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-16700 | Lindelöf's Lemma/Lemma 1 | Let $C$ be a set of open real sets.
Then there is a countable subset $D$ of $C$ such that:
:$\ds \bigcup_{O \mathop \in D} O = \bigcup_{O \mathop \in C} O$ | === Lemma $2$ ===
{{:Lindelöf's Lemma/Lemma 2}}{{qed|lemma}}
Let $\ds U = \bigcup_{O \mathop \in C} O$.
Let $x$ be an arbitrary point in $U$.
Since $U$ is the union of the sets in $C$, the point $x$ belongs to a set in $C$.
Name such a set $O_x$.
Since $O_x$ is open, $O_x$ contains an open interval $I_x$ that contains ... | Let $C$ be a [[Definition:Set|set]] of [[Definition:Open Set of Real Numbers|open real sets]].
Then there is a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] $D$ of $C$ such that:
:$\ds \bigcup_{O \mathop \in D} O = \bigcup_{O \mathop \in C} O$ | === [[Lindelöf's Lemma/Lemma 2|Lemma $2$]] ===
{{:Lindelöf's Lemma/Lemma 2}}{{qed|lemma}}
Let $\ds U = \bigcup_{O \mathop \in C} O$.
Let $x$ be an arbitrary point in $U$.
Since $U$ is the [[Definition:Set Union|union]] of the [[Definition:Set|set]]s in $C$, the point $x$ belongs to a [[Definition:Set|set]] in $C$.
... | Lindelöf's Lemma/Lemma 1 | https://proofwiki.org/wiki/Lindelöf's_Lemma/Lemma_1 | https://proofwiki.org/wiki/Lindelöf's_Lemma/Lemma_1 | [
"Lindelöf's Lemma"
] | [
"Definition:Set",
"Definition:Open Set/Real Analysis/Real Numbers",
"Definition:Countable Set",
"Definition:Subset"
] | [
"Lindelöf's Lemma/Lemma 2",
"Definition:Set Union",
"Definition:Set",
"Definition:Set",
"Definition:Open Set/Real Analysis/Real Numbers",
"Definition:Real Interval/Open",
"Between two Real Numbers exists Rational Number",
"Definition:Rational Number",
"Definition:Real Interval/Endpoints",
"Definit... |
proofwiki-16701 | Linear Second Order ODE/y'' - 4 y = x^2 - 3 x - 4 | The second order ODE:
:$(1): \quad y'' - 4 y = x^2 - 3 x - 4$
has the general solution:
:$y = C_1 e^{2 x} + C_2 e^{-2 x} - \dfrac {x^2} 4 + \dfrac {3 x} 4 + \dfrac 7 8$ | It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:
:$y'' + p y' + q y = \map R x$
where:
:$p = 0$
:$q = -4$
:$\map R x = x^2 - 3 x - 4$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y'' - 4 y = ... | The [[Definition:Second Order ODE|second order ODE]]:
:$(1): \quad y'' - 4 y = x^2 - 3 x - 4$
has the [[Definition:General Solution to Differential Equation|general solution]]:
:$y = C_1 e^{2 x} + C_2 e^{-2 x} - \dfrac {x^2} 4 + \dfrac {3 x} 4 + \dfrac 7 8$ | It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
:$y'' + p y' + q y = \map R x$
where:
:$p = 0$
:$q = -4$
:$\map R x = x^2 - 3 x - 4$
First we establish... | Linear Second Order ODE/y'' - 4 y = x^2 - 3 x - 4 | https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_-_4_y_=_x^2_-_3_x_-_4 | https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_-_4_y_=_x^2_-_3_x_-_4 | [
"Examples of Constant Coefficient LSOODEs",
"Examples of Method of Undetermined Coefficients"
] | [
"Definition:Second Order Ordinary Differential Equation",
"Definition:Differential Equation/Solution/General Solution"
] | [
"Definition:Nonhomogeneous Linear Second Order ODE",
"Definition:Constant",
"Definition:Coefficient",
"Definition:Homogeneous Linear Second Order ODE with Constant Coefficients",
"Linear Second Order ODE/y'' - 4 y = 0",
"Definition:Differential Equation/Solution/General Solution",
"Method of Undetermine... |
proofwiki-16702 | Linear Second Order ODE/y'' + 2 y' + y = x exp -x | The second order ODE:
:$(1): \quad y'' + 2 y' + y = x e^{-x}$
has the general solution:
:$y = e^{-x} \paren {C_1 + C_2 x + \dfrac {x^3} 6}$ | It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
:$y'' + p y' + q y = \map R x$
where:
:$p = 2$
:$q = 1$
:$\map R x = x e^{-x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y'' + 2 y' + y = 0$
From Linear Second Order ... | The [[Definition:Second Order ODE|second order ODE]]:
:$(1): \quad y'' + 2 y' + y = x e^{-x}$
has the [[Definition:General Solution to Differential Equation|general solution]]:
:$y = e^{-x} \paren {C_1 + C_2 x + \dfrac {x^3} 6}$ | It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] in the form:
:$y'' + p y' + q y = \map R x$
where:
:$p = 2$
:$q = 1$
:$\map R x = x e^{-x}$
First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Lin... | Linear Second Order ODE/y'' + 2 y' + y = x exp -x | https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_+_2_y'_+_y_=_x_exp_-x | https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_+_2_y'_+_y_=_x_exp_-x | [
"Examples of Constant Coefficient LSOODEs",
"Examples of Method of Undetermined Coefficients"
] | [
"Definition:Second Order Ordinary Differential Equation",
"Definition:Differential Equation/Solution/General Solution"
] | [
"Definition:Nonhomogeneous Linear Second Order ODE",
"Definition:Homogeneous Linear Second Order ODE with Constant Coefficients",
"Linear Second Order ODE/y'' + 2 y' + y = 0",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Differential Equation/Solution/Particular Solution",
"Me... |
proofwiki-16703 | Linear Second Order ODE/y'' + 2 y' + 5 y = x sin x | The second order ODE:
:$(1): \quad y'' + 2 y' + 5 y = x \sin x$
has the general solution:
:$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \ldots$ | It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
:$y'' + p y' + q y = \map R x$
where:
:$p = 2$
:$q = 5$
:$\map R x = x \sin x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
:$y'' + 2 y' + 5 y = 0$
From Linear Second Orde... | The [[Definition:Second Order ODE|second order ODE]]:
:$(1): \quad y'' + 2 y' + 5 y = x \sin x$
has the [[Definition:General Solution to Differential Equation|general solution]]:
:$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \ldots$ | It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] in the form:
:$y'' + p y' + q y = \map R x$
where:
:$p = 2$
:$q = 5$
:$\map R x = x \sin x$
First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Lin... | Linear Second Order ODE/y'' + 2 y' + 5 y = x sin x | https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_+_2_y'_+_5_y_=_x_sin_x | https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_+_2_y'_+_5_y_=_x_sin_x | [
"Examples of Constant Coefficient LSOODEs",
"Examples of Method of Undetermined Coefficients"
] | [
"Definition:Second Order Ordinary Differential Equation",
"Definition:Differential Equation/Solution/General Solution"
] | [
"Definition:Nonhomogeneous Linear Second Order ODE",
"Definition:Homogeneous Linear Second Order ODE with Constant Coefficients",
"Linear Second Order ODE/y'' + 2 y' + 5 y = 0",
"Definition:Differential Equation/Solution/General Solution",
"Definition:Differential Equation/Solution/Particular Solution",
"... |
proofwiki-16704 | Angular Momentum Commutation Rules | Let $J_x$, $J_y$ and $J_z$ denote the angular momentum operators.
Then:
{{begin-eqn}}
{{eqn | l = \sqbrk {J_x, J_y}
| r = i J_z
}}
{{eqn | l = \sqbrk {J_y, J_z}
| r = i J_x
}}
{{eqn | l = \sqbrk {J_z, J_x}
| r = i J_y
}}
{{end-eqn}}
where $\sqbrk {\, \cdot, \cdot \,}$ denotes the commutator operator. | {{ProofWanted|Got a lot to work through before we get this far}} | Let $J_x$, $J_y$ and $J_z$ denote the [[Definition:Angular Momentum Operator|angular momentum operators]].
Then:
{{begin-eqn}}
{{eqn | l = \sqbrk {J_x, J_y}
| r = i J_z
}}
{{eqn | l = \sqbrk {J_y, J_z}
| r = i J_x
}}
{{eqn | l = \sqbrk {J_z, J_x}
| r = i J_y
}}
{{end-eqn}}
where $\sqbrk {\, \cdot, \... | {{ProofWanted|Got a lot to work through before we get this far}} | Angular Momentum Commutation Rules | https://proofwiki.org/wiki/Angular_Momentum_Commutation_Rules | https://proofwiki.org/wiki/Angular_Momentum_Commutation_Rules | [
"Physics"
] | [
"Definition:Angular Momentum Operator",
"Definition:Commutator"
] | [] |
proofwiki-16705 | P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient | Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\ds \sum_{i \mathop = m}^\infty d_i p^i$ be a $p$-adic expansion.
Let $a$ be the $p$-adic number, that is left coset, in $\Q_p$ containing $\ds \sum_{i \mathop = m}^\infty d_i p^i$.
Let $l$ be the index of the first non-z... | For all $n \ge m$, let:
:$\alpha_n = \ds \sum_{i \mathop = m}^n d_i p^i$
By assumption:
:$\sequence{\alpha_n}$ is a representative of $a$
By definition of the induced norm:
:$\norm a_p = \ds \lim_{n \mathop \to \infty} \norm {\alpha_n}_p$
From Eventually Constant Sequence Converges to Constant it is sufficient to show:... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:Valued Field of P-adic Numbers|$p$-adic numbers]].
Let $\ds \sum_{i \mathop = m}^\infty d_i p^i$ be a [[Definition:P-adic Expansion|$p$-adic expansion]].
Let $a$ be the [[Definition:P-adic Number|$p$-... | For all $n \ge m$, let:
:$\alpha_n = \ds \sum_{i \mathop = m}^n d_i p^i$
By assumption:
:$\sequence{\alpha_n}$ is a [[Definition:Representative of P-adic Number|representative]] of $a$
By definition of the [[Definition:Induced Norm on Quotient of Cauchy Sequences|induced norm]]:
:$\norm a_p = \ds \lim_{n \mathop \to... | P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient | https://proofwiki.org/wiki/P-adic_Norm_of_P-adic_Expansion_is_determined_by_First_Nonzero_Coefficient | https://proofwiki.org/wiki/P-adic_Norm_of_P-adic_Expansion_is_determined_by_First_Nonzero_Coefficient | [
"P-adic Number Theory"
] | [
"Definition:Prime Number",
"Definition:Valued Field of P-adic Numbers",
"Definition:P-adic Expansion",
"Definition:P-adic Number",
"Definition:Coset/Left Coset",
"Definition:Subset",
"Definition:Summation/Index Variable",
"Definition:Zero (Number)",
"Definition:Power Series/Coefficient",
"Definiti... | [
"Definition:P-adic Number/Representative",
"Definition:Induced Norm on Quotient of Cauchy Sequences",
"Eventually Constant Sequence Converges to Constant",
"Definition:Divisor",
"Power Function on Base between Zero and One is Strictly Decreasing",
"Definition:P-adic Expansion",
"Definition:P-adic Norm"
... |
proofwiki-16706 | Eventually Constant Sequence Converges to Constant | :$\ds \lim_{n \mathop \to \infty} x_n = \lambda$ | Let $\sequence {y_n}$ be the subsequence of $\sequence {\norm {x_n} }$ defined as:
:$\forall n: y_n = x_{N + n}$
The $\sequence {y_n}$ is the constant sequence $\tuple {\lambda, \lambda, \lambda, \dotsc}$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} x_n
| r = \lim_{n \mathop \to \infty} y_n
... | :$\ds \lim_{n \mathop \to \infty} x_n = \lambda$ | Let $\sequence {y_n}$ be the [[Definition:Subsequence|subsequence]] of $\sequence {\norm {x_n} }$ defined as:
:$\forall n: y_n = x_{N + n}$
The $\sequence {y_n}$ is the constant [[Definition:Sequence|sequence]] $\tuple {\lambda, \lambda, \lambda, \dotsc}$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} ... | Eventually Constant Sequence Converges to Constant | https://proofwiki.org/wiki/Eventually_Constant_Sequence_Converges_to_Constant | https://proofwiki.org/wiki/Eventually_Constant_Sequence_Converges_to_Constant | [
"Sequences",
"Normed Division Rings"
] | [] | [
"Definition:Subsequence",
"Definition:Sequence",
"Limit of Subsequence equals Limit of Sequence/Normed Division Ring",
"Constant Sequence Converges to Constant in Normed Division Ring",
"Category:Sequences",
"Category:Normed Division Rings"
] |
proofwiki-16707 | Non-Zero Complex Numbers under Multiplication form Group | Let $\C_{\ne 0}$ be the set of complex numbers without zero, that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The structure $\struct {\C_{\ne 0}, \times}$ is a group. | Taking the group axioms in turn: | Let $\C_{\ne 0}$ be the set of [[Definition:Complex Number|complex numbers]] without [[Definition:Zero (Number)|zero]], that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\C_{\ne 0}, \times}$ is a [[Definition:Group|group]]. | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Non-Zero Complex Numbers under Multiplication form Group | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_under_Multiplication_form_Group | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_under_Multiplication_form_Group | [
"Complex Multiplication",
"Examples of Groups"
] | [
"Definition:Complex Number",
"Definition:Zero (Number)",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] | [
"Axiom:Group Axioms"
] |
proofwiki-16708 | Real Numbers under Addition form Group | Let $\R$ be the set of real numbers.
The structure $\struct {\R, +}$ is a group. | Taking the group axioms in turn: | Let $\R$ be the set of [[Definition:Real Number|real numbers]].
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\R, +}$ is a [[Definition:Group|group]]. | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Real Numbers under Addition form Group | https://proofwiki.org/wiki/Real_Numbers_under_Addition_form_Group | https://proofwiki.org/wiki/Real_Numbers_under_Addition_form_Group | [
"Additive Group of Real Numbers"
] | [
"Definition:Real Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] | [
"Axiom:Group Axioms"
] |
proofwiki-16709 | Complex Numbers under Addition form Group | Let $\C$ be the set of complex numbers.
The structure $\struct {\C, +}$ is a group. | Taking the group axioms in turn: | Let $\C$ be the set of [[Definition:Complex Number|complex numbers]].
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\C, +}$ is a [[Definition:Group|group]]. | Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Complex Numbers under Addition form Group | https://proofwiki.org/wiki/Complex_Numbers_under_Addition_form_Group | https://proofwiki.org/wiki/Complex_Numbers_under_Addition_form_Group | [
"Additive Group of Complex Numbers",
"Complex Addition",
"Examples of Groups"
] | [
"Definition:Complex Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] | [
"Axiom:Group Axioms"
] |
proofwiki-16710 | Unitary Group is Group | The set of unitary $n \times n$ matrices forms a group under (conventional) matrix multiplication. | Let $\mathbf A$ and $\mathbf B$ be $n \times n$ unitary matrices. | The [[Definition:Set|set]] of [[Definition:Unitary Matrix|unitary $n \times n$ matrices]] forms a [[Definition:Group|group]] under [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]]. | Let $\mathbf A$ and $\mathbf B$ be $n \times n$ [[Definition:Unitary Matrix|unitary matrices]]. | Unitary Group is Group | https://proofwiki.org/wiki/Unitary_Group_is_Group | https://proofwiki.org/wiki/Unitary_Group_is_Group | [
"Unitary Groups",
"Unitary Matrices",
"Examples of Groups"
] | [
"Definition:Set",
"Definition:Unitary Matrix",
"Definition:Group",
"Definition:Matrix Product (Conventional)"
] | [
"Definition:Unitary Matrix",
"Definition:Unitary Matrix",
"Definition:Unitary Matrix",
"Definition:Unitary Matrix"
] |
proofwiki-16711 | General Linear Group is not Abelian | Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\GL {n, K}$ be the general linear group of order $n$ over $K$.
Then $\GL {n, K}$ is not an abelian group. | Let:
{{begin-eqn}}
{{eqn | l = A
| r = \begin {pmatrix} 1 & 1 \\ 0 & 1 \end {pmatrix}
}}
{{eqn | l = B
| r = \begin {pmatrix} 1 & 1 \\ 1 & 1 \end {pmatrix}
}}
{{end-eqn}}
Both $A$ and $B$ are elements of the general linear group.
Then:
{{begin-eqn}}
{{eqn | l = A B
| r = \begin {pmatrix} 1 & 1 \\ 0 & ... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Ring Zero|zero]] is $0_K$ and [[Definition:Unity of Field|unity]] is $1_K$.
Let $\GL {n, K}$ be the [[Definition:General Linear Group|general linear group of order $n$ over $K$]].
Then $\GL {n, K}$ is not an [[Definition:Abelian Group|abel... | Let:
{{begin-eqn}}
{{eqn | l = A
| r = \begin {pmatrix} 1 & 1 \\ 0 & 1 \end {pmatrix}
}}
{{eqn | l = B
| r = \begin {pmatrix} 1 & 1 \\ 1 & 1 \end {pmatrix}
}}
{{end-eqn}}
Both $A$ and $B$ are [[Definition:Element|elements]] of the [[Definition:General Linear Group|general linear group]].
Then:
{{begin-e... | General Linear Group is not Abelian/Proof 2 | https://proofwiki.org/wiki/General_Linear_Group_is_not_Abelian | https://proofwiki.org/wiki/General_Linear_Group_is_not_Abelian/Proof_2 | [
"General Linear Group is not Abelian",
"General Linear Group",
"Abelian Groups"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Multiplicative Identity",
"Definition:General Linear Group",
"Definition:Abelian Group"
] | [
"Definition:Element",
"Definition:General Linear Group",
"Definition:General Linear Group",
"Definition:Abelian Group"
] |
proofwiki-16712 | Group of Unitary Matrices under Multiplication is not Abelian | Let $n > 1$ be a natural number.
Let $\map U n$ denote the unitary group.
Then $\map U n$ is not abelian. | ;Proof by Counterexample
{{Recall|Unitary Group}}
{{:Definition:Unitary Group}}
The matrices:
:$A = \begin {pmatrix} 1 & 0 \\ 0 & -1 \end {pmatrix}$
and
:$B = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix}$
are both real and symmetric.
Hence, by Real Symmetric Matrix is Hermitian, both are Hermitian.
Furthermore, both ... | Let $n > 1$ be a [[Definition:Natural Number|natural number]].
Let $\map U n$ denote the [[Definition:Unitary Group|unitary group]].
Then $\map U n$ is not [[Definition:Abelian Group|abelian]]. | ;[[Proof by Counterexample]]
{{Recall|Unitary Group}}
{{:Definition:Unitary Group}}
The [[Definition:Matrix|matrices]]:
:$A = \begin {pmatrix} 1 & 0 \\ 0 & -1 \end {pmatrix}$
and
:$B = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix}$
are both [[Definition:Real Matrix|real]] and [[Definition:Symmetric Matrix|symmetric]... | Group of Unitary Matrices under Multiplication is not Abelian | https://proofwiki.org/wiki/Group_of_Unitary_Matrices_under_Multiplication_is_not_Abelian | https://proofwiki.org/wiki/Group_of_Unitary_Matrices_under_Multiplication_is_not_Abelian | [
"Unitary Groups",
"Abelian Groups"
] | [
"Definition:Natural Numbers",
"Definition:Unitary Group",
"Definition:Abelian Group"
] | [
"Proof by Counterexample",
"Definition:Matrix",
"Definition:Real Matrix",
"Definition:Symmetric Matrix",
"Real Symmetric Matrix is Hermitian",
"Definition:Hermitian Matrix",
"Definition:Inverse Matrix",
"Definition:Unitary Matrix",
"Definition:Matrix Product",
"Definition:Commutative/Operation",
... |
proofwiki-16713 | Complex Numbers form Vector Space over Themselves | The set of complex numbers $\C$, with the operations of addition and multiplication, forms a vector space. | Let the field of complex numbers be denoted $\struct {\C, +, \times}$.
By Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is an abelian group.
From Complex Multiplication Distributes over Addition:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in \C
| l = x \times \paren {y + z}
|... | The [[Definition:Complex Number|set of complex numbers]] $\C$, with the operations of [[Definition:Complex Addition|addition]] and [[Definition:Complex Multiplication|multiplication]], forms a [[Definition:Vector Space|vector space]]. | Let the [[Definition:Field of Complex Numbers|field of complex numbers]] be denoted $\struct {\C, +, \times}$.
By [[Complex Numbers under Addition form Infinite Abelian Group]], $\struct {\C, +}$ is an [[Definition:Abelian Group|abelian group]].
From [[Complex Multiplication Distributes over Addition]]:
{{begin-eqn}... | Complex Numbers form Vector Space over Themselves | https://proofwiki.org/wiki/Complex_Numbers_form_Vector_Space_over_Themselves | https://proofwiki.org/wiki/Complex_Numbers_form_Vector_Space_over_Themselves | [
"Linear Algebra",
"Complex Numbers",
"Examples of Vector Spaces"
] | [
"Definition:Complex Number",
"Definition:Addition/Complex Numbers",
"Definition:Multiplication/Complex Numbers",
"Definition:Vector Space"
] | [
"Definition:Field of Complex Numbers",
"Complex Numbers under Addition form Infinite Abelian Group",
"Definition:Abelian Group",
"Complex Multiplication Distributes over Addition",
"Complex Multiplication is Associative",
"Complex Multiplication Identity is One",
"Definition:Vector Space"
] |
proofwiki-16714 | Quaternions form Vector Space over Reals | Let $\R$ be the set of real numbers.
