner Product Norm|inner product norm]] on $\KK$.
We show that:
:if $T$ is not [[Definition:Bo... | Compact Linear Transformation is Bounded | https://proofwiki.org/wiki/Compact_Linear_Transformation_is_Bounded | https://proofwiki.org/wiki/Compact_Linear_Transformation_is_Bounded | [
"Linear Transformations on Hilbert Spaces",
"Compact Linear Transformations"
] | [
"Definition:Hilbert Space",
"Definition:Space of Compact Linear Transformations",
"Definition:Space of Bounded Linear Transformations"
] | [
"Definition:Linear Transformation",
"Definition:Inner Product Norm",
"Definition:Inner Product Norm",
"Definition:Bounded Linear Transformation",
"Definition:Compact Linear Transformation",
"Definition:Bounded Linear Transformation",
"Definition:Real Number",
"Definition:Sequence",
"Definition:Bound... |
proofwiki-4977 | Compact Linear Transformations Composed with Bounded Linear Operator | Let $H, K$ be Hilbert spaces.
Let $T \in \map {B_0} {H, K}$ be a compact linear transformation.
Let $A \in \map B H, B \in \map B K$ be bounded linear operators.
Then the compositions $T A$ and $B T$ are also compact linear transformations. | Let $\sequence {h_n}_{n \in \N}$ is a bounded sequence in $H$.
That is, there exists a $M > 0$ such that:
:$\forall n \in \N : \norm {h_n}_H \le M$
Then $\sequence {A h_n}_{n \in \N}$ is also bounded, since:
:$\norm {A h_n}_H \le \norm {A}_{\map B H} \norm {h_n}_H \le \norm {A}_{\map B H} M$
As $T$ is compact,
the sequ... | Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $T \in \map {B_0} {H, K}$ be a [[Definition:Compact Linear Transformation|compact linear transformation]].
Let $A \in \map B H, B \in \map B K$ be [[Definition:Bounded Linear Operator|bounded linear operators]].
Then the [[Definition:Composition of Map... | Let $\sequence {h_n}_{n \in \N}$ is a [[Definition:Bounded Sequence in Normed Vector Space|bounded sequence]] in $H$.
That is, there exists a $M > 0$ such that:
:$\forall n \in \N : \norm {h_n}_H \le M$
Then $\sequence {A h_n}_{n \in \N}$ is also [[Definition:Bounded Sequence in Normed Vector Space|bounded]], since:
... | Compact Linear Transformations Composed with Bounded Linear Operator | https://proofwiki.org/wiki/Compact_Linear_Transformations_Composed_with_Bounded_Linear_Operator | https://proofwiki.org/wiki/Compact_Linear_Transformations_Composed_with_Bounded_Linear_Operator | [
"Compact Linear Transformations",
"Linear Transformations on Hilbert Spaces",
"Compact Linear Transformations"
] | [
"Definition:Hilbert Space",
"Definition:Compact Linear Transformation",
"Definition:Bounded Linear Operator",
"Definition:Composition of Mappings",
"Definition:Compact Linear Transformation"
] | [
"Definition:Bounded Sequence/Normed Vector Space",
"Definition:Bounded Sequence/Normed Vector Space",
"Definition:Compact Linear Transformation/Inner Product Space/Definition 2",
"Definition:Sequence",
"Definition:Subsequence",
"Definition:Convergent Sequence/Normed Vector Space",
"Definition:Compact Li... |
proofwiki-4978 | Expansion of Included Set Topology | Let $S$ be a set.
Let $A_1 \subseteq S$ and $A_2 \subseteq S$.
Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be included set spaces on $S$.
Then:
:$(1): \quad T_1 \ge T_2 \iff A_1 \subseteq A_2$
:$(2): \quad T_1 > T_2 \iff A_1 \subsetneq A_2$
where:
:$T_1 \ge T_2$ denotes that $T_1$ is finer... | Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be included set spaces on $S$. | Let $S$ be a [[Definition:Set|set]].
Let $A_1 \subseteq S$ and $A_2 \subseteq S$.
Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be [[Definition:Included Set Space|included set spaces]] on $S$.
Then:
:$(1): \quad T_1 \ge T_2 \iff A_1 \subseteq A_2$
:$(2): \quad T_1 > T_2 \iff A_1 \subsetn... | Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be [[Definition:Included Set Space|included set spaces]] on $S$. | Expansion of Included Set Topology | https://proofwiki.org/wiki/Expansion_of_Included_Set_Topology | https://proofwiki.org/wiki/Expansion_of_Included_Set_Topology | [
"Included Set Topology"
] | [
"Definition:Set",
"Definition:Included Set Topology",
"Definition:Finer Topology",
"Definition:Finer Topology/Strictly Finer"
] | [
"Definition:Included Set Topology"
] |
proofwiki-4979 | Linear Transformation Compact iff Adjoint Compact | Let $H, K$ be Hilbert spaces.
Let $T: H \to K$ be a linear transformation.
Then $T$ is compact {{iff}} its adjoint $T^*$ is. | First suppose that $T$ is compact.
From the construction of the adjoint, $T^\ast$ is bounded.
From Right Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact, $T T^\ast$ is compact.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence in $K$.
Then there exists a convergen... | Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $T: H \to K$ be a [[Definition:Linear Transformation|linear transformation]].
Then $T$ is [[Definition:Compact Linear Transformation|compact]] {{iff}} its [[Definition:Adjoint Linear Transformation|adjoint]] $T^*$ is. | First suppose that $T$ is [[Definition:Compact Linear Transformation|compact]].
From the construction of the [[Definition:Adjoint Linear Transformation|adjoint]], $T^\ast$ is [[Definition:Bounded Linear Transformation|bounded]].
