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-5700 | Product of Powers in B-Algebra | Let $\struct {X, \circ}$ be a $B$-algebra.
Let $x \in X$ and $m, n \in \N$.
Then:
:$x^m \circ x^n = \begin {cases}
x^{m - n} & : m \ge n \\
0 \circ x^{n - m} & : n > m
\end {cases}$ | For $m \ge n$ the result follows from $B$-Algebra Power Law.
For $n > m$ the result follows from $B$-Algebra Power Law with Zero.
{{qed}}
Category:B-Algebras
80evozxwp6k58y1u28pq15t2boz2ulb | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Let $x \in X$ and $m, n \in \N$.
Then:
:$x^m \circ x^n = \begin {cases}
x^{m - n} & : m \ge n \\
0 \circ x^{n - m} & : n > m
\end {cases}$ | For $m \ge n$ the result follows from [[B-Algebra Power Law|$B$-Algebra Power Law]].
For $n > m$ the result follows from [[B-Algebra Power Law with Zero|$B$-Algebra Power Law with Zero]].
{{qed}}
[[Category:B-Algebras]]
80evozxwp6k58y1u28pq15t2boz2ulb | Product of Powers in B-Algebra | https://proofwiki.org/wiki/Product_of_Powers_in_B-Algebra | https://proofwiki.org/wiki/Product_of_Powers_in_B-Algebra | [
"B-Algebras"
] | [
"Definition:B-Algebra"
] | [
"B-Algebra Power Law",
"B-Algebra Power Law with Zero",
"Category:B-Algebras"
] |
proofwiki-5701 | Superset of Co-Countable Set | Every superset of a co-countable set is co-countable. | Let $S$ be a set.
Let $A$ be co-countable in $S$, and let $B$ be such that $A \subseteq B \subseteq S$.
From Relative Complement inverts Subsets, it follows that:
:$\complement_S \left({B}\right) \subseteq \complement_S \left({A}\right)$
As $A$ is co-countable, $\complement_S \left({A}\right)$ is countable.
By Subset o... | Every [[Definition:Superset|superset]] of a [[Definition:Co-Countable Set|co-countable set]] is [[Definition:Co-Countable Set|co-countable]]. | Let $S$ be a [[Definition:Set|set]].
Let $A$ be [[Definition:Co-Countable Set|co-countable]] in $S$, and let $B$ be such that $A \subseteq B \subseteq S$.
From [[Relative Complement inverts Subsets]], it follows that:
:$\complement_S \left({B}\right) \subseteq \complement_S \left({A}\right)$
As $A$ is [[Definition... | Superset of Co-Countable Set | https://proofwiki.org/wiki/Superset_of_Co-Countable_Set | https://proofwiki.org/wiki/Superset_of_Co-Countable_Set | [
"Set Theory"
] | [
"Definition:Subset/Superset",
"Definition:Co-Countable Set",
"Definition:Co-Countable Set"
] | [
"Definition:Set",
"Definition:Co-Countable Set",
"Relative Complement inverts Subsets",
"Definition:Co-Countable Set",
"Definition:Countable Set",
"Subset of Countably Infinite Set is Countable",
"Definition:Countable Set",
"Definition:Co-Countable Set"
] |
proofwiki-5702 | Group Induces B-Algebra | Let $\struct {G, \circ}$ be a group whose identity element is $e$.
Let $*$ be the product inverse operation on $G$:
:$\forall a, b \in G: a * b = a \circ b^{-1}$
where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$.
Then the algebraic structure $\struct {G, *}$ is a $B$-algebra. | First note that by {{Group-axiom|0}}:
:$\forall a, b \in G: a * b \in G$
and so {{B-algebra-axiom|C}} holds.
We have that:
:$\forall x \in G: x * x = x \circ x^{-1} = e$
by definition of inverse element.
Let $0 := e$.
Then:
:{{B-algebra-axiom|0}}: $\quad \exists 0 \in G$
:{{B-algebra-axiom|1}}: $\quad \forall x \in G: ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e$.
Let $*$ be the [[Definition:Product Inverse Operation|product inverse operation]] on $G$:
:$\forall a, b \in G: a * b = a \circ b^{-1}$
where $b^{-1}$ is the [[Definition:Inverse Element|inverse elem... | First note that by {{Group-axiom|0}}:
:$\forall a, b \in G: a * b \in G$
and so {{B-algebra-axiom|C}} holds.
We have that:
:$\forall x \in G: x * x = x \circ x^{-1} = e$
by definition of [[Definition:Inverse Element|inverse element]].
Let $0 := e$.
Then:
:{{B-algebra-axiom|0}}: $\quad \exists 0 \in G$
:{{B-algebra... | Group Induces B-Algebra | https://proofwiki.org/wiki/Group_Induces_B-Algebra | https://proofwiki.org/wiki/Group_Induces_B-Algebra | [
"B-Algebras",
"Group Theory"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Product Inverse Operation",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Operation/Binary Operation",
"Definition:Algebraic Structure",
"Definition:B-Algebra"
] | [
"Definition:Inverse (Abstract Algebra)/Inverse",
"Inverse of Identity Element is Itself",
"Inverse of Group Product",
"Inverse of Group Inverse",
"Definition:B-Algebra",
"Category:B-Algebras",
"Category:Group Theory"
] |
proofwiki-5703 | Abelian Group Induces Commutative B-Algebra | Let $\struct {G, \circ}$ be an abelian group whose identity element is $e$.
Let $*$ be the product inverse operation on $G$:
:$\forall a, b \in G: a * b = a \circ b^{-1}$
where $b^{-1}$ is the inverse element of $b$ under the operation $\circ$.
Then the algebraic structure $\struct {G, *}$ is a commutative $B$-algebra.... | From Group Induces $B$-Algebra, $\struct {G, *}$ is a $B$-Algebra.
As in the proof Group Induces $B$-Algebra, we let:
:$0 := e$
Now we demonstrate $0$-commutativity.
Let $x, y \in G$:
{{begin-eqn}}
{{eqn | l = x * \paren {0 * y}
| r = x \circ \paren {e \circ y^{-1} }^{-1}
| c = Definition of $*$ and $0$
}}... | Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity element]] is $e$.
Let $*$ be the [[Definition:Product Inverse Operation|product inverse operation]] on $G$:
:$\forall a, b \in G: a * b = a \circ b^{-1}$
where $b^{-1}$ is the [[Definition:Inverse Ele... | From [[Group Induces B-Algebra|Group Induces $B$-Algebra]], $\struct {G, *}$ is a [[Definition:B-Algebra|$B$-Algebra]].
As in the proof [[Group Induces B-Algebra|Group Induces $B$-Algebra]], we let:
:$0 := e$
Now we demonstrate $0$-commutativity.
Let $x, y \in G$:
{{begin-eqn}}
{{eqn | l = x * \paren {0 * y}
... | Abelian Group Induces Commutative B-Algebra | https://proofwiki.org/wiki/Abelian_Group_Induces_Commutative_B-Algebra | https://proofwiki.org/wiki/Abelian_Group_Induces_Commutative_B-Algebra | [
"B-Algebras",
"Abelian Groups"
] | [
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Product Inverse Operation",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Operation/Binary Operation",
"Definition:Algebraic Structure",
"Definition:Commutative B-Algebra"
] | [
"Group Induces B-Algebra",
"Definition:B-Algebra",
"Group Induces B-Algebra",
"Inverse of Group Inverse",
"Inverse of Group Inverse",
"Category:B-Algebras",
"Category:Abelian Groups"
] |
proofwiki-5704 | Identity of Cardinal Product is One | Let $\mathbf a$ be a cardinal.
Then:
: $\bsone \mathbf a = \mathbf a$
where $\bsone \mathbf a$ denotes the product of the (cardinal) one and $\mathbf a$.
That is, $\bsone$ is the identity element of the product operation on cardinals. | Let $\mathbf a = \card A$ for some set $A$.
From the definition of (cardinal) one, $\bsone$ is the cardinal associated with a singleton set, say, $\set \O$.
We have by definition of product of cardinals that $\bsone \mathbf a$ is the cardinal associated with $\set \O \times A$.
Consider the mapping $f: \set \O \times A... | Let $\mathbf a$ be a [[Definition:Cardinal|cardinal]].
Then:
: $\bsone \mathbf a = \mathbf a$
where $\bsone \mathbf a$ denotes the [[Definition:Product of Cardinals|product]] of the [[Definition:One (Cardinal)|(cardinal) one]] and $\mathbf a$.
That is, $\bsone$ is the [[Definition:Identity Element|identity element]... | Let $\mathbf a = \card A$ for some [[Definition:Set|set]] $A$.
From the definition of [[Definition:One (Cardinal)|(cardinal) one]], $\bsone$ is the [[Definition:Cardinal|cardinal]] associated with a [[Definition:Singleton|singleton set]], say, $\set \O$.
We have by definition of [[Definition:Product of Cardinals|prod... | Identity of Cardinal Product is One | https://proofwiki.org/wiki/Identity_of_Cardinal_Product_is_One | https://proofwiki.org/wiki/Identity_of_Cardinal_Product_is_One | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Product of Cardinals",
"Definition:One (Cardinal)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Product of Cardinals",
"Definition:Cardinal"
] | [
"Definition:Set",
"Definition:One (Cardinal)",
"Definition:Cardinal",
"Definition:Singleton",
"Definition:Product of Cardinals",
"Definition:Cardinal",
"Definition:Mapping",
"Equality of Ordered Pairs",
"Definition:Injection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjec... |
proofwiki-5705 | Zero of Cardinal Product is Zero | Let $\mathbf a$ be a cardinal.
Then:
: $\mathbf 0 \mathbf a = \mathbf 0$
where $\mathbf 0 \mathbf a$ denotes the product of the (cardinal) zero and $\mathbf a$.
That is, $\mathbf 0$ is the zero element of the product operation on cardinals. | Let $\mathbf a = \card A$ for some set $A$.
From the definition of (cardinal) zero, $\mathbf 0$ is the cardinal associated with the empty set $\O$.
We have by definition of product of cardinals that $\mathbf 0 \mathbf a$ is the cardinal associated with $\O \times A$.
But from Cartesian Product is Empty iff Factor is Em... | Let $\mathbf a$ be a [[Definition:Cardinal|cardinal]].
Then:
: $\mathbf 0 \mathbf a = \mathbf 0$
where $\mathbf 0 \mathbf a$ denotes the [[Definition:Product of Cardinals|product]] of the [[Definition:Zero (Cardinal)|(cardinal) zero]] and $\mathbf a$.
That is, $\mathbf 0$ is the [[Definition:Zero Element|zero eleme... | Let $\mathbf a = \card A$ for some [[Definition:Set|set]] $A$.
From the definition of [[Definition:Zero (Cardinal)|(cardinal) zero]], $\mathbf 0$ is the [[Definition:Cardinal|cardinal]] associated with the [[Definition:Empty Set|empty set]] $\O$.
We have by definition of [[Definition:Product of Cardinals|product of c... | Zero of Cardinal Product is Zero | https://proofwiki.org/wiki/Zero_of_Cardinal_Product_is_Zero | https://proofwiki.org/wiki/Zero_of_Cardinal_Product_is_Zero | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Product of Cardinals",
"Definition:Zero (Cardinal)",
"Definition:Zero Element",
"Definition:Product of Cardinals",
"Definition:Cardinal"
] | [
"Definition:Set",
"Definition:Zero (Cardinal)",
"Definition:Cardinal",
"Definition:Empty Set",
"Definition:Product of Cardinals",
"Definition:Cardinal",
"Cartesian Product is Empty iff Factor is Empty"
] |
proofwiki-5706 | Sum of Cardinals is Commutative | Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
:$\mathbf a + \mathbf b = \mathbf b + \mathbf a$
where $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \map \Card A$ and $\mathbf b = \map \Card B$ for some sets $A$ and $B$ such that $A \cap B = \O$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a + \mathbf b
| r = \map \Card {A \cup B}
| c = {{Defof|Sum of Cardinals}}
}}
{{eqn | r = \map \Card {B \cup A}
| c = Union is Commutative
}}
{{eqn... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Cardinal|cardinals]].