Let $\H$ be the set of quaternions.
Then the $\R$-module $\H$ is a vector space. | Recall that Real Numbers form Field.
Thus $\R$ is a division ring {{afortiori}}.
Thus we only need to show that $\R$-module $\H$ is a unitary module, by demonstrating the unitary (left) module axioms:
{{begin-axiom}}
{{axiom | n = \text {UM} 1
| q = x, y \in \H: \forall \lambda \in \R
| ml= \lambda \par... | Let $\R$ be the set of [[Definition:Real Number|real numbers]].
Let $\H$ be the set of [[Definition:Quaternion|quaternions]].
Then the [[Definition:Module over Ring|$\R$-module]] $\H$ is a [[Definition:Vector Space|vector space]]. | Recall that [[Real Numbers form Field]].
Thus $\R$ is a [[Definition:Division Ring|division ring]] {{afortiori}}.
Thus we only need to show that [[Definition:Module over Ring|$\R$-module]] $\H$ is a [[Definition:Unitary Module over Ring|unitary module]], by demonstrating the [[Axiom:Unitary Left Module Axioms|unitary... | Quaternions form Vector Space over Reals | https://proofwiki.org/wiki/Quaternions_form_Vector_Space_over_Reals | https://proofwiki.org/wiki/Quaternions_form_Vector_Space_over_Reals | [
"Quaternions",
"Examples of Vector Spaces"
] | [
"Definition:Real Number",
"Definition:Quaternion",
"Definition:Module over Ring",
"Definition:Vector Space"
] | [
"Real Numbers form Field",
"Definition:Division Ring",
"Definition:Module over Ring",
"Definition:Unitary Module over Ring",
"Axiom:Unitary Left Module Axioms",
"Quaternion Multiplication Distributes over Addition",
"Quaternion Multiplication is Associative",
"Multiplicative Identity for Quaternions",... |
proofwiki-16715 | Quaternions form Vector Space over Themselves | The set of quaternions $\H$, with the operations of addition and multiplication, forms a vector space. | Let the set of quaternions be denoted $\struct {\H, +, \times}$.
From Quaternions form Skew Field, the algebraic structure $\struct {\H, +, \times}$ is a skew field.
By definition, a skew field is a division ring.
{{ProofWanted}} | The [[Definition:Quaternion|set of quaternions]] $\H$, with the operations of [[Definition:Quaternion Addition|addition]] and [[Definition:Quaternion Multiplication|multiplication]], forms a [[Definition:Vector Space|vector space]]. | Let the [[Definition:Quaternion|set of quaternions]] be denoted $\struct {\H, +, \times}$.
From [[Quaternions form Skew Field]], the [[Definition:Algebraic Structure with Two Operations|algebraic structure]] $\struct {\H, +, \times}$ is a [[Definition:Skew Field|skew field]].
By definition, a [[Definition:Skew Field|... | Quaternions form Vector Space over Themselves | https://proofwiki.org/wiki/Quaternions_form_Vector_Space_over_Themselves | https://proofwiki.org/wiki/Quaternions_form_Vector_Space_over_Themselves | [
"Linear Algebra",
"Quaternions",
"Examples of Vector Spaces"
] | [
"Definition:Quaternion",
"Definition:Quaternion/Addition",
"Definition:Quaternion/Multiplication",
"Definition:Vector Space"
] | [
"Definition:Quaternion",
"Quaternions form Skew Field",
"Definition:Algebraic Structure/Two Operations",
"Definition:Skew Field",
"Definition:Skew Field",
"Definition:Division Ring"
] |
proofwiki-16716 | Set of Matrices under Entrywise Addition forms Vector Space | Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let $\map {\MM_\GF} {m, n}$ be the $m \times n$ matrix space over a field $\GF$.
Let $\struct {\map {\MM_\GF} {m, n}, +}$ denote the abelian group formed from $\map {\MM_\GF} {m, n}$ under matrix entrywise addition.
Let $\struct {\map {\MM_\GF} {m, n}, +, \times}_... | It is to be demonstrated that $\map {\MM_\GF} {m, n}$ satisfies all the vector space axioms.
First we note that from Matrix Entrywise Addition forms Abelian Group, $\struct {\map {\MM_\GF} {m, n}, +}$ is indeed an abelian group.
Hence the underlying group axioms $\text G 0$ to $\text G 3$ and $\text C$ are satisfied. | Let $m, n \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
Let $\map {\MM_\GF} {m, n}$ be the [[Definition:Matrix Space|$m \times n$ matrix space]] over a [[Definition:Field (Abstract Algebra)|field]] $\GF$.
Let $\struct {\map {\MM_\GF} {m, n}, +}$ denote the [[Definition:Abelian... | It is to be demonstrated that $\map {\MM_\GF} {m, n}$ satisfies all the [[Axiom:Vector Space Axioms|vector space axioms]].
First we note that from [[Matrix Entrywise Addition forms Abelian Group]], $\struct {\map {\MM_\GF} {m, n}, +}$ is indeed an [[Definition:Abelian Group|abelian group]].
Hence the [[Axiom:Underly... | Set of Matrices under Entrywise Addition forms Vector Space | https://proofwiki.org/wiki/Set_of_Matrices_under_Entrywise_Addition_forms_Vector_Space | https://proofwiki.org/wiki/Set_of_Matrices_under_Entrywise_Addition_forms_Vector_Space | [
"Set of Matrices under Entrywise Addition forms Vector Space",
"Vector Spaces of Matrices",
"Examples of Vector Spaces",
"Matrix Entrywise Addition"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Matrix Space",
"Definition:Field (Abstract Algebra)",
"Definition:Abelian Group",
"Definition:Matrix Entrywise Addition",
"Definition:Unitary Module over Ring",
"Definition:Matrix Scalar Product",
"Definition:Vector Space"
] | [
"Axiom:Vector Space Axioms",
"Matrix Entrywise Addition forms Abelian Group",
"Definition:Abelian Group",
"Axiom:Underlying Group Axioms",
"Axiom:Vector Space Axioms"
] |
proofwiki-16717 | Simple Events are Mutually Exclusive | Let $\EE$ be an experiment.
Let $e_1$ and $e_2$ be distinct simple events in $\EE$.
Then $e_1$ and $e_2$ are mutually exclusive. | By definition of simple event:
{{begin-eqn}}
{{eqn | l = e_1
| r = \set {s_1}
}}
{{eqn | l = e_2
| r = \set {s_2}
}}
{{end-eqn}}
for some elementary events $s_1$ and $s_2$ of $\EE$ such that $s_1 \ne s_2$.
It follows that:
{{begin-eqn}}
{{eqn | l = e_1 \cap e_2
| r = \set {s_1} \cap \set {s_2}
|... | Let $\EE$ be an [[Definition:Experiment|experiment]].
Let $e_1$ and $e_2$ be [[Definition:Distinct Objects|distinct]] [[Definition:Simple Event|simple events]] in $\EE$.
Then $e_1$ and $e_2$ are [[Definition:Mutually Exclusive Events|mutually exclusive]]. | By definition of [[Definition:Simple Event|simple event]]:
{{begin-eqn}}
{{eqn | l = e_1
| r = \set {s_1}
}}
{{eqn | l = e_2
| r = \set {s_2}
}}
{{end-eqn}}
for some [[Definition:Elementary Event|elementary events]] $s_1$ and $s_2$ of $\EE$ such that $s_1 \ne s_2$.
It follows that:
{{begin-eqn}}
{{eqn ... | Simple Events are Mutually Exclusive | https://proofwiki.org/wiki/Simple_Events_are_Mutually_Exclusive | https://proofwiki.org/wiki/Simple_Events_are_Mutually_Exclusive | [
"Events"
] | [
"Definition:Experiment",
"Definition:Distinct/Plural",
"Definition:Event/Simple Event",
"Definition:Disjoint Events"
] | [
"Definition:Event/Simple Event",
"Definition:Elementary Event",
"Definition:Disjoint Events"
] |
proofwiki-16718 | Non-Trivial Event is Union of Simple Events | Let $\EE$ be an experiment.
Let $e$ be an event in $\EE$ such that $e \ne \O$.
That is, such that $e$ is non-trivial.
Then $e$ can be expressed as the union of a set of simple events in $\EE$. | By definition of event, $e$ is a subset of the sample space $\Omega$ of $\EE$.
By hypothesis:
:$e \ne \O$
and so:
:$\exists s \in \Omega: s \in e$
Let $S$ be the set defined as:
:$S = \set {\set s: s \in e}$
By Union is Smallest Superset: Set of Sets it follows that:
:$\ds \bigcup S \subseteq e$
Let $x \in e$.
Then by ... | Let $\EE$ be an [[Definition:Experiment|experiment]].
Let $e$ be an [[Definition:Event|event]] in $\EE$ such that $e \ne \O$.
That is, such that $e$ is [[Definition:Non-Trivial Event|non-trivial]].
Then $e$ can be expressed as the [[Definition:Set Union|union]] of a [[Definition:Set|set]] of [[Definition:Simple Eve... | By definition of [[Definition:Event|event]], $e$ is a [[Definition:Subset|subset]] of the [[Definition:Sample Space|sample space]] $\Omega$ of $\EE$.
[[Definition:By Hypothesis|By hypothesis]]:
:$e \ne \O$
and so:
:$\exists s \in \Omega: s \in e$
Let $S$ be the [[Definition:Set|set]] defined as:
:$S = \set {\set s: ... | Non-Trivial Event is Union of Simple Events | https://proofwiki.org/wiki/Non-Trivial_Event_is_Union_of_Simple_Events | https://proofwiki.org/wiki/Non-Trivial_Event_is_Union_of_Simple_Events | [
"Events"
] | [
"Definition:Experiment",
"Definition:Event",
"Definition:Non-Trivial Event",
"Definition:Set Union",
"Definition:Set",
"Definition:Event/Simple Event"
] | [
"Definition:Event",
"Definition:Subset",
"Definition:Sample Space",
"Definition:By Hypothesis",
"Definition:Set",
"Union is Smallest Superset/Set of Sets",
"Singleton of Element is Subset",
"Definition:Set Union",
"Definition:Set Equality"
] |
proofwiki-16719 | Sample Space is Union of All Distinct Simple Events | Let $\EE$ be an experiment.
Let $\Omega$ denote the sample space of $\EE$.
Then $\Omega$ is the union of the set of simple events in $\EE$. | By Set is Subset of Itself:
:$\Omega \subseteq \Omega$
That is, $\Omega$ is itself an event in $\EE$.
The result as an application of Non-Trivial Event is Union of Simple Events.
{{qed}} | Let $\EE$ be an [[Definition:Experiment|experiment]].
Let $\Omega$ denote the [[Definition:Sample Space|sample space]] of $\EE$.
Then $\Omega$ is the [[Definition:Set Union|union]] of the [[Definition:Set|set]] of [[Definition:Simple Event|simple events]] in $\EE$. | By [[Set is Subset of Itself]]:
:$\Omega \subseteq \Omega$
That is, $\Omega$ is itself an [[Definition:Event|event]] in $\EE$.
The result as an application of [[Non-Trivial Event is Union of Simple Events]].
{{qed}} | Sample Space is Union of All Distinct Simple Events | https://proofwiki.org/wiki/Sample_Space_is_Union_of_All_Distinct_Simple_Events | https://proofwiki.org/wiki/Sample_Space_is_Union_of_All_Distinct_Simple_Events | [
"Events"
] | [
"Definition:Experiment",
"Definition:Sample Space",
"Definition:Set Union",
"Definition:Set",
"Definition:Event/Simple Event"
] | [
"Set is Subset of Itself",
"Definition:Event",
"Non-Trivial Event is Union of Simple Events"
] |
proofwiki-16720 | Equivalence of Definitions of Probability Measure | Let $\EE$ be an experiment.
{{TFAE|def = Probability Measure}} | {{ProofWanted}}
Category:Probability Measures
3h2zo3dlivx9v7189tggd120brw05hd | Let $\EE$ be an [[Definition:Experiment|experiment]].
{{TFAE|def = Probability Measure}} | {{ProofWanted}}
[[Category:Probability Measures]]
3h2zo3dlivx9v7189tggd120brw05hd | Equivalence of Definitions of Probability Measure | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Probability_Measure | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Probability_Measure | [
"Probability Measures"
] | [
"Definition:Experiment"
] | [
"Category:Probability Measures"
] |
proofwiki-16721 | Total Probability Theorem/Conditional Probabilities | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\set {B_1, B_2, \ldots}$ be a partition of $\Omega$ such that $\forall i: \map \Pr {B_i} > 0$.
Let $C \in \Sigma$ be an event independent to any of the $B_i$.
Let $\map \Pr C > 0$.
Then:
:$\ds \forall A \in \Sigma: \condprob A C = \sum_i \condprob A {C \c... | First define $Q_C := \condprob {\, \cdot} C$.
Then, from Conditional Probability Defines Probability Space, $\struct {\Omega, \Sigma, Q_C}$ is a probability space.
Moreover:
{{begin-eqn}}
{{eqn | q = \forall i
| l = \map {Q_C} {B_i}
| r = \map \Pr {B_i \mid C}
}}
{{eqn | r = \map \Pr {B_i}
| c = $C$ a... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $\set {B_1, B_2, \ldots}$ be a [[Definition:Partition (Probability Theory)|partition of $\Omega$]] such that $\forall i: \map \Pr {B_i} > 0$.
Let $C \in \Sigma$ be an event independent to any of the $B_i$.
Let $\map \Pr ... | First define $Q_C := \condprob {\, \cdot} C$.
Then, from [[Conditional Probability Defines Probability Space]], $\struct {\Omega, \Sigma, Q_C}$ is a [[Definition:Probability Space|probability space]].
Moreover:
{{begin-eqn}}
{{eqn | q = \forall i
| l = \map {Q_C} {B_i}
| r = \map \Pr {B_i \mid C}
}}
{{eqn... | Total Probability Theorem/Conditional Probabilities | https://proofwiki.org/wiki/Total_Probability_Theorem/Conditional_Probabilities | https://proofwiki.org/wiki/Total_Probability_Theorem/Conditional_Probabilities | [
"Total Probability Theorem"
] | [
"Definition:Probability Space",
"Definition:Partition (Probability Theory)"
] | [
"Conditional Probability Defines Probability Space",
"Definition:Probability Space",
"Definition:Independent Events/Definition 1",
"Total Probability Theorem",
"Total Probability Theorem",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Definition:Independent Events/Definition 1",... |
proofwiki-16722 | Product Space is Product in Category of Topological Spaces | Let $\mathbf {Top}$ be the category of topological spaces.
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\struct {\XX, \tau}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Then $\struct {\XX, ... | {{ProofWanted|Use Continuous Mapping to Product Space/General Result}}
Category:Category of Topological Spaces
Category:Product Topology
Category:Products (Category Theory)
auofr2hdeswk67x1womd9iyj6pl1yki | Let $\mathbf {Top}$ be the [[Definition:Category of Topological Spaces|category of topological spaces]].
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set... | {{ProofWanted|Use [[Continuous Mapping to Product Space/General Result]]}}
[[Category:Category of Topological Spaces]]
[[Category:Product Topology]]
[[Category:Products (Category Theory)]]
auofr2hdeswk67x1womd9iyj6pl1yki | Product Space is Product in Category of Topological Spaces | https://proofwiki.org/wiki/Product_Space_is_Product_in_Category_of_Topological_Spaces | https://proofwiki.org/wiki/Product_Space_is_Product_in_Category_of_Topological_Spaces | [
"Category of Topological Spaces",
"Product Topology",
"Products (Category Theory)"
] | [
"Definition:Category of Topological Spaces",
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Product (Category Theory)"
] | [
"Continuous Mapping to Product Space/General Result",
"Category:Category of Topological Spaces",
"Category:Product Topology",
"Category:Products (Category Theory)"
] |
proofwiki-16723 | Strictly Positive Rational Numbers are Closed under Addition | Let $\Q_{>0}$ denote the set of strictly positive rational numbers:
:$\Q_{>0} := \set {x \in \Q: x > 0}$
where $\Q$ denotes the set of rational numbers.
The algebraic structure $\struct {\Q_{>0}, +}$ is closed in the sense that:
:$\forall a, b \in \Q_{>0}: a + b \in \Q_{>0}$
where $+$ denotes rational addition. | Let $a$ and $b$ be expressed in canonical form:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of rational addition:
:$\dfrac {p_1} {q_1} + \dfrac {p_2} {q_2} = \dfrac {p_1 q_2 + p_... | Let $\Q_{>0}$ denote the [[Definition:Set|set]] of [[Definition:Strictly Positive Rational Number|strictly positive rational numbers]]:
:$\Q_{>0} := \set {x \in \Q: x > 0}$
where $\Q$ denotes the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]].
The [[Definition:Algebraic Structure|algebraic ... | Let $a$ and $b$ be expressed in [[Definition:Canonical Form of Rational Number|canonical form]]:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of [[Definition:Rational Addition|... | Strictly Positive Rational Numbers are Closed under Addition | https://proofwiki.org/wiki/Strictly_Positive_Rational_Numbers_are_Closed_under_Addition | https://proofwiki.org/wiki/Strictly_Positive_Rational_Numbers_are_Closed_under_Addition | [
"Rational Addition"
] | [
"Definition:Set",
"Definition:Strictly Positive/Rational Number",
"Definition:Set",
"Definition:Rational Number",
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Addition/Rational Numbers"
] | [
"Definition:Rational Number/Canonical Form",
"Definition:Addition/Rational Numbers",
"Integers form Ordered Integral Domain"
] |
proofwiki-16724 | Strictly Positive Rational Numbers are Closed under Multiplication | :$\forall a, b \in \Q_{>0}: a b \in \Q_{>0}$ | Let $a$ and $b$ be expressed in canonical form:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of rational multiplication:
:$\dfrac {p_1} {q_1} \times \dfrac {p_2} {q_2} = \dfrac {p... | :$\forall a, b \in \Q_{>0}: a b \in \Q_{>0}$ | Let $a$ and $b$ be expressed in [[Definition:Canonical Form of Rational Number|canonical form]]:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of [[Definition:Rational Multiplic... | Strictly Positive Rational Numbers are Closed under Multiplication | https://proofwiki.org/wiki/Strictly_Positive_Rational_Numbers_are_Closed_under_Multiplication | https://proofwiki.org/wiki/Strictly_Positive_Rational_Numbers_are_Closed_under_Multiplication | [
"Rational Multiplication",
"Algebraic Closure"
] | [] | [
"Definition:Rational Number/Canonical Form",
"Definition:Multiplication/Rational Numbers",
"Integers form Ordered Integral Domain"
] |
proofwiki-16725 | Set of Rationals Less than Root 2 has no Greatest Element | Let $A$ be the set of all positive rational numbers $p$ such that $p^2 < 2$.
Then $A$ has no greatest element. | {{AimForCont}} $p \in A$ is the greatest element of $A$.
Then by definition of $A$:
:$p^2 < 2$
Let $h \in \Q$ be a rational number such that $0 < h < 1$ such that:
:$h < \dfrac {2 - p^2} {2 p + 1}$
This is always possible, because by definition $2 - p^2 > 0$ and $2 p + 1 > 0$.
Let $q = p + h$.
Then $q > p$, and:
{{begi... | Let $A$ be the [[Definition:Set|set]] of all [[Definition:Positive Rational Number|positive rational numbers]] $p$ such that $p^2 < 2$.
Then $A$ has no [[Definition:Greatest Element|greatest element]]. | {{AimForCont}} $p \in A$ is the [[Definition:Greatest Element|greatest element]] of $A$.
Then by definition of $A$:
:$p^2 < 2$
Let $h \in \Q$ be a [[Definition:Rational Number|rational number]] such that $0 < h < 1$ such that:
:$h < \dfrac {2 - p^2} {2 p + 1}$
This is always possible, because by definition $2 - p^2 ... | Set of Rationals Less than Root 2 has no Greatest Element | https://proofwiki.org/wiki/Set_of_Rationals_Less_than_Root_2_has_no_Greatest_Element | https://proofwiki.org/wiki/Set_of_Rationals_Less_than_Root_2_has_no_Greatest_Element | [
"Rational Numbers"
] | [
"Definition:Set",
"Definition:Positive/Rational Number",
"Definition:Greatest Element"
] | [
"Definition:Greatest Element",
"Definition:Rational Number",
"Definition:Contradiction",
"Definition:Greatest Element",
"Definition:Greatest Element"
] |
proofwiki-16726 | Set of Rationals Greater than Root 2 has no Smallest Element | Let $B$ be the set of all positive rational numbers $p$ such that $p^2 > 2$.
Then $B$ has no smallest element. | {{AimForCont}} $p \in B$ is the smallest element of $B$.
Then by definition of $B$:
:$p^2 > 2$
Let $q = p - \dfrac {p_2 - 2} {2 p}$.
Then $q > p$, and:
{{begin-eqn}}
{{eqn | l = q
| r = p - \dfrac {p_2 - 2} {2 p}
| c =
}}
{{eqn | r = \dfrac p 2 + \dfrac 1 p
| c =
}}
{{end-eqn}}
Hence:
:$0 < q < p$
a... | Let $B$ be the [[Definition:Set|set]] of all [[Definition:Positive Rational Number|positive rational numbers]] $p$ such that $p^2 > 2$.