From [[Right Composition of Compact Linear Transformation with Bounded Linear Transformat... | Linear Transformation Compact iff Adjoint Compact | https://proofwiki.org/wiki/Linear_Transformation_Compact_iff_Adjoint_Compact | https://proofwiki.org/wiki/Linear_Transformation_Compact_iff_Adjoint_Compact | [
"Adjoints",
"Compact Linear Transformations"
] | [
"Definition:Hilbert Space",
"Definition:Linear Transformation",
"Definition:Compact Linear Transformation",
"Definition:Adjoint Linear Transformation"
] | [
"Definition:Compact Linear Transformation",
"Definition:Adjoint Linear Transformation",
"Definition:Bounded Linear Transformation",
"Compact Linear Transformations Composed with Bounded Linear Operator",
"Definition:Compact Linear Transformation",
"Definition:Bounded Sequence/Normed Vector Space",
"Defi... |
proofwiki-4980 | Finite Rank Operators Dense in Compact Linear Transformations | Let $H, K$ be Hilbert spaces.
Then:
:$\map {B_{00} } {H, K}$ is everywhere dense in $\map {B_0} {H, K}$
where:
:$\map {B_{00} } {H, K}$ is the space of continuous finite rank operators from $H$ to $K$
:$\map {B_0} {H, K}$ is the space of compact linear transformations from $H$ to $K$.
That is, for every $T \in \map {B_... | {{proof wanted|Proceeds by sequence definition of limit point}} | Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
Then:
:$\map {B_{00} } {H, K}$ is [[Definition:Everywhere Dense|everywhere dense]] in $\map {B_0} {H, K}$
where:
:$\map {B_{00} } {H, K}$ is the [[Definition:Space of Continuous Finite Rank Operators|space of continuous finite rank operators from $H$ to $K$]... | {{proof wanted|Proceeds by sequence definition of limit point}} | Finite Rank Operators Dense in Compact Linear Transformations | https://proofwiki.org/wiki/Finite_Rank_Operators_Dense_in_Compact_Linear_Transformations | https://proofwiki.org/wiki/Finite_Rank_Operators_Dense_in_Compact_Linear_Transformations | [
"Linear Transformations on Hilbert Spaces",
"Compact Linear Transformations",
"Compact Linear Transformations"
] | [
"Definition:Hilbert Space",
"Definition:Everywhere Dense",
"Definition:Space of Continuous Finite Rank Operators",
"Definition:Space of Compact Linear Transformations",
"Definition:Sequence",
"Definition:Norm/Bounded Linear Transformation"
] | [] |
proofwiki-4981 | Inverse of Matrix Product | Let $\mathbf {A, B}$ be square matrices of order $n$
Let $\mathbf I$ be the $n \times n$ unit matrix.
Let $\mathbf A$ and $\mathbf B$ be nonsingular.
Then the matrix product $\mathbf {AB}$ is also nonsingular, and:
:$\paren {\mathbf A \mathbf B}^{-1} = \mathbf B^{-1} \mathbf A^{-1}$ | We are given that $\mathbf A$ and $\mathbf B$ are nonsingular.
From Product of Matrices is Nonsingular iff Matrices are Nonsingular, $\mathbf A \mathbf B$ is also nonsingular.
By the definition of inverse matrix:
:$\mathbf A \mathbf A^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I$
and
:$\mathbf B \mathbf B^{-1} = \mathb... | Let $\mathbf {A, B}$ be [[Definition:Square Matrix|square matrices of order $n$]]
Let $\mathbf I$ be the $n \times n$ [[Definition:Unit Matrix|unit matrix]].
Let $\mathbf A$ and $\mathbf B$ be [[Definition:Nonsingular Matrix|nonsingular]].
Then the [[Definition:Matrix Product (Conventional)|matrix product]] $\mathb... | We are given that $\mathbf A$ and $\mathbf B$ are [[Definition:Nonsingular Matrix|nonsingular]].
From [[Product of Matrices is Nonsingular iff Matrices are Nonsingular]], $\mathbf A \mathbf B$ is also [[Definition:Nonsingular Matrix|nonsingular]].
By the definition of [[Definition:Inverse Matrix|inverse matrix]]:
:... | Inverse of Matrix Product | https://proofwiki.org/wiki/Inverse_of_Matrix_Product | https://proofwiki.org/wiki/Inverse_of_Matrix_Product | [
"Inverse Matrices",
"Conventional Matrix Multiplication"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Unit Matrix",
"Definition:Nonsingular Matrix",
"Definition:Matrix Product (Conventional)",
"Definition:Nonsingular Matrix"
] | [
"Definition:Nonsingular Matrix",
"Product of Matrices is Nonsingular iff Matrices are Nonsingular",
"Definition:Nonsingular Matrix",
"Definition:Inverse Matrix",
"Matrix Multiplication is Associative",
"Matrix Multiplication is Associative",
"Definition:Inverse Matrix"
] |
proofwiki-4982 | Elementary Row Operations as Matrix Multiplications | Let $e$ be an elementary row operation.
Let $\mathbf E$ be the elementary row matrix of order $m$ defined as:
:$\mathbf E = \map e {\mathbf I}$
where $\mathbf I$ is the unit matrix.
Then for every $m \times n$ matrix $\mathbf A$:
:$\map e {\mathbf A} = \mathbf {E A}$
where $\mathbf {E A}$ denotes the conventional matri... | Let $s, t \in \closedint 1 m$ such that $s \ne t$. | Let $e$ be an [[Definition:Elementary Row Operation|elementary row operation]].
Let $\mathbf E$ be the [[Definition:Elementary Row Matrix|elementary row matrix]] of [[Definition:Order of Square Matrix|order]] $m$ defined as:
:$\mathbf E = \map e {\mathbf I}$
where $\mathbf I$ is the [[Definition:Unit Matrix|unit matri... | Let $s, t \in \closedint 1 m$ such that $s \ne t$. | Elementary Row Operations as Matrix Multiplications | https://proofwiki.org/wiki/Elementary_Row_Operations_as_Matrix_Multiplications | https://proofwiki.org/wiki/Elementary_Row_Operations_as_Matrix_Multiplications | [
"Conventional Matrix Multiplication",
"Elementary Row Operations",
"Elementary Matrices"
] | [
"Definition:Elementary Operation/Row",
"Definition:Elementary Matrix/Row Operation",
"Definition:Matrix/Square Matrix/Order",
"Definition:Unit Matrix",
"Definition:Matrix",
"Definition:Matrix Product (Conventional)"
] | [] |
proofwiki-4983 | Included Set Topology on Finite Intersection | Let $T = \struct {S, \tau}$ be a topological space on a set $S$.