Then:
:$\mathbf a + \mathbf b = \mathbf b + \mathbf a$
where $\mathbf a + \mathbf b$ denotes the [[Definition:Sum of Cardinals|sum]] of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \map \Card A$ and $\mathbf b = \map \Card B$ for some [[Definition:Set|sets]] $A$ and $B$ such that $A \cap B = \O$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf a + \mathbf b
| r = \map \Card {A \cup B}
| c = {{Defof|Sum of Cardinals}}
}}
{{eqn | r = \map \Card {B \cup A}
| c = [[Union ... | Sum of Cardinals is Commutative | https://proofwiki.org/wiki/Sum_of_Cardinals_is_Commutative | https://proofwiki.org/wiki/Sum_of_Cardinals_is_Commutative | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Sum of Cardinals"
] | [
"Definition:Set",
"Union is Commutative"
] |
proofwiki-5707 | Sum of Cardinals is Associative | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
: $\mathbf a + \paren {\mathbf b + \mathbf c} = \paren {\mathbf a + \mathbf b} + \mathbf c$
where $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \card A, \mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.
Let $A, B, C$ be pairwise disjoint, that is:
:$A \cap B = \O$
:$B \cap C = \O$
:$A \cap C = \O$
Then we can define:
:$A \sqcup B := A \cup B$
:$B \sqcup C := B \cup C$
:$A \sqcup C := A \cup C$
where $A \sqcup B$ de... | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be [[Definition:Cardinal|cardinals]].
Then:
: $\mathbf a + \paren {\mathbf b + \mathbf c} = \paren {\mathbf a + \mathbf b} + \mathbf c$
where $\mathbf a + \mathbf b$ denotes the [[Definition:Sum of Cardinals|sum]] of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \card A, \mathbf b = \card B$ and $\mathbf c = \card C$ for some [[Definition:Set|sets]] $A$, $B$ and $C$.
Let $A, B, C$ be [[Definition:Pairwise Disjoint|pairwise disjoint]], that is:
:$A \cap B = \O$
:$B \cap C = \O$
:$A \cap C = \O$
Then we can define:
:$A \sqcup B := A \cup B$
:$B \sqcup C := B ... | Sum of Cardinals is Associative | https://proofwiki.org/wiki/Sum_of_Cardinals_is_Associative | https://proofwiki.org/wiki/Sum_of_Cardinals_is_Associative | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Sum of Cardinals"
] | [
"Definition:Set",
"Definition:Pairwise Disjoint",
"Definition:Disjoint Union (Set Theory)",
"Intersection Distributes over Union",
"Union with Empty Set",
"Intersection Distributes over Union",
"Union with Empty Set",
"Union is Associative"
] |
proofwiki-5708 | B-Algebra is Commutative iff x(xy)=y | Let $\struct {X, \circ}$ be a $B$-algebra.
Then $\struct {X, \circ}$ is commutative {{iff}}:
:$\forall x, y \in X: x \circ \paren {x \circ y} = y$ | === Necessary Condition ===
Let $x, y \in X$:
{{begin-eqn}}
{{eqn | l = x \circ \paren {x \circ y}
| r = \paren {x \circ \paren {0 \circ y} } \circ x
| c = Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$
}}
{{eqn | r = \paren {y \circ \paren {0 \circ x} } \circ x
... | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then $\struct {X, \circ}$ is [[Definition:Commutative B-Algebra|commutative]] {{iff}}:
:$\forall x, y \in X: x \circ \paren {x \circ y} = y$ | === Necessary Condition ===
Let $x, y \in X$:
{{begin-eqn}}
{{eqn | l = x \circ \paren {x \circ y}
| r = \paren {x \circ \paren {0 \circ y} } \circ x
| c = [[B-Algebra Identity: x (y z) = (x (0 z)) y|Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$]]
}}
{{eqn | r = \p... | B-Algebra is Commutative iff x(xy)=y | https://proofwiki.org/wiki/B-Algebra_is_Commutative_iff_x(xy)=y | https://proofwiki.org/wiki/B-Algebra_is_Commutative_iff_x(xy)=y | [
" B-Algebras"
] | [
"Definition:B-Algebra",
"Definition:Commutative B-Algebra"
] | [
"B-Algebra Identity: x (y z) = (x (0 z)) y",
"Definition:Commutative B-Algebra",
"Definition:Commutative B-Algebra"
] |
proofwiki-5709 | Identity of Cardinal Sum is Zero | Let $\mathbf a$ be a cardinal.
Then:
: $\mathbf a + \mathbf 0 = \mathbf a$
where $\mathbf a + \mathbf 0$ denotes the sum of the zero cardinal and $\mathbf a$.
That is, $\mathbf 0$ is the identity element of the sum operation on cardinals. | Let $\mathbf a = \card A$ for some set $A$.
From Union with Empty Set we have $A \cup \O = A$.
From Intersection with Empty Set we have $A \cap \O = \O$.
So $A$ and $\O$ are disjoint and so:
{{begin-eqn}}
{{eqn | l=\mathbf a + \mathbf 0
| r=\card {A \cup \O}
| c={{Defof|Sum of Cardinals}}
}}
{{eqn | r=\card... | Let $\mathbf a$ be a [[Definition:Cardinal|cardinal]].
Then:
: $\mathbf a + \mathbf 0 = \mathbf a$
where $\mathbf a + \mathbf 0$ denotes the [[Definition:Sum of Cardinals|sum]] of the [[Definition:Zero (Cardinal)|zero cardinal]] and $\mathbf a$.
That is, $\mathbf 0$ is the [[Definition:Identity Element|identity ele... | Let $\mathbf a = \card A$ for some [[Definition:Set|set]] $A$.
From [[Union with Empty Set]] we have $A \cup \O = A$.
From [[Intersection with Empty Set]] we have $A \cap \O = \O$.
So $A$ and $\O$ are [[Definition:Disjoint Sets|disjoint]] and so:
{{begin-eqn}}
{{eqn | l=\mathbf a + \mathbf 0
| r=\card {A \cup... | Identity of Cardinal Sum is Zero | https://proofwiki.org/wiki/Identity_of_Cardinal_Sum_is_Zero | https://proofwiki.org/wiki/Identity_of_Cardinal_Sum_is_Zero | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Sum of Cardinals",
"Definition:Zero (Cardinal)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Sum of Cardinals",
"Definition:Cardinal"
] | [
"Definition:Set",
"Union with Empty Set",
"Intersection with Empty Set",
"Definition:Disjoint Sets",
"Cardinality of Set Union/Corollary",
"Cardinality of Empty Set"
] |
proofwiki-5710 | Commutative B-Algebra Implies (zy)(zx)=xy | Let $\struct {B, \circ}$ be a commutative $B$-algebra.
Then:
:$\forall x, y, z \in X: \paren {z \circ y} \circ \paren {z \circ x} = x \circ y$ | Let $x, y, z \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {z \circ y} \circ \paren {z \circ x}
| r = \paren {\paren {z \circ y} \circ \paren {0 \circ x} } \circ z
| c = Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$
}}
{{eqn | r = \paren {x \circ \paren {0 \circ \par... | Let $\struct {B, \circ}$ be a [[Definition:Commutative B-Algebra|commutative $B$-algebra]].
Then:
:$\forall x, y, z \in X: \paren {z \circ y} \circ \paren {z \circ x} = x \circ y$ | Let $x, y, z \in X$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {z \circ y} \circ \paren {z \circ x}
| r = \paren {\paren {z \circ y} \circ \paren {0 \circ x} } \circ z
| c = [[B-Algebra Identity: x (y z) = (x (0 z)) y|Identity: $x \circ \paren {y \circ z} = \paren {x \circ \paren {0 \circ z} } \circ y$]]
}}
{... | Commutative B-Algebra Implies (zy)(zx)=xy | https://proofwiki.org/wiki/Commutative_B-Algebra_Implies_(zy)(zx)=xy | https://proofwiki.org/wiki/Commutative_B-Algebra_Implies_(zy)(zx)=xy | [
"B-Algebras"
] | [
"Definition:Commutative B-Algebra"
] | [
"B-Algebra Identity: x (y z) = (x (0 z)) y",
"B-Algebra Identity: x (y z) = (x (0 z)) y",
"B-Algebra is Commutative iff x(xy)=y",
"Category:B-Algebras"
] |
proofwiki-5711 | Cardinal Product Distributes over Cardinal Sum | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
: $\mathbf a \paren {\mathbf b + \mathbf c} = \mathbf a \mathbf b + \mathbf a \mathbf c$
where:
: $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.
: $\mathbf a \mathbf b$ denotes the product of $\mathbf a$ and $\mathbf b$. | Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.
Let $B$ and $C$ be pairwise disjoint, that is:
:$B \cap C = \O$
Then we can define:
:$B \sqcup C := B \cup C$
where $B \sqcup C$ denotes the disjoint union of $B$ and $C$.
Then we have:
:$\mathbf b + \mathbf c = \... | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be [[Definition:Cardinal|cardinals]].
Then:
: $\mathbf a \paren {\mathbf b + \mathbf c} = \mathbf a \mathbf b + \mathbf a \mathbf c$
where:
: $\mathbf a + \mathbf b$ denotes the [[Definition:Sum of Cardinals|sum]] of $\mathbf a$ and $\mathbf b$.
: $\mathbf a \mathbf b$ den... | Let $\mathbf a = \card A$, $\mathbf b = \card B$ and $\mathbf c = \card C$ for some [[Definition:Set|sets]] $A$, $B$ and $C$.
Let $B$ and $C$ be [[Definition:Pairwise Disjoint|pairwise disjoint]], that is:
:$B \cap C = \O$
Then we can define:
:$B \sqcup C := B \cup C$
where $B \sqcup C$ denotes the [[Definition:Disjo... | Cardinal Product Distributes over Cardinal Sum | https://proofwiki.org/wiki/Cardinal_Product_Distributes_over_Cardinal_Sum | https://proofwiki.org/wiki/Cardinal_Product_Distributes_over_Cardinal_Sum | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Sum of Cardinals",
"Definition:Product of Cardinals"
] | [
"Definition:Set",
"Definition:Pairwise Disjoint",
"Definition:Disjoint Union (Set Theory)",
"Cartesian Product of Intersections/Corollary 1",
"Cartesian Product is Empty iff Factor is Empty",
"Definition:Disjoint Sets",
"Cartesian Product Distributes over Union"
] |
proofwiki-5712 | Ordering of Cardinals Compatible with Cardinal Product | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
:$\mathbf a \le \mathbf b \implies \mathbf a \mathbf c \le \mathbf b \mathbf c$
where $\mathbf a \mathbf c$ denotes the product of $\mathbf a$ and $\mathbf c$. | Let $\mathbf a = \map \Card A$, $\mathbf b = \map \Card B$ and $\mathbf c = \map \Card C$ for some sets $A$, $B$ and $C$.
Let $\mathbf a \le \mathbf b$.
Then by definition of cardinal, there exists an injection $f: A \to B$.
Then the mapping $g: A \times C \to B \times C$ defined as:
:$\forall \tuple {a, c} \in A \time... | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be [[Definition:Cardinal|cardinals]].
Then:
:$\mathbf a \le \mathbf b \implies \mathbf a \mathbf c \le \mathbf b \mathbf c$
where $\mathbf a \mathbf c$ denotes the [[Definition:Product of Cardinals|product]] of $\mathbf a$ and $\mathbf c$. | Let $\mathbf a = \map \Card A$, $\mathbf b = \map \Card B$ and $\mathbf c = \map \Card C$ for some [[Definition:Set|sets]] $A$, $B$ and $C$.
Let $\mathbf a \le \mathbf b$.
Then by definition of [[Definition:Cardinal|cardinal]], there exists an [[Definition:Injection|injection]] $f: A \to B$.
Then the [[Definition:Ma... | Ordering of Cardinals Compatible with Cardinal Product | https://proofwiki.org/wiki/Ordering_of_Cardinals_Compatible_with_Cardinal_Product | https://proofwiki.org/wiki/Ordering_of_Cardinals_Compatible_with_Cardinal_Product | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Product of Cardinals"
] | [
"Definition:Set",
"Definition:Cardinal",
"Definition:Injection",
"Definition:Mapping",
"Equality of Ordered Pairs",
"Definition:Injection",
"Equality of Ordered Pairs",
"Definition:Injection",
"Definition:Product of Cardinals"
] |
proofwiki-5713 | Ordering of Cardinals Compatible with Cardinal Sum | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
:$\mathbf a \le \mathbf b \implies \mathbf a + \mathbf c \le \mathbf b + \mathbf c$
where $\mathbf a \mathbf c$ denotes the sum of $\mathbf a$ and $\mathbf c$. | Let $\mathbf a = \map \Card A$, $\mathbf b = \map \Card B$ and $\mathbf c = \map \Card C$ for some sets $A$, $B$ and $C$.
Let $C$ be chosen such that $A \cap C = \O = B \cap C$.
Let $\mathbf a \le \mathbf b$.