Then $B$ has no [[Definition:Smallest Element|smallest element]]. | {{AimForCont}} $p \in B$ is the [[Definition:Smallest Element|smallest element]] of $B$.
Then by definition of $B$:
:$p^2 > 2$
Let $q = p - \dfrac {p_2 - 2} {2 p}$.
Then $q > p$, and:
{{begin-eqn}}
{{eqn | l = q
| r = p - \dfrac {p_2 - 2} {2 p}
| c =
}}
{{eqn | r = \dfrac p 2 + \dfrac 1 p
| c = ... | Set of Rationals Greater than Root 2 has no Smallest Element | https://proofwiki.org/wiki/Set_of_Rationals_Greater_than_Root_2_has_no_Smallest_Element | https://proofwiki.org/wiki/Set_of_Rationals_Greater_than_Root_2_has_no_Smallest_Element | [
"Rational Numbers"
] | [
"Definition:Set",
"Definition:Positive/Rational Number",
"Definition:Smallest Element"
] | [
"Definition:Smallest Element",
"Definition:Contradiction",
"Definition:Smallest Element",
"Definition:Smallest Element"
] |
proofwiki-16727 | Rational Number Not in Cut is Greater than Element of Cut | Let $\alpha$ be a cut.
Let $p \in \alpha$.
Let $q \in \Q$ such that $q \notin \alpha$.
Then $q > p$. | We have been given that $p \in \alpha$ while $q \notin \alpha$.
{{AimForCont}} $q \le p$.
If $q = p$ then $q \in \alpha$ immediately.
If $q < p$ then $q \in \alpha$ by property $(2)$ of the definition of cut.
In either case $q \in \alpha$ which contradicts $q \notin \alpha$.
Hence by Proof by Contradiction it follows t... | Let $\alpha$ be a [[Definition:Cut (Analysis)|cut]].
Let $p \in \alpha$.
Let $q \in \Q$ such that $q \notin \alpha$.
Then $q > p$. | We have been given that $p \in \alpha$ while $q \notin \alpha$.
{{AimForCont}} $q \le p$.
If $q = p$ then $q \in \alpha$ immediately.
If $q < p$ then $q \in \alpha$ by property $(2)$ of the definition of [[Definition:Cut (Analysis)|cut]].
In either case $q \in \alpha$ which [[Definition:Contradiction|contradicts]] ... | Rational Number Not in Cut is Greater than Element of Cut | https://proofwiki.org/wiki/Rational_Number_Not_in_Cut_is_Greater_than_Element_of_Cut | https://proofwiki.org/wiki/Rational_Number_Not_in_Cut_is_Greater_than_Element_of_Cut | [
"Cuts"
] | [
"Definition:Cut (Analysis)"
] | [
"Definition:Cut (Analysis)",
"Definition:Contradiction",
"Proof by Contradiction"
] |
proofwiki-16728 | Rational Cut has Smallest Upper Number | Let $r \in \Q$ be rational.
Let $\alpha$ be the rational cut consisting of all rational numbers $p$ such that $p < r$.
Then $\alpha$ is indeed a cut, and has a smallest upper number that is $r$. | That $\alpha$ fulfils conditions $(1)$ and $(2)$ of the definition of a cut follows directly from that definition.
Then it is noted that if $p \in \alpha$ then from Mediant is Between:
:$p < \dfrac {p + r} 2 < r$
and so $p$ cannot be the greatest element of $\alpha$.
Hence it is seen that $\alpha$ fulfils all condition... | Let $r \in \Q$ be [[Definition:Rational Number|rational]].
Let $\alpha$ be the [[Definition:Rational Cut|rational cut]] consisting of all [[Definition:Rational Number|rational numbers]] $p$ such that $p < r$.
Then $\alpha$ is indeed a [[Definition:Cut (Analysis)|cut]], and has a [[Definition:Smallest Element|smalles... | That $\alpha$ fulfils conditions $(1)$ and $(2)$ of the definition of a [[Definition:Cut (Analysis)|cut]] follows directly from that definition.
Then it is noted that if $p \in \alpha$ then from [[Mediant is Between]]:
:$p < \dfrac {p + r} 2 < r$
and so $p$ cannot be the [[Definition:Greatest Element|greatest element]... | Rational Cut has Smallest Upper Number | https://proofwiki.org/wiki/Rational_Cut_has_Smallest_Upper_Number | https://proofwiki.org/wiki/Rational_Cut_has_Smallest_Upper_Number | [
"Cuts"
] | [
"Definition:Rational Number",
"Definition:Cut (Analysis)/Rational",
"Definition:Rational Number",
"Definition:Cut (Analysis)",
"Definition:Smallest Element",
"Definition:Cut (Analysis)/Upper Number"
] | [
"Definition:Cut (Analysis)",
"Mediant is Between",
"Definition:Greatest Element",
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)/Upper Number",
"Definition:Cut (Analysis)/Upper Number",
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)/Upper Number",
"Proof by Contradiction",
"Defin... |
proofwiki-16729 | Natural Basis of Product Topology | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Then the natural basis on $X$ is the set $\BB$ of cartesian products of ... | Let $\NN$ denote the natural basis on $X$.
By definition of the natural basis, $\NN$ is the basis generated by:
:$\SS = \set {\pr_i^{-1} \sqbrk U: i \in I, \, U \in \tau_i}$
where for each $i \in I$, $\pr_i: X \to X_i$ denotes the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\... | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $X$ be the [[Definition:Cartesian Product of Family|cartesian product]] of $\family {X_... | Let $\NN$ denote the [[Definition:Natural Basis of Product Topology|natural basis]] on $X$.
By definition of the [[Definition:Natural Basis of Product Topology|natural basis]], $\NN$ is the [[Synthetic Basis formed from Synthetic Sub-Basis|basis generated]] by:
:$\SS = \set {\pr_i^{-1} \sqbrk U: i \in I, \, U \in \tau... | Natural Basis of Product Topology | https://proofwiki.org/wiki/Natural_Basis_of_Product_Topology | https://proofwiki.org/wiki/Natural_Basis_of_Product_Topology | [
"Product Topology",
"Natural Basis of Product Topology"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Cartesian Product/Family of Sets",
"Definition:Product Topology/Natural Basis",
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Index"
] | [
"Definition:Product Topology/Natural Basis",
"Definition:Product Topology/Natural Basis",
"Synthetic Basis formed from Synthetic Sub-Basis",
"Definition:Projection (Mapping Theory)/Family of Sets",
"Synthetic Basis formed from Synthetic Sub-Basis",
"Synthetic Basis formed from Synthetic Sub-Basis"
] |
proofwiki-16730 | Natural Basis of Product Topology/Finite Product | Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $\struct {X_k, \tau_k}$ be topological spaces.
Let $\ds X = \prod_{k \mathop = 1}^n X_k$ be the (finite) cartesian product of $X_1, \ldots, X_n$.
Then the natural basis on $X$ is:
:$\BB = \set {\ds \prod_{k \mathop = 1}^n U_k : \forall k : U_k \in \tau_k}$ | From Natural Basis of Product Topology, the natural basis on $X$ is the set $\BB$ of cartesian products of the form:
:$\ds \prod_{i \mathop \in I} U_i$
where:
:for all $k = 1, \dots, n: U_k \in \tau_i$
:for all but finitely many indices $k : U_k = X_k$
Let:
:$\BB' = \set {\ds \prod_{k \mathop = 1}^n U_k : \forall k : U... | Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $\struct {X_k, \tau_k}$ be [[Definition:Topological Space|topological spaces]].
Let $\ds X = \prod_{k \mathop = 1}^n X_k$ be the [[Definition:Finite Cartesian Product|(finite) cartesian product]] of $X_1, \ldots, X_n$.
Then the [[Definition:Natural Basis of P... | From [[Natural Basis of Product Topology]], the [[Definition:Natural Basis of Product Topology|natural basis]] on $X$ is the [[Definition:Set|set]] $\BB$ of [[Definition:Cartesian Product|cartesian products]] of the form:
:$\ds \prod_{i \mathop \in I} U_i$
where:
:for all $k = 1, \dots, n: U_k \in \tau_i$
:for all but ... | Natural Basis of Product Topology/Finite Product | https://proofwiki.org/wiki/Natural_Basis_of_Product_Topology/Finite_Product | https://proofwiki.org/wiki/Natural_Basis_of_Product_Topology/Finite_Product | [
"Natural Basis of Product Topology"
] | [
"Definition:Topological Space",
"Definition:Cartesian Product/Finite",
"Definition:Product Topology/Natural Basis"
] | [
"Natural Basis of Product Topology",
"Definition:Product Topology/Natural Basis",
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Index",
"Definition:Finite Set",
"Definition:Index",
"Definition:Set Equality",
"Category:Natural Basis of Product Topology"
] |
proofwiki-16731 | Box Topology may not be Coarsest Topology such that Projections are Continuous | Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the box topology on $X$.
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$t... | From Projection from Box Topology is Continuous, the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous.
From Product Topology is Coarsest Topology such that Projections are Continuous, the product topology is the coarsest topology such that the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous... | Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Topological Space|topological spaces]].
Let $X$ be the [[Definition:Cartesian Product of Family|cartesian product]] of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \i... | From [[Projection from Box Topology is Continuous]], the [[Definition:Projection|projections]] $\family{\pr_i} _{i \mathop \in I}$ are [[Definition:Continuous|continuous]].
From [[Product Topology is Coarsest Topology such that Projections are Continuous]], the [[Definition:Product Topology|product topology]] is the [... | Box Topology may not be Coarsest Topology such that Projections are Continuous | https://proofwiki.org/wiki/Box_Topology_may_not_be_Coarsest_Topology_such_that_Projections_are_Continuous | https://proofwiki.org/wiki/Box_Topology_may_not_be_Coarsest_Topology_such_that_Projections_are_Continuous | [
"Box Topology"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Cartesian Product/Family of Sets",
"Definition:Box Topology",
"Definition:Projection (Mapping Theory)/Family of Sets",
"Definition:Coarser Topology",
"Definition:Projection",
"Definition:Continuous"
] | [
"Projection from Box Topology is Continuous",
"Definition:Projection",
"Definition:Continuous",
"Product Topology is Coarsest Topology such that Projections are Continuous",
"Definition:Product Topology",
"Definition:Coarser Topology",
"Definition:Projection",
"Definition:Continuous",
"Definition:Pr... |
proofwiki-16732 | Box Topology may not form Categorical Product in the Category of Topological Spaces | Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the box topology on $X$.
Then $\struct{X, \tau}$ may not be the categorical product... | There exists such a $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ that the box topology $\tau$ on $X$ differs from the product topology, since:
:Box Topology may not be Coarsest Topology such that Projections are Continuous
:Product Topology is Coarsest Topology such that Projections are Continuous
Then $\struct{X... | Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Topological Space|topological spaces]].
Let $X$ be the [[Definition:Cartesian Product of Family|cartesian product]] of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \i... | There exists such a $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ that the [[Definition:Box Topology|box topology]] $\tau$ on $X$ differs from the [[Definition:Product Topology|product topology]], since:
:[[Box Topology may not be Coarsest Topology such that Projections are Continuous]]
:[[Product Topology is Coar... | Box Topology may not form Categorical Product in the Category of Topological Spaces | https://proofwiki.org/wiki/Box_Topology_may_not_form_Categorical_Product_in_the_Category_of_Topological_Spaces | https://proofwiki.org/wiki/Box_Topology_may_not_form_Categorical_Product_in_the_Category_of_Topological_Spaces | [
"Box Topology",
"Products (Category Theory)"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Cartesian Product/Family of Sets",
"Definition:Box Topology",
"Definition:Product (Category Theory)",
"Definition:Category of Topological Spaces"
] | [
"Definition:Box Topology",
"Definition:Product Topology",
"Box Topology may not be Coarsest Topology such that Projections are Continuous",
"Product Topology is Coarsest Topology such that Projections are Continuous",
"Definition:Product (Category Theory)",
"Product Space is Product in Category of Topolog... |
proofwiki-16733 | Ordering on Cuts satisfies Trichotomy Law | Let $\alpha$ and $\beta$ be cuts.
Then exactly one of the following applies:
{{begin-eqn}}
{{eqn | n = 1
| l = \alpha
| o = <
| r = \beta
}}
{{eqn | n = 2
| l = \alpha
| r = \beta
}}
{{eqn | n = 3
| l = \alpha
| o = >
| r = \beta
}}
{{end-eqn}}
where $<$ and so $>$ denote... | Let $\alpha = \beta$.
By definition of equality of cuts:
:$p \in \alpha \iff p \in \beta$
By definition of strict ordering of cuts it follows that neither $\alpha < \beta$ or $\alpha > \beta$.
{{AimForCont}} both $\alpha < \beta$ and $\alpha > \beta$.
Because $\alpha < \beta$ there exists $p \in \Q$ such that:
:$p \in ... | Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]].
Then exactly one of the following applies:
{{begin-eqn}}
{{eqn | n = 1
| l = \alpha
| o = <
| r = \beta
}}
{{eqn | n = 2
| l = \alpha
| r = \beta
}}
{{eqn | n = 3
| l = \alpha
| o = >
| r = \beta
}}
{{end-e... | Let $\alpha = \beta$.
By definition of [[Definition:Equality of Cuts|equality of cuts]]:
:$p \in \alpha \iff p \in \beta$
By definition of [[Definition:Strict Ordering of Cuts|strict ordering of cuts]] it follows that neither $\alpha < \beta$ or $\alpha > \beta$.
{{AimForCont}} both $\alpha < \beta$ and $\alpha > \b... | Ordering on Cuts satisfies Trichotomy Law | https://proofwiki.org/wiki/Ordering_on_Cuts_satisfies_Trichotomy_Law | https://proofwiki.org/wiki/Ordering_on_Cuts_satisfies_Trichotomy_Law | [
"Cuts"
] | [
"Definition:Cut (Analysis)",
"Definition:Ordering of Cuts/Strict",
"Definition:Ordering of Cuts",
"Definition:Total Ordering"
] | [
"Definition:Equality of Cuts",
"Definition:Ordering of Cuts/Strict",
"Rational Number Not in Cut is Greater than Element of Cut",
"Rational Numbers form Totally Ordered Field",
"Proof by Contradiction",
"Definition:Ordering of Cuts/Strict",
"Definition:Equality of Cuts"
] |
proofwiki-16734 | Ordering on Cuts is Transitive | Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let:
{{begin-eqn}}
{{eqn | n = 1
| l = \alpha
| o = <
| r = \beta
}}
{{eqn | n = 2
| l = \beta
| o = <
| r = \gamma
}}
{{end-eqn}}
where $<$ denotes the strict ordering of cuts:
:$\alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin... | We have that:
:$\alpha < \beta$
By definition of strict ordering of cuts:
:$\exists p \in \Q: p \in \beta, p \notin \alpha$
Similarly, we have that:
:$\beta < \gamma$
and so:
:$\exists q \in \Q: q \in \gamma, q \notin \beta$
We have by definition of a cut that:
:$p \in \beta$ and $q \notin \beta$ implies that $p < q$
T... | Let $\alpha$, $\beta$ and $\gamma$ be [[Definition:Cut (Analysis)|cuts]].
Let:
{{begin-eqn}}
{{eqn | n = 1
| l = \alpha
| o = <
| r = \beta
}}
{{eqn | n = 2
| l = \beta
| o = <
| r = \gamma
}}
{{end-eqn}}
where $<$ denotes the [[Definition:Strict Ordering of Cuts|strict ordering o... | We have that:
:$\alpha < \beta$
By definition of [[Definition:Strict Ordering of Cuts|strict ordering of cuts]]:
:$\exists p \in \Q: p \in \beta, p \notin \alpha$
Similarly, we have that:
:$\beta < \gamma$
and so:
:$\exists q \in \Q: q \in \gamma, q \notin \beta$
We have by definition of a [[Definition:Cut (Analys... | Ordering on Cuts is Transitive | https://proofwiki.org/wiki/Ordering_on_Cuts_is_Transitive | https://proofwiki.org/wiki/Ordering_on_Cuts_is_Transitive | [
"Cuts",
"Examples of Transitive Relations"
] | [
"Definition:Cut (Analysis)",
"Definition:Ordering of Cuts/Strict",
"Definition:Ordering of Cuts",
"Definition:Transitive Relation"
] | [
"Definition:Ordering of Cuts/Strict",
"Definition:Cut (Analysis)",
"Definition:Ordering of Cuts/Strict"
] |
proofwiki-16735 | Ordering on Cuts is Total | Let $\CC$ denote the set of cuts.
Let $<$ denote the strict ordering on cuts defined as:
:$\forall \alpha, \beta \in \CC: \alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin \beta$
Then $<$ is a (strict) total ordering on $\CC$. | Let $\struct {\CC, <}$ denote the relational structure defined from the above.
From Ordering on Cuts is Transitive, $<$ is a transitive relation on $\CC$.
From Ordering on Cuts satisfies Trichotomy Law, we have that:
:$\alpha < \beta \implies \lnot \paren {\beta < \alpha}$
demonstrating that $<$ is asymmetric.
Hence, b... | Let $\CC$ denote the [[Definition:Set|set]] of [[Definition:Cut (Analysis)|cuts]].
Let $<$ denote the [[Definition:Strict Ordering of Cuts|strict ordering on cuts]] defined as:
:$\forall \alpha, \beta \in \CC: \alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin \beta$
Then $<$ is a [[Definition:Strict Total... | Let $\struct {\CC, <}$ denote the [[Definition:Relational Structure|relational structure]] defined from the above.
From [[Ordering on Cuts is Transitive]], $<$ is a [[Definition:Transitive Relation|transitive relation]] on $\CC$.
From [[Ordering on Cuts satisfies Trichotomy Law]], we have that:
:$\alpha < \beta \impl... | Ordering on Cuts is Total | https://proofwiki.org/wiki/Ordering_on_Cuts_is_Total | https://proofwiki.org/wiki/Ordering_on_Cuts_is_Total | [
"Cuts"
] | [
"Definition:Set",
"Definition:Cut (Analysis)",
"Definition:Ordering of Cuts/Strict",
"Definition:Strict Total Ordering"
] | [
"Definition:Relational Structure",
"Ordering on Cuts is Transitive",
"Definition:Transitive Relation",
"Ordering on Cuts satisfies Trichotomy Law",
"Definition:Asymmetric Relation",
"Definition:Strict Ordering",
"Ordering on Cuts satisfies Trichotomy Law",
"Definition:Non-Comparable Elements",
"Defi... |
proofwiki-16736 | Sum of Cuts is Cut | Let $\alpha$ and $\beta$ be cuts.
Let $\gamma$ be the set of all rational numbers $r$ such that:
:$\exists p \in \alpha, q \in \beta: r = p + q$
Then $\gamma$ is also a cut.
Thus the operation of addition on the set of cuts is closed. | By definition of cut, neither $\alpha$ nor $\beta$ are empty.
Hence there exist $p \in \alpha$ and $q \in \beta$.
Hence there exists $r = p + q$ and so $\gamma$ is likewise not empty.
Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers.
Such $s$ and $t$ are... | Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]].
Let $\gamma$ be the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] $r$ such that:
:$\exists p \in \alpha, q \in \beta: r = p + q$
Then $\gamma$ is also a [[Definition:Cut (Analysis)|cut]].
Thus the operation of [[Definiti... | By definition of [[Definition:Cut (Analysis)|cut]], neither $\alpha$ nor $\beta$ are [[Definition:Empty Set|empty]].
Hence there exist $p \in \alpha$ and $q \in \beta$.
Hence there exists $r = p + q$ and so $\gamma$ is likewise not [[Definition:Empty Set|empty]].
Let $s, t \in \Q$ such that $s \notin \alpha$ and $t... | Sum of Cuts is Cut | https://proofwiki.org/wiki/Sum_of_Cuts_is_Cut | https://proofwiki.org/wiki/Sum_of_Cuts_is_Cut | [
"Cuts",
"Addition"
] | [
"Definition:Cut (Analysis)",
"Definition:Set",
"Definition:Rational Number",
"Definition:Cut (Analysis)",
"Definition:Addition of Cuts",
"Definition:Set",
"Definition:Cut (Analysis)",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Cut (Analysis)",
"Definition:Empty Set",
"Definition:Empty Set",
"Definition:Rational Number",
"Definition:Cut (Analysis)",
"Rational Numbers form Totally Ordered Field",
"Definition:Rational Number",
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)",
"Definition:Greatest Elemen... |
proofwiki-16737 | Addition of Cuts is Commutative | Let $\alpha$ and $\beta$ be cuts.
Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.
Then:
:$\alpha + \beta = \beta + \alpha$ | $\alpha + \beta$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$ and $q \in \beta$.
Similarly, $\beta + \alpha$ is the set of all rational numbers of the form $q + p$ such that $p \in \alpha$ and $q \in \beta$.
From Rational Addition is Commutative we have that:
:$p + q = q + p$
The resu... | Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]].
Let the operation of $\alpha + \beta$ be the [[Definition:Addition of Cuts|sum]] of $\alpha$ and $\beta$.
Then:
:$\alpha + \beta = \beta + \alpha$ | $\alpha + \beta$ is the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] of the form $p + q$ such that $p \in \alpha$ and $q \in \beta$.
Similarly, $\beta + \alpha$ is the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] of the form $q + p$ such that $p \in \al... | Addition of Cuts is Commutative | https://proofwiki.org/wiki/Addition_of_Cuts_is_Commutative | https://proofwiki.org/wiki/Addition_of_Cuts_is_Commutative | [
"Cuts",
"Addition",
"Examples of Commutative Operations"
] | [
"Definition:Cut (Analysis)",
"Definition:Addition of Cuts"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Set",
"Definition:Rational Number",
"Rational Addition is Commutative"
] |
proofwiki-16738 | Addition of Cuts is Associative | Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.