Let $A_1, A_2, \ldots, A_n$ be a finite set of subsets of $S$:
:$\forall i \in \closedint 1 n: A_i \subseteq S$
Let $\forall i \in \closedint 1 n: \map T {A_i} = \struct {S, \tau_{A_i} }$ be the included set spaces on $S$ by $A_i$.
Let:
:$\forall i \in \c... | For ease of notation, define:
:$A := \ds \bigcap_{i \mathop = 1}^n A_i$
and let $\tau_A$ denote the included set topology on $S$ by $A$.
Let $U \in \tau_A$ be nonempty.
Then by definition, $A \subseteq U$.
Hence there is a subset $Z \subseteq S$ of $S$, such that $U = A \cup Z$; that is:
:$U = \ds \paren {\bigcap_{i \m... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] on a [[Definition:Set|set]] $S$.
Let $A_1, A_2, \ldots, A_n$ be a [[Definition:Finite Set|finite set]] of [[Definition:Subset|subsets]] of $S$:
:$\forall i \in \closedint 1 n: A_i \subseteq S$
Let $\forall i \in \closedint 1 n: \map T... | For ease of notation, define:
:$A := \ds \bigcap_{i \mathop = 1}^n A_i$
and let $\tau_A$ denote the [[Definition:Included Set Topology|included set topology]] on $S$ by $A$.
Let $U \in \tau_A$ be [[Definition:Empty Set|nonempty]].
Then by definition, $A \subseteq U$.
Hence there is a [[Definition:Subset|subset]] $... | Included Set Topology on Finite Intersection | https://proofwiki.org/wiki/Included_Set_Topology_on_Finite_Intersection | https://proofwiki.org/wiki/Included_Set_Topology_on_Finite_Intersection | [
"Included Set Topology"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Included Set Topology",
"Definition:Coarser Topology",
"Definition:Included Set Topology"
] | [
"Definition:Included Set Topology",
"Definition:Empty Set",
"Definition:Subset",
"Union Distributes over Intersection",
"Definition:Included Set Topology",
"Definition:Topology",
"Definition:Coarser Topology"
] |
proofwiki-4984 | Transformation of Unit Matrix into Inverse | Let $\mathbf A$ be a square matrix of order $n$ of the matrix space $\map {\MM_\R} n$.
Let $\mathbf I$ be the unit matrix of order $n$.
Suppose there exists a sequence of elementary row operations that reduces $\mathbf A$ to $\mathbf I$.
Then $\mathbf A$ is nonsingular.
Futhermore, the same sequence, when performed on ... | For ease of presentation, let $\breve {\mathbf X}$ be the inverse of $\mathbf X$.
We have that $\mathbf A$ can be transformed into $\mathbf I$ by a sequence of elementary row operations.
By repeated application of Elementary Row Operations as Matrix Multiplications, we can write this assertion as:
{{begin-eqn}}
{{eqn |... | Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix of order $n$]] of the [[Definition:Matrix Space|matrix space]] $\map {\MM_\R} n$.
Let $\mathbf I$ be the [[Definition:Unit Matrix|unit matrix]] of order $n$.
Suppose there exists a [[Definition:Sequence|sequence]] of [[Definition:Elementary Row Operation|... | For ease of presentation, let $\breve {\mathbf X}$ be the inverse of $\mathbf X$.
We have that $\mathbf A$ can be transformed into $\mathbf I$ by a sequence of elementary row operations.
By repeated application of [[Elementary Row Operations as Matrix Multiplications]], we can write this assertion as:
{{begin-eqn}}
... | Transformation of Unit Matrix into Inverse | https://proofwiki.org/wiki/Transformation_of_Unit_Matrix_into_Inverse | https://proofwiki.org/wiki/Transformation_of_Unit_Matrix_into_Inverse | [
"Unit Matrices",
"Inverse Matrices",
"Elementary Row Operations"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Matrix Space",
"Definition:Unit Matrix",
"Definition:Sequence",
"Definition:Elementary Operation/Row",
"Definition:Nonsingular Matrix",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Elementary Row Operations as Matrix Multiplications",
"Elementary Row Matrix is Nonsingular",
"Inverse of Matrix Product",
"Axiom:Leibniz's Law",
"Inverse of Group Inverse",
"Elementary Row Operations as Matrix Multiplications",
"Category:Unit Matrices",
"Category:Inverse Matrices",
"Category:Eleme... |
proofwiki-4985 | Included Set Topology on Union | Let $T = \struct {S, \tau}$ be a topological space on a set $S$.
Let $\family {A_i}_{i \mathop \in I}$ be a family of subsets of $S$ indexed by the indexing set $I$:
:$\forall i \in I: A_i \subseteq S$
Let $\forall i \in I: \map T {A_i} = \struct {S, \tau_{A_i} }$ be the included set spaces on $S$ by $A_i$.
Let:
:$\for... | For ease of notation, define:
:$A := \ds \bigcup_{i \mathop \in I} A_i$
and let $\tau_A$ denote the included set topology on $S$ by $A$.
Let $U \in \tau$ be nonempty.
As $\map T {A_i}$ is finer than $T$ it follows by definition that:
:$\forall i \in I: \tau \subseteq \tau_{A_i}$
Thus:
:$\forall i \in I: U \in \tau_{A_i... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] on a [[Definition:Set|set]] $S$.
Let $\family {A_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Subsets|family of subsets]] of $S$ indexed by the [[Definition:Indexing Set|indexing set]] $I$:
:$\forall i \in I: A_i \subseteq... | For ease of notation, define:
:$A := \ds \bigcup_{i \mathop \in I} A_i$
and let $\tau_A$ denote the [[Definition:Included Set Topology|included set topology]] on $S$ by $A$.
Let $U \in \tau$ be [[Definition:Non-Empty Set|nonempty]].