Then by definition of cardinal, there exists an injection $f: A \to B$.
Then the mapping $h: A \cup C \to B \cu... | Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be [[Definition:Cardinal|cardinals]].
Then:
:$\mathbf a \le \mathbf b \implies \mathbf a + \mathbf c \le \mathbf b + \mathbf c$
where $\mathbf a \mathbf c$ denotes the [[Definition:Sum of Cardinals|sum]] of $\mathbf a$ and $\mathbf c$. | Let $\mathbf a = \map \Card A$, $\mathbf b = \map \Card B$ and $\mathbf c = \map \Card C$ for some [[Definition:Set|sets]] $A$, $B$ and $C$.
Let $C$ be chosen such that $A \cap C = \O = B \cap C$.
Let $\mathbf a \le \mathbf b$.
Then by definition of [[Definition:Cardinal|cardinal]], there exists an [[Definition:Inj... | Ordering of Cardinals Compatible with Cardinal Sum | https://proofwiki.org/wiki/Ordering_of_Cardinals_Compatible_with_Cardinal_Sum | https://proofwiki.org/wiki/Ordering_of_Cardinals_Compatible_with_Cardinal_Sum | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Sum of Cardinals"
] | [
"Definition:Set",
"Definition:Cardinal",
"Definition:Injection",
"Definition:Mapping",
"Definition:Injection",
"Definition:Injection",
"Definition:Sum of Cardinals"
] |
proofwiki-5714 | Group is B-Algebra Iff All Elements Self-Inverse | Let $\struct {G, \circ}$ be a group whose identity element is $e$.
Then $\struct {G, \circ}$ is also a $B$-algebra {{iff}}:
:$\forall g \in G: g = g^{-1}$
That is, {{iff}} all elements of $G$ are self-inverse. | Let $\struct {G, \circ}$ be a group such that $\forall g \in G: g = g^{-1}$.
From Group Induces B-Algebra, the algebraic structure $\struct {G, *}$ where $*$ is defined as:
:$\forall a, b \in G: a * b := a \circ b^{-1}$
is a $B$-algebra.
But as $b = b^{-1}$, it follows that:
:$\forall a, b \in G: a * b := a \circ b$
an... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e$.
Then $\struct {G, \circ}$ is also a [[Definition:B-Algebra|$B$-algebra]] {{iff}}:
:$\forall g \in G: g = g^{-1}$
That is, {{iff}} all [[Definition:Element|elements]] of $G$ are [[Definition:Self-In... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] such that $\forall g \in G: g = g^{-1}$.
From [[Group Induces B-Algebra]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {G, *}$ where $*$ is defined as:
:$\forall a, b \in G: a * b := a \circ b^{-1}$
is a [[Definition:B-Algebra|$B$-algebra... | Group is B-Algebra Iff All Elements Self-Inverse | https://proofwiki.org/wiki/Group_is_B-Algebra_Iff_All_Elements_Self-Inverse | https://proofwiki.org/wiki/Group_is_B-Algebra_Iff_All_Elements_Self-Inverse | [
"Group Theory",
"B-Algebras"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:B-Algebra",
"Definition:Element",
"Definition:Self-Inverse Element"
] | [
"Definition:Group",
"Group Induces B-Algebra",
"Definition:Algebraic Structure",
"Definition:B-Algebra",
"Definition:B-Algebra",
"Definition:B-Algebra",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Identity is Unique",
"Definition:B-Algebra",
"Definition:Self-Inverse Element",
"... |
proofwiki-5715 | Condition for Existence of Cardinal Sum | Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
: $\mathbf a \le \mathbf b \iff \exists \mathbf c: \mathbf a + \mathbf c = \mathbf b$
where $\mathbf c$ is also a cardinal. | Let $\mathbf a = \card A$ and $\mathbf b = \card B$ for some sets $A$ and $B$.
From Equivalence of Definitions of Dominate (Set Theory), there exists a bijection from $A$ onto a subset $E$ of $B$.
Thus:
: $\mathop a = \card E$
Let $F = \relcomp B E$
From Set with Relative Complement forms Partition:
: $B = E \cup F$
: ... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Cardinal|cardinals]].
Then:
: $\mathbf a \le \mathbf b \iff \exists \mathbf c: \mathbf a + \mathbf c = \mathbf b$
where $\mathbf c$ is also a [[Definition:Cardinal|cardinal]]. | Let $\mathbf a = \card A$ and $\mathbf b = \card B$ for some [[Definition:Set|sets]] $A$ and $B$.
From [[Equivalence of Definitions of Dominate (Set Theory)]], there exists a [[Definition:Bijection|bijection]] from $A$ onto a [[Definition:Subset|subset]] $E$ of $B$.
Thus:
: $\mathop a = \card E$
Let $F = \relcomp B... | Condition for Existence of Cardinal Sum | https://proofwiki.org/wiki/Condition_for_Existence_of_Cardinal_Sum | https://proofwiki.org/wiki/Condition_for_Existence_of_Cardinal_Sum | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:Cardinal"
] | [
"Definition:Set",
"Equivalence of Definitions of Dominate (Set Theory)",
"Definition:Bijection",
"Definition:Subset",
"Set Difference and Intersection form Partition/Corollary 2",
"Definition:Sum of Cardinals",
"Definition:Cardinal",
"Definition:Sum of Cardinals",
"Definition:Set",
"Definition:Inj... |
proofwiki-5716 | Cardinal One is Cancellable for Cardinal Sum | Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
:$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 \implies \mathbf a = \mathbf b$
where $\mathbf 1$ is (cardinal) one. | Suppose $\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1$.
Then from Condition for Existence of Cardinal Sum there exists some cardinal $\mathbf c$ such that
:$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 = \mathbf c$
By definition of cardinal there exists a set $C$ such that $\mathbf c = \map \Card C$.
$C$ also has su... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Cardinal|cardinals]].
Then:
:$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 \implies \mathbf a = \mathbf b$
where $\mathbf 1$ is [[Definition:One (Cardinal)|(cardinal) one]]. | Suppose $\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1$.
Then from [[Condition for Existence of Cardinal Sum]] there exists some [[Definition:Cardinal|cardinal]] $\mathbf c$ such that
:$\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 = \mathbf c$
By definition of [[Definition:Cardinal|cardinal]] there exists a [[Defin... | Cardinal One is Cancellable for Cardinal Sum | https://proofwiki.org/wiki/Cardinal_One_is_Cancellable_for_Cardinal_Sum | https://proofwiki.org/wiki/Cardinal_One_is_Cancellable_for_Cardinal_Sum | [
"Cardinals"
] | [
"Definition:Cardinal",
"Definition:One (Cardinal)"
] | [
"Condition for Existence of Cardinal Sum",
"Definition:Cardinal",
"Definition:Cardinal",
"Definition:Set",
"Definition:Subset",
"Definition:Element"
] |
proofwiki-5717 | Mapping from Finite Set to Itself is Injection iff Surjection | Let $S$ be a finite set.
Let $f: S \to S$ be a mapping.
Then $f$ is injective {{iff}} $f$ is surjective. | Let $f$ be an injection.
From Injection to Image is Bijection, $S$ is equivalent to the image $\Img f$ of $f$.
We are given that $S$ is finite.
It follows from Infinite Set is Equivalent to Proper Subset that $\Img f = S$.
It follows by definition that $f$ is surjective.
{{qed|lemma}}
Let $f$ be a surjection.
Then by S... | Let $S$ be a [[Definition:Finite Set|finite set]].
Let $f: S \to S$ be a [[Definition:Mapping|mapping]].
Then $f$ is [[Definition:Injection|injective]] {{iff}} $f$ is [[Definition:Surjection|surjective]]. | Let $f$ be an [[Definition:Injection|injection]].
From [[Injection to Image is Bijection]], $S$ is [[Definition:Set Equivalence|equivalent]] to the [[Definition:Image of Mapping|image]] $\Img f$ of $f$.
We are given that $S$ is [[Definition:Finite Set|finite]].
It follows from [[Infinite Set is Equivalent to Proper ... | Mapping from Finite Set to Itself is Injection iff Surjection | https://proofwiki.org/wiki/Mapping_from_Finite_Set_to_Itself_is_Injection_iff_Surjection | https://proofwiki.org/wiki/Mapping_from_Finite_Set_to_Itself_is_Injection_iff_Surjection | [
"Injections",
"Surjections"
] | [
"Definition:Finite Set",
"Definition:Mapping",
"Definition:Injection",
"Definition:Surjection"
] | [
"Definition:Injection",
"Injection to Image is Bijection",
"Definition:Set Equivalence",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Finite Set",
"Infinite Set is Equivalent to Proper Subset",
"Definition:Surjection",
"Definition:Surjection",
"Surjection iff Right Inverse",
"Defin... |
proofwiki-5718 | Finite Cardinals form Infinite Set | The finite cardinals form a set which is infinite. | {{proof wanted|This proof depends upon the Axiom of Infinity.}} | The [[Definition:Finite Cardinal|finite cardinals]] form a [[Definition:Set|set]] which is [[Definition:Infinite Set|infinite]]. | {{proof wanted|This proof depends upon the [[Axiom:Axiom of Infinity|Axiom of Infinity]].}} | Finite Cardinals form Infinite Set | https://proofwiki.org/wiki/Finite_Cardinals_form_Infinite_Set | https://proofwiki.org/wiki/Finite_Cardinals_form_Infinite_Set | [
"Cardinals",
"Infinite Sets"
] | [
"Definition:Finite Cardinal",
"Definition:Set",
"Definition:Infinite Set"
] | [
"Axiom:Axiom of Infinity"
] |
proofwiki-5719 | Natural Numbers as Cardinals | The natural numbers $\N = \set {0, 1, 2, 3, \ldots}$ can be defined as the set of cardinals. | From Finite Cardinals form Infinite Set we have that the cardinals form a set which is infinite.
{{proof wanted|Show that the set of finite cardinals satisfy Peano's axioms.}} | The [[Definition:Natural Numbers|natural numbers]] $\N = \set {0, 1, 2, 3, \ldots}$ can be defined as the [[Definition:Set|set]] of [[Definition:Cardinal|cardinals]]. | From [[Finite Cardinals form Infinite Set]] we have that the [[Definition:Cardinal|cardinals]] form a [[Definition:Set|set]] which is [[Definition:Infinite Set|infinite]].
{{proof wanted|Show that the set of finite cardinals satisfy Peano's axioms.}} | Natural Numbers as Cardinals | https://proofwiki.org/wiki/Natural_Numbers_as_Cardinals | https://proofwiki.org/wiki/Natural_Numbers_as_Cardinals | [
"Cardinals",
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Set",
"Definition:Cardinal"
] | [
"Finite Cardinals form Infinite Set",
"Definition:Cardinal",
"Definition:Set",
"Definition:Infinite Set"
] |
proofwiki-5720 | Increasing Real Function has Countably Many Discontinuities | Let $f: \R \to \R$ be an increasing real function.
Then $f$ has at most countably many discontinuities. | Let $D \subseteq \R$ be the set of discontinuities of $f$.
From Limit of Monotone Real Function: Increasing: Corollary, we have that:
:$\ds \lim_{x \mathop \to b^-} \map f x = l_x$ and $\ds \lim_{x \mathop \to b^+} \map f x = r_x$
both exist for each $x \in \R$.
Further:
:$l_x \le r_x$
Let $x \in D$.
Then since:
:$f... | Let $f: \R \to \R$ be an [[Definition:Increasing Real Function|increasing real function]].
Then $f$ has at most [[Definition:Countable Set|countably many]] [[Definition:Discontinuity (Real Analysis)|discontinuities]]. | Let $D \subseteq \R$ be the [[Definition:Set|set]] of [[Definition:Discontinuity (Real Analysis)|discontinuities]] of $f$.
From [[Limit of Monotone Real Function/Increasing/Corollary|Limit of Monotone Real Function: Increasing: Corollary]], we have that:
:$\ds \lim_{x \mathop \to b^-} \map f x = l_x$ and $\ds \lim_{x... | Increasing Real Function has Countably Many Discontinuities | https://proofwiki.org/wiki/Increasing_Real_Function_has_Countably_Many_Discontinuities | https://proofwiki.org/wiki/Increasing_Real_Function_has_Countably_Many_Discontinuities | [
"Increasing Real Function has Countably Many Discontinuities",
"Monotone Real Functions",
"Real Analysis"
] | [
"Definition:Increasing/Real Function",
"Definition:Countable Set",
"Definition:Discontinuity (Real Analysis)"
] | [
"Definition:Set",
"Definition:Discontinuity (Real Analysis)",
"Limit of Increasing Function/Corollary",
"Definition:Discontinuity (Real Analysis)",
"Definition:Limit of Real Function/Left",
"Definition:Limit of Real Function/Right",
"Definition:Interval/Ordered Set/Open",
"Definition:Disjoint Sets",
... |
proofwiki-5721 | Integral with respect to Pushforward Measure | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\struct {X', \Sigma'}$ be a measurable space.
Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.
Let $f: X' \to \overline \R$ be a positive $\Sigma'$-measurable function.
Let $\map T \mu$ be the pushforward measure of $\mu$ under $T$.
Then $f \cir... | From Composition of Measurable Mappings is Measurable:
:$f \circ T$ is $\Sigma$-measurable.
Clearly $f \circ T \ge 0$, so $f \circ T$ is a positive $\Sigma$-measurable function.
We first show the proposition for characteristic functions $f$.
Suppose $f = \chi_A$ for $A \in \Sigma'$.
Then, we have:
{{begin-eqn}}
{{eqn ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\struct {X', \Sigma'}$ be a [[Definition:Measurable Space|measurable space]].
Let $T: X \to X'$ be a [[Definition:Measurable Mapping|$\Sigma \, / \, \Sigma'$-measurable mapping]].
Let $f: X' \to \overline \R$ be a [[Definition:Posit... | From [[Composition of Measurable Mappings is Measurable]]:
:$f \circ T$ is [[Definition:Measurable Function|$\Sigma$-measurable]].
Clearly $f \circ T \ge 0$, so $f \circ T$ is a [[Definition:Measurable Function|positive $\Sigma$-measurable function]].
We first show the proposition for [[Definition:Characteristic Fun... | Integral with respect to Pushforward Measure | https://proofwiki.org/wiki/Integral_with_respect_to_Pushforward_Measure | https://proofwiki.org/wiki/Integral_with_respect_to_Pushforward_Measure | [
"Integral with respect to Pushforward Measure",
"Integrals of Integrable Functions",
"Pushforward Measures"
] | [
"Definition:Measure Space",
"Definition:Measurable Space",
"Definition:Measurable Mapping",
"Definition:Measurable Function/Positive",
"Definition:Pushforward Measure",
"Definition:Measurable Function/Positive"
] | [
"Composition of Measurable Mappings is Measurable",
"Definition:Measurable Function",
"Definition:Measurable Function",
"Definition:Characteristic Function",
"Characteristic Function of Preimage",
"Definition:Simple Function",
"Measurable Function is Simple Function iff Finite Image Set/Corollary",
"D... |
proofwiki-5722 | Tartaglia's Formula | Let $T$ be a tetrahedron with vertices $\mathbf d_1, \mathbf d_2, \mathbf d_3$ and $\mathbf d_4$.
For all $i$ and $j$, let the distance between $\mathbf d_i$ and $\mathbf d_j$ be denoted $d_{ij}$.
Then the volume $V_T$ of $T$ satisfies:
$\quad V_T^2 = \dfrac 1 {288} \det \ \begin{vmatrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 &... | A proof of Tartaglia's Formula will be found in a proof of the Value of Cayley-Menger Determinant as a tetrahedron is a $3$-simplex.
{{proof wanted}} | Let $T$ be a [[Definition:Tetrahedron|tetrahedron]] with [[Definition:Vertex of Polyhedron|vertices]] $\mathbf d_1, \mathbf d_2, \mathbf d_3$ and $\mathbf d_4$.
For all $i$ and $j$, let the [[Definition:Length (Linear Measure)|distance]] between $\mathbf d_i$ and $\mathbf d_j$ be denoted $d_{ij}$.
Then the [[Definit... | A proof of [[Tartaglia's Formula]] will be found in a proof of the [[Value of Cayley-Menger Determinant]] as a [[Definition:Tetrahedron|tetrahedron]] is a [[Definition:Simplex|$3$-simplex]].
{{proof wanted}} | Tartaglia's Formula | https://proofwiki.org/wiki/Tartaglia's_Formula | https://proofwiki.org/wiki/Tartaglia's_Formula | [
"Tartaglia's Formula",
"Tetrahedra"
] | [
"Definition:Tetrahedron",
"Definition:Polyhedron/Vertex",
"Definition:Linear Measure/Length",
"Definition:Volume"
] | [
"Tartaglia's Formula",
"Value of Cayley-Menger Determinant",
"Definition:Tetrahedron",
"Definition:Simplex"
] |
proofwiki-5723 | 0 in B-Algebra is Left Cancellable Element | Let $\struct {X, \circ}$ be a $B$-Algebra.
Then:
:$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$ | Let $x, y \in X$ and let $0 \circ x = 0 \circ y$.
Then:
{{begin-eqn}}
{{eqn | l = 0
| r = x \circ x
| c = {{B-algebra-axiom|1}}
}}
{{eqn | r = \paren {x \circ x} \circ 0
| c = {{B-algebra-axiom|2}}
}}
{{eqn | r = x \circ \paren {0 \circ \paren {0 \circ x} }
| c = {{B-algebra-axiom|3}}
}}
{{eqn |... | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-Algebra]].
Then:
:$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$ | Let $x, y \in X$ and let $0 \circ x = 0 \circ y$.
Then:
{{begin-eqn}}
{{eqn | l = 0
| r = x \circ x
| c = {{B-algebra-axiom|1}}
}}
{{eqn | r = \paren {x \circ x} \circ 0
| c = {{B-algebra-axiom|2}}
}}
{{eqn | r = x \circ \paren {0 \circ \paren {0 \circ x} }
| c = {{B-algebra-axiom|3}}
}}
{{eqn... | 0 in B-Algebra is Left Cancellable Element | https://proofwiki.org/wiki/0_in_B-Algebra_is_Left_Cancellable_Element | https://proofwiki.org/wiki/0_in_B-Algebra_is_Left_Cancellable_Element | [
"B-Algebras"
] | [
"Definition:B-Algebra"
] | [
"B-Algebra Identity: xy = 0 iff x = y",
"Category:B-Algebras"
] |
proofwiki-5724 | B-Algebra Identity: 0(0x)=x | Let $\struct {X, \circ}$ be a $B$-algebra.
Then:
:$\forall x \in X: 0 \circ \paren {0 \circ x} = x$ | Let $x \in X$.
Then:
{{begin-eqn}}
{{eqn | l = 0 \circ x
| r = \paren {0 \circ x} \circ 0
| c = {{B-algebra-axiom|2}}
}}
{{eqn | r = 0 \circ \paren {0 \circ \paren {0 \circ x} }
| c = {{B-algebra-axiom|3}}
}}
{{eqn | ll= \leadsto
| l = x
| r = 0 \circ \paren {0 \circ x}
| c = $0$ in ... | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Then:
:$\forall x \in X: 0 \circ \paren {0 \circ x} = x$ | Let $x \in X$.
Then:
{{begin-eqn}}
{{eqn | l = 0 \circ x
| r = \paren {0 \circ x} \circ 0
| c = {{B-algebra-axiom|2}}
}}
{{eqn | r = 0 \circ \paren {0 \circ \paren {0 \circ x} }
| c = {{B-algebra-axiom|3}}
}}
{{eqn | ll= \leadsto
| l = x
| r = 0 \circ \paren {0 \circ x}
| c = [[0 i... | B-Algebra Identity: 0(0x)=x | https://proofwiki.org/wiki/B-Algebra_Identity:_0(0x)=x | https://proofwiki.org/wiki/B-Algebra_Identity:_0(0x)=x | [
"B-Algebras"
] | [
"Definition:B-Algebra"
] | [
"0 in B-Algebra is Left Cancellable Element",
"Category:B-Algebras"
] |
proofwiki-5725 | B-Algebra Induces Group | Let $\struct {X, \circ}$ be a $B$-algebra.
Let $*$ be the binary operation on $X$ defined as:
:$\forall a, b \in X: a * b := a \circ \paren {0 \circ b}$
Then the algebraic structure $\struct {X, *}$ is a group such that:
:$\forall x \in X: 0 \circ x$ is the inverse element of $x$ under $*$.
That is:
:$\forall a, b \in ... | Let $x, y, z \in X$:
We will show that $\struct {X, *}$ satisfies each of the group axioms in turn: | Let $\struct {X, \circ}$ be a [[Definition:B-Algebra|$B$-algebra]].
Let $*$ be the [[Definition:Binary Operation|binary operation]] on $X$ defined as:
:$\forall a, b \in X: a * b := a \circ \paren {0 \circ b}$
Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {X, *}$ is a [[Definition:Group|g... | Let $x, y, z \in X$:
We will show that $\struct {X, *}$ satisfies each of the [[Axiom:Group Axioms|group axioms]] in turn: | B-Algebra Induces Group | https://proofwiki.org/wiki/B-Algebra_Induces_Group | https://proofwiki.org/wiki/B-Algebra_Induces_Group | [
"B-Algebras",
"Group Theory"
] | [
"Definition:B-Algebra",
"Definition:Operation/Binary Operation",
"Definition:Algebraic Structure",
"Definition:Group",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Axiom:Group Axioms"
] |
proofwiki-5726 | Klein Four-Group as Order 2 Matrices | Let $G$ be the set of order $2$ square matrices:
:$G = \set {I, A, B, C}$
where:
:$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$
Then the algebr... | From Unit Matrix is Unity of Ring of Square Matrices, $I$ can be identified as the unit matrix of order $2$.
Then:
:$A^2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
:$A B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatr... | Let $G$ be the [[Definition:Set|set]] of [[Definition:Order of Square Matrix|order $2$]] [[Definition:Square Matrix|square matrices]]:
:$G = \set {I, A, B, C}$
where:
:$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 0 \\ 0 & 1 \... | From [[Unit Matrix is Unity of Ring of Square Matrices]], $I$ can be identified as the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order $2$]].
Then:
:$A^2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & ... | Klein Four-Group as Order 2 Matrices | https://proofwiki.org/wiki/Klein_Four-Group_as_Order_2_Matrices | https://proofwiki.org/wiki/Klein_Four-Group_as_Order_2_Matrices | [
"Matrix Algebra",
"Klein Four-Group"
] | [
"Definition:Set",
"Definition:Matrix/Square Matrix/Order",
"Definition:Matrix/Square Matrix",
"Definition:Algebraic Structure",
"Definition:Matrix Product (Conventional)",
"Definition:Klein Four-Group"
] | [
"Unit Matrix is Unity of Ring of Square Matrices",
"Definition:Unit Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Cayley Table",
"Definition:Klein Four-Group"
] |
proofwiki-5727 | Young's Inequality for Convolutions | Let $p, q, r \in \R_{\ge 1}$ satisfy:
:$1 + \dfrac 1 r = \dfrac 1 p + \dfrac 1 q$
Let $\map {L^p} {\R^n}$, $\map {L^q} {\R^n}$, and $\map {L^r} {\R^n}$ be Lebesgue spaces with seminorms $\norm {\, \cdot \,}_p$, $\norm {\, \cdot \,}_q$, and $\norm {\, \cdot \,}_r$ respectively.
Let $f \in \map {L^p} {\R^n}$ and $g \in \... | We begin by seeking to bound $\size {\map {\paren {f * g} } x}$:
{{begin-eqn}}
{{eqn | l = \map {\paren {f * g} } x
| r = \int \map f {x - y} \map g y \rd y
}}
{{eqn | ll= \leadsto
| l = \size {\map {\paren {f * g} } x}
| o = \le
| r = \int \size {\map f {x - y} } \cdot \size {\map g y} \rd y
}... | Let $p, q, r \in \R_{\ge 1}$ satisfy:
:$1 + \dfrac 1 r = \dfrac 1 p + \dfrac 1 q$
Let $\map {L^p} {\R^n}$, $\map {L^q} {\R^n}$, and $\map {L^r} {\R^n}$ be [[Definition:Lebesgue Space|Lebesgue spaces]] with [[Definition:P-Seminorm|seminorms]] $\norm {\, \cdot \,}_p$, $\norm {\, \cdot \,}_q$, and $\norm {\, \cdot \,}_r$... | We begin by seeking to bound $\size {\map {\paren {f * g} } x}$:
{{begin-eqn}}
{{eqn | l = \map {\paren {f * g} } x
| r = \int \map f {x - y} \map g y \rd y
}}
{{eqn | ll= \leadsto
| l = \size {\map {\paren {f * g} } x}
| o = \le
| r = \int \size {\map f {x - y} } \cdot \size {\map g y} \rd y
... | Young's Inequality for Convolutions | https://proofwiki.org/wiki/Young's_Inequality_for_Convolutions | https://proofwiki.org/wiki/Young's_Inequality_for_Convolutions | [
"Young's Inequality for Convolutions",
"Measure Theory",
"Inequalities"
] | [
"Definition:Lebesgue Space",
"Definition:P-Seminorm",
"Definition:Convolution of Measurable Functions"
] | [
"Hölder's Inequality for Integrals/General",
"Definition:Conjugate Exponents",
"Hölder's Inequality for Integrals/General",
"Fubini's Theorem",
"Category:Young's Inequality for Convolutions",
"Category:Measure Theory",
"Category:Inequalities"
] |
proofwiki-5728 | F-Sigma Sets Closed under Union | Let $T = \struct {S, \tau}$ be a topological space.