Then:
:$\paren {\alpha + \beta} + \gamma = \alpha + \paren {\beta + \gamma}$ | $\paren {\alpha + \beta} + \gamma$ is the set of all rational numbers of the form $\paren {p + q} + r$ such that $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
Similarly, $\alpha + \paren {\beta + \gamma}$ is the set of all rational numbers of the form $p + \paren {q + r}$ such that $p \in \alpha$, $q \in \beta$ and... | Let $\alpha$, $\beta$ and $\gamma$ be [[Definition:Cut (Analysis)|cuts]].
Let the operation of $\alpha + \beta$ be the [[Definition:Addition of Cuts|sum]] of $\alpha$ and $\beta$.
Then:
:$\paren {\alpha + \beta} + \gamma = \alpha + \paren {\beta + \gamma}$ | $\paren {\alpha + \beta} + \gamma$ is the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] of the form $\paren {p + q} + r$ such that $p \in \alpha$, $q \in \beta$ and $r \in \gamma$.
Similarly, $\alpha + \paren {\beta + \gamma}$ is the [[Definition:Set|set]] of all [[Definition:Rational N... | Addition of Cuts is Associative | https://proofwiki.org/wiki/Addition_of_Cuts_is_Associative | https://proofwiki.org/wiki/Addition_of_Cuts_is_Associative | [
"Cuts",
"Addition",
"Examples of Associative Operations"
] | [
"Definition:Cut (Analysis)",
"Definition:Addition of Cuts"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Set",
"Definition:Rational Number",
"Rational Addition is Associative"
] |
proofwiki-16739 | Identity Element for Addition of Cuts | Let $\alpha$ be a cut.
Let $0^*$ be the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
Then:
:$\alpha + 0^* = \alpha$
where $+$ denotes the operation of addition of cuts. | Let $r \in \alpha + 0^*$.
$\alpha + 0^*$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$, $q \in 0^*$, that is, $q < 0$.
It follows that:
:$p + q < p$
and so:
:$p + q \in \alpha$
that is:
:$r \in \alpha$
Let $r \in \alpha$.
Let $s \in \Q$ be a rational number such that $s > r$ and $s \in... | Let $\alpha$ be a [[Definition:Cut (Analysis)|cut]].
Let $0^*$ be the [[Definition:Rational Cut|rational cut]] associated with the [[Definition:Zero (Number)|(rational) number $0$]]:
:$0^* = \set {r \in \Q: r < 0}$
Then:
:$\alpha + 0^* = \alpha$
where $+$ denotes the operation of [[Definition:Addition of Cuts|additi... | Let $r \in \alpha + 0^*$.
$\alpha + 0^*$ is the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] of the form $p + q$ such that $p \in \alpha$, $q \in 0^*$, that is, $q < 0$.
It follows that:
:$p + q < p$
and so:
:$p + q \in \alpha$
that is:
:$r \in \alpha$
Let $r \in \alpha$.
Let $s \i... | Identity Element for Addition of Cuts | https://proofwiki.org/wiki/Identity_Element_for_Addition_of_Cuts | https://proofwiki.org/wiki/Identity_Element_for_Addition_of_Cuts | [
"Cuts",
"Addition"
] | [
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Addition of Cuts"
] | [
"Definition:Set",
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Greatest Element",
"Definition:Cut (Analysis)",
"Definition:Set Equality"
] |
proofwiki-16740 | Existence of Upper and Lower Numbers of Cut whose Difference equal Given Rational | Let $\alpha$ be a cut.
Let $r \in \Q_{>0}$ be a (strictly) positive rational number.
Then there exist rational numbers $p$ and $q$ such that:
:$p \in \alpha, q \notin \alpha$
:$q - p = r$
such that $q$ is not the smallest upper number of $\alpha$. | Let $s \in \alpha$ be a rational number.
For $n = 0, 1, 2, \ldots$ let $s_n = s + n r$.
Then there exists a unique integer $m$ such that:
:$s_m \in \alpha$
and:
:$s_{m + 1} \notin \alpha$
If $s_{m + 1}$ is not the smallest upper number of $\alpha$, take:
:$p = s_m$
:$q = s_{m + 1}$
If $s_{m + 1}$ ''is'' the smallest up... | Let $\alpha$ be a [[Definition:Cut (Analysis)|cut]].
Let $r \in \Q_{>0}$ be a [[Definition:Strictly Positive Rational Number|(strictly) positive rational number]].
Then there exist [[Definition:Rational Number|rational numbers]] $p$ and $q$ such that:
:$p \in \alpha, q \notin \alpha$
:$q - p = r$
such that $q$ is no... | Let $s \in \alpha$ be a [[Definition:Rational Number|rational number]].
For $n = 0, 1, 2, \ldots$ let $s_n = s + n r$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Integer|integer]] $m$ such that:
:$s_m \in \alpha$
and:
:$s_{m + 1} \notin \alpha$
If $s_{m + 1}$ is not the [[Definition:Smallest Elem... | Existence of Upper and Lower Numbers of Cut whose Difference equal Given Rational | https://proofwiki.org/wiki/Existence_of_Upper_and_Lower_Numbers_of_Cut_whose_Difference_equal_Given_Rational | https://proofwiki.org/wiki/Existence_of_Upper_and_Lower_Numbers_of_Cut_whose_Difference_equal_Given_Rational | [
"Cuts"
] | [
"Definition:Cut (Analysis)",
"Definition:Strictly Positive/Rational Number",
"Definition:Rational Number",
"Definition:Smallest Element",
"Definition:Cut (Analysis)/Upper Number"
] | [
"Definition:Rational Number",
"Definition:Unique",
"Definition:Integer",
"Definition:Smallest Element",
"Definition:Cut (Analysis)/Upper Number",
"Definition:Smallest Element",
"Definition:Cut (Analysis)/Upper Number"
] |
proofwiki-16741 | Existence of Unique Inverse Element for Addition of Cuts | Let $\alpha$ be a cut.
Let $0^*$ be the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
Then there exists a unique cut $\beta$ such that:
:$\alpha + \beta = 0^*$
where $+$ denotes the operation of addition of cuts. | === Proof of Uniqueness ===
Suppose $\alpha + \beta_1 = \alpha + \beta_2 = 0^*$.
We have:
{{begin-eqn}}
{{eqn | l = \beta_2
| r = 0^* + \beta_2
| c = Identity Element for Addition of Cuts
}}
{{eqn | r = \paren {\alpha + \beta_1} + \beta_2
| c = Definition of $\beta_1$
}}
{{eqn | r = \paren {\alpha + \... | Let $\alpha$ be a [[Definition:Cut (Analysis)|cut]].
Let $0^*$ be the [[Definition:Rational Cut|rational cut]] associated with the [[Definition:Zero (Number)|(rational) number $0$]]:
:$0^* = \set {r \in \Q: r < 0}$
Then there exists a [[Definition:Unique|unique]] [[Definition:Cut (Analysis)|cut]] $\beta$ such that:
... | === Proof of Uniqueness ===
Suppose $\alpha + \beta_1 = \alpha + \beta_2 = 0^*$.
We have:
{{begin-eqn}}
{{eqn | l = \beta_2
| r = 0^* + \beta_2
| c = [[Identity Element for Addition of Cuts]]
}}
{{eqn | r = \paren {\alpha + \beta_1} + \beta_2
| c = Definition of $\beta_1$
}}
{{eqn | r = \paren {\al... | Existence of Unique Inverse Element for Addition of Cuts | https://proofwiki.org/wiki/Existence_of_Unique_Inverse_Element_for_Addition_of_Cuts | https://proofwiki.org/wiki/Existence_of_Unique_Inverse_Element_for_Addition_of_Cuts | [
"Cuts",
"Addition"
] | [
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Unique",
"Definition:Cut (Analysis)",
"Definition:Addition of Cuts"
] | [
"Identity Element for Addition of Cuts",
"Addition of Cuts is Commutative",
"Addition of Cuts is Associative",
"Identity Element for Addition of Cuts",
"Definition:Unique"
] |
proofwiki-16742 | Ordering on Cuts is Compatible with Addition of Cuts | Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.
Let $\beta < \gamma$ denotes the strict ordering on cuts defined as:
:$\beta < \gamma \iff \exists p \in \Q: p \in \beta, p \notin \gamma$
Then:
:$\beta < \gamma \implies \alpha + \beta < \alpha + \gamm... | By definition of the strict ordering on cuts and addition of cuts:
:$\alpha + \beta \le \alpha + \gamma$
{{explain|Details needed for the above}}
Suppose $\alpha + \beta = \alpha + \gamma$.
Then:
{{begin-eqn}}
{{eqn | l = \beta
| r = 0^* + \beta
| c = Identity Element for Addition of Cuts
}}
{{eqn | r = \pa... | Let $\alpha$, $\beta$ and $\gamma$ be [[Definition:Cut (Analysis)|cuts]].
Let the operation of $\alpha + \beta$ be the [[Definition:Addition of Cuts|sum]] of $\alpha$ and $\beta$.
Let $\beta < \gamma$ denotes the [[Definition:Strict Ordering of Cuts|strict ordering on cuts]] defined as:
:$\beta < \gamma \iff \exists ... | By definition of the [[Definition:Strict Ordering of Cuts|strict ordering on cuts]] and [[Definition:Addition of Cuts|addition of cuts]]:
:$\alpha + \beta \le \alpha + \gamma$
{{explain|Details needed for the above}}
Suppose $\alpha + \beta = \alpha + \gamma$.
Then:
{{begin-eqn}}
{{eqn | l = \beta
| r = 0^*... | Ordering on Cuts is Compatible with Addition of Cuts | https://proofwiki.org/wiki/Ordering_on_Cuts_is_Compatible_with_Addition_of_Cuts | https://proofwiki.org/wiki/Ordering_on_Cuts_is_Compatible_with_Addition_of_Cuts | [
"Cuts",
"Addition"
] | [
"Definition:Cut (Analysis)",
"Definition:Addition of Cuts",
"Definition:Ordering of Cuts/Strict"
] | [
"Definition:Ordering of Cuts/Strict",
"Definition:Addition of Cuts",
"Identity Element for Addition of Cuts"
] |
proofwiki-16743 | Ordering on Cuts is Compatible with Addition of Cuts/Corollary | Let $0^*$ denote the rational cut associated with the (rational) number $0$.
If:
:$\alpha > 0^*$ and $\gamma > 0^*$
then:
:$\alpha + \gamma > 0^*$ | From Ordering on Cuts is Compatible with Addition of Cuts
:$0^* + 0^* < 0^* + \alpha$
:$\alpha + 0^* < \alpha + \gamma$
The result follows from Ordering on Cuts is Transitive.
{{qed}} | Let $0^*$ denote the [[Definition:Rational Cut|rational cut]] associated with the [[Definition:Zero (Number)|(rational) number $0$]].
If:
:$\alpha > 0^*$ and $\gamma > 0^*$
then:
:$\alpha + \gamma > 0^*$ | From [[Ordering on Cuts is Compatible with Addition of Cuts]]
:$0^* + 0^* < 0^* + \alpha$
:$\alpha + 0^* < \alpha + \gamma$
The result follows from [[Ordering on Cuts is Transitive]].
{{qed}} | Ordering on Cuts is Compatible with Addition of Cuts/Corollary | https://proofwiki.org/wiki/Ordering_on_Cuts_is_Compatible_with_Addition_of_Cuts/Corollary | https://proofwiki.org/wiki/Ordering_on_Cuts_is_Compatible_with_Addition_of_Cuts/Corollary | [
"Cuts",
"Addition"
] | [
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)"
] | [
"Ordering on Cuts is Compatible with Addition of Cuts",
"Ordering on Cuts is Transitive"
] |
proofwiki-16744 | Existence of Unique Difference between Cuts | Let $\alpha$ and $\beta$ be cuts.
Then there exists exactly one cut $\gamma$ such that:
:$\alpha + \gamma = \beta$ | From Ordering on Cuts is Compatible with Addition of Cuts:
:$\gamma_1 \ne \gamma_2 \implies \alpha + \gamma_1 \ne \alpha + \gamma_2$
That demonstrates uniqueness.
Let:
:$\gamma = \beta + \paren {-\alpha}$
where $-\alpha$ is the negative of $\alpha$.
Then by Identity Element for Addition of Cuts:
{{begin-eqn}}
{{eqn | l... | Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]].
Then there exists [[Definition:Unique|exactly one]] [[Definition:Cut (Analysis)|cut]] $\gamma$ such that:
:$\alpha + \gamma = \beta$ | From [[Ordering on Cuts is Compatible with Addition of Cuts]]:
:$\gamma_1 \ne \gamma_2 \implies \alpha + \gamma_1 \ne \alpha + \gamma_2$
That demonstrates [[Definition:Unique|uniqueness]].
Let:
:$\gamma = \beta + \paren {-\alpha}$
where $-\alpha$ is the [[Definition:Negative of Cut|negative of $\alpha$]].
Then by [... | Existence of Unique Difference between Cuts | https://proofwiki.org/wiki/Existence_of_Unique_Difference_between_Cuts | https://proofwiki.org/wiki/Existence_of_Unique_Difference_between_Cuts | [
"Cuts",
"Subtraction"
] | [
"Definition:Cut (Analysis)",
"Definition:Unique",
"Definition:Cut (Analysis)"
] | [
"Ordering on Cuts is Compatible with Addition of Cuts",
"Definition:Unique",
"Definition:Negative of Cut",
"Identity Element for Addition of Cuts",
"Addition of Cuts is Commutative",
"Addition of Cuts is Associative",
"Existence of Unique Inverse Element for Addition of Cuts",
"Identity Element for Ad... |
proofwiki-16745 | Set of Cuts under Addition forms Abelian Group | Let $\CC$ denote the set of cuts.
Let $\struct {\CC, +}$ denote the algebraic structure formed from $\CC$ and the operation $+$ of addition of cuts.
Then $\struct {\CC, +}$ forms an abelian group. | In the below, $\alpha$, $\beta$ and $\gamma$ denote arbitrary cuts.
Taking the abelian group axioms in turn: | Let $\CC$ denote the [[Definition:Set|set]] of [[Definition:Cut (Analysis)|cuts]].
Let $\struct {\CC, +}$ denote the [[Definition:Algebraic Structure with One Operation|algebraic structure]] formed from $\CC$ and the [[Definition:Addition of Cuts|operation $+$ of addition of cuts]].
Then $\struct {\CC, +}$ forms an ... | In the below, $\alpha$, $\beta$ and $\gamma$ denote arbitrary [[Definition:Cut (Analysis)|cuts]].
Taking the [[Axiom:Abelian Group Axioms|abelian group axioms]] in turn: | Set of Cuts under Addition forms Abelian Group | https://proofwiki.org/wiki/Set_of_Cuts_under_Addition_forms_Abelian_Group | https://proofwiki.org/wiki/Set_of_Cuts_under_Addition_forms_Abelian_Group | [
"Cuts",
"Examples of Abelian Groups"
] | [
"Definition:Set",
"Definition:Cut (Analysis)",
"Definition:Algebraic Structure/One Operation",
"Definition:Addition of Cuts",
"Definition:Abelian Group"
] | [
"Definition:Cut (Analysis)",
"Axiom:Abelian Group Axioms",
"Definition:Cut (Analysis)",
"Axiom:Abelian Group Axioms"
] |
proofwiki-16746 | Product of Positive Cuts is Positive Cut | Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Let $\alpha$ and $\beta$ be cuts such that $\alpha \ge 0^*$ and $\beta \ge 0^*$, where $\ge$ denotes the ordering on cuts.
Let $\gamma$ be the set of all rational numbers $r$ such that either:
:$r < 0$
or:
:$\exists p \in \alpha, q \in \beta: ... | By definition of $\gamma$, we have that $r < 0 \implies r \in \gamma$.
Hence $\gamma$ is not empty.
First suppose that either $\alpha = 0^*$ or $\beta = 0^*$.
Then by definition of cut:
:$p \in \alpha \implies p < 0$
:$q \in \beta \implies q < 0$
and so there exist no $r \in \gamma$ such that $r = p q$.
Thus $\gamma$ c... | Let $0^*$ denote the [[Definition:Rational Cut|rational cut]] associated with the [[Definition:Zero (Number)|(rational) number $0$]].
Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]] such that $\alpha \ge 0^*$ and $\beta \ge 0^*$, where $\ge$ denotes the [[Definition:Ordering of Cuts|ordering on cuts]].
... | By definition of $\gamma$, we have that $r < 0 \implies r \in \gamma$.
Hence $\gamma$ is not [[Definition:Empty Set|empty]].
First suppose that either $\alpha = 0^*$ or $\beta = 0^*$.
Then by definition of [[Definition:Cut (Analysis)|cut]]:
:$p \in \alpha \implies p < 0$
:$q \in \beta \implies q < 0$
and so there ... | Product of Positive Cuts is Positive Cut | https://proofwiki.org/wiki/Product_of_Positive_Cuts_is_Positive_Cut | https://proofwiki.org/wiki/Product_of_Positive_Cuts_is_Positive_Cut | [
"Cuts",
"Multiplication"
] | [
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Cut (Analysis)",
"Definition:Ordering of Cuts",
"Definition:Set",
"Definition:Rational Number",
"Definition:Cut (Analysis)",
"Definition:Multiplication of Positive Cuts",
"Definition:Set",
"Definition:Positive Cut",
"D... | [
"Definition:Empty Set",
"Definition:Cut (Analysis)",
"Definition:Rational Number",
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)",
"Definition:Greatest Element",
"Definition:Rational Number",
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)",
"Ration... |
proofwiki-16747 | Product of Cuts is Cut | Let $\alpha$ and $\beta$ be cuts.
Let $\alpha \beta$ denote the product of cuts.
Then $\alpha \beta$ is also a cut.
Thus the operation of multiplication on the set of cuts is closed. | {{ProofWanted}}
Category:Cuts
Category:Multiplication
m4f4p5wp4vizxgz4w7o28tg6sd15z9z | Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]].
Let $\alpha \beta$ denote the [[Definition:Multiplication of Cuts|product]] of [[Definition:Cut (Analysis)|cuts]].
Then $\alpha \beta$ is also a [[Definition:Cut (Analysis)|cut]].
Thus the operation of [[Definition:Multiplication of Cuts|multiplicatio... | {{ProofWanted}}
[[Category:Cuts]]
[[Category:Multiplication]]
m4f4p5wp4vizxgz4w7o28tg6sd15z9z | Product of Cuts is Cut | https://proofwiki.org/wiki/Product_of_Cuts_is_Cut | https://proofwiki.org/wiki/Product_of_Cuts_is_Cut | [
"Cuts",
"Multiplication"
] | [
"Definition:Cut (Analysis)",
"Definition:Multiplication of Cuts",
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)",
"Definition:Multiplication of Cuts",
"Definition:Set",
"Definition:Cut (Analysis)",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Category:Cuts",
"Category:Multiplication"
] |
proofwiki-16748 | Factorisation of z^n+a | Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a \in \C$ be a complex number.
Then:
:$z^n + a = \ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha_k b}$
where:
:$\alpha_k$ are the complex $n$th roots of negative unity
:$b$ is any complex number such that $b^n = a$. | From $z^n + a = 0$ we have that:
:$z^n = - a$
Let $b = a^{1 / n}$, hence $b^n = a$, with $a, b \in \C$.
From Roots of Complex Number:
{{begin-eqn}}
{{eqn | l = z^{1 / n}
| r = \set {a^{1 / n} e^{i \paren {\theta + 2 k \pi} / n}: k \in \set {0, 1, 2, \ldots, n - 1}, \theta = \arg -a}
| c = $z^n = -a$ so we n... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $a \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$z^n + a = \ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \alpha_k b}$
where:
:$\alpha_k$ are the complex $n$th roots of negative unity
:$b$ is any ... | From $z^n + a = 0$ we have that:
:$z^n = - a$
Let $b = a^{1 / n}$, hence $b^n = a$, with $a, b \in \C$.
From [[Roots of Complex Number]]:
{{begin-eqn}}
{{eqn | l = z^{1 / n}
| r = \set {a^{1 / n} e^{i \paren {\theta + 2 k \pi} / n}: k \in \set {0, 1, 2, \ldots, n - 1}, \theta = \arg -a}
| c = $z^n = -a$ s... | Factorisation of z^n+a | https://proofwiki.org/wiki/Factorisation_of_z^n+a | https://proofwiki.org/wiki/Factorisation_of_z^n+a | [
"Algebra",
"Complex Roots"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Complex Number",
"Definition:Complex Number"
] | [
"Roots of Complex Number",
"Definition:Root of Polynomial",
"Polynomial Factor Theorem/Corollary/Complex Numbers",
"Definition:Monic Polynomial",
"Category:Algebra",
"Category:Complex Roots"
] |
proofwiki-16749 | Multiplication of Cuts is Commutative | Let $\alpha$ and $\beta$ be cuts.
Let the $\alpha \beta$ be the product of $\alpha$ and $\beta$.
Then:
:$\alpha \beta = \beta \alpha$ | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \... | Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]].
Let the $\alpha \beta$ be the [[Definition:Multiplication of Cuts|product]] of $\alpha$ and $\beta$.