As $\map T {A_i}$ is [[Definition:Finer Topology|finer]] than $T$ it follows by def... | Included Set Topology on Union | https://proofwiki.org/wiki/Included_Set_Topology_on_Union | https://proofwiki.org/wiki/Included_Set_Topology_on_Union | [
"Included Set Topology"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Indexing Set",
"Definition:Included Set Topology",
"Definition:Finer Topology",
"Definition:Included Set Topology"
] | [
"Definition:Included Set Topology",
"Definition:Non-Empty Set",
"Definition:Finer Topology",
"Definition:Subset",
"Union is Associative",
"Union is Commutative",
"Definition:Set Union",
"Definition:Included Set Topology",
"Definition:Finer Topology"
] |
proofwiki-4986 | Nonzero Eigenvalue of Compact Operator has Finite Dimensional Eigenspace | Let $H$ be a Hilbert space.
Let $T \in \map {B_0} H$ be a compact operator.
Let $\lambda \in \map {\sigma_p} T, \lambda \ne 0$ be a nonzero eigenvalue of $T$.
Then the eigenspace for $\lambda$ has finite dimension. | {{tidy}}
{{MissingLinks}}
$\def \sequence#1{\left({#1}\right)}$
Note that in the following, the notation for a sequence of terms in the form $e_n$ has been changed for this particular page to be $\sequence {e_n}$ rather than the {{ProofWiki}} standard $\left\langle{e_n}\right\rangle$.
This is in order to avoid potentia... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $T \in \map {B_0} H$ be a [[Definition:Compact Operator|compact operator]].
Let $\lambda \in \map {\sigma_p} T, \lambda \ne 0$ be a nonzero [[Definition:Eigenvalue of Linear Operator|eigenvalue]] of $T$.
Then the [[Definition:Eigenspace of Linear Operator... | {{tidy}}
{{MissingLinks}}
$\def \sequence#1{\left({#1}\right)}$
Note that in the following, the notation for a [[Definition:Sequence|sequence]] of [[Definition:Term of Sequence|terms]] in the form $e_n$ has been changed for this particular page to be $\sequence {e_n}$ rather than the {{ProofWiki}} standard $\left\lan... | Nonzero Eigenvalue of Compact Operator has Finite Dimensional Eigenspace | https://proofwiki.org/wiki/Nonzero_Eigenvalue_of_Compact_Operator_has_Finite_Dimensional_Eigenspace | https://proofwiki.org/wiki/Nonzero_Eigenvalue_of_Compact_Operator_has_Finite_Dimensional_Eigenspace | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Compact Linear Transformation",
"Definition:Eigenvalue/Linear Operator",
"Definition:Eigenspace/Linear Operator",
"Definition:Finite",
"Definition:Dimension (Hilbert Space)"
] | [
"Definition:Sequence",
"Definition:Term of Sequence",
"Definition:Inner Product"
] |
proofwiki-4987 | Condition for Nonzero Eigenvalue of Compact Operator | Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.
Let $T \in \map {B_0} H$ be a compact operator.
Let $\lambda \in \Bbb F, \lambda \ne 0$ be a nonzero scalar.
Suppose that the following holds:
:$\inf \set {\norm {\paren {T - \lambda I} h}_H: \norm h_H = 1} = 0$
Then $\lambda \in \map {\sigma_p} T$, that is, $... | {{proof wanted}}
By hypothesis, there is a sequence of unit vectors $\sequence{h_n}$ such that $\cmod{(T-\lambda) h_n} \to 0$.
Since $T$ is compact, by definition 2 there is a $f\in H$ and a subsequence $\sequence{h_{n_k}}$ such that $T h_{n_k}\to f$.
But $h_{n_k}=\lambda^{-1}\left[(\lambda-T) h_{n_k}+T h_{n_k}\right]... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] over $\Bbb F \in \set {\R, \C}$.
Let $T \in \map {B_0} H$ be a [[Definition:Compact Operator|compact operator]].
Let $\lambda \in \Bbb F, \lambda \ne 0$ be a nonzero [[Definition:Scalar (Vector Space)|scalar]].
Suppose that the following holds:
:$\inf \set {\... | {{proof wanted}}
By hypothesis, there is a sequence of unit vectors $\sequence{h_n}$ such that $\cmod{(T-\lambda) h_n} \to 0$.
Since $T$ is compact, by [[Definition:Compact Linear Transformation/Normed Vector Space/Definition 2|definition 2]] there is a $f\in H$ and a subsequence $\sequence{h_{n_k}}$ such that $T h_{n... | Condition for Nonzero Eigenvalue of Compact Operator | https://proofwiki.org/wiki/Condition_for_Nonzero_Eigenvalue_of_Compact_Operator | https://proofwiki.org/wiki/Condition_for_Nonzero_Eigenvalue_of_Compact_Operator | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Compact Linear Transformation",
"Definition:Scalar/Vector Space",
"Definition:Eigenvalue/Linear Operator"
] | [
"Definition:Compact Linear Transformation/Normed Vector Space/Definition 2"
] |
proofwiki-4988 | Finite Rank Operator is Compact | Let $H, K$ be Hilbert spaces.
Let $T \in \map {B_{00} } {H, K}$ be a bounded finite rank operator.
Then $T \in \map {B_0} {H, K}$, that is, $T$ is compact. | {{tidy}}
{{MissingLinks}}
Let $A$ be a bounded set in $H$.
Since $T$ is bounded, the image $T \sqbrk A$ is a bounded set by definition.
Since $T$ is finite rank operator, its range $\Rng T$ is finite dimensional.
The closure of a bounded set in $\Rng T$ is compact, by Heine-Borel Theorem.
So the closure of $T \sqbrk A$... | Let $H, K$ be [[Definition: Hilbert Space|Hilbert spaces]].
Let $T \in \map {B_{00} } {H, K}$ be a [[Definition:Bounded Linear Transformation|bounded]] [[Definition:Finite Rank Operator|finite rank operator]].
Then $T \in \map {B_0} {H, K}$, that is, $T$ is [[Definition:Compact Linear Operator|compact]]. | {{tidy}}
{{MissingLinks}}
Let $A$ be a bounded set in $H$.
Since $T$ is [[Definition:Bounded Linear Transformation|bounded]], the image $T \sqbrk A$ is a bounded set by definition.