Let $F, F'$ be $F_\sigma$ sets of $T$.
Then their union $F \cup F'$ is also an $F_\sigma$ set of $T$. | By definition of $F_\sigma$ set, there exist sequences $\sequence {C_n}_{n \mathop \in \N}$ and $\sequence {C'_n}_{n \mathop \in \N}$ of closed sets of $T$ such that:
:$F = \ds \bigcup_{n \mathop \in \N} C_n$
:$F' = \ds \bigcup_{n \mathop \in \N} C'_n$
By General Self-Distributivity of Set Union, we have:
:$F \cup F' =... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $F, F'$ be [[Definition:F-Sigma Set|$F_\sigma$ sets]] of $T$.
Then their [[Definition:Set Union|union]] $F \cup F'$ is also an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$. | By definition of [[Definition:F-Sigma Set|$F_\sigma$ set]], there exist [[Definition:Sequence|sequences]] $\sequence {C_n}_{n \mathop \in \N}$ and $\sequence {C'_n}_{n \mathop \in \N}$ of [[Definition:Closed Set (Topology)|closed sets]] of $T$ such that:
:$F = \ds \bigcup_{n \mathop \in \N} C_n$
:$F' = \ds \bigcup_{n ... | F-Sigma Sets Closed under Union | https://proofwiki.org/wiki/F-Sigma_Sets_Closed_under_Union | https://proofwiki.org/wiki/F-Sigma_Sets_Closed_under_Union | [
"F-Sigma Sets"
] | [
"Definition:Topological Space",
"Definition:F-Sigma Set",
"Definition:Set Union",
"Definition:F-Sigma Set"
] | [
"Definition:F-Sigma Set",
"Definition:Sequence",
"Definition:Closed Set/Topology",
"Set Union is Self-Distributive/General Result",
"Finite Union of Closed Sets is Closed/Topology",
"Definition:Closed Set/Topology",
"Definition:F-Sigma Set"
] |
proofwiki-5729 | F-Sigma Sets Closed under Intersection | Let $T = \struct {S, \tau}$ be a topological space.
Let $F, F'$ be $F_\sigma$ sets of $T$.
Then their intersection $F \cap F'$ is also a $F_\sigma$ set of $T$. | By definition of $F_\sigma$ set, there exist sequences $\sequence {C_n}_{n \mathop \in \N}$ and $\sequence {C'_n}_{n \mathop \in \N}$ of closed sets of $T$ such that:
:$F = \ds \bigcup_{n \mathop \in \N} C_n$
:$F' = \ds \bigcup_{n \mathop \in \N} C'_n$
Now compute:
{{begin-eqn}}
{{eqn | l = F \cap F'
| r = \bigcu... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $F, F'$ be [[Definition:F-Sigma Set|$F_\sigma$ sets]] of $T$.
Then their [[Definition:Set Intersection|intersection]] $F \cap F'$ is also a [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$. | By definition of [[Definition:F-Sigma Set|$F_\sigma$ set]], there exist [[Definition:Sequence|sequences]] $\sequence {C_n}_{n \mathop \in \N}$ and $\sequence {C'_n}_{n \mathop \in \N}$ of [[Definition:Closed Set (Topology)|closed sets]] of $T$ such that:
:$F = \ds \bigcup_{n \mathop \in \N} C_n$
:$F' = \ds \bigcup_{n ... | F-Sigma Sets Closed under Intersection | https://proofwiki.org/wiki/F-Sigma_Sets_Closed_under_Intersection | https://proofwiki.org/wiki/F-Sigma_Sets_Closed_under_Intersection | [
"F-Sigma Sets"
] | [
"Definition:Topological Space",
"Definition:F-Sigma Set",
"Definition:Set Intersection",
"Definition:F-Sigma Set"
] | [
"Definition:F-Sigma Set",
"Definition:Sequence",
"Definition:Closed Set/Topology",
"Intersection Distributes over Union/General Result",
"Intersection Distributes over Union/General Result",
"Intersection of Closed Sets is Closed/Topology",
"Definition:Closed Set/Topology",
"Cartesian Product of Count... |
proofwiki-5730 | G-Delta Sets Closed under Union | Let $T = \struct {S, \tau}$ be a topological space.
Let $G, G'$ be $G_\delta$ sets of $T$.
Then their union $G \cup G'$ is also a $G_\delta$ set of $T$. | By definition of $G_\delta$ set, there exist sequences $\sequence {U_n}_{n \mathop \in \N}$ and $\sequence {U'_n}_{n \mathop \in \N}$ of open sets of $T$ such that:
:$G = \ds \bigcap_{n \mathop \in \N} U_n$
:$G' = \ds \bigcap_{n \mathop \in \N} U'_n$
Now compute:
{{begin-eqn}}
{{eqn | l = G \cup G'
| r = \bigcap_... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $G, G'$ be [[Definition:G-Delta Set|$G_\delta$ sets]] of $T$.
Then their [[Definition:Set Union|union]] $G \cup G'$ is also a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$. | By definition of [[Definition:G-Delta Set|$G_\delta$ set]], there exist [[Definition:Sequence|sequences]] $\sequence {U_n}_{n \mathop \in \N}$ and $\sequence {U'_n}_{n \mathop \in \N}$ of [[Definition:Open Set (Topology)|open sets]] of $T$ such that:
:$G = \ds \bigcap_{n \mathop \in \N} U_n$
:$G' = \ds \bigcap_{n \mat... | G-Delta Sets Closed under Union | https://proofwiki.org/wiki/G-Delta_Sets_Closed_under_Union | https://proofwiki.org/wiki/G-Delta_Sets_Closed_under_Union | [
"G-Delta Sets"
] | [
"Definition:Topological Space",
"Definition:G-Delta Set",
"Definition:Set Union",
"Definition:G-Delta Set"
] | [
"Definition:G-Delta Set",
"Definition:Sequence",
"Definition:Open Set/Topology",
"Intersection Distributes over Union/General Result",
"Intersection Distributes over Union/General Result",
"Definition:Topology",
"Definition:Open Set/Topology",
"Cartesian Product of Countable Sets is Countable",
"Def... |
proofwiki-5731 | G-Delta Sets Closed under Intersection | Let $T = \struct {S, \tau}$ be a topological space.
Let $G, G'$ be $G_\delta$ sets of $T$.
Then their intersection $G \cap G'$ is also a $G_\delta$ set of $T$. | By definition of $G_\delta$ set, there exist sequences $\sequence {U_n}_{n \mathop \in \N}$ and $\sequence {U'_n}_{n \mathop \in \N}$ of open sets of $T$ such that:
:$G = \ds \bigcap_{n \mathop \in \N} U_n$
:$G' = \ds \bigcap_{n \mathop \in \N} U'_n$
By General Self-Distributivity of Intersection, we have:
:$G \cap G' ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $G, G'$ be [[Definition:G-Delta Set|$G_\delta$ sets]] of $T$.
Then their [[Definition:Set Intersection|intersection]] $G \cap G'$ is also a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$. | By definition of [[Definition:G-Delta Set|$G_\delta$ set]], there exist [[Definition:Sequence|sequences]] $\sequence {U_n}_{n \mathop \in \N}$ and $\sequence {U'_n}_{n \mathop \in \N}$ of [[Definition:Open Set (Topology)|open sets]] of $T$ such that:
:$G = \ds \bigcap_{n \mathop \in \N} U_n$
:$G' = \ds \bigcap_{n \mat... | G-Delta Sets Closed under Intersection | https://proofwiki.org/wiki/G-Delta_Sets_Closed_under_Intersection | https://proofwiki.org/wiki/G-Delta_Sets_Closed_under_Intersection | [
"G-Delta Sets"
] | [
"Definition:Topological Space",
"Definition:G-Delta Set",
"Definition:Set Intersection",
"Definition:G-Delta Set"
] | [
"Definition:G-Delta Set",
"Definition:Sequence",
"Definition:Open Set/Topology",
"Set Intersection is Self-Distributive/General Result",
"Intersection of Closed Sets is Closed/Topology",
"Definition:Closed Set/Topology",
"Definition:G-Delta Set"
] |
proofwiki-5732 | F-Sigma Sets form Lattice | Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be the collection of all $F_\sigma$ sets of $T$.
Then $\struct {\FF, \subseteq}$ is a lattice, where $\subseteq$ denotes the subset relation. | From Subset Relation is Ordering, $\subseteq$ is an ordering on $\FF$.
Let $F, F'$ be $F_\sigma$ sets of $T$.
We have $F_\sigma$ Sets Closed under Union, so that $F \cup F' \in \FF$.
From Union is Smallest Superset and Subset of Union, it follows that $F \cup F'$ is the supremum of $F$ and $F'$.
Similarly, we have $F_\... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\FF$ be the collection of all [[Definition:F-Sigma Set|$F_\sigma$ sets]] of $T$.
Then $\struct {\FF, \subseteq}$ is a [[Definition:Lattice (Order Theory)|lattice]], where $\subseteq$ denotes the [[Definition:Subset|subset relat... | From [[Subset Relation is Ordering]], $\subseteq$ is an [[Definition:Ordering|ordering]] on $\FF$.
Let $F, F'$ be [[Definition:F-Sigma Set|$F_\sigma$ sets]] of $T$.
We have [[F-Sigma Sets Closed under Union|$F_\sigma$ Sets Closed under Union]], so that $F \cup F' \in \FF$.
From [[Union is Smallest Superset]] and [[... | F-Sigma Sets form Lattice | https://proofwiki.org/wiki/F-Sigma_Sets_form_Lattice | https://proofwiki.org/wiki/F-Sigma_Sets_form_Lattice | [
"F-Sigma Sets",
"Lattice Theory"
] | [
"Definition:Topological Space",
"Definition:F-Sigma Set",
"Definition:Lattice (Order Theory)",
"Definition:Subset"
] | [
"Subset Relation is Ordering",
"Definition:Ordering",
"Definition:F-Sigma Set",
"F-Sigma Sets Closed under Union",
"Union is Smallest Superset",
"Set is Subset of Union",
"Definition:Supremum of Set",
"F-Sigma Sets Closed under Intersection",
"Intersection is Largest Subset",
"Intersection is Subs... |
proofwiki-5733 | G-Delta Sets form Lattice | Let $T = \struct {S, \tau}$ be a topological space.
Let $\GG$ be the collection of all $G_\delta$ sets of $T$.
Then $\struct {\GG, \subseteq}$ is a lattice, where $\subseteq$ denotes the subset relation. | From Subset Relation is Ordering, $\subseteq$ is an ordering on $\GG$.
Let $G, G'$ be $G_\delta$ sets of $T$.
We have $G_\delta$ Sets Closed under Union, so that $G \cup G' \in \GG$.
From Union is Smallest Superset and Subset of Union, it follows that $G \cup G'$ is the supremum of $G$ and $G'$.
Similarly, we have $G_\... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\GG$ be the collection of all [[Definition:G-Delta Set|$G_\delta$ sets]] of $T$.
Then $\struct {\GG, \subseteq}$ is a [[Definition:Lattice (Order Theory)|lattice]], where $\subseteq$ denotes the [[Definition:Subset|subset relat... | From [[Subset Relation is Ordering]], $\subseteq$ is an [[Definition:Ordering|ordering]] on $\GG$.
Let $G, G'$ be [[Definition:G-Delta Set|$G_\delta$ sets]] of $T$.
We have [[G-Delta Sets Closed under Union|$G_\delta$ Sets Closed under Union]], so that $G \cup G' \in \GG$.