Then:
:$\alpha \beta = \beta \alpha$ | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, ... | Multiplication of Cuts is Commutative | https://proofwiki.org/wiki/Multiplication_of_Cuts_is_Commutative | https://proofwiki.org/wiki/Multiplication_of_Cuts_is_Commutative | [
"Cuts",
"Multiplication",
"Examples of Commutative Operations"
] | [
"Definition:Cut (Analysis)",
"Definition:Multiplication of Cuts"
] | [
"Definition:Absolute Value of Cut",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Ordering of Cuts",
"Definition:Set",
"Definition:Rational Number",
"Definition:Set",
"Definition:Rational Number",
"Definition:Set",
"Definition:Rational Number",
"Rational Multiplic... |
proofwiki-16750 | Multiplication of Cuts is Associative | Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let $\alpha \beta$ denote the product of $\alpha$ and $\beta$.
Then:
:$\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$ | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \... | Let $\alpha$, $\beta$ and $\gamma$ be [[Definition:Cut (Analysis)|cuts]].
Let $\alpha \beta$ denote the [[Definition:Multiplication of Cuts|product]] of $\alpha$ and $\beta$.
Then:
:$\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$ | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, ... | Multiplication of Cuts is Associative | https://proofwiki.org/wiki/Multiplication_of_Cuts_is_Associative | https://proofwiki.org/wiki/Multiplication_of_Cuts_is_Associative | [
"Cuts",
"Multiplication",
"Examples of Associative Operations"
] | [
"Definition:Cut (Analysis)",
"Definition:Multiplication of Cuts"
] | [
"Definition:Absolute Value of Cut",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Ordering of Cuts",
"Definition:Set",
"Definition:Rational Number",
"Definition:Set",
"Definition:Rational Number",
"Rational Multiplication is Associative"
] |
proofwiki-16751 | Multiplication of Cuts Distributes over Addition | Let $\alpha$, $\beta$ and $\gamma$ be cuts.
Let:
:$\alpha + \beta$ denote the sum of $\alpha$ and $\beta$.
:$\alpha \beta$ denote the product of $\alpha$ and $\beta$.
Then:
:$\alpha \paren {\beta + \gamma} = \alpha \beta + \alpha \gamma$ | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \... | Let $\alpha$, $\beta$ and $\gamma$ be [[Definition:Cut (Analysis)|cuts]].
Let:
:$\alpha + \beta$ denote the [[Definition:Addition of Cuts|sum]] of $\alpha$ and $\beta$.
:$\alpha \beta$ denote the [[Definition:Multiplication of Cuts|product]] of $\alpha$ and $\beta$.
Then:
:$\alpha \paren {\beta + \gamma} = \alpha \b... | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, ... | Multiplication of Cuts Distributes over Addition | https://proofwiki.org/wiki/Multiplication_of_Cuts_Distributes_over_Addition | https://proofwiki.org/wiki/Multiplication_of_Cuts_Distributes_over_Addition | [
"Cuts",
"Addition",
"Multiplication",
"Examples of Distributive Operations"
] | [
"Definition:Cut (Analysis)",
"Definition:Addition of Cuts",
"Definition:Multiplication of Cuts"
] | [
"Definition:Absolute Value of Cut",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Ordering of Cuts",
"Definition:Set",
"Definition:Rational Number",
"Definition:Set",
"Definition:Rational Number",
"Rational Multiplication Distributes over Addition"
] |
proofwiki-16752 | Product of Cut with Zero Cut equals Zero Cut | Let $\alpha$ be a cut.
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Then:
:$\alpha 0^* = 0^*$
where $\alpha 0^*$ denotes the product of $\alpha$ and $0^*$. | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \... | Let $\alpha$ be a [[Definition:Cut (Analysis)|cut]].
Let $0^*$ denote the [[Definition:Rational Cut|rational cut]] associated with the [[Definition:Zero (Number)|(rational) number $0$]].
Then:
:$\alpha 0^* = 0^*$
where $\alpha 0^*$ denotes the [[Definition:Multiplication of Cuts|product]] of $\alpha$ and $0^*$. | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, ... | Product of Cut with Zero Cut equals Zero Cut | https://proofwiki.org/wiki/Product_of_Cut_with_Zero_Cut_equals_Zero_Cut | https://proofwiki.org/wiki/Product_of_Cut_with_Zero_Cut_equals_Zero_Cut | [
"Cuts",
"Multiplication"
] | [
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Multiplication of Cuts"
] | [
"Definition:Absolute Value of Cut",
"Definition:Multiplication of Positive Cuts",
"Definition:Ordering of Cuts",
"Absolute Value of Cut is Zero iff Cut is Zero",
"Absolute Value of Cut is Greater Than or Equal To Zero Cut",
"Definition:Multiplication of Positive Cuts",
"Definition:Set",
"Definition:Ra... |
proofwiki-16753 | Product of Cuts is Zero Cut iff Either Factor equals Zero Cut | Let $\alpha$ and $\beta$ be cuts.
Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Then:
:$\alpha \beta = 0^*$
{{iff}}:
:$\alpha = 0^*$ or $\beta = 0^*$
where $\alpha \beta$ denotes the product of $\alpha$ and $\beta$. | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, \... | Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]].
Let $0^*$ denote the [[Definition:Rational Cut|rational cut]] associated with the [[Definition:Zero (Number)|(rational) number $0$]].
Then:
:$\alpha \beta = 0^*$
{{iff}}:
:$\alpha = 0^*$ or $\beta = 0^*$
where $\alpha \beta$ denotes the [[Definition:Mu... | By definition, we have that:
:$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha < 0^*, ... | Product of Cuts is Zero Cut iff Either Factor equals Zero Cut | https://proofwiki.org/wiki/Product_of_Cuts_is_Zero_Cut_iff_Either_Factor_equals_Zero_Cut | https://proofwiki.org/wiki/Product_of_Cuts_is_Zero_Cut_iff_Either_Factor_equals_Zero_Cut | [
"Cuts",
"Multiplication"
] | [
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Multiplication of Cuts"
] | [
"Definition:Absolute Value of Cut",
"Definition:Multiplication of Positive Cuts",
"Definition:Ordering of Cuts",
"Absolute Value of Cut is Zero iff Cut is Zero",
"Definition:Multiplication of Positive Cuts"
] |
proofwiki-16754 | Cut Associated with 1 is Identity for Multiplication of Cuts | Let $\alpha$ be a cut.
Let $1^*$ denote the rational cut (rational) number $1$.
Then:
:$\alpha 1^* = \alpha$
where $\alpha 1^*$ denote the product of $\alpha$ and $1^*$. | By definition, we have that:
:<nowiki>$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha ... | Let $\alpha$ be a [[Definition:Cut (Analysis)|cut]].
Let $1^*$ denote the [[Definition:Rational Cut|rational cut]] [[Definition:Rational Number|(rational) number]] $1$.
Then:
:$\alpha 1^* = \alpha$
where $\alpha 1^*$ denote the [[Definition:Multiplication of Cuts|product]] of $\alpha$ and $1^*$. | By definition, we have that:
:<nowiki>$\alpha \beta := \begin {cases}
\size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\
-\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\
\size \alpha \, \size \beta & : \alpha... | Cut Associated with 1 is Identity for Multiplication of Cuts | https://proofwiki.org/wiki/Cut_Associated_with_1_is_Identity_for_Multiplication_of_Cuts | https://proofwiki.org/wiki/Cut_Associated_with_1_is_Identity_for_Multiplication_of_Cuts | [
"Cuts",
"Multiplication"
] | [
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)/Rational",
"Definition:Rational Number",
"Definition:Multiplication of Cuts"
] | [
"Definition:Absolute Value of Cut",
"Definition:Multiplication of Positive Cuts",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Ordering of Cuts",
"Definition:Ordering of Cuts/Strict",
"Definition:Absolute Value of Cut",
"Definition:Multiplication of Positive Cuts"
] |
proofwiki-16755 | Multiplication of Positive Cuts preserves Ordering | Let $0^*$ denote the rational cut associated with the (rational) number $0$.
Let $\alpha$, $\beta$ and $\gamma$ be cuts such that:
:$0^* < \alpha < \beta$
:$0^* < \gamma$
where $<$ denotes the strict ordering on cuts.
Then
:$\alpha \gamma < \beta \gamma$
where $\alpha \gamma$ denotes the product of $\alpha$ and $\gamma... | {{ProofWanted|bored with this}} | Let $0^*$ denote the [[Definition:Rational Cut|rational cut]] associated with the [[Definition:Zero (Number)|(rational) number $0$]].
Let $\alpha$, $\beta$ and $\gamma$ be [[Definition:Cut (Analysis)|cuts]] such that:
:$0^* < \alpha < \beta$
:$0^* < \gamma$
where $<$ denotes the [[Definition:Strict Ordering of Cuts|... | {{ProofWanted|bored with this}} | Multiplication of Positive Cuts preserves Ordering | https://proofwiki.org/wiki/Multiplication_of_Positive_Cuts_preserves_Ordering | https://proofwiki.org/wiki/Multiplication_of_Positive_Cuts_preserves_Ordering | [
"Cuts",
"Multiplication"
] | [
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)",
"Definition:Cut (Analysis)",
"Definition:Ordering of Cuts/Strict",
"Definition:Multiplication of Cuts"
] | [] |
proofwiki-16756 | Sum of Rational Cuts is Rational Cut | Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* + q^* = \paren {p + q}^*$
Thus the operation of addition on the set of rational cuts is closed. | From Sum of Cuts is Cut, $p^* + q^*$ is a cut.
Let $r \in p^* + q^*$.
Then:
:$r = s + t$
where $s < p$ and $t < q$
Thus:
:$r < p + q$
and so:
:$r \in \paren {p + q}^*$
Hence:
:$p^* + q^* \subseteq \paren {p + q}^*$
{{qed|lemma}}
Let $r \in \paren {p + q}^*$.
Then:
:$r < p + q$
Let:
:$h = p + q - r$
:$s = p - \dfrac h 2... | Let $p \in\ Q$ and $q \in \Q$ be [[Definition:Rational Number|rational numbers]].
Let $p^*$ and $q^*$ denote the [[Definition:Rational Cut|rational cuts]] associated with $p$ and $q$.
Then:
:$p^* + q^* = \paren {p + q}^*$
Thus the operation of [[Definition:Addition of Cuts|addition]] on the [[Definition:Set|set]] ... | From [[Sum of Cuts is Cut]], $p^* + q^*$ is a [[Definition:Cut (Analysis)|cut]].
Let $r \in p^* + q^*$.
Then:
:$r = s + t$
where $s < p$ and $t < q$
Thus:
:$r < p + q$
and so:
:$r \in \paren {p + q}^*$
Hence:
:$p^* + q^* \subseteq \paren {p + q}^*$
{{qed|lemma}}
Let $r \in \paren {p + q}^*$.
Then:
:$r < p + q$
... | Sum of Rational Cuts is Rational Cut | https://proofwiki.org/wiki/Sum_of_Rational_Cuts_is_Rational_Cut | https://proofwiki.org/wiki/Sum_of_Rational_Cuts_is_Rational_Cut | [
"Cuts",
"Addition"
] | [
"Definition:Rational Number",
"Definition:Cut (Analysis)/Rational",
"Definition:Addition of Cuts",
"Definition:Set",
"Definition:Cut (Analysis)/Rational",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Sum of Cuts is Cut",
"Definition:Cut (Analysis)",
"Definition:Set Equality"
] |
proofwiki-16757 | Product of Rational Cuts is Rational Cut | Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* q^* = \paren {p q}^*$
Thus the operation of multiplication on the set of rational cuts is closed. | From Product of Cuts is Cut, $p^* q^*$ is a cut.
{{ProofWanted}} | Let $p \in\ Q$ and $q \in \Q$ be [[Definition:Rational Number|rational numbers]].
Let $p^*$ and $q^*$ denote the [[Definition:Rational Cut|rational cuts]] associated with $p$ and $q$.
Then:
:$p^* q^* = \paren {p q}^*$
Thus the operation of [[Definition:Multiplication of Cuts|multiplication]] on the [[Definition:Se... | From [[Product of Cuts is Cut]], $p^* q^*$ is a [[Definition:Cut (Analysis)|cut]].
{{ProofWanted}} | Product of Rational Cuts is Rational Cut | https://proofwiki.org/wiki/Product_of_Rational_Cuts_is_Rational_Cut | https://proofwiki.org/wiki/Product_of_Rational_Cuts_is_Rational_Cut | [
"Cuts",
"Multiplication"
] | [
"Definition:Rational Number",
"Definition:Cut (Analysis)/Rational",
"Definition:Multiplication of Cuts",
"Definition:Set",
"Definition:Cut (Analysis)/Rational",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Product of Cuts is Cut",
"Definition:Cut (Analysis)"
] |
proofwiki-16758 | Ordering of Rational Cuts preserves Ordering of Associated Rational Numbers | Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* < q^* \iff p < q$
where $p^* < q^*$ denotes the strict ordering on cuts defined as:
:$\beta < \gamma \iff \exists p \in \Q: p \in \beta, p \notin \gamma$ | Let $p < q$.
Then $p \notin p^*$ but $q \in q^*$.
Thus $p^* < q^*$ by definition of the strict ordering on cuts .
Let $p^* < q^*$.
Then:
:$\exists r \in \Q: r \notin p^*, r \in q^*$
Hence:
:$p \le r < q$
and so:
:$p < q$
{{qed}} | Let $p \in\ Q$ and $q \in \Q$ be [[Definition:Rational Number|rational numbers]].
Let $p^*$ and $q^*$ denote the [[Definition:Rational Cut|rational cuts]] associated with $p$ and $q$.
Then:
:$p^* < q^* \iff p < q$
where $p^* < q^*$ denotes the [[Definition:Strict Ordering of Cuts|strict ordering on cuts]] defined a... | Let $p < q$.
Then $p \notin p^*$ but $q \in q^*$.
Thus $p^* < q^*$ by definition of the [[Definition:Strict Ordering of Cuts|strict ordering on cuts]] .
Let $p^* < q^*$.
Then:
:$\exists r \in \Q: r \notin p^*, r \in q^*$
Hence:
:$p \le r < q$
and so:
:$p < q$
{{qed}} | Ordering of Rational Cuts preserves Ordering of Associated Rational Numbers | https://proofwiki.org/wiki/Ordering_of_Rational_Cuts_preserves_Ordering_of_Associated_Rational_Numbers | https://proofwiki.org/wiki/Ordering_of_Rational_Cuts_preserves_Ordering_of_Associated_Rational_Numbers | [
"Cuts"
] | [
"Definition:Rational Number",
"Definition:Cut (Analysis)/Rational",
"Definition:Ordering of Cuts/Strict"
] | [
"Definition:Ordering of Cuts/Strict"
] |
proofwiki-16759 | Exists Rational Cut Between two Cuts | Let $\alpha$ and $\beta$ be cuts.
Let $\alpha < \beta$, where $<$ denotes the strict ordering on cuts.
Then there exists a rational cut $r^*$ associated with the rational number $r$ such that:
:$\alpha < r^* < \beta$ | Let $\alpha < \beta$.
Then there exists a rational number $p$ such that $p \in \beta$ and $p \notin \alpha$.
Let $r \in \Q$ such that $r > p$ and $r \in \beta$.
Because $r \in \beta$ and $r \notin r^*$, we have that $r^* < \beta$.
Because $p \in r^*$ and $p \notin \alpha$, we have that $\alpha < r^*$.
{{qed}} | Let $\alpha$ and $\beta$ be [[Definition:Cut (Analysis)|cuts]].
Let $\alpha < \beta$, where $<$ denotes the [[Definition:Strict Ordering of Cuts|strict ordering on cuts]].
Then there exists a [[Definition:Rational Cut|rational cut]] $r^*$ associated with the [[Definition:Rational Number|rational number]] $r$ such th... | Let $\alpha < \beta$.
Then there exists a [[Definition:Rational Number|rational number]] $p$ such that $p \in \beta$ and $p \notin \alpha$.
Let $r \in \Q$ such that $r > p$ and $r \in \beta$.
Because $r \in \beta$ and $r \notin r^*$, we have that $r^* < \beta$.
Because $p \in r^*$ and $p \notin \alpha$, we have tha... | Exists Rational Cut Between two Cuts | https://proofwiki.org/wiki/Exists_Rational_Cut_Between_two_Cuts | https://proofwiki.org/wiki/Exists_Rational_Cut_Between_two_Cuts | [
"Cuts"
] | [
"Definition:Cut (Analysis)",
"Definition:Ordering of Cuts/Strict",
"Definition:Cut (Analysis)/Rational",
"Definition:Rational Number"
] | [
"Definition:Rational Number"
] |
proofwiki-16760 | Condition for Rational Cut to be Less than Given Cut | Let $\alpha$ be a cut.
Let $p^*$ be the rational cut associated with a rational number $p$.
Then:
:$p \in \alpha$
{{iff}}:
:$p^* < \alpha$
where $<$ denotes the strict ordering on cuts. | Let $p$ be a rational number such that $p \in \alpha$.
Then by definition of rational cut:
:$p \notin p^*$
Thus:
:$p \in \alpha \implies p^* < \alpha$
{{qed|lemma}}
Let $p^* < \alpha$.
Then there exists a rational number $q$ such that $q \in \alpha$ and $q \notin p$.
Thus $q \ge p$.
But as $q \in \alpha$ it follows tha... | Let $\alpha$ be a [[Definition:Cut (Analysis)|cut]].
Let $p^*$ be the [[Definition:Rational Cut|rational cut]] associated with a [[Definition:Rational Number|rational number]] $p$.
Then:
:$p \in \alpha$
{{iff}}:
:$p^* < \alpha$
where $<$ denotes the [[Definition:Strict Ordering of Cuts|strict ordering on cuts]]. | Let $p$ be a [[Definition:Rational Number|rational number]] such that $p \in \alpha$.
Then by definition of [[Definition:Rational Cut|rational cut]]:
:$p \notin p^*$
Thus:
:$p \in \alpha \implies p^* < \alpha$
{{qed|lemma}}
Let $p^* < \alpha$.
Then there exists a [[Definition:Rational Number|rational number]] $q$ ... | Condition for Rational Cut to be Less than Given Cut | https://proofwiki.org/wiki/Condition_for_Rational_Cut_to_be_Less_than_Given_Cut | https://proofwiki.org/wiki/Condition_for_Rational_Cut_to_be_Less_than_Given_Cut | [
"Cuts"
] | [
"Definition:Cut (Analysis)",
"Definition:Cut (Analysis)/Rational",
"Definition:Rational Number",
"Definition:Ordering of Cuts/Strict"
] | [
"Definition:Rational Number",
"Definition:Cut (Analysis)/Rational",
"Definition:Rational Number"
] |
proofwiki-16761 | Continuous Mapping to Product Space/General Result | Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$.
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:... | Suppose $f$ is continuous.
From Projection from Product Topology is Continuous $\pr_i$ is continuous for each $i \in I$.
By Composite of Continuous Mappings is Continuous it follows that $\pr_i \circ f$ is continuous for each $i \in I$.
Conversely, suppose that each $\pr_i \circ f$ is continuous.
Let $U = \pr_{i_1}^{-... | Let $X$ be a [[Definition:Topological Space|topological space]].
Let $\family {Y_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] for some [[Definition:Indexing Set|indexing set]] $I$.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the [[Def... | Suppose $f$ is [[Definition:Continuous Mapping (Topology)|continuous]].
From [[Projection from Product Topology is Continuous]] $\pr_i$ is [[Definition:Continuous Mapping (Topology)|continuous]] for each $i \in I$.
By [[Composite of Continuous Mappings is Continuous]] it follows that $\pr_i \circ f$ is [[Definition:C... | Continuous Mapping to Product Space/General Result | https://proofwiki.org/wiki/Continuous_Mapping_to_Product_Space/General_Result | https://proofwiki.org/wiki/Continuous_Mapping_to_Product_Space/General_Result | [
"Product Topology",
"Continuous Mappings"
] | [
"Definition:Topological Space",
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Projection (Mapping Theory)/Family of Sets",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:C... | [
"Definition:Continuous Mapping (Topology)",
"Projection from Product Topology is Continuous",
"Definition:Continuous Mapping (Topology)",
"Composite of Continuous Mappings is Continuous",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Definition:Open Set/Topolog... |
proofwiki-16762 | Set of Cuts forms Ordered Field | Let $\CC$ denote the set of cuts.
Let $\struct {\CC, +, \times, \le}$ denote the ordered structure formed from $\CC$ and:
:the operation $+$ of addition of cuts
:the operation $\times$ of multiplication of cuts
:the ordering $\le$ of cuts.
Then $\struct {\CC, +, \times, \le}$ is an ordered field. | First we show that $\struct {\CC, +, \times}$ is a field by demonstrating that it fulfils the field axioms:
{{:Axiom:Field Axioms}}
It has been established from Set of Cuts under Addition forms Abelian Group that $\struct {\CC, +}$ forms an abelian group.
Thus $\text A 0$ through to $\text A 4$ are fulfilled.
It remai... | Let $\CC$ denote the [[Definition:Set|set]] of [[Definition:Cut (Analysis)|cuts]].