Since $T$ is [[Definition:Finite Rank Operator|finite rank operator]], its range $\Rng T$ is [[Definition:Finite Dimensional Vector Spac... | Finite Rank Operator is Compact | https://proofwiki.org/wiki/Finite_Rank_Operator_is_Compact | https://proofwiki.org/wiki/Finite_Rank_Operator_is_Compact | [
"Linear Transformations on Hilbert Spaces",
"Compact Linear Transformations"
] | [
"Definition: Hilbert Space",
"Definition:Bounded Linear Transformation",
"Definition:Finite Rank Operator",
"Definition:Compact Linear Operator"
] | [
"Definition:Bounded Linear Transformation",
"Definition:Finite Rank Operator",
"Definition:Dimension of Vector Space/Finite",
"Heine-Borel Theorem/Normed Vector Space",
"Definition:Compact Linear Operator"
] |
proofwiki-4989 | Adjoint of Finite Rank Operator | Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\GF$.
Let $T \in \map {B_{00} } {\HH, \KK}$ be a bounded finite rank operator.
Then:
:$T^* \in \map {B_{00} } {\KK, \HH}$
that is, the adjoint of $T$ is also a bounded fi... | From Characterization of Finite Rank Operators, there exists $e_1, e_2, \ldots, e_n \in \HH$ and $g_1, g_2, \ldots, g_n \in \KK$ such that:
:$\ds T x = \sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH g_i$
for each $x \in \HH$.
Then for each $x \in \HH$, $y \in \KK$ we have:
{{begin-eqn}}
{{eqn | l = \innerprod {T x} ... | Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be [[Definition:Hilbert Space|Hilbert spaces]] over $\GF$.
Let $T \in \map {B_{00} } {\HH, \KK}$ be a [[Definition:Bounded Linear Transformation|bounded]] [[Definition:Finite Rank Operator|fin... | From [[Characterization of Finite Rank Operators]], there exists $e_1, e_2, \ldots, e_n \in \HH$ and $g_1, g_2, \ldots, g_n \in \KK$ such that:
:$\ds T x = \sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH g_i$
for each $x \in \HH$.
Then for each $x \in \HH$, $y \in \KK$ we have:
{{begin-eqn}}
{{eqn | l = \innerpro... | Adjoint of Finite Rank Operator | https://proofwiki.org/wiki/Adjoint_of_Finite_Rank_Operator | https://proofwiki.org/wiki/Adjoint_of_Finite_Rank_Operator | [
"Finite Rank Operators",
"Adjoints"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Transformation",
"Definition:Finite Rank Operator",
"Definition:Adjoint Linear Transformation",
"Definition:Bounded Linear Transformation",
"Definition:Finite Rank Operator"
] | [
"Characterization of Finite Rank Operators",
"Inner Product is Sesquilinear",
"Inner Product is Sesquilinear",
"Definition:Inner Product",
"Definition:Conjugate Symmetric Mapping",
"Inner Product is Sesquilinear",
"Existence and Uniqueness of Adjoint/Lemma 1",
"Definition:Finite Rank Operator"
] |
proofwiki-4990 | Compact Idempotent is of Finite Rank | Let $H$ be a Hilbert space.
Let $T \in \map {B_0} H$ be a compact linear operator.
Let $T$ be idempotent.
Then:
:$T \in \map {B_{00} } H$
that is, $T$ is a bounded finite rank operator. | {{MissingLinks}}
{{tidy}}
Let $\sequence {y_n}$ be a bounded sequence in the range of $T$.
Since $T$ is idempotent:
:$\forall n \in \N: T y_n = y_n$
Since $T$ is compact, $\sequence {T y_n}$ and hence $\sequence {y_n}$, contains a convergent subsequence.
Thus the range of $T$ has the Bolzano-Weierstrass property.
By N... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $T \in \map {B_0} H$ be a [[Definition:Compact Linear Operator|compact linear operator]].
Let $T$ be [[Definition:Idempotent Operator|idempotent]].
Then:
:$T \in \map {B_{00} } H$
that is, $T$ is a [[Definition:Bounded Linear Transformation|bounded]] [[D... | {{MissingLinks}}
{{tidy}}
Let $\sequence {y_n}$ be a bounded sequence in the range of $T$.
Since $T$ is idempotent:
:$\forall n \in \N: T y_n = y_n$
Since $T$ is compact, $\sequence {T y_n}$ and hence $\sequence {y_n}$, contains a convergent subsequence.
Thus the range of $T$ has the Bolzano-Weierstrass property.
... | Compact Idempotent is of Finite Rank | https://proofwiki.org/wiki/Compact_Idempotent_is_of_Finite_Rank | https://proofwiki.org/wiki/Compact_Idempotent_is_of_Finite_Rank | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Compact Linear Operator",
"Definition:Idempotent Operator",
"Definition:Bounded Linear Transformation",
"Definition:Finite Rank Operator"
] | [
"Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Sufficient Condition"
] |
proofwiki-4991 | Characterization of Finite Rank Operators | Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\GF$.
Let $T \in \map {B_{00} } {\HH, \KK}$ be a bounded finite rank operator.
Let $n = \map \dim {\operatorname {ran} T}$ be the rank of $T$.
Then there are orthonormal ... | Let:
:$\set {e_1, e_2, \ldots, e_n}$
be a basis of $T \sqbrk \HH$.
By Gram-Schmidt Orthogonalization, there exists an orthonormal subset such that:
:$\set {g_1, g_2, \ldots, g_n}$
that is a basis of $T \sqbrk \HH$.
So we have $T x \in \map \span {g_1, g_2, \ldots, g_n}$ for each $x \in \HH$.
So, we can write:
:$\ds... | Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be [[Definition:Hilbert Space|Hilbert spaces]] over $\GF$.
Let $T \in \map {B_{00} } {\HH, \KK}$ be a [[Definition:Bounded Linear Operator|bounded]] [[Definition:Finite Rank Operator|finite ra... | Let:
:$\set {e_1, e_2, \ldots, e_n}$
be a [[Definition:Basis of Vector Space|basis]] of $T \sqbrk \HH$.