From [[Union is Smallest Superset]] and [[... | G-Delta Sets form Lattice | https://proofwiki.org/wiki/G-Delta_Sets_form_Lattice | https://proofwiki.org/wiki/G-Delta_Sets_form_Lattice | [
"G-Delta Sets",
"Lattice Theory"
] | [
"Definition:Topological Space",
"Definition:G-Delta Set",
"Definition:Lattice (Order Theory)",
"Definition:Subset"
] | [
"Subset Relation is Ordering",
"Definition:Ordering",
"Definition:G-Delta Set",
"G-Delta Sets Closed under Union",
"Union is Smallest Superset",
"Set is Subset of Union",
"Definition:Supremum of Set",
"G-Delta Sets Closed under Intersection",
"Intersection is Largest Subset",
"Intersection is Subs... |
proofwiki-5734 | Bounded Sequence in Euclidean Space has Convergent Subsequence | Let $\sequence {x_i}_{i \mathop \in \N}$ be a bounded sequence in the Euclidean space $\R^n$.
Then some subsequence of $\sequence {x_i}_{i \mathop \in \N}$ converges to a limit. | Denote with $d$ the Euclidean metric on $\R^n$.
Because $\sequence {x_i}_{i \mathop \in \N}$ is bounded, we find $v \in \R^n$ and $\epsilon \in \R_{>0}$ such that:
:$\forall i \in \N: \map d {v, x_i} < \epsilon$
Therefore, all $x_i$ are members of the closed $\epsilon$-ball $S = \map {B_\epsilon^-} v$.
By Closed Ball i... | Let $\sequence {x_i}_{i \mathop \in \N}$ be a [[Definition:Bounded Sequence|bounded sequence]] in the [[Definition:Euclidean Space|Euclidean space]] $\R^n$.
Then some [[Definition:Subsequence|subsequence]] of $\sequence {x_i}_{i \mathop \in \N}$ converges to a [[Definition:Limit of Sequence in Metric Space|limit]]. | Denote with $d$ the [[Definition:Euclidean Metric on Real Vector Space|Euclidean metric]] on $\R^n$.
Because $\sequence {x_i}_{i \mathop \in \N}$ is [[Definition:Bounded Metric Space|bounded]], we find $v \in \R^n$ and $\epsilon \in \R_{>0}$ such that:
:$\forall i \in \N: \map d {v, x_i} < \epsilon$
Therefore, all $... | Bounded Sequence in Euclidean Space has Convergent Subsequence/Proof 1 | https://proofwiki.org/wiki/Bounded_Sequence_in_Euclidean_Space_has_Convergent_Subsequence | https://proofwiki.org/wiki/Bounded_Sequence_in_Euclidean_Space_has_Convergent_Subsequence/Proof_1 | [
"Bounded Sequence in Euclidean Space has Convergent Subsequence",
"Limits of Sequences",
"Euclidean Spaces"
] | [
"Definition:Bounded Sequence",
"Definition:Euclidean Space",
"Definition:Subsequence",
"Definition:Limit of Sequence/Metric Space"
] | [
"Definition:Euclidean Metric/Real Vector Space",
"Definition:Bounded Metric Space",
"Definition:Closed Ball",
"Closed Ball in Euclidean Space is Compact",
"Definition:Compact Topological Space/Subspace",
"Definition:Sequence",
"Definition:Compact Space/Metric Space",
"Compact Subspace of Metric Space ... |
proofwiki-5735 | Bounded Sequence in Euclidean Space has Convergent Subsequence | Let $\sequence {x_i}_{i \mathop \in \N}$ be a bounded sequence in the Euclidean space $\R^n$.
Then some subsequence of $\sequence {x_i}_{i \mathop \in \N}$ converges to a limit. | Let the range of $\sequence {x_i}$ be $S$.
By Closure of Bounded Subset of Metric Space is Bounded $\map \cl S$ is bounded in $\R^n$.
By Topological Closure is Closed, $\map \cl S$ is closed in $\R^n$.
By the Heine-Borel Theorem, $S$ is compact.
The result follows from Compact Subspace of Metric Space is Sequentially C... | Let $\sequence {x_i}_{i \mathop \in \N}$ be a [[Definition:Bounded Sequence|bounded sequence]] in the [[Definition:Euclidean Space|Euclidean space]] $\R^n$.
Then some [[Definition:Subsequence|subsequence]] of $\sequence {x_i}_{i \mathop \in \N}$ converges to a [[Definition:Limit of Sequence in Metric Space|limit]]. | Let the [[Definition:Range of Sequence|range of $\sequence {x_i}$]] be $S$.
By [[Closure of Bounded Subset of Metric Space is Bounded]] $\map \cl S$ is [[Definition:Bounded Metric Space|bounded]] in $\R^n$.
By [[Topological Closure is Closed]], $\map \cl S$ is [[Definition:Closed Set (Metric Space)|closed]] in $\R^n$... | Bounded Sequence in Euclidean Space has Convergent Subsequence/Proof 2 | https://proofwiki.org/wiki/Bounded_Sequence_in_Euclidean_Space_has_Convergent_Subsequence | https://proofwiki.org/wiki/Bounded_Sequence_in_Euclidean_Space_has_Convergent_Subsequence/Proof_2 | [
"Bounded Sequence in Euclidean Space has Convergent Subsequence",
"Limits of Sequences",
"Euclidean Spaces"
] | [
"Definition:Bounded Sequence",
"Definition:Euclidean Space",
"Definition:Subsequence",
"Definition:Limit of Sequence/Metric Space"
] | [
"Definition:Range of Sequence",
"Closure of Bounded Subset of Metric Space is Bounded",
"Definition:Bounded Metric Space",
"Topological Closure is Closed",
"Definition:Closed Set/Metric Space",
"Heine-Borel Theorem/Euclidean Space",
"Definition:Compact Space/Metric Space",
"Compact Subspace of Metric ... |
proofwiki-5736 | Bounded Sequence in Euclidean Space has Convergent Subsequence | Let $\sequence {x_i}_{i \mathop \in \N}$ be a bounded sequence in the Euclidean space $\R^n$.
Then some subsequence of $\sequence {x_i}_{i \mathop \in \N}$ converges to a limit. | We have that all norms on $\R^n$ are equivalent.
Choose the Euclidean norm $\norm {\, \cdot \,}_2$.
Proof by induction will be used.
=== Basis for the induction ===
Let $n = 1$.
Then the proof is given by Bolzano-Weierstrass theorem.
=== Induction hypothesis ===
Suppose, a bounded sequence $\sequence {\boldsymbol x_i}_... | Let $\sequence {x_i}_{i \mathop \in \N}$ be a [[Definition:Bounded Sequence|bounded sequence]] in the [[Definition:Euclidean Space|Euclidean space]] $\R^n$.
Then some [[Definition:Subsequence|subsequence]] of $\sequence {x_i}_{i \mathop \in \N}$ converges to a [[Definition:Limit of Sequence in Metric Space|limit]]. | We have that [[Norms on Finite-Dimensional Real Vector Space are Equivalent|all norms on $\R^n$ are equivalent]].
Choose the [[Definition:Euclidean Norm|Euclidean norm]] [[Definition:P-Norm|$\norm {\, \cdot \,}_2$]].
[[Principle of Mathematical Induction|Proof by induction]] will be used.
=== Basis for the inductio... | Bounded Sequence in Euclidean Space has Convergent Subsequence/Proof 3 | https://proofwiki.org/wiki/Bounded_Sequence_in_Euclidean_Space_has_Convergent_Subsequence | https://proofwiki.org/wiki/Bounded_Sequence_in_Euclidean_Space_has_Convergent_Subsequence/Proof_3 | [
"Bounded Sequence in Euclidean Space has Convergent Subsequence",
"Limits of Sequences",
"Euclidean Spaces"
] | [
"Definition:Bounded Sequence",
"Definition:Euclidean Space",
"Definition:Subsequence",
"Definition:Limit of Sequence/Metric Space"
] | [
"Norms on Finite-Dimensional Real Vector Space are Equivalent",
"Definition:Euclidean Norm",
"Definition:P-Norm",
"Principle of Mathematical Induction",
"Bolzano-Weierstrass Theorem/Proof 1",
"Definition:Bounded Sequence/Normed Vector Space",
"Definition:Convergent Sequence/Normed Vector Space",
"Defi... |
proofwiki-5737 | Exponential Growth Equation/Special Case | All solutions of the differential equation $y' = y$ take the form $y = C e^x$. | Let $\map f x = C e^x$.
Then by Derivative of Exponential Function:
:$\map {f'} x = \map f x$
From Exponential of Zero:
:$\map f 0 = C$
Hence $C e^x$ is a solution of $y' = y$.
Now suppose that a function $f$ satisfies $\map {f'} x = \map f x$.
Consider $\map h x = \map f x e^{-x}$.
By the Product Rule for Derivatives... | All [[Definition:Solution of Differential Equation|solutions]] of the [[Definition:Differential Equation|differential equation]] $y' = y$ take the form $y = C e^x$. | Let $\map f x = C e^x$.
Then by [[Derivative of Exponential Function]]:
:$\map {f'} x = \map f x$
From [[Exponential of Zero]]:
:$\map f 0 = C$
Hence $C e^x$ is a [[Definition:Solution of Differential Equation|solution]] of $y' = y$.
Now suppose that a function $f$ satisfies $\map {f'} x = \map f x$.
Consider ... | Exponential Growth Equation/Special Case | https://proofwiki.org/wiki/Exponential_Growth_Equation/Special_Case | https://proofwiki.org/wiki/Exponential_Growth_Equation/Special_Case | [
"First Order ODEs"
] | [
"Definition:Differential Equation/Solution",
"Definition:Differential Equation"
] | [
"Derivative of Exponential Function",
"Exponential of Zero",
"Definition:Differential Equation/Solution",
"Product Rule for Derivatives",
"Zero Derivative implies Constant Function",
"Definition:Constant Mapping"
] |
proofwiki-5738 | Existence-Uniqueness Theorem for Homogeneous First-Order Differential Equation | Let $\map P x$ be a continuous function on an open interval $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
Let $\map f x = y$ be a function satisfying the differential equation:
:$y' + \map P x y = 0$
and the initial condition:
:$\map f a = b$
Then there exists a unique function satisfying these initial conditions on... | === Existence ===
Differentiating $\map f x = b e^{-\map A x}$ {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | l = \map {f'} x
| r = b e^{-\map A x} \cdot \paren {-\map {A'} x}
| c =
}}
{{eqn | r = -b \map P x e^{-\map A x}
| c =
}}
{{eqn | r = -\map P x \map f x
| c =
}}
{{end-eqn}}
So th... | Let $\map P x$ be a [[Definition:Continuous Real Function|continuous function]] on an [[Definition:Open Real Interval|open interval]] $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
Let $\map f x = y$ be a [[Definition:Real Function|function]] satisfying the [[Definition:Differential Equation|differential equation]... | === Existence ===
[[Definition:Differentiation|Differentiating]] $\map f x = b e^{-\map A x}$ {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | l = \map {f'} x
| r = b e^{-\map A x} \cdot \paren {-\map {A'} x}
| c =
}}
{{eqn | r = -b \map P x e^{-\map A x}
| c =
}}
{{eqn | r = -\map P x \map f x
... | Existence-Uniqueness Theorem for Homogeneous First-Order Differential Equation | https://proofwiki.org/wiki/Existence-Uniqueness_Theorem_for_Homogeneous_First-Order_Differential_Equation | https://proofwiki.org/wiki/Existence-Uniqueness_Theorem_for_Homogeneous_First-Order_Differential_Equation | [
"Calculus",
"Differential Equations"
] | [
"Definition:Continuous Real Function",
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Differential Equation",
"Definition:Initial Condition",
"Definition:Unique",
"Definition:Real Function",
"Definition:Initial Condition",
"Definition:Real Interval/Open",
"Definition:Real... | [
"Definition:Differentiation",
"Definition:Differential Equation",
"Definition:Initial Condition",
"Definition:Real Function",
"Definition:Real Function",
"Definition:Initial Condition"
] |
proofwiki-5739 | Existence-Uniqueness Theorem for First-Order Differential Equation | Let $P$ and $Q$ be continuous real functions on some open interval $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
There exists a unique function $\map f x = y$ on $I$ that satisfies the linear first order ordinary differential equation:
:$(1): \quad y' + \map P x y = \map Q x$
along with the initial condition:
:$\ma... | === Existence ===
Because $P$ and $Q$ are continuous, they are integrable.
Hence we may use the Fundamental Theorem of Calculus.
{{begin-eqn}}
{{eqn | l = \map {f'} x
| r = -b \map P x e^{-\map A x} - \map P x e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t + e^{-\map A x} \map Q x e^{\map A x}
| c = Pro... | Let $P$ and $Q$ be [[Definition:Continuous Real Function|continuous real functions]] on some [[Definition:Open Real Interval|open interval]] $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
There exists a unique [[Definition:Real Function|function]] $\map f x = y$ on $I$ that satisfies the [[Definition:Linear First... | === Existence ===
Because $P$ and $Q$ are [[Definition:Continuous Real Function|continuous]], they are [[Definition:Darboux Integrable Function|integrable]].