Let $\struct {\CC, +, \times, \le}$ denote the [[Definition:Ordered Structure|ordered structure]] formed from $\CC$ and:
:the [[Definition:Addition of Cuts|operation $+$ of addition of cuts]]
:the [[Definition:Addition of Cuts|operation... | First we show that $\struct {\CC, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]] by demonstrating that it fulfils the [[Axiom:Field Axioms|field axioms]]:
{{:Axiom:Field Axioms}}
It has been established from [[Set of Cuts under Addition forms Abelian Group]] that $\struct {\CC, +}$ forms an [[Definiti... | Set of Cuts forms Ordered Field | https://proofwiki.org/wiki/Set_of_Cuts_forms_Ordered_Field | https://proofwiki.org/wiki/Set_of_Cuts_forms_Ordered_Field | [
"Cuts",
"Totally Ordered Fields"
] | [
"Definition:Set",
"Definition:Cut (Analysis)",
"Definition:Ordered Structure",
"Definition:Addition of Cuts",
"Definition:Addition of Cuts",
"Definition:Ordering of Cuts",
"Definition:Ordered Field"
] | [
"Definition:Field (Abstract Algebra)",
"Axiom:Field Axioms",
"Set of Cuts under Addition forms Abelian Group",
"Definition:Abelian Group",
"Axiom:Field Axioms",
"Axiom:Field Axioms",
"Definition:Field (Abstract Algebra)"
] |
proofwiki-16763 | Set of Rational Cuts forms Ordered Field | Let $\RR$ denote the set of rational cuts.
Let $\struct {\RR, +, \times, \le}$ denote the ordered structure formed from $\RR$ and:
:the operation $+$ of addition of cuts
:the operation $\times$ of multiplication of cuts
:the ordering $\le$ of cuts.
Then $\struct {\RR, + \times, \le}$ is an ordered field. | We demonstrate that $\struct {\RR, +, \times}$ is a field by showing it is a subfield of the structure $\struct {\CC, +, \times}$, where $\CC$ denotes the set of all cuts.
We do this by establishing that all $4$ criteria of the Subfield Test are satisfied.
We note that $0^* \in \RR$, where $0^*$ is the rational cut ass... | Let $\RR$ denote the [[Definition:Set|set]] of [[Definition:Rational Cut|rational cuts]].
Let $\struct {\RR, +, \times, \le}$ denote the [[Definition:Ordered Structure|ordered structure]] formed from $\RR$ and:
:the [[Definition:Addition of Cuts|operation $+$ of addition of cuts]]
:the [[Definition:Addition of Cuts|op... | We demonstrate that $\struct {\RR, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]] by showing it is a [[Definition:Subfield|subfield]] of the [[Definition:Algebraic Structure with Two Operations|structure]] $\struct {\CC, +, \times}$, where $\CC$ denotes the [[Definition:Set|set]] of all [[Definition:Cut... | Set of Rational Cuts forms Ordered Field | https://proofwiki.org/wiki/Set_of_Rational_Cuts_forms_Ordered_Field | https://proofwiki.org/wiki/Set_of_Rational_Cuts_forms_Ordered_Field | [
"Cuts",
"Totally Ordered Fields"
] | [
"Definition:Set",
"Definition:Cut (Analysis)/Rational",
"Definition:Ordered Structure",
"Definition:Addition of Cuts",
"Definition:Addition of Cuts",
"Definition:Ordering of Cuts",
"Definition:Ordered Field"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Subfield",
"Definition:Algebraic Structure/Two Operations",
"Definition:Set",
"Definition:Cut (Analysis)",
"Subfield Test",
"Definition:Cut (Analysis)/Rational",
"Definition:Zero (Number)"
] |
proofwiki-16764 | Box Topology contains Product Topology | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the product topology on $X$.
Let $\tau'$ be the box topolo... | From Natural Basis of Product Topology and Basis for Box Topology, it follows immediately that the natural basis $\BB$ for the product topology is contained in the basis $\BB'$ for the box topology.
From Corollary of Basis Condition for Coarser Topology, it follows that $\tau \subseteq \tau'$.
{{qed}}
Category:Product ... | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $X$ be the [[Definition:Cartesian Product of Family|cartesian product]] of $\family {X_... | From [[Natural Basis of Product Topology]] and [[Basis for Box Topology]], it follows immediately that the [[Definition:Natural Basis of Product Topology|natural basis]] $\BB$ for the [[Definition:Product Topology|product topology]] is contained in the [[Definition:Synthetic Basis|basis]] $\BB'$ for the [[Definition:Bo... | Box Topology contains Product Topology | https://proofwiki.org/wiki/Box_Topology_contains_Product_Topology | https://proofwiki.org/wiki/Box_Topology_contains_Product_Topology | [
"Product Topology",
"Box Topology"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Cartesian Product/Family of Sets",
"Definition:Product Topology",
"Definition:Box Topology"
] | [
"Natural Basis of Product Topology",
"Basis for Box Topology",
"Definition:Product Topology/Natural Basis",
"Definition:Product Topology",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Box Topology",
"Basis Condition for Coarser Topology/Corollary 2",
"Category:Product Topology",
"Categ... |
proofwiki-16765 | Ordered Field of Rational Cuts is Isomorphic to Rational Numbers | Let $\struct {\RR, +, \times, \le}$ denote the ordered field of rational cuts.
Let $\struct {\Q, +, \times, \le}$ denote the field of rational numbers.
Then $\struct {\RR, +, \times, \le}$ and $\struct {\Q, +, \times, \le}$ are isomorphic. | {{ProofWanted|Straightforward but tedious}} | Let $\struct {\RR, +, \times, \le}$ denote the [[Definition:Ordered Field|ordered field]] of [[Definition:Rational Cut|rational cuts]].
Let $\struct {\Q, +, \times, \le}$ denote the [[Definition:Field of Rational Numbers|field of rational numbers]].
Then $\struct {\RR, +, \times, \le}$ and $\struct {\Q, +, \times, \... | {{ProofWanted|Straightforward but tedious}} | Ordered Field of Rational Cuts is Isomorphic to Rational Numbers | https://proofwiki.org/wiki/Ordered_Field_of_Rational_Cuts_is_Isomorphic_to_Rational_Numbers | https://proofwiki.org/wiki/Ordered_Field_of_Rational_Cuts_is_Isomorphic_to_Rational_Numbers | [
"Cuts",
"Rational Numbers"
] | [
"Definition:Ordered Field",
"Definition:Cut (Analysis)/Rational",
"Definition:Field of Rational Numbers",
"Definition:Ordered Field Isomorphism"
] | [] |
proofwiki-16766 | Supremum of Subset of Real Numbers May or May Not be in Subset | Let $S \subset \R$ be a proper subset of the set $\R$ of real numbers.
Let $S$ admit a supremum $M$.
Then $M$ may or may not be an element of $S$. | Consider the subset $S$ of the real numbers $\R$ defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
It is seen that:
:$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$
and hence $\sup S = 1$.
Thus $\sup S \in S$.
Consider the subset $T$ of the real numbers $\R$ defined as:
:$T = \set {-\dfrac 1 n: n \in \Z_{>0} }$
It is... | Let $S \subset \R$ be a [[Definition:Proper Subset|proper subset]] of the [[Definition:Real Number|set $\R$ of real numbers]].
Let $S$ admit a [[Definition:Supremum of Subset of Real Numbers|supremum]] $M$.
Then $M$ may or may not be an [[Definition:Element|element]] of $S$. | Consider the [[Definition:Subset|subset]] $S$ of the [[Definition:Real Number|real numbers]] $\R$ defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
It is seen that:
:$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$
and hence $\sup S = 1$.
Thus $\sup S \in S$.
Consider the [[Definition:Subset|subset]] $T$ of the [... | Supremum of Subset of Real Numbers May or May Not be in Subset | https://proofwiki.org/wiki/Supremum_of_Subset_of_Real_Numbers_May_or_May_Not_be_in_Subset | https://proofwiki.org/wiki/Supremum_of_Subset_of_Real_Numbers_May_or_May_Not_be_in_Subset | [
"Suprema"
] | [
"Definition:Proper Subset",
"Definition:Real Number",
"Definition:Supremum of Set/Real Numbers",
"Definition:Element"
] | [
"Definition:Subset",
"Definition:Real Number",
"Definition:Subset",
"Definition:Real Number"
] |
proofwiki-16767 | Infimum of Subset of Real Numbers May or May Not be in Subset | Let $S \subset \R$ be a proper subset of the set $\R$ of real numbers.
Let $S$ admit an infimum $m$.
Then $m$ may or may not be an element of $S$. | Consider the subset $S$ of the real numbers $\R$ defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
It is seen that:
:$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$
and hence $\inf S = 0$.
Thus $\inf S \notin S$.
Consider the subset $T$ of the real numbers $\R$ defined as:
:$T = \set {-\dfrac 1 n: n \in \Z_{>0} }$
It... | Let $S \subset \R$ be a [[Definition:Proper Subset|proper subset]] of the [[Definition:Real Number|set $\R$ of real numbers]].
Let $S$ admit an [[Definition:Infimum of Subset of Real Numbers|infimum]] $m$.
Then $m$ may or may not be an [[Definition:Element|element]] of $S$. | Consider the [[Definition:Subset|subset]] $S$ of the [[Definition:Real Number|real numbers]] $\R$ defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
It is seen that:
:$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$
and hence $\inf S = 0$.
Thus $\inf S \notin S$.
Consider the [[Definition:Subset|subset]] $T$ of th... | Infimum of Subset of Real Numbers May or May Not be in Subset | https://proofwiki.org/wiki/Infimum_of_Subset_of_Real_Numbers_May_or_May_Not_be_in_Subset | https://proofwiki.org/wiki/Infimum_of_Subset_of_Real_Numbers_May_or_May_Not_be_in_Subset | [
"Suprema"
] | [
"Definition:Proper Subset",
"Definition:Real Number",
"Definition:Infimum of Set/Real Numbers",
"Definition:Element"
] | [
"Definition:Subset",
"Definition:Real Number",
"Definition:Subset",
"Definition:Real Number"
] |
proofwiki-16768 | Uniqueness of Positive Root of Positive Real Number/Positive Exponent | Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$. | We have that:
:$0 < y_1 < y_2 \implies y_1^n < y_2^n$
so there exists at most one $y \in \R: y \ge 0$ such that $y^n = x$.
{{Qed}} | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$. | We have that:
:$0 < y_1 < y_2 \implies y_1^n < y_2^n$
so there exists at most one $y \in \R: y \ge 0$ such that $y^n = x$.
{{Qed}} | Uniqueness of Positive Root of Positive Real Number/Positive Exponent/Proof 2 | https://proofwiki.org/wiki/Uniqueness_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent | https://proofwiki.org/wiki/Uniqueness_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent/Proof_2 | [
"Uniqueness of Positive Root of Positive Real Number"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [] |
proofwiki-16769 | Uniqueness of Positive Root of Positive Real Number/Positive Exponent | Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$. | To prove uniqueness, we must show that:
:$y_1^n = x = y_2^n$ implies $y_1 = y_2$
{{AimForCont}} that $y_1 \ne y_2$.
Then $y_1 < y_2$ or $y_2 < y_1$.
{{WLOG}}, assume that $y_1 < y_2$.
We will show by induction that $y_1^n < y_2^n$, contradicting the assumption that $y_1^n = x = y_2^n$.
=== Basis for the Induction ===
B... | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$. | To prove [[Definition:Unique|uniqueness]], we must show that:
:$y_1^n = x = y_2^n$ implies $y_1 = y_2$
{{AimForCont}} that $y_1 \ne y_2$.
Then $y_1 < y_2$ or $y_2 < y_1$.
{{WLOG}}, assume that $y_1 < y_2$.
We will show by [[Principle of Mathematical Induction|induction]] that $y_1^n < y_2^n$, contradicting the assu... | Uniqueness of Positive Root of Positive Real Number/Positive Exponent/Proof 3 | https://proofwiki.org/wiki/Uniqueness_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent | https://proofwiki.org/wiki/Uniqueness_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent/Proof_3 | [
"Uniqueness of Positive Root of Positive Real Number"
] | [
"Definition:Real Number",
"Definition:Integer"
] | [
"Definition:Unique",
"Principle of Mathematical Induction",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Uniqueness of Positive Root of Positive Real Number/Positive Exponent/Proof 3",
"Definition:Positive",
"Axiom:Real Number/Axioms",
"Principle of Mathematical Induction"... |
proofwiki-16770 | Existence and Uniqueness of Positive Root of Positive Real Number/Positive Exponent | Let $x \in \R$ be a real number such that $x \ge 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there always exists a unique $y \in \R: \paren {y \ge 0} \land \paren {y^n = x}$. | === Proof of Existence ===
{{:Existence of Positive Root of Positive Real Number/Positive Exponent}} | Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x \ge 0$.
Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
Then there always exists a [[Definition:Unique|unique]] $y \in \R: \paren {y \ge 0} \land \paren {y^n = x}$. | === [[Existence of Positive Root of Positive Real Number/Positive Exponent|Proof of Existence]] ===
{{:Existence of Positive Root of Positive Real Number/Positive Exponent}} | Existence and Uniqueness of Positive Root of Positive Real Number/Positive Exponent | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Positive_Root_of_Positive_Real_Number/Positive_Exponent | [
"Existence and Uniqueness of Positive Root of Positive Real Number"
] | [
"Definition:Real Number",
"Definition:Integer",
"Definition:Unique"
] | [
"Existence of Positive Root of Positive Real Number/Positive Exponent"
] |
proofwiki-16771 | Basis Condition for Coarser Topology | Let $S$ be a set.
Let $\BB_1$ and $\BB_2$ be two bases on $S$.
Let $\tau_1$ and $\tau_2$ be the topologies generated by $\BB_1$ and $\BB_2$ respectively.
If $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
then $\tau_1$ is coarser than $\tau_2$. | Let $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
Let $U \in \BB_1$.
Then there exists $\AA_2 \subseteq \BB_2$ such that $U = \bigcup \AA_2$.
By definition of the topology generated by $\BB_2$, it follows that $U \in \tau_2$.
Since $U$ was arbitrary, it follows that ... | Let $S$ be a [[Definition:Set|set]].
Let $\BB_1$ and $\BB_2$ be two [[Definition:Synthetic Basis|bases]] on $S$.
Let $\tau_1$ and $\tau_2$ be the [[Definition:Topology|topologies]] [[Definition:Topology Generated by Synthetic Basis|generated]] by $\BB_1$ and $\BB_2$ respectively.
If $\BB_1$ and $\BB_2$ satisfy:
:$\... | Let $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
Let $U \in \BB_1$.
Then there exists $\AA_2 \subseteq \BB_2$ such that $U = \bigcup \AA_2$.
By definition of the [[Definition:Topology Generated by Synthetic Basis|topology generated]] by $\BB_2$, it follows that ... | Basis Condition for Coarser Topology | https://proofwiki.org/wiki/Basis_Condition_for_Coarser_Topology | https://proofwiki.org/wiki/Basis_Condition_for_Coarser_Topology | [
"Topology"
] | [
"Definition:Set",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Topology",
"Definition:Topology Generated by Synthetic Basis",
"Definition:Coarser Topology"
] | [
"Definition:Topology Generated by Synthetic Basis",
"Definition:Topology Generated by Synthetic Basis",
"Definition:Topology",
"Category:Topology"
] |
proofwiki-16772 | Basis Condition for Coarser Topology/Corollary 1 | If $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$
then $\tau_1$ is coarser than $\tau_2$. | Let $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$
Let $U \in \BB_1$.
Let $\AA = \set {V \in \BB_2 : V \subseteq U}$
From Union of Family of Sets is Smallest Superset:
:$\bigcup \AA \subseteq V$
Let $x \in U$.
Then:
:$\exists V_x \in \BB_2 : x \in V_x \sub... | If $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$
then $\tau_1$ is [[Definition:Coarser Topology|coarser]] than $\tau_2$. | Let $\BB_1$ and $\BB_2$ satisfy:
:$\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$
Let $U \in \BB_1$.
Let $\AA = \set {V \in \BB_2 : V \subseteq U}$
From [[Union is Smallest Superset/Family of Sets|Union of Family of Sets is Smallest Superset]]:
:$\bigcup \AA \subseteq V$
Let $x \i... | Basis Condition for Coarser Topology/Corollary 1 | https://proofwiki.org/wiki/Basis_Condition_for_Coarser_Topology/Corollary_1 | https://proofwiki.org/wiki/Basis_Condition_for_Coarser_Topology/Corollary_1 | [
"Topology"
] | [
"Definition:Coarser Topology"
] | [
"Union is Smallest Superset/Family of Sets",
"Definition:Set Equality",
"Basis Condition for Coarser Topology",
"Definition:Coarser Topology",
"Category:Topology"
] |
proofwiki-16773 | Basis Condition for Coarser Topology/Corollary 2 | If $\BB_1 \subseteq \BB_2$ then $\tau_1$ is coarser than $\tau_2$. | Let $\BB_1 \subseteq \BB_2$.
Let $U \in \BB_1$.
Let $\AA = \set U$.
Then:
:$\AA \subseteq \BB_2$
and
:$U = \bigcup \AA$
So:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
From Basis Condition for Coarser Topology:
:$\tau_1$ is coarser than $\tau_2$
{{qed}}
Category:Topology
gmwsn3hqv1yry5z79x0w8t... | If $\BB_1 \subseteq \BB_2$ then $\tau_1$ is [[Definition:Coarser Topology|coarser]] than $\tau_2$. | Let $\BB_1 \subseteq \BB_2$.
Let $U \in \BB_1$.
Let $\AA = \set U$.
Then:
:$\AA \subseteq \BB_2$
and
:$U = \bigcup \AA$
So:
:$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$
From [[Basis Condition for Coarser Topology]]:
:$\tau_1$ is [[Definition:Coarser Topology|coarser]] than $\tau_2$
{{qed}}... | Basis Condition for Coarser Topology/Corollary 2 | https://proofwiki.org/wiki/Basis_Condition_for_Coarser_Topology/Corollary_2 | https://proofwiki.org/wiki/Basis_Condition_for_Coarser_Topology/Corollary_2 | [
"Topology"
] | [
"Definition:Coarser Topology"
] | [
"Basis Condition for Coarser Topology",
"Definition:Coarser Topology",
"Category:Topology"
] |
proofwiki-16774 | Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous | Let $\struct {X, \tau_1}$ and $\struct {Y, \tau_2}$ be topological spaces.
Let $f: \struct {X, \tau_1} \to \struct {Y, \tau_2}$ be a continuous mapping.
Let $\tau'_1$ be a finer topology on $X$ than $\tau_1$, that is, $\tau_1 \subseteq \tau'_1$.
Let $\tau'_2$ be a coarser topology on $Y$ than $\tau_2$, that is, $\tau'_... | Let $U \in \tau'_2$.
Since $\tau'_2$ is a coarser topology than $\tau_2$:
:$U \in \tau_2$
By definition of continuity:
:$\map {f^{-1}} U \in \tau_1$
Since $\tau'_1$ is a finer topology than $\tau_1$:
:$\map {f^{-1}} U \in \tau'_1$
Since $U$ was arbitrary, by definition of continuity:
:$f: \struct {X, \tau'_1} \to \stru... | Let $\struct {X, \tau_1}$ and $\struct {Y, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $f: \struct {X, \tau_1} \to \struct {Y, \tau_2}$ be a [[Definition:Continuous Mapping|continuous mapping]].
Let $\tau'_1$ be a [[Definition:Finer Topology|finer topology]] on $X$ than $\tau_1$, that is, $\... | Let $U \in \tau'_2$.
Since $\tau'_2$ is a [[Definition:Coarser Topology|coarser topology]] than $\tau_2$:
:$U \in \tau_2$
By definition of [[Definition:Continuous Mapping|continuity]]:
:$\map {f^{-1}} U \in \tau_1$
Since $\tau'_1$ is a [[Definition:Finer Topology|finer topology]] than $\tau_1$:
:$\map {f^{-1}} U \i... | Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous | https://proofwiki.org/wiki/Continuous_Mapping_on_Finer_Domain_and_Coarser_Codomain_Topologies_is_Continuous | https://proofwiki.org/wiki/Continuous_Mapping_on_Finer_Domain_and_Coarser_Codomain_Topologies_is_Continuous | [
"Topology"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping",
"Definition:Finer Topology",
"Definition:Coarser Topology",
"Definition:Continuous Mapping"
] | [
"Definition:Coarser Topology",
"Definition:Continuous Mapping",
"Definition:Finer Topology",
"Definition:Continuous Mapping",
"Definition:Continuous Mapping",
"Category:Topology"
] |
proofwiki-16775 | Projection from Box Topology is Continuous | Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the box topology on $X$.
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$t... | Let $\tau'$ be the product topology on $X$.
From Projection from Product Topology is Continuous:
:$\forall i \in I : \pr_i : \struct{X, \tau'} \to \struct{X_i, \tau_i}$ is continuous
From Box Topology contains Product Topology, $\tau$ is a finer topology than $\tau'$ on $X$.
From Continuous Mapping on Finer Domain and ... | Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Topological Space|topological spaces]].
Let $X$ be the [[Definition:Cartesian Product of Family|cartesian product]] of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \i... | Let $\tau'$ be the [[Definition:Product Topology|product topology]] on $X$.
From [[Projection from Product Topology is Continuous]]:
:$\forall i \in I : \pr_i : \struct{X, \tau'} \to \struct{X_i, \tau_i}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]
From [[Box Topology contains Product Topolo... | Projection from Box Topology is Continuous | https://proofwiki.org/wiki/Projection_from_Box_Topology_is_Continuous | https://proofwiki.org/wiki/Projection_from_Box_Topology_is_Continuous | [
"Box Topology"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Cartesian Product/Family of Sets",
"Definition:Box Topology",
"Definition:Projection (Mapping Theory)/Family of Sets",
"Definition:Continuous Mapping (Topology)/Everywhere"
] | [
"Definition:Product Topology",
"Projection from Product Topology is Continuous",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Box Topology contains Product Topology",
"Definition:Finer Topology",
"Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous",
"Definition:C... |
proofwiki-16776 | Domain Topology Contains Initial Topology iff Mappings are Continuous | Let $\struct{Y, \tau}$ be a topological space.
Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be a family of topological spaces.
Let $\family {f_i}_{i \mathop \in I}$ be a family of mappings $f_i : Y \to X_i$.
Let $\tau'$ be the initial topology on $Y$ with respect to $\family {f_i}_{i \mathop \in I}$.
Then:
:$... | === Necessary Condition ===
Let $\tau' \subseteq \tau$.
From Equivalence of Definitions of Initial Topology:
:for each $i \in I$, $f_i: \struct{Y, \tau'} \to \struct{X_i, \tau_i}$ is $\tuple{\tau', \tau_i}$-continuous
From Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous:
:for each $i \i... | Let $\struct{Y, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]].
Let $\family {f_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[... | === Necessary Condition ===
Let $\tau' \subseteq \tau$.
From [[Equivalence of Definitions of Initial Topology]]:
:for each $i \in I$, $f_i: \struct{Y, \tau'} \to \struct{X_i, \tau_i}$ is [[Definition:Continuous Mapping (Topology)|$\tuple{\tau', \tau_i}$-continuous]]
From [[Continuous Mapping on Finer Domain and Coar... | Domain Topology Contains Initial Topology iff Mappings are Continuous | https://proofwiki.org/wiki/Domain_Topology_Contains_Initial_Topology_iff_Mappings_are_Continuous | https://proofwiki.org/wiki/Domain_Topology_Contains_Initial_Topology_iff_Mappings_are_Continuous | [
"Initial Topology"
] | [
"Definition:Topological Space",
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set/Family",
"Definition:Mapping",
"Definition:Initial Topology",
"Definition:Continuous Mapping (Topology)"
] | [
"Equivalence of Definitions of Initial Topology",
"Definition:Continuous Mapping (Topology)",
"Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Equivalence of Definitions of Initial T... |
proofwiki-16777 | Final Topology Contains Codomain Topology iff Mappings are Continuous | Let $\struct{Y, \tau}$ be a topological space.
Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be a family of topological spaces.
Let $\family {f_i}_{i \mathop \in I}$ be a family of mappings $f_i : X_i \to Y$.
Let $\tau'$ be the final topology on $Y$ with respect to $\family {f_i}_{i \mathop \in I}$.
Then:
:$\t... | === Necessary Condition ===
Let $\tau \subseteq \tau'$.
From Equivalence of Definitions of Final Topology:
:for each $i \in I$, $f_i: \struct{X_i, \tau_i} \to \struct{Y, \tau'}$ is $\tuple{\tau_i, \tau'}$-continuous
From Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous:
:for each $i \in ... | Let $\struct{Y, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]].
Let $\family {f_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[... | === Necessary Condition ===
Let $\tau \subseteq \tau'$.
From [[Equivalence of Definitions of Final Topology]]:
:for each $i \in I$, $f_i: \struct{X_i, \tau_i} \to \struct{Y, \tau'}$ is [[Definition:Continuous Mapping (Topology)|$\tuple{\tau_i, \tau'}$-continuous]]
From [[Continuous Mapping on Finer Domain and Coarse... | Final Topology Contains Codomain Topology iff Mappings are Continuous | https://proofwiki.org/wiki/Final_Topology_Contains_Codomain_Topology_iff_Mappings_are_Continuous | https://proofwiki.org/wiki/Final_Topology_Contains_Codomain_Topology_iff_Mappings_are_Continuous | [
"Final Topology"
] | [
"Definition:Topological Space",
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set/Family",
"Definition:Mapping",
"Definition:Final Topology",
"Definition:Continuous"
] | [
"Equivalence of Definitions of Final Topology",
"Definition:Continuous Mapping (Topology)",
"Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Equivalence of Definitions of Final Topol... |
proofwiki-16778 | Equivalence of Definitions of Final Topology | Let $X$ be a set.
Let $I$ be an indexing set.
Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$.
Let $\family {f_i: Y_i \to X}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$.
{{TFAE|def = Final Topology}} | === Definition 1 Implies Definition 2 ===
{{:Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2}}{{qed|lemma}} | Let $X$ be a [[Definition:Set|set]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] [[Definition:Indexing Set|indexed]] by $I$.
Let $\family {f_... | === [[Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2|Definition 1 Implies Definition 2]] ===
{{:Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2}}{{qed|lemma}} | Equivalence of Definitions of Final Topology | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Final_Topology | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Final_Topology | [
"Final Topology"
] | [
"Definition:Set",
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Mapping",
"Definition:Indexing Set"
] | [
"Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2"
] |
proofwiki-16779 | Condition for Trivial Relation to be Mapping | Let $S$ and $T$ be sets.
Let $\RR = S \times T$ be the trivial relation in $S$ to $T$.
Then $\RR$ is a mapping {{iff}} either:
:$(2): \card S = 0$
or:
:$(1): \card T = 1$
where $\card {\, \cdot \,}$ denotes cardinality. | By definition, the trivial relation in $S$ to $T$ is the set:
:$\RR = \set {\tuple {s, t}: s \in S, t \in T}$
Proof by Cases:
:$(1)$: Let $\card S = 0$.
That is, $S = \O$.
By definition of the trivial relation, in this case $\RR = \O$.
This is the empty mapping.
From Empty Mapping is Mapping, this is a mapping.
:$(2)$:... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR = S \times T$ be the [[Definition:Trivial Relation|trivial relation]] in $S$ to $T$.
Then $\RR$ is a [[Definition:Mapping|mapping]] {{iff}} either:
:$(2): \card S = 0$
or:
:$(1): \card T = 1$
where $\card {\, \cdot \,}$ denotes [[Definition:Cardinality|cardinalit... | By definition, the [[Definition:Trivial Relation|trivial relation]] in $S$ to $T$ is the [[Definition:Set|set]]:
:$\RR = \set {\tuple {s, t}: s \in S, t \in T}$
[[Proof by Cases]]:
:$(1)$: Let $\card S = 0$.
That is, $S = \O$.
By definition of the [[Definition:Trivial Relation|trivial relation]], in this case $\R... | Condition for Trivial Relation to be Mapping | https://proofwiki.org/wiki/Condition_for_Trivial_Relation_to_be_Mapping | https://proofwiki.org/wiki/Condition_for_Trivial_Relation_to_be_Mapping | [
"Trivial Relation"
] | [
"Definition:Set",
"Definition:Trivial Relation",
"Definition:Mapping",
"Definition:Cardinality"
] | [
"Definition:Trivial Relation",
"Definition:Set",
"Proof by Cases",
"Definition:Trivial Relation",
"Definition:Empty Mapping",
"Empty Mapping is Mapping",
"Definition:Mapping",
"Relation to Empty Set is Mapping iff Domain is Empty",
"Definition:Mapping",
"Definition:Trivial Relation",
"Definition... |
proofwiki-16780 | Relation to Empty Set is Mapping iff Domain is Empty | Let $S$ be a set
Let $S \times \O$ denote the cartesian product of $S$ with the empty set $\O$.
Let $\RR \subseteq S \times \O$ be a relation in $S$ to $\O$.
Then $\RR$ is a mapping {{iff}} $S = \O$. | Let $S \ne \O$.
Then $\exists s \in S$.
But there exists no $t \in \O$.
Hence there is no $\tuple {s, t} \in \RR$.
So $\RR$ is not a mapping by definition.
Let $S = \O$.
Then $\RR$ is the empty mapping by definition.
From Empty Mapping is Mapping, it is demonstrated that $\RR$ is indeed a mapping.
Hence the result.
{{q... | Let $S$ be a [[Definition:Set|set]]
Let $S \times \O$ denote the [[Definition:Cartesian Product|cartesian product]] of $S$ with the [[Definition:Empty Set|empty set]] $\O$.
Let $\RR \subseteq S \times \O$ be a [[Definition:Relation|relation]] in $S$ to $\O$.
Then $\RR$ is a [[Definition:Mapping|mapping]] {{iff}} $S... | Let $S \ne \O$.
Then $\exists s \in S$.
But there exists no $t \in \O$.
Hence there is no $\tuple {s, t} \in \RR$.
So $\RR$ is not a [[Definition:Mapping|mapping]] by definition.
Let $S = \O$.
Then $\RR$ is the [[Definition:Empty Mapping|empty mapping]] by definition.
From [[Empty Mapping is Mapping]], it is de... | Relation to Empty Set is Mapping iff Domain is Empty | https://proofwiki.org/wiki/Relation_to_Empty_Set_is_Mapping_iff_Domain_is_Empty | https://proofwiki.org/wiki/Relation_to_Empty_Set_is_Mapping_iff_Domain_is_Empty | [
"Empty Mapping",
"Relation Theory"
] | [
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Empty Set",
"Definition:Relation",
"Definition:Mapping"
] | [
"Definition:Mapping",
"Definition:Empty Mapping",
"Empty Mapping is Mapping",
"Definition:Mapping",
"Category:Empty Mapping",
"Category:Relation Theory"
] |
proofwiki-16781 | Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2 | Let:
:$\SS = \set{\map {f_i^{-1}} U: i \in I, U \in \tau_i} \subseteq \powerset X$
where $\map {f_i^{-1}} U$ denotes the preimage of $U$ under $f_i$.
Let $\tau$ be the topology on $X$ generated by the subbase $\SS$. | ==== Mappings are Continuous ====
Let $i \in I$.
Let $U \in \tau_i$.
Then $\map {f_i^{-1} } U$ is an element of the subbase $\SS$ of $X$, and is therefore trivially in $\tau$.
{{qed|lemma}}
==== $\tau$ is the Coarsest such Topology ====
Let $\struct {X, \vartheta}$ be a topological space.
Let the mappings $\family {f_i... | Let:
:$\SS = \set{\map {f_i^{-1}} U: i \in I, U \in \tau_i} \subseteq \powerset X$
where $\map {f_i^{-1}} U$ denotes the [[Definition:Preimage of Subset under Mapping|preimage]] of $U$ under $f_i$.
Let $\tau$ be the [[Definition:Topology Generated by Synthetic Sub-Basis|topology on $X$ generated]] by the [[Definition:... | ==== Mappings are Continuous ====
Let $i \in I$.
Let $U \in \tau_i$.
Then $\map {f_i^{-1} } U$ is an [[Definition:Element|element]] of the [[Definition:Subbase|subbase]] $\SS$ of $X$, and is therefore trivially in $\tau$.
{{qed|lemma}}
==== $\tau$ is the Coarsest such Topology ====
Let $\struct {X, \vartheta}$ be... | Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Initial_Topology/Definition_1_Implies_Definition_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Initial_Topology/Definition_1_Implies_Definition_2 | [
"Initial Topology"
] | [
"Definition:Preimage/Mapping/Subset",
"Definition:Topology Generated by Synthetic Sub-Basis",
"Definition:Sub-Basis"
] | [
"Definition:Element",
"Definition:Sub-Basis",
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis",
"Definition:Coarser Topology"
] |
proofwiki-16782 | Equivalence of Definitions of Initial Topology/Definition 2 Implies Definition 1 | Let $\tau$ be the coarsest topology on $X$ such that each $f_i: X \to Y_i$ is $\tuple{\tau, \tau_i}$-continuous.
Let:
:$\SS = \set {\map {f_i^{-1} } U: i \in I, U \in \tau_i} \subseteq \map \PP X$
where $\map {f_i^{-1} } U$ denotes the preimage of $U$ under $f_i$. | Let $\map \tau \SS$ be the topology on $X$ generated by the subbase $\SS$.
==== $\tau$ contains Topology Generated by $\SS$ ====
Let $i \in I$ and $U \in \tau_i$.
By definition of $\tuple{\tau, \tau_i}$-continuity:
:$\map {f_i^{-1} } U \in \tau$
Since $i \in I$ and $U \in \tau_i$ were arbitrary, then $\SS \subseteq \t... | Let $\tau$ be the [[Definition:Coarser Topology|coarsest topology]] on $X$ such that each $f_i: X \to Y_i$ is [[Definition:Continuous Mapping (Topology)|$\tuple{\tau, \tau_i}$-continuous]].
Let:
:$\SS = \set {\map {f_i^{-1} } U: i \in I, U \in \tau_i} \subseteq \map \PP X$
where $\map {f_i^{-1} } U$ denotes the [[Defi... | Let $\map \tau \SS$ be the [[Definition:Topology Generated by Synthetic Sub-Basis|topology on $X$ generated ]] by the [[Definition:Subbase|subbase]] $\SS$.
==== $\tau$ contains Topology Generated by $\SS$ ====
Let $i \in I$ and $U \in \tau_i$.
By definition of [[Definition:Continuous Mapping (Topology)|$\tuple{\tau... | Equivalence of Definitions of Initial Topology/Definition 2 Implies Definition 1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Initial_Topology/Definition_2_Implies_Definition_1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Initial_Topology/Definition_2_Implies_Definition_1 | [
"Initial Topology"
] | [
"Definition:Coarser Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Preimage/Mapping/Subset"
] | [
"Definition:Topology Generated by Synthetic Sub-Basis",
"Definition:Sub-Basis",
"Definition:Continuous Mapping (Topology)",
"Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis",
"Definition:Element",
"Definition:Sub-Basis",
"Definition:Continuous Mapping (Topology)",
"Definition:C... |
proofwiki-16783 | Even Natural Numbers are Infinite | The set of even natural numbers is infinite. | Let $E$ denote the set of even natural numbers.
{{AimForCont}} $E$ is finite.
Then there exists $n \in \N$ such that $E$ has $n$ elements.
Let $m$ be the greatest element of $E$.
But then $m + 2$ is an even natural number.
But $m + 2 > m$, and $m$ is the greatest element of $E$.
Therefore $m + 2$ is an even natural num... | The [[Definition:Set|set]] of [[Definition:Even Integer|even]] [[Definition:Natural Numbers|natural numbers]] is [[Definition:Infinite Set|infinite]]. | Let $E$ denote the [[Definition:Set|set]] of [[Definition:Even Integer|even]] [[Definition:Natural Numbers|natural numbers]].
{{AimForCont}} $E$ is [[Definition:Finite Set|finite]].
Then there exists $n \in \N$ such that $E$ has $n$ [[Definition:Element|elements]].
Let $m$ be the [[Definition:Greatest Element|greate... | Even Natural Numbers are Infinite | https://proofwiki.org/wiki/Even_Natural_Numbers_are_Infinite | https://proofwiki.org/wiki/Even_Natural_Numbers_are_Infinite | [
"Even Integers",
"Infinite Sets"
] | [
"Definition:Set",
"Definition:Even Integer",
"Definition:Natural Numbers",
"Definition:Infinite Set"
] | [
"Definition:Set",
"Definition:Even Integer",
"Definition:Natural Numbers",
"Definition:Finite Set",
"Definition:Element",
"Definition:Greatest Element",
"Definition:Even Integer",
"Definition:Natural Numbers",
"Definition:Greatest Element",
"Definition:Even Integer",
"Definition:Natural Numbers"... |
proofwiki-16784 | Set of Points on Line Segment is Infinite | The set of points on a line segment is infinite. | Let $S$ denote the set of points on a line segment.
{{AimForCont}} $S$ is finite.
Then there exists $n \in \N$ such that $S$ has $n$ elements.
Let $s_1$ and $s_2$ be two arbitrary adjacent points in $S$.
That is, such that there are no points in $S$ between $s_1$ and $s_2$.
But there exists (at least) one point on the ... | The [[Definition:Set|set]] of [[Definition:Point|points]] on a [[Definition:Line Segment|line segment]] is [[Definition:Infinite Set|infinite]]. | Let $S$ denote the [[Definition:Set|set]] of [[Definition:Point|points]] on a [[Definition:Line Segment|line segment]].
{{AimForCont}} $S$ is [[Definition:Finite Set|finite]].
Then there exists $n \in \N$ such that $S$ has $n$ [[Definition:Element|elements]].
Let $s_1$ and $s_2$ be two arbitrary adjacent [[Definitio... | Set of Points on Line Segment is Infinite | https://proofwiki.org/wiki/Set_of_Points_on_Line_Segment_is_Infinite | https://proofwiki.org/wiki/Set_of_Points_on_Line_Segment_is_Infinite | [
"Lines",
"Infinite Sets"
] | [
"Definition:Set",
"Definition:Point",
"Definition:Line/Segment",
"Definition:Infinite Set"
] | [
"Definition:Set",
"Definition:Point",
"Definition:Line/Segment",
"Definition:Finite Set",
"Definition:Element",
"Definition:Point",
"Definition:Point",
"Definition:Point",
"Definition:Line/Segment",
"Definition:Element",
"Definition:Contradiction",
"Proof by Contradiction",
"Definition:Finit... |
proofwiki-16785 | Set of Doubletons of Natural Numbers is Countable | Let $S$ be the set defined as:
:$S = \set {\set {n_1, n_2}: n_1, n_2 \in \N, n_1 \ne n_2}$
where $\N$ denotes the set of natural numbers.
Then $S$ is countably infinite. | We can write the elements of $S \times T$ in the form of an infinite table:
:$\begin{array} {*{4}c}
\tuple {1, 0} & & & \\
\tuple {2, 0} & \tuple {2, 1} & & \\
\tuple {3, 0} & \tuple {3, 1} & \tuple {3, 2} & \\
\vdots & \vdots & \vdots & \ddots \\
\end{array}$
This table clearly contains all the elements of $... | Let $S$ be the [[Definition:Set|set]] defined as:
:$S = \set {\set {n_1, n_2}: n_1, n_2 \in \N, n_1 \ne n_2}$
where $\N$ denotes the [[Definition:Set|set]] of [[Definition:Natural Number|natural numbers]].
Then $S$ is [[Definition:Countably Infinite Set|countably infinite]]. | We can write the [[Definition:Element|elements]] of $S \times T$ in the form of an [[Definition:Infinite Set|infinite]] table:
:$\begin{array} {*{4}c}
\tuple {1, 0} & & & \\
\tuple {2, 0} & \tuple {2, 1} & & \\
\tuple {3, 0} & \tuple {3, 1} & \tuple {3, 2} & \\
\vdots & \vdots & \vdots & \ddots \\
\end{array... | Set of Doubletons of Natural Numbers is Countable | https://proofwiki.org/wiki/Set_of_Doubletons_of_Natural_Numbers_is_Countable | https://proofwiki.org/wiki/Set_of_Doubletons_of_Natural_Numbers_is_Countable | [
"Cartesian Product of Countable Sets is Countable"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Countably Infinite/Set"
] | [
"Definition:Element",
"Definition:Infinite Set",
"Definition:Element",
"Definition:Triangular Number",
"Definition:Injection",
"Definition:Bijection"
] |
proofwiki-16786 | Finite Sets are Comparable | Let $S$ and $T$ be finite sets.
Then $S$ and $T$ are comparable by size. | By definition of finite set, there exist $m, n \in \N$ such that:
:$S \sim \N_{<n}$
:$T \sim \N_{<m}$
That is, there are bijections $f, g$:
:$f: S \to \N_{<n}$
:$g: T \to \N_{<m}$
{{WLOG}}, suppose that $m \le n$.
Then $\N_{<m} \subseteq \N_{<n}$, and so we can define:
:$h: T \to S, \map h t = \map {f^{-1}} { \map g t ... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
Then $S$ and $T$ are [[Definition:Comparable Sets by Size|comparable by size]]. | By definition of [[Definition:Finite Set|finite set]], there exist $m, n \in \N$ such that:
:$S \sim \N_{<n}$
:$T \sim \N_{<m}$
That is, there are [[Definition:Bijection|bijections]] $f, g$:
:$f: S \to \N_{<n}$
:$g: T \to \N_{<m}$
{{WLOG}}, suppose that $m \le n$.
Then $\N_{<m} \subseteq \N_{<n}$, and so we can de... | Finite Sets are Comparable | https://proofwiki.org/wiki/Finite_Sets_are_Comparable | https://proofwiki.org/wiki/Finite_Sets_are_Comparable | [
"Comparable Sets"
] | [
"Definition:Finite Set",
"Definition:Comparable Sets/Cardinality"
] | [
"Definition:Finite Set",
"Definition:Bijection",
"Definition:Inverse Mapping",
"Inverse of Bijection is Bijection",
"Composite of Injections is Injection",
"Definition:Injection",
"Injection to Image is Bijection",
"Definition:Bijection",
"Definition:Comparable Sets/Cardinality"
] |
proofwiki-16787 | Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2 | Let:
:$\tau = \set{U \subseteq X: \forall i \in I: \map {f_i^{-1}} U \in \tau_i} \subseteq \powerset X$ | From Final Topology is Topology, $\tau$ is a topology.
==== Mappings are continuous ====
Let $U \in \tau$.
Let $i \in I$.
Then $\map {f_i^{-1}} {U} \in \tau_i$ by definition of $\tau$.
It follows that for each $i \in I$, $f_i: Y_i \to X$ is $\tuple {\tau_i, \tau}$-continuous.
{{qed|lemma}}
==== $\tau$ is the finest suc... | Let:
:$\tau = \set{U \subseteq X: \forall i \in I: \map {f_i^{-1}} U \in \tau_i} \subseteq \powerset X$ | From [[Final Topology is Topology]], $\tau$ is a [[Definition:Topology|topology]].
==== Mappings are continuous ====
Let $U \in \tau$.
Let $i \in I$.
Then $\map {f_i^{-1}} {U} \in \tau_i$ by definition of $\tau$.