By [[Gram-Schmidt Orthogonalization]], there exists an [[Definition:Orthonormal Subset|orthonormal subset]] such that:
:$\set {g_1, g_2, \ldots, g_n}$
that is a [[Definition:Basis of Vector Space|basis]] of $... | Characterization of Finite Rank Operators | https://proofwiki.org/wiki/Characterization_of_Finite_Rank_Operators | https://proofwiki.org/wiki/Characterization_of_Finite_Rank_Operators | [
"Linear Transformations on Hilbert Spaces",
"Finite Rank Operators"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator",
"Definition:Finite Rank Operator",
"Definition:Rank (Linear Algebra)",
"Definition:Orthonormal Subset",
"Definition:Vector/Linear Algebra",
"Definition:Vector/Linear Algebra"
] | [
"Definition:Basis of Vector Space",
"Gram-Schmidt Orthogonalization",
"Definition:Orthonormal Subset",
"Definition:Basis of Vector Space",
"Definition:Function",
"Definition:Bounded Linear Functional",
"Orthogonal Set is Linearly Independent Set",
"Definition:Linearly Independent/Set",
"Definition:L... |
proofwiki-4992 | Compact Operator on Hilbert Space Direct Sum | Let $\sequence {\HH_n}_{n \mathop \in \N}$ be a sequence of Hilbert spaces.
Denote by $\HH = \bigoplus_{n \mathop \in \N} \HH_n$ their Hilbert space direct sum.
For each $n \in \N$, let $T_n \in \map B {\HH_n}$ be a bounded linear operator.
Suppose that:
:$\ds \sup_{n \mathop \in \N} \norm {T_n} < \infty$
where $\norm ... | Assume:
:$(1):$ For each $n \in \N$, $T_n$ is compact
:$(2): \ds \lim_{n \mathop \to \infty} \norm {T_n} = 0$
We show $\map \cl {T \sqbrk {\operatorname {ball} \HH } }$ is compact in $\HH$.
By $(1)$, $K_k := \map \cl {T_k \sqbrk {\operatorname {ball} \HH_k } }$ is compact in $\HH_k$ for each $k\in\N$.
Further, each $K_... | Let $\sequence {\HH_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Hilbert Space|Hilbert spaces]].
Denote by $\HH = \bigoplus_{n \mathop \in \N} \HH_n$ their [[Definition:Hilbert Space Direct Sum|Hilbert space direct sum]].
For each $n \in \N$, let $T_n \in \map B {\HH_n}$ be a [[Defini... | Assume:
:$(1):$ For each $n \in \N$, $T_n$ is [[Definition:Compact Linear Operator|compact]]
:$(2): \ds \lim_{n \mathop \to \infty} \norm {T_n} = 0$
We show $\map \cl {T \sqbrk {\operatorname {ball} \HH } }$ is [[Definition:Compact Subset of Normed Vector Space|compact]] in $\HH$.
By $(1)$, $K_k := \map \cl {T_k \sq... | Compact Operator on Hilbert Space Direct Sum | https://proofwiki.org/wiki/Compact_Operator_on_Hilbert_Space_Direct_Sum | https://proofwiki.org/wiki/Compact_Operator_on_Hilbert_Space_Direct_Sum | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Sequence",
"Definition:Hilbert Space",
"Definition:Hilbert Space Direct Sum",
"Definition:Bounded Linear Operator",
"Definition:Norm/Bounded Linear Transformation",
"Definition:Bounded Linear Operator",
"Bounded Linear Operator on Hilbert Space Direct Sum",
"Definition:Compact Linear Opera... | [
"Definition:Compact Linear Operator",
"Definition:Compact Space/Normed Vector Space",
"Definition:Compact Space/Normed Vector Space",
"Definition:Sequentially Compact Space",
"Definition:Subset",
"Compact Metric Space is Totally Bounded",
"Complete and Totally Bounded Metric Space is Sequentially Compac... |
proofwiki-4993 | Bounded Linear Operator on Hilbert Space Direct Sum | Let $\GF \in \set {\R, \C}$.
Let $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ be a $I$-indexed family of Hilbert spaces over $\GF$.
For each $i \in I$, let $T_i : \HH_i \to \HH_i$ be a bounded linear operator.
Suppose that:
:$\ds \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} } < \inf... | We first verify that $T$ is well-defined.
Let:
:$\ds M = \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} }$
For this we want to verify that for each $f \in \HH$ we have:
:$\ds \sum \set {\norm {\map {T_i} {\map f i} }_i^2 : i \in I} < \infty$
From Fundamental Property of Norm on Bounded Linear Transformation, we hav... | Let $\GF \in \set {\R, \C}$.
Let $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ be a [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Hilbert Space|Hilbert spaces]] over $\GF$.
For each $i \in I$, let $T_i : \HH_i \to \HH_i$ be a [[Definition:Bounded Linear Operator|bounde... | We first verify that $T$ is well-defined.
Let:
:$\ds M = \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} }$
For this we want to verify that for each $f \in \HH$ we have:
:$\ds \sum \set {\norm {\map {T_i} {\map f i} }_i^2 : i \in I} < \infty$
From [[Fundamental Property of Norm on Bounded Linear Transformation]],... | Bounded Linear Operator on Hilbert Space Direct Sum | https://proofwiki.org/wiki/Bounded_Linear_Operator_on_Hilbert_Space_Direct_Sum | https://proofwiki.org/wiki/Bounded_Linear_Operator_on_Hilbert_Space_Direct_Sum | [
"Direct Sums of Hilbert Spaces"
] | [
"Definition:Indexing Set/Family",
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator",
"Definition:Norm/Bounded Linear Transformation",
"Definition:Hilbert Space Direct Sum",
"Definition:Norm/Vector Space",
"Definition:Bounded Linear Operator"
] | [
"Fundamental Property of Norm on Bounded Linear Transformation",
"Generalized Sum Preserves Inequality",
"Generalized Sum Preserves Inequality",
"Generalized Sum is Linear",
"Definition:Linear Operator",
"Definition:Linear Transformation",
"Definition:Norm/Bounded Linear Transformation",
"Definition:S... |
proofwiki-4994 | Compact Hermitian Operator has Countable Point Spectrum | Let $\HH$ be a Hilbert space.