Hence we may use the [[Fundamental Theorem of Calculus]].
{{begin-eqn}}
{{eqn | l = \map {f'} x
| r = -b \map P x e^{-\map A x} - \map P x e^{-\map A x} ... | Existence-Uniqueness Theorem for First-Order Differential Equation | https://proofwiki.org/wiki/Existence-Uniqueness_Theorem_for_First-Order_Differential_Equation | https://proofwiki.org/wiki/Existence-Uniqueness_Theorem_for_First-Order_Differential_Equation | [
"Calculus",
"Differential Equations"
] | [
"Definition:Continuous Real Function",
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Linear First Order Ordinary Differential Equation",
"Definition:Initial Condition",
"Definition:Real Function"
] | [
"Definition:Continuous Real Function",
"Definition:Darboux Integrable Function",
"Fundamental Theorem of Calculus",
"Product Rule for Derivatives",
"Definition:Initial Condition",
"Definite Integral on Zero Interval",
"Definition:Initial Condition",
"Product Rule for Derivatives",
"Fundamental Theor... |
proofwiki-5740 | Proper Well-Ordering determines Smallest Elements | Let $S$ be a class.
Let $\preceq$ be a proper well-ordering on $S$.
Let $B$ be a nonempty subclass of $S$.
Then $B$ has a $\preceq$-smallest element. | Because $B \ne \O$ it follows that $B$ has an element $x$.
If $x$ is the smallest element of $B$ then the theorem holds.
Otherwise, there is an element $y \in B$ such that $x \npreceq y$.
By Well-Ordering is Total Ordering, $y \prec x$.
Thus $S_x$, the $\preceq$-initial segment of $x$, is not empty.
By the hypothesis, ... | Let $S$ be a [[Definition:Class (Class Theory)|class]].
Let $\preceq$ be a [[Definition:Proper Well-Ordering|proper well-ordering]] on $S$.
Let $B$ be a [[Definition:Non-Empty Set|nonempty]] [[Definition:Subclass|subclass]] of $S$.
Then $B$ has a $\preceq$-[[Definition:Smallest Element|smallest element]]. | Because $B \ne \O$ it follows that $B$ has an element $x$.
If $x$ is the [[Definition:Smallest Element|smallest element]] of $B$ then the theorem holds.
Otherwise, there is an element $y \in B$ such that $x \npreceq y$.
By [[Well-Ordering is Total Ordering]], $y \prec x$.
Thus $S_x$, the [[Definition:Initial Segmen... | Proper Well-Ordering determines Smallest Elements | https://proofwiki.org/wiki/Proper_Well-Ordering_determines_Smallest_Elements | https://proofwiki.org/wiki/Proper_Well-Ordering_determines_Smallest_Elements | [
"Class Theory",
"Well-Orderings"
] | [
"Definition:Class (Class Theory)",
"Definition:Proper Well-Ordering",
"Definition:Non-Empty Set",
"Definition:Subclass",
"Definition:Smallest Element"
] | [
"Definition:Smallest Element",
"Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2",
"Definition:Initial Segment",
"Intersection is Subset",
"Definition:Proper Well-Ordering",
"Definition:Smallest Element",
"Extended Transitivity",
"Definition:Contradiction",
"Definition:S... |
proofwiki-5741 | Well-Ordered Induction | Let $\struct {A, \prec}$ be a strict well-ordering.
For all $x \in A$, let the $\prec$-initial segment of $x$ be a small class.
Let $B$ be a class such that $B \subseteq A$.
Let:
:$(1): \quad \forall x \in A: \paren {\paren {A \mathop \cap \map {\prec^{-1} } x} \subseteq B \implies x \in B}$
Then:
:$A = B$
That is, if ... | {{AimForCont}} that $A \nsubseteq B$.
Then:
:$A \setminus B \ne 0$.
By Proper Well-Ordering Determines Smallest Elements, $A \setminus B$ must have some $\prec$-minimal element.
Thus:
:$\ds \exists x \in \paren {A \setminus B}: \paren {A \setminus B} \cap \map {\prec^{-1} } x = \O$
implies that:
:$A \cap \map {\prec^{-... | Let $\struct {A, \prec}$ be a [[Definition:Strict Well-Ordering|strict well-ordering]].
For all $x \in A$, let the $\prec$-[[Definition:Initial Segment|initial segment]] of $x$ be a [[Definition:Small Class|small class]].
Let $B$ be a [[Definition:Class (Class Theory)|class]] such that $B \subseteq A$.
Let:
:$(1): \... | {{AimForCont}} that $A \nsubseteq B$.
Then:
:$A \setminus B \ne 0$.
By [[Proper Well-Ordering Determines Smallest Elements]], $A \setminus B$ must have some $\prec$-minimal element.
Thus:
:$\ds \exists x \in \paren {A \setminus B}: \paren {A \setminus B} \cap \map {\prec^{-1} } x = \O$
implies that:
:$A \cap \map {\... | Well-Ordered Induction | https://proofwiki.org/wiki/Well-Ordered_Induction | https://proofwiki.org/wiki/Well-Ordered_Induction | [
"Well-Orderings",
"Class Theory",
"Mathematical Induction"
] | [
"Definition:Strict Well-Ordering",
"Definition:Initial Segment",
"Definition:Small Class",
"Definition:Class (Class Theory)"
] | [
"Proper Well-Ordering determines Smallest Elements",
"Definition:Contradiction",
"Proof by Contradiction",
"Definition:Set Equality"
] |
proofwiki-5742 | Subset of Ordinals has Minimal Element | Let $A$ be an ordinal (we shall allow $A$ to be a proper class).
Let $B$ be a nonempty subset of $A$.
Then $B$ has an $\Epsilon$-minimal element.
{{explain|$\Epsilon$}}
That is:
:$\exists x \in B: B \cap x = \O$ | We have that $\Epsilon$ creates a well-ordering on any ordinal.
Also, the initial segments of $x$ are sets.
Therefore from Proper Well-Ordering Determines Smallest Elements:
:$B$ has an $\Epsilon$-minimal element. | Let $A$ be an [[Definition:Ordinal|ordinal]] (we shall allow $A$ to be a [[Definition:Proper Class|proper class]]).
Let $B$ be a [[Definition:Non-Empty Set|nonempty]] [[Definition:Subset|subset]] of $A$.
Then $B$ has an $\Epsilon$-[[Definition:Minimal Element|minimal element]].
{{explain|$\Epsilon$}}
That is:
:$\e... | We have that $\Epsilon$ creates a [[Definition:Well-Ordering|well-ordering]] on any [[Alternative Definition of Ordinal|ordinal]].
Also, the [[Definition:Initial Segment|initial segments]] of $x$ are [[Definition:Set|sets]].
Therefore from [[Proper Well-Ordering Determines Smallest Elements]]:
:$B$ has an $\Epsilon$-... | Subset of Ordinals has Minimal Element | https://proofwiki.org/wiki/Subset_of_Ordinals_has_Minimal_Element | https://proofwiki.org/wiki/Subset_of_Ordinals_has_Minimal_Element | [
"Class Theory",
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Minimal/Element"
] | [
"Definition:Well-Ordering",
"Alternative Definition of Ordinal",
"Definition:Initial Segment",
"Definition:Set",
"Proper Well-Ordering determines Smallest Elements"
] |
proofwiki-5743 | Union of Set of Ordinals is Ordinal | Let $A$ be a set of ordinals.
Then $\bigcup A$ is an ordinal. | By the Axiom of Replacement, $\map F y$ is a set.
Thus by the Axiom of Unions, $\bigcup \map F y$ is a set.
By Union of Set of Ordinals is Ordinal, $\bigcup \map F y$ is transitive.
{{mistake|This link is the result which needs to be proved. Something with transitive instead?}}
By the epsilon relation $\bigcup \map F y... | Let $A$ be a [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]].
Then $\bigcup A$ is an [[Definition:Ordinal|ordinal]]. | By the [[Axiom:Axiom of Replacement|Axiom of Replacement]], $\map F y$ is a [[Definition:Set|set]].
Thus by the [[Axiom:Axiom of Unions (Set Theory)|Axiom of Unions]], $\bigcup \map F y$ is a [[Definition:Set|set]].
By [[Union of Set of Ordinals is Ordinal]], $\bigcup \map F y$ is [[Definition:Transitive Class|transi... | Union of Set of Ordinals is Ordinal/Corollary/Proof 1 | https://proofwiki.org/wiki/Union_of_Set_of_Ordinals_is_Ordinal | https://proofwiki.org/wiki/Union_of_Set_of_Ordinals_is_Ordinal/Corollary/Proof_1 | [
"Union of Set of Ordinals is Ordinal",
"Ordinals",
"Set Union"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Ordinal"
] | [
"Axiom:Axiom of Replacement",
"Definition:Set",
"Axiom:Axiom of Unions/Set Theory",
"Definition:Set",
"Union of Set of Ordinals is Ordinal",
"Definition:Transitive Class",
"Definition:Epsilon Relation",
"Definition:Well-Ordering",
"Definition:Class of All Ordinals"
] |
proofwiki-5744 | Union of Set of Ordinals is Ordinal | Let $A$ be a set of ordinals.
Then $\bigcup A$ is an ordinal. | By Class of All Ordinals is Well-Ordered by Subset Relation, a set of ordinals forms a chain.
The result then follows from Union of Chain of Ordinals is Ordinal. | Let $A$ be a [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]].
Then $\bigcup A$ is an [[Definition:Ordinal|ordinal]]. | By [[Class of All Ordinals is Well-Ordered by Subset Relation]], a [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]] forms a [[Definition:Chain of Sets|chain]].
The result then follows from [[Union of Chain of Ordinals is Ordinal]]. | Union of Set of Ordinals is Ordinal/Proof 1 | https://proofwiki.org/wiki/Union_of_Set_of_Ordinals_is_Ordinal | https://proofwiki.org/wiki/Union_of_Set_of_Ordinals_is_Ordinal/Proof_1 | [
"Union of Set of Ordinals is Ordinal",
"Ordinals",
"Set Union"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Ordinal"
] | [
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Definition:Set",
"Definition:Ordinal",
"Definition:Chain (Order Theory)/Subset Relation",
"Properties of Class of All Ordinals/Union of Chain of Ordinals is Ordinal"
] |
proofwiki-5745 | Union of Set of Ordinals is Ordinal | Let $A$ be a set of ordinals.
Then $\bigcup A$ is an ordinal. | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup A
| c =
}}
{{eqn | ll= \leadsto
| q = \exists y \in A
| l = x
| o = \in
| r = y
| c =
}}
{{eqn | ll= \leadsto
| q = \exists y \subseteq \bigcup A
| l = x
| o = \subseteq
| r = y
| c =
}... | Let $A$ be a [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]].
Then $\bigcup A$ is an [[Definition:Ordinal|ordinal]]. | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup A
| c =
}}
{{eqn | ll= \leadsto
| q = \exists y \in A
| l = x
| o = \in
| r = y
| c =
}}
{{eqn | ll= \leadsto
| q = \exists y \subseteq \bigcup A
| l = x
| o = \subseteq
| r = y
| c =
}... | Union of Set of Ordinals is Ordinal/Proof 2 | https://proofwiki.org/wiki/Union_of_Set_of_Ordinals_is_Ordinal | https://proofwiki.org/wiki/Union_of_Set_of_Ordinals_is_Ordinal/Proof_2 | [
"Union of Set of Ordinals is Ordinal",
"Ordinals",
"Set Union"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Ordinal"
] | [
"Definition:Transitive Class",
"Class is Transitive iff Union is Subclass",
"Definition:Set",
"Definition:Class (Class Theory)",
"Subset of Well-Ordered Set is Well-Ordered",
"Definition:Strict Well-Ordering",
"Alternative Definition of Ordinal",
"Definition:Ordinal"
] |
proofwiki-5746 | Union of Ordinals is Least Upper Bound | Let $A \subset \On$.
That is, let $A$ be a class of ordinals (every member of $A$ is an ordinal).
Then $\bigcup A$, the union of $A$, is the least upper bound of $A$:
:$\ds \forall x \in A: x \le A$
:$\ds \forall y \in A: y \le x \implies \bigcup A \le x$ | First, we must show that $\ds \bigcup A$ is an upper bound.
Take any member $a \in A$.