It follows that for each $i \in I$, $f_i: Y_i \to X$ is [[Definition:Continuous Mapping (Topology)|$\... | Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Final_Topology/Definition_1_Implies_Definition_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Final_Topology/Definition_1_Implies_Definition_2 | [
"Final Topology"
] | [] | [
"Final Topology is Topology",
"Definition:Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Definition:Finer Topology"
] |
proofwiki-16788 | Equivalence of Definitions of Final Topology/Definition 2 Implies Definition 1 | Let $\tau$ be the finest topology on $X$ such that each $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau}$-continuous. | Let $\tau' = \set{U \subseteq X: \forall i \in I: \map {f_i^{-1} } U \in \tau_i}$.
==== $\tau'$ contains $\tau$ ====
Let $U \in \tau$.
By definition of $\tuple {\tau_i, \tau}$-continuity for each $i \in I$:
:$\forall i \in I : \map {f_i^{-1} } U \in \tau_i$
So:
:$U \in \tau'$
Since $U$ was arbitrary:
:$\tau \subseteq... | Let $\tau$ be the [[Definition:Finer Topology|finest topology]] on $X$ such that each $f_i: Y_i \to X$ is [[Definition:Continuous Mapping (Topology)|$\tuple{\tau_i, \tau}$-continuous]]. | Let $\tau' = \set{U \subseteq X: \forall i \in I: \map {f_i^{-1} } U \in \tau_i}$.
==== $\tau'$ contains $\tau$ ====
Let $U \in \tau$.
By definition of [[Definition:Continuous Mapping (Topology)| $\tuple {\tau_i, \tau}$-continuity]] for each $i \in I$:
:$\forall i \in I : \map {f_i^{-1} } U \in \tau_i$
So:
:$U \i... | Equivalence of Definitions of Final Topology/Definition 2 Implies Definition 1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Final_Topology/Definition_2_Implies_Definition_1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Final_Topology/Definition_2_Implies_Definition_1 | [
"Final Topology"
] | [
"Definition:Finer Topology",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Continuous Mapping (Topology)",
"Final Topology is Topology",
"Definition:Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Finer Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Set Equ... |
proofwiki-16789 | Strictly Positive Integers have same Cardinality as Natural Numbers | Let $\Z_{>0} := \set {1, 2, 3, \ldots}$ denote the set of strictly positive integers.
Let $\N := \set {0, 1, 2, \ldots}$ denote the set of natural numbers.
Then $\Z_{>0}$ has the same cardinality as $\N$. | Consider the mapping $f: \N \to \Z_{>0}$ defined as:
:$\forall x \in \N: \map f x = x + 1$
Then $f$ is trivially seen to be a bijection.
The result follows by definition of cardinality. | Let $\Z_{>0} := \set {1, 2, 3, \ldots}$ denote the [[Definition:Set|set]] of [[Definition:Strictly Positive Integer|strictly positive integers]].
Let $\N := \set {0, 1, 2, \ldots}$ denote the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]].
Then $\Z_{>0}$ has the same [[Definition:Cardinalit... | Consider the [[Definition:Mapping|mapping]] $f: \N \to \Z_{>0}$ defined as:
:$\forall x \in \N: \map f x = x + 1$
Then $f$ is trivially seen to be a [[Definition:Bijection|bijection]].
The result follows by definition of [[Definition:Cardinality|cardinality]]. | Strictly Positive Integers have same Cardinality as Natural Numbers | https://proofwiki.org/wiki/Strictly_Positive_Integers_have_same_Cardinality_as_Natural_Numbers | https://proofwiki.org/wiki/Strictly_Positive_Integers_have_same_Cardinality_as_Natural_Numbers | [
"Cardinality",
"Natural Numbers",
"Integers"
] | [
"Definition:Set",
"Definition:Strictly Positive/Integer",
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Cardinality"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Definition:Cardinality"
] |
proofwiki-16790 | Power Set of Natural Numbers has Cardinality of Continuum | Let $\N$ denote the set of natural numbers.
Let $\powerset \N$ denote the power set of $\N$.
Let $\card {\powerset \N}$ denote the cardinality of $\powerset \N$.
Let $\mathfrak c = \card \R$ denote the cardinality of the continuum.
Then:
:$\mathfrak c = \card {\powerset \N}$ | === Outline ===
$\powerset \N$ is demonstrated to have the same cardinality as the set of real numbers.
This is done by identifying a real number with its basis expansion in binary notation.
Such a basis expansion is a sequence of $0$s and $1$s.
Each $1$ of a real number $x$ expressed in such a way can be identified wi... | Let $\N$ denote the [[Definition:Set|set]] of [[Definition:Natural Number|natural numbers]].
Let $\powerset \N$ denote the [[Definition:Power Set|power set]] of $\N$.
Let $\card {\powerset \N}$ denote the [[Definition:Cardinality|cardinality]] of $\powerset \N$.
Let $\mathfrak c = \card \R$ denote the [[Definition:C... | === Outline ===
$\powerset \N$ is demonstrated to have the same [[Definition:Cardinality|cardinality]] as the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]].
This is done by identifying a [[Definition:Real Number|real number]] with its [[Definition:Basis Expansion/Positive Real Numbers|basis expans... | Power Set of Natural Numbers has Cardinality of Continuum/Proof 1 | https://proofwiki.org/wiki/Power_Set_of_Natural_Numbers_has_Cardinality_of_Continuum | https://proofwiki.org/wiki/Power_Set_of_Natural_Numbers_has_Cardinality_of_Continuum/Proof_1 | [
"Power Set of Natural Numbers has Cardinality of Continuum",
"Cardinality of Continuum",
"Power Set",
"Natural Numbers",
"Infinite Sets"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Power Set",
"Definition:Cardinality",
"Definition:Cardinality of Continuum",
"Definition:Continuum"
] | [
"Definition:Cardinality",
"Definition:Set",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Basis Expansion/Positive Real Numbers",
"Definition:Binary Notation",
"Definition:Basis Expansion/Positive Real Numbers",
"Definition:Sequence",
"Definition:Real Number",
"Definition:Integer... |
proofwiki-16791 | Power Set of Natural Numbers has Cardinality of Continuum | Let $\N$ denote the set of natural numbers.
Let $\powerset \N$ denote the power set of $\N$.
Let $\card {\powerset \N}$ denote the cardinality of $\powerset \N$.
Let $\mathfrak c = \card \R$ denote the cardinality of the continuum.
Then:
:$\mathfrak c = \card {\powerset \N}$ | By Reals are Isomorphic to Dedekind Cuts there exists bijection:
:$f: \R \to \mathscr D$
where:
:$\mathscr D$ denotes the set of all Dedekind cuts of $\struct {\Q, \le}$.
Dedekind's cuts are subsets of $\Q$.
Therefore by definition of power set:
:$\mathscr D \subseteq \powerset \Q$
By Subset implies Cardinal Inequality... | Let $\N$ denote the [[Definition:Set|set]] of [[Definition:Natural Number|natural numbers]].
Let $\powerset \N$ denote the [[Definition:Power Set|power set]] of $\N$.
Let $\card {\powerset \N}$ denote the [[Definition:Cardinality|cardinality]] of $\powerset \N$.
Let $\mathfrak c = \card \R$ denote the [[Definition:C... | By [[Reals are Isomorphic to Dedekind Cuts]] there exists [[Definition:Bijection|bijection]]:
:$f: \R \to \mathscr D$
where:
:$\mathscr D$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Dedekind Cut|Dedekind cuts]] of $\struct {\Q, \le}$.
Dedekind's cuts are [[Definition:Subset|subsets]] of $\Q$.
Ther... | Power Set of Natural Numbers has Cardinality of Continuum/Proof 2 | https://proofwiki.org/wiki/Power_Set_of_Natural_Numbers_has_Cardinality_of_Continuum | https://proofwiki.org/wiki/Power_Set_of_Natural_Numbers_has_Cardinality_of_Continuum/Proof_2 | [
"Power Set of Natural Numbers has Cardinality of Continuum",
"Cardinality of Continuum",
"Power Set",
"Natural Numbers",
"Infinite Sets"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Power Set",
"Definition:Cardinality",
"Definition:Cardinality of Continuum",
"Definition:Continuum"
] | [
"Reals are Isomorphic to Dedekind Cuts",
"Definition:Bijection",
"Definition:Set of Sets",
"Definition:Dedekind Cut",
"Definition:Subset",
"Definition:Power Set",
"Subset implies Cardinal Inequality",
"Rational Numbers are Countably Infinite",
"Definition:Countably Infinite/Set",
"Definition:Count... |
proofwiki-16792 | Empty Set can be Derived from Axiom of Abstraction | The '''empty set''' can be formed by application of the {{axiom-link|Abstraction}}.
Hence the '''empty set''' can be derived as a valid object in Frege set theory. | Let $P$ be the property defined as:
:$\forall x: \map P x := \neg \paren {x = x}$
Hence, using the {{axiom-link|Abstraction}}, we form the set:
:$\O := \set {x: \neg \paren {x = x} }$
where the property ${\map P x}$ is:
:$\neg \paren {x = x}$
Since we have that:
:$\forall x: x = x$
it is seen that $\O$ as defined here ... | The '''[[Definition:Empty Set|empty set]]''' can be formed by application of the {{axiom-link|Abstraction}}.
Hence the '''[[Definition:Empty Set|empty set]]''' can be derived as a valid [[Definition:Object|object]] in [[Definition:Frege Set Theory|Frege set theory]]. | Let $P$ be the [[Definition:Property|property]] defined as:
:$\forall x: \map P x := \neg \paren {x = x}$
Hence, using the {{axiom-link|Abstraction}}, we form the set:
:$\O := \set {x: \neg \paren {x = x} }$
where the [[Definition:Property|property]] ${\map P x}$ is:
:$\neg \paren {x = x}$
Since we have that:
:$\f... | Empty Set can be Derived from Axiom of Abstraction | https://proofwiki.org/wiki/Empty_Set_can_be_Derived_from_Axiom_of_Abstraction | https://proofwiki.org/wiki/Empty_Set_can_be_Derived_from_Axiom_of_Abstraction | [
"Empty Set",
"Axiom of Abstraction"
] | [
"Definition:Empty Set",
"Definition:Empty Set",
"Definition:Object",
"Definition:Frege Set Theory"
] | [
"Definition:Property",
"Definition:Property",
"Definition:Element",
"Definition:Frege Set Theory",
"Definition:Property",
"Definition:Unique",
"Definition:Set",
"Definition:Object",
"Definition:Property",
"Definition:Element",
"Definition:Empty Set",
"Definition:Object",
"Definition:Frege Se... |
proofwiki-16793 | Doubleton of Sets can be Derived using Axiom of Abstraction | Let $a$ and $b$ be sets.
By application of the {{axiom-link|Abstraction}}, the set $\set {a, b}$ can be formed.
Hence the '''doubleton''' $\set {a, b}$ can be derived as a valid object in Frege set theory. | Let $P$ be the property defined as:
:$\forall x: \map P x := \paren {x = a \lor x = b}$
where $\lor$ is the disjunction operator.
Hence, using the {{axiom-link|Abstraction}}, we form the set:
:$\set {a, b} := \set {x: x = a \lor x = b}$
{{qed}} | Let $a$ and $b$ be [[Definition:Set|sets]].
By application of the {{axiom-link|Abstraction}}, the [[Definition:Set|set]] $\set {a, b}$ can be formed.
Hence the '''[[Definition:Doubleton|doubleton]]''' $\set {a, b}$ can be derived as a valid [[Definition:Object|object]] in [[Definition:Frege Set Theory|Frege set theo... | Let $P$ be the [[Definition:Property|property]] defined as:
:$\forall x: \map P x := \paren {x = a \lor x = b}$
where $\lor$ is the [[Definition:Disjunction|disjunction operator]].
Hence, using the {{axiom-link|Abstraction}}, we form the [[Definition:Set|set]]:
:$\set {a, b} := \set {x: x = a \lor x = b}$
{{qed}} | Doubleton of Sets can be Derived using Axiom of Abstraction | https://proofwiki.org/wiki/Doubleton_of_Sets_can_be_Derived_using_Axiom_of_Abstraction | https://proofwiki.org/wiki/Doubleton_of_Sets_can_be_Derived_using_Axiom_of_Abstraction | [
"Doubletons",
"Axiom of Abstraction"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Doubleton",
"Definition:Object",
"Definition:Frege Set Theory"
] | [
"Definition:Property",
"Definition:Disjunction",
"Definition:Set"
] |
proofwiki-16794 | Power Set can be Derived using Axiom of Abstraction | Let $a$ be a set.
By application of the {{axiom-link|Abstraction}}, the power set $\powerset a$ can be formed.
Hence the '''power set''' $\powerset a$ can be derived as a valid object in Frege set theory. | Let $P$ be the property defined as:
:$\forall x: \map P x := \paren {x \subseteq a}$
where $\lor$ is the disjunction operator.
Hence, using the {{axiom-link|Abstraction}}, we form the set:
:$\powerset a := \set {x: x \subseteq a}$
{{qed}} | Let $a$ be a [[Definition:Set|set]].
By application of the {{axiom-link|Abstraction}}, the [[Definition:Power Set|power set]] $\powerset a$ can be formed.
Hence the '''[[Definition:Power Set|power set]]''' $\powerset a$ can be derived as a valid [[Definition:Object|object]] in [[Definition:Frege Set Theory|Frege set... | Let $P$ be the [[Definition:Property|property]] defined as:
:$\forall x: \map P x := \paren {x \subseteq a}$
where $\lor$ is the [[Definition:Disjunction|disjunction operator]].
Hence, using the {{axiom-link|Abstraction}}, we form the [[Definition:Set|set]]:
:$\powerset a := \set {x: x \subseteq a}$
{{qed}} | Power Set can be Derived using Axiom of Abstraction | https://proofwiki.org/wiki/Power_Set_can_be_Derived_using_Axiom_of_Abstraction | https://proofwiki.org/wiki/Power_Set_can_be_Derived_using_Axiom_of_Abstraction | [
"Power Set",
"Axiom of Abstraction"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Power Set",
"Definition:Object",
"Definition:Frege Set Theory"
] | [
"Definition:Property",
"Definition:Disjunction",
"Definition:Set"
] |
proofwiki-16795 | Set Union can be Derived using Axiom of Abstraction | Let $a$ be a set of sets.
By application of the {{axiom-link|Abstraction}}, the union $\bigcup a$ can be formed.
Hence the '''union''' $\bigcup a$ can be derived as a valid object in Frege set theory. | Let $P$ be the property defined as:
:$\forall x: \map P x := \paren {\exists y: y \in a \land x \in y}$
where $\land$ is the conjunction operator.
That is, $\map P x$ {{iff}}:
:$x$ is an element of some set $y$, where $y$ is one of the sets that comprise the elements of $a$.
Hence, using the {{axiom-link|Abstraction}},... | Let $a$ be a [[Definition:Set of Sets|set of sets]].
By application of the {{axiom-link|Abstraction}}, the [[Definition:Set Union|union]] $\bigcup a$ can be formed.
Hence the '''[[Definition:Set Union|union]]''' $\bigcup a$ can be derived as a valid [[Definition:Object|object]] in [[Definition:Frege Set Theory|Frege... | Let $P$ be the [[Definition:Property|property]] defined as:
:$\forall x: \map P x := \paren {\exists y: y \in a \land x \in y}$
where $\land$ is the [[Definition:Conjunction|conjunction operator]].
That is, $\map P x$ {{iff}}:
:$x$ is an [[Definition:Element|element]] of some [[Definition:Set|set]] $y$, where $y$ i... | Set Union can be Derived using Axiom of Abstraction | https://proofwiki.org/wiki/Set_Union_can_be_Derived_using_Axiom_of_Abstraction | https://proofwiki.org/wiki/Set_Union_can_be_Derived_using_Axiom_of_Abstraction | [
"Set Union",
"Axiom of Abstraction"
] | [
"Definition:Set of Sets",
"Definition:Set Union",
"Definition:Set Union",
"Definition:Object",
"Definition:Frege Set Theory"
] | [
"Definition:Property",
"Definition:Conjunction",
"Definition:Element",
"Definition:Set",
"Definition:Set",
"Definition:Element",
"Definition:Set"
] |
proofwiki-16796 | Set of Natural Numbers can be Derived using Axiom of Abstraction | Let $\N$ denote the set of natural numbers.
By application of the {{axiom-link|Abstraction}}, $\N$ can be derived as a valid object in Frege set theory. | {{ProofWanted|Use the same construction as in ZF}} | Let $\N$ denote the [[Definition:Natural Numbers|set of natural numbers]].
By application of the {{axiom-link|Abstraction}}, $\N$ can be derived as a valid [[Definition:Object|object]] in [[Definition:Frege Set Theory|Frege set theory]]. | {{ProofWanted|Use the same construction as in [[Definition:Zermelo-Fraenkel Set Theory|ZF]]}} | Set of Natural Numbers can be Derived using Axiom of Abstraction | https://proofwiki.org/wiki/Set_of_Natural_Numbers_can_be_Derived_using_Axiom_of_Abstraction | https://proofwiki.org/wiki/Set_of_Natural_Numbers_can_be_Derived_using_Axiom_of_Abstraction | [
"Natural Numbers",
"Axiom of Abstraction"
] | [
"Definition:Natural Numbers",
"Definition:Object",
"Definition:Frege Set Theory"
] | [
"Definition:Zermelo-Fraenkel Set Theory"
] |
proofwiki-16797 | Frege Set Theory is Logically Inconsistent | The system of axiomatic set theory that is Frege set theory is inconsistent. | From Russell's Paradox, the {{axiom-link|Abstraction}} leads to a contradiction.
Let $q$ be such a contradiction:
:$q = p \land \neg p$
for some statement $p$.
From the Rule of Explosion it then follows that every logical formula is a provable consequence of $q$.
Hence the result, by definition of inconsistent.
{{qed}} | The [[Definition:Axiomatic Set Theory|system of axiomatic set theory]] that is [[Definition:Frege Set Theory|Frege set theory]] is [[Definition:Inconsistent (Logic)|inconsistent]]. | From [[Russell's Paradox]], the {{axiom-link|Abstraction}} leads to a [[Definition:Contradiction|contradiction]].
Let $q$ be such a [[Definition:Contradiction|contradiction]]:
:$q = p \land \neg p$
for some [[Definition:Statement|statement]] $p$.
From the [[Rule of Explosion]] it then follows that every [[Definition:... | Frege Set Theory is Logically Inconsistent | https://proofwiki.org/wiki/Frege_Set_Theory_is_Logically_Inconsistent | https://proofwiki.org/wiki/Frege_Set_Theory_is_Logically_Inconsistent | [
"Frege Set Theory"
] | [
"Definition:Axiomatic Set Theory",
"Definition:Frege Set Theory",
"Definition:Inconsistent (Logic)"
] | [
"Russell's Paradox",
"Definition:Contradiction",
"Definition:Contradiction",
"Definition:Statement",
"Rule of Explosion",
"Definition:Logical Formula",
"Definition:Provable Consequence",
"Definition:Inconsistent (Logic)"
] |
proofwiki-16798 | Exists Subset which is not Element | Let $S$ be a set.
Then there exists at least one subset of $S$ which is not an element of $S$. | Let $S$ be a set.
Let $T$ be the set of all elements of $S$ which do not contain $S$ as elements.
Then by {{Corollary|Russell's Paradox}}, $T$ itself cannot be an element of $S$.
This $T$ is the required subset. | Let $S$ be a [[Definition:Set|set]].
Then there exists at least one [[Definition:Subset|subset]] of $S$ which is not an [[Definition:Element|element]] of $S$. | Let $S$ be a [[Definition:Set|set]].
Let $T$ be the [[Definition:Set|set]] of all [[Definition:Element|elements]] of $S$ which do not contain $S$ as [[Definition:Element|elements]].
Then by {{Corollary|Russell's Paradox}}, $T$ itself cannot be an [[Definition:Element|element]] of $S$.
This $T$ is the required [[Defi... | Exists Subset which is not Element/Proof 1 | https://proofwiki.org/wiki/Exists_Subset_which_is_not_Element | https://proofwiki.org/wiki/Exists_Subset_which_is_not_Element/Proof_1 | [
"Subsets",
"Exists Subset which is not Element"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Element"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Subset"
] |
proofwiki-16799 | Exists Subset which is not Element | Let $S$ be a set.
Then there exists at least one subset of $S$ which is not an element of $S$. | Consider the power set $\powerset S$ of $S$.
From Cantor's Theorem, there is no surjection $f: S \to \powerset S$.
That is, there are more subsets of $S$ than there are elements of $S$.
So there must be at least one subset of $S$ which is not an element of $S$.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Then there exists at least one [[Definition:Subset|subset]] of $S$ which is not an [[Definition:Element|element]] of $S$. | Consider the [[Definition:Power Set|power set]] $\powerset S$ of $S$.
From [[Cantor's Theorem]], there is no [[Definition:Surjection|surjection]] $f: S \to \powerset S$.
That is, there are more [[Definition:Subset|subsets]] of $S$ than there are [[Definition:Element|elements]] of $S$.
So there must be at least one [... | Exists Subset which is not Element/Proof 2 | https://proofwiki.org/wiki/Exists_Subset_which_is_not_Element | https://proofwiki.org/wiki/Exists_Subset_which_is_not_Element/Proof_2 | [
"Subsets",
"Exists Subset which is not Element"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Element"
] | [
"Definition:Power Set",
"Cantor's Theorem",
"Definition:Surjection",
"Definition:Subset",
"Definition:Element",
"Definition:Subset",
"Definition:Element"
] |
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