Let $T \in \map {B_0} \HH$ be a compact Hermitian operator.
{{explain|What is $\map {B_0} \HH$?}}
Then its point spectrum $\map {\sigma_p} T$ is countable. | {{proof wanted|Use the next result, the spectral theorem}} | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $T \in \map {B_0} \HH$ be a [[Definition:Compact Linear Operator|compact]] [[Definition:Hermitian Operator|Hermitian operator]].
{{explain|What is $\map {B_0} \HH$?}}
Then its [[Definition:Point Spectrum of Linear Operator|point spectrum]] $\map {\sigma_... | {{proof wanted|Use the next result, the spectral theorem}} | Compact Hermitian Operator has Countable Point Spectrum | https://proofwiki.org/wiki/Compact_Hermitian_Operator_has_Countable_Point_Spectrum | https://proofwiki.org/wiki/Compact_Hermitian_Operator_has_Countable_Point_Spectrum | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Compact Linear Operator",
"Definition:Hermitian Operator",
"Definition:Point Spectrum of Linear Operator",
"Definition:Countable Set"
] | [] |
proofwiki-4995 | Eigenspace for Normal Operator is Reducing Subspace | Let $\HH$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.
Let $A \in \map B \HH$ be a normal operator.
Let $\lambda \in \Bbb F$.
Let $I \in \map B \HH$ be the identity operator.
Then $\map \ker {A - \lambda I}$ is a reducing subspace for $A$.
Here $\ker$ denotes kernel. | We are given that $A$ is normal.
Hence by {{Corollary|Kernel of Linear Transformation is Orthocomplement of Image of Adjoint}}:
:$\ker A = \Rng A^\perp$
and in particular, that:
:$\ker A \subseteq \Rng A^\perp$
Now, by Orthocomplement of Subset of Orthocomplement is Superset:
:$\Rng A \subseteq \ker A^\perp$
From Opera... | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]] over $\Bbb F \in \set {\R, \C}$.
Let $A \in \map B \HH$ be a [[Definition:Normal Operator|normal operator]].
Let $\lambda \in \Bbb F$.
Let $I \in \map B \HH$ be the [[Definition:Identity Operator|identity operator]].
Then $\map \ker {A - \lambda I}$ is a [[... | We are given that $A$ is [[Definition:Normal Operator|normal]].
Hence by {{Corollary|Kernel of Linear Transformation is Orthocomplement of Image of Adjoint}}:
:$\ker A = \Rng A^\perp$
and in particular, that:
:$\ker A \subseteq \Rng A^\perp$
Now, by [[Orthocomplement of Subset of Orthocomplement is Superset]]:
:$... | Eigenspace for Normal Operator is Reducing Subspace | https://proofwiki.org/wiki/Eigenspace_for_Normal_Operator_is_Reducing_Subspace | https://proofwiki.org/wiki/Eigenspace_for_Normal_Operator_is_Reducing_Subspace | [
"Normal Operators",
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Normal Operator",
"Definition:Identity Mapping",
"Definition:Reducing Subspace",
"Definition:Kernel of Linear Transformation"
] | [
"Definition:Normal Operator",
"Orthocomplement of Subset of Orthocomplement is Superset",
"Operator is Normal iff Operator Minus Multiple of Identity Operator is Normal",
"Definition:Normal Operator",
"Orthocomplement of Subset of Orthocomplement is Superset",
"Definition:Reducing Subspace",
"Definition... |
proofwiki-4996 | Eigenvalues of Normal Operator have Orthogonal Eigenspaces | Let $\HH$ be a Hilbert space.
Let $\mathbf T: \HH \to \HH$ be a normal operator.
Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $\mathbf T$.
Then, the eigenspaces of $\lambda_1$ and $\lambda_2$ are orthogonal:
:$\map \ker {\mathbf T - \lambda_1 \mathbf I} \perp \map \ker {\mathbf T - \lambda_2 \mathbf I}$
where:... | Let $\mathbf v_1$ and $\mathbf v_2$ be eigenvectors of $\mathbf T$ with corresponding eigenvalues $\lambda_1$ and $\lambda_2$, respectively.
Then:
{{begin-eqn}}
{{eqn | l = \lambda_1 \innerprod {\mathbf v_1} {\mathbf v_2}
| r = \innerprod {\lambda_1 \mathbf v_1} {\mathbf v_2}
| c = Inner Product is Sesquili... | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $\mathbf T: \HH \to \HH$ be a [[Definition:Normal Operator|normal operator]].
Let $\lambda_1, \lambda_2$ be distinct [[Definition:Eigenvalue of Linear Operator|eigenvalues]] of $\mathbf T$.
Then, the [[Definition:Eigenspace of Linear Operator|eigenspace... | Let $\mathbf v_1$ and $\mathbf v_2$ be [[Definition:Eigenvector of Linear Operator|eigenvectors]] of $\mathbf T$ with corresponding [[Definition:Eigenvalue of Linear Operator|eigenvalues]] $\lambda_1$ and $\lambda_2$, respectively.
Then:
{{begin-eqn}}
{{eqn | l = \lambda_1 \innerprod {\mathbf v_1} {\mathbf v_2}
... | Eigenvalues of Normal Operator have Orthogonal Eigenspaces | https://proofwiki.org/wiki/Eigenvalues_of_Normal_Operator_have_Orthogonal_Eigenspaces | https://proofwiki.org/wiki/Eigenvalues_of_Normal_Operator_have_Orthogonal_Eigenspaces | [
"Normal Operators",
"Linear Transformations on Hilbert Spaces",
"Eigenvectors of Linear Operators"
] | [
"Definition:Hilbert Space",
"Definition:Normal Operator",
"Definition:Eigenvalue/Linear Operator",
"Definition:Eigenspace/Linear Operator",
"Definition:Orthogonal (Linear Algebra)/Sets",
"Definition:Kernel of Linear Transformation",
"Definition:Identity Mapping",
"Definition:Orthogonal (Linear Algebra... | [
"Definition:Eigenvector/Linear Operator",
"Definition:Eigenvalue/Linear Operator",
"Inner Product is Sesquilinear",
"Adjoint of Normal Operator has Same Eigenvectors and Complex Conjugated Eigenvalues",
"Inner Product is Sesquilinear",
"Definition:Eigenvector/Linear Operator",
"Definition:Normal Operato... |
proofwiki-4997 | Eigenvalues of Hermitian Operator are Real | Let $\HH$ be a Hilbert space.