Then by Subset of Union:
:$\ds a \subseteq \bigcup A$
By Ordering on Ordinal is Subset Relation:
:$a \le A$
By generalizing for all $a \in A$:
:$\ds \forall x \in A: x \le \bigcup A$
Similarly, suppose now that $x$ is an upper bound ... | Let $A \subset \On$.
That is, let $A$ be a [[Definition:Class (Class Theory)|class]] of [[Definition:Ordinal|ordinals]] (every [[Definition:Element of Class|member]] of $A$ is an [[Definition:Ordinal|ordinal]]).
Then $\bigcup A$, the [[Definition:Union of Set of Sets|union]] of $A$, is the [[Definition:Supremum of S... | First, we must show that $\ds \bigcup A$ is an [[Definition:Upper Bound of Set|upper bound]].
Take any [[Definition:Element of Class|member]] $a \in A$.
Then by [[Subset of Union]]:
:$\ds a \subseteq \bigcup A$
By [[Ordering on Ordinal is Subset Relation]]:
:$a \le A$
By [[Universal Generalization|generalizing]] [[... | Union of Ordinals is Least Upper Bound | https://proofwiki.org/wiki/Union_of_Ordinals_is_Least_Upper_Bound | https://proofwiki.org/wiki/Union_of_Ordinals_is_Least_Upper_Bound | [
"Ordinals"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordinal",
"Definition:Element/Class",
"Definition:Ordinal",
"Definition:Set Union/Set of Sets",
"Definition:Supremum of Set"
] | [
"Definition:Upper Bound of Set",
"Definition:Element/Class",
"Set is Subset of Union",
"Ordering on Ordinal is Subset Relation",
"Universal Generalisation",
"Definition:Universal Quantifier",
"Definition:Upper Bound of Set",
"Definition:Ordering",
"Definition:Ordinal",
"Ordering on Ordinal is Subs... |
proofwiki-5747 | Ordinal is Less than Successor | Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Then:
:$x \in x^+$
:$x \subset x^+$ | $x$ is an ordinal and so by definition is also a set.
:$x \in \paren {x \cup \set x} \land x \subset \paren {x \cup \set x}$
so by applying the definition of a successor set:
:$x \in x^+ \land x \subset x^+$
{{qed}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $x^+$ denote the [[Definition:Successor Set|successor]] of $x$.
Then:
:$x \in x^+$
:$x \subset x^+$ | $x$ is an [[Definition:Ordinal|ordinal]] and so by definition is also a [[Definition:Set|set]].
:$x \in \paren {x \cup \set x} \land x \subset \paren {x \cup \set x}$
so by applying the definition of a [[Definition:Successor Set|successor set]]:
:$x \in x^+ \land x \subset x^+$
{{qed}} | Ordinal is Less than Successor | https://proofwiki.org/wiki/Ordinal_is_Less_than_Successor | https://proofwiki.org/wiki/Ordinal_is_Less_than_Successor | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Definition:Ordinal",
"Definition:Set",
"Definition:Successor Mapping/Successor Set"
] |
proofwiki-5748 | No Natural Number between Number and Successor | Let $\N$ be the natural numbers.
Let $n \in \N$.
Then no natural number $m$ exists strictly between $n$ and its successor $n + 1$:
:$\neg \exists m \in \N: \paren {n < m < n + 1}$
That is:
:If $n \le m \le n + 1$, then $m = n$ or $m = n + 1$. | We use the model that $\N \cong \omega$ where $\omega$ is the minimally inductive set.
{{AimForCont}} such an ordinal $m$ exists.
Then, by Ordering on Ordinal is Subset Relation:
:$n \in m$
and from Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
:$m \in n^+$
Applying the definition of a successor se... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $n \in \N$.
Then no [[Definition:Natural Numbers|natural number]] $m$ exists strictly between $n$ and its [[Definition:Successor Mapping on Natural Numbers|successor]] $n + 1$:
:$\neg \exists m \in \N: \paren {n < m < n + 1}$
That is:
:If $n \le m... | We use the model that $\N \cong \omega$ where $\omega$ is the [[Definition:Minimally Inductive Set|minimally inductive set]].
{{AimForCont}} such an [[Definition:Ordinal|ordinal]] $m$ exists.
Then, by [[Ordering on Ordinal is Subset Relation]]:
:$n \in m$
and from [[Transitive Set is Proper Subset of Ordinal iff E... | No Natural Number between Number and Successor/Proof using Minimally Inductive Set | https://proofwiki.org/wiki/No_Natural_Number_between_Number_and_Successor | https://proofwiki.org/wiki/No_Natural_Number_between_Number_and_Successor/Proof_using_Minimally_Inductive_Set | [
"Ordering on Natural Numbers",
"No Natural Number between Number and Successor"
] | [
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Successor Mapping on Natural Numbers"
] | [
"Definition:Minimally Inductive Set",
"Definition:Ordinal",
"Ordering on Ordinal is Subset Relation",
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal",
"Definition:Successor Mapping/Successor Set",
"Definition:Membership Loop",
"No Membership Loops",
"Definition:Contradiction",
"P... |
proofwiki-5749 | No Natural Number between Number and Successor | Let $\N$ be the natural numbers.
Let $n \in \N$.
Then no natural number $m$ exists strictly between $n$ and its successor $n + 1$:
:$\neg \exists m \in \N: \paren {n < m < n + 1}$
That is:
:If $n \le m \le n + 1$, then $m = n$ or $m = n + 1$. | Let $\N$ be defined as the von Neumann construction $\omega$.
By definition of the ordering on von Neumann construction:
:$m \le n \iff m \subseteq n$
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping.
The result follows from the S... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $n \in \N$.
Then no [[Definition:Natural Numbers|natural number]] $m$ exists strictly between $n$ and its [[Definition:Successor Mapping on Natural Numbers|successor]] $n + 1$:
:$\neg \exists m \in \N: \paren {n < m < n + 1}$
That is:
:If $n \le m... | Let $\N$ be defined as the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]] $\omega$.
By definition of the [[Definition:Ordering on Von Neumann Construction of Natural Numbers|ordering on von Neumann construction]]:
:$m \le n \iff m \subseteq n$
From [[Von Neumann Construction of N... | No Natural Number between Number and Successor/Proof using Von Neumann Construction | https://proofwiki.org/wiki/No_Natural_Number_between_Number_and_Successor | https://proofwiki.org/wiki/No_Natural_Number_between_Number_and_Successor/Proof_using_Von_Neumann_Construction | [
"Ordering on Natural Numbers",
"No Natural Number between Number and Successor"
] | [
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Successor Mapping on Natural Numbers"
] | [
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Ordering on Natural Numbers/Von Neumann Construction",
"Von Neumann Construction of Natural Numbers is Minimally Inductive",
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Natural Numbers/Von Neumann Construction/S... |
proofwiki-5750 | No Largest Ordinal | Let $a$ be a set of ordinals.
Then:
:$\forall x \in a: x \prec \paren {\bigcup a}^+$ | For this proof, we shall use $\prec$, $\in$, and $\subset$ interchangeably.
We are justified in doing this because of Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = a
| c = {{Hypothesis}}
}}
{{eq... | Let $a$ be a [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]].
Then:
:$\forall x \in a: x \prec \paren {\bigcup a}^+$ | For this proof, we shall use $\prec$, $\in$, and $\subset$ interchangeably.
We are justified in doing this because of [[Ordering on Ordinal is Subset Relation]] and [[Transitive Set is Proper Subset of Ordinal iff Element of Ordinal]].
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = a
| c = {{Hypothesis... | No Largest Ordinal | https://proofwiki.org/wiki/No_Largest_Ordinal | https://proofwiki.org/wiki/No_Largest_Ordinal | [
"Ordinals"
] | [
"Definition:Set",
"Definition:Ordinal"
] | [
"Ordering on Ordinal is Subset Relation",
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal",
"Ordinal is Less than Successor"
] |
proofwiki-5751 | Minimally Inductive Set is Ordinal | Let $\omega$ denote the minimally inductive set.
Then $\omega$ is an ordinal. | The minimally inductive set is a set of ordinals by definition.
From {{Corollary|Ordinals are Well-Ordered}}, it is seen that $\struct {\omega, \Epsilon \restriction_\omega}$ is a strictly well-ordered set.
It is to be shown by induction on minimally inductive set that $\forall n \in \omega: \omega_n = n$
=== Basis for... | Let $\omega$ denote the [[Definition:Minimally Inductive Set|minimally inductive set]].
Then $\omega$ is an [[Definition:Ordinal|ordinal]]. | The [[Definition:Minimally Inductive Set|minimally inductive set]] is a [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]] by definition.
From {{Corollary|Ordinals are Well-Ordered}}, it is seen that $\struct {\omega, \Epsilon \restriction_\omega}$ is a [[Definition:Strictly Well-Ordered Set|strictly well-order... | Minimally Inductive Set is Ordinal/Proof 1 | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Ordinal | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Ordinal/Proof_1 | [
"Ordinals",
"Minimally Inductive Set",
"Minimally Inductive Set is Ordinal"
] | [
"Definition:Minimally Inductive Set",
"Definition:Ordinal"
] | [
"Definition:Minimally Inductive Set",
"Definition:Set",
"Definition:Ordinal",
"Definition:Strictly Well-Ordered Set",
"Principle of Mathematical Induction for Minimally Inductive Set",
"Minimally Inductive Set is Ordinal/Proof 1",
"Definition:Ordinal"
] |
proofwiki-5752 | Minimally Inductive Set is Ordinal | Let $\omega$ denote the minimally inductive set.
Then $\omega$ is an ordinal. | Let $K_I$ denote the set of all nonlimit ordinals.
Let $\On$ denote the set of all ordinals.
Let $a \in \omega$.
It follows that $a^+ \subseteq K_I$, so $a \in K_I$.
Thus:
:$\omega \subseteq K_I \subseteq \On$
We now must prove that $\omega$ is a transitive set, at which point it will satisfy the Alternative Definition... | Let $\omega$ denote the [[Definition:Minimally Inductive Set|minimally inductive set]].
Then $\omega$ is an [[Definition:Ordinal|ordinal]]. | Let $K_I$ denote the [[Definition:Set|set]] of all [[Definition:Limit Ordinal|nonlimit ordinals]].
Let $\On$ denote the [[Definition:Set|set]] of all [[Definition:Ordinal|ordinals]].
Let $a \in \omega$.
It follows that $a^+ \subseteq K_I$, so $a \in K_I$.
Thus:
:$\omega \subseteq K_I \subseteq \On$
We now must pr... | Minimally Inductive Set is Ordinal/Proof 2 | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Ordinal | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Ordinal/Proof_2 | [
"Ordinals",
"Minimally Inductive Set",
"Minimally Inductive Set is Ordinal"
] | [
"Definition:Minimally Inductive Set",
"Definition:Ordinal"
] | [
"Definition:Set",
"Definition:Limit Ordinal",
"Definition:Set",
"Definition:Ordinal",
"Definition:Transitive Class",
"Alternative Definition of Ordinal",
"Definition:Ordinal",
"Definition:Transitive Class",
"Definition:Minimally Inductive Set/Definition 3",
"Definition:Transitive Class"
] |
proofwiki-5753 | Minimally Inductive Set is Limit Ordinal | Let $\omega$ denote the minimally inductive set.
Then $\omega$ is a limit ordinal. | Let $K_I$ denote the class of all nonlimit ordinals.
{{AimForCont}} $\omega$ is not a limit ordinal.
Every element of $\omega$ is a nonlimit ordinal.
So, if $\omega$ is also a nonlimit ordinal, $\omega + 1 \subseteq K_I$.
By the definition of $\omega$:
:$\omega \in \omega$
which violates No Membership Loops and is thus... | Let $\omega$ denote the [[Definition:Minimally Inductive Set|minimally inductive set]].
Then $\omega$ is a [[Definition:Limit Ordinal|limit ordinal]]. | Let $K_I$ denote the class of all nonlimit [[Definition:Ordinal|ordinals]].
{{AimForCont}} $\omega$ is not a [[Definition:Limit Ordinal|limit ordinal]].
Every [[Definition:Element|element]] of $\omega$ is a [[Definition:Limit Ordinal|nonlimit ordinal]].
So, if $\omega$ is also a nonlimit ordinal, $\omega + 1 \subset... | Minimally Inductive Set is Limit Ordinal | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Limit_Ordinal | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Limit_Ordinal | [
"Ordinals",
"Minimally Inductive Set",
"Limit Ordinals"
] | [
"Definition:Minimally Inductive Set",
"Definition:Limit Ordinal"
] | [
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Element",
"Definition