Let $A \in \map B \HH$ be a Hermitian operator.
Then all eigenvalues of $A$ are real. | Let $\lambda$ be an eigenvalue of $A$.
Let $v \in \HH$ be an eigenvector for $\lambda$.
That is:
:$A v = \lambda v$
Now compute:
{{begin-eqn}}
{{eqn | l = \lambda \innerprod v v
| r = \innerprod {\lambda v} v
| c = Property $(2)$ of an inner product
}}
{{eqn | r = \innerprod {A v} v
| c = $v$ is an ei... | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $A \in \map B \HH$ be a [[Definition:Hermitian Operator|Hermitian operator]].
Then all [[Definition:Eigenvalue of Linear Operator|eigenvalues]] of $A$ are [[Definition:Real Number|real]]. | Let $\lambda$ be an [[Definition:Eigenvalue of Linear Operator|eigenvalue]] of $A$.
Let $v \in \HH$ be an [[Definition:Eigenvector of Linear Operator|eigenvector]] for $\lambda$.
That is:
:$A v = \lambda v$
Now compute:
{{begin-eqn}}
{{eqn | l = \lambda \innerprod v v
| r = \innerprod {\lambda v} v
| c ... | Eigenvalues of Hermitian Operator are Real/Proof 1 | https://proofwiki.org/wiki/Eigenvalues_of_Hermitian_Operator_are_Real | https://proofwiki.org/wiki/Eigenvalues_of_Hermitian_Operator_are_Real/Proof_1 | [
"Eigenvalues of Hermitian Operator are Real",
"Hermitian Operators",
"Eigenvalues of Linear Operators"
] | [
"Definition:Hilbert Space",
"Definition:Hermitian Operator",
"Definition:Eigenvalue/Linear Operator",
"Definition:Real Number"
] | [
"Definition:Eigenvalue/Linear Operator",
"Definition:Eigenvector/Linear Operator",
"Definition:Inner Product",
"Definition:Eigenvector/Linear Operator",
"Definition:Hermitian Operator",
"Definition:Inner Product",
"Definition:Eigenvector/Linear Operator",
"Definition:Inner Product",
"Definition:Eige... |
proofwiki-4998 | Eigenvalues of Hermitian Operator are Real | Let $\HH$ be a Hilbert space.
Let $A \in \map B \HH$ be a Hermitian operator.
Then all eigenvalues of $A$ are real. | We use Dirac notation in the following:
Let $\ket x \in \HH$ be an eigenvector of $A$.
Let $\lambda \in \C$ be the associated eigenvalue:
That is:
:$A \ket x = \lambda \ket x$
For a general operator $T$ in $\map B \HH$ and $\ket x, \ket y \in \HH$:
{{begin-eqn}}
{{eqn | l = \bra x T \ket y
| r = \braket x {T y}
... | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $A \in \map B \HH$ be a [[Definition:Hermitian Operator|Hermitian operator]].
Then all [[Definition:Eigenvalue of Linear Operator|eigenvalues]] of $A$ are [[Definition:Real Number|real]]. | We use [[Definition:Dirac Notation|Dirac notation]] in the following:
Let $\ket x \in \HH$ be an [[Definition:Eigenvector of Linear Operator|eigenvector]] of $A$.
Let $\lambda \in \C$ be the associated [[Definition:Eigenvalue of Linear Operator|eigenvalue]]:
That is:
:$A \ket x = \lambda \ket x$
For a general [[D... | Eigenvalues of Hermitian Operator are Real/Proof 2 | https://proofwiki.org/wiki/Eigenvalues_of_Hermitian_Operator_are_Real | https://proofwiki.org/wiki/Eigenvalues_of_Hermitian_Operator_are_Real/Proof_2 | [
"Eigenvalues of Hermitian Operator are Real",
"Hermitian Operators",
"Eigenvalues of Linear Operators"
] | [
"Definition:Hilbert Space",
"Definition:Hermitian Operator",
"Definition:Eigenvalue/Linear Operator",
"Definition:Real Number"
] | [
"Definition:Dirac Notation",
"Definition:Eigenvector/Linear Operator",
"Definition:Eigenvalue/Linear Operator",
"Definition:Bounded Linear Operator",
"Definition:Inner Product/Complex Field",
"Definition:Complex Conjugate",
"Definition:Adjoint Linear Transformation",
"Definition:Dirac Notation",
"De... |
proofwiki-4999 | Existence of Hartogs Number | Let $S$ be a set.
Then $S$ has a Hartogs number. | From Hartogs' lemma there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$.
We also have that Ordinals are Well-Ordered.
It follows from the definition of well-ordering that there exists a smallest such ordinal.
Hence the result.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Then $S$ has a [[Definition:Hartogs Number|Hartogs number]]. | From [[Hartogs' Lemma (Set Theory)|Hartogs' lemma]] there exists an [[Definition:Ordinal|ordinal]] $\alpha$ such that there is no [[Definition:Injection|injection]] from $\alpha$ to $S$.
We also have that [[Ordinals are Well-Ordered]].
It follows from the definition of [[Definition:Well-Ordering|well-ordering]] that ... | Existence of Hartogs Number/Proof 1 | https://proofwiki.org/wiki/Existence_of_Hartogs_Number | https://proofwiki.org/wiki/Existence_of_Hartogs_Number/Proof_1 | [
"Ordinals",
"Existence of Hartogs Number"
] | [
"Definition:Set",
"Definition:Hartogs Number"
] | [
"Hartogs' Lemma (Set Theory)",
"Definition:Ordinal",
"Definition:Injection",
"Ordinals are Well-Ordered",
"Definition:Well-Ordering",
"Definition:Smallest Element",
"Definition:Ordinal"
] |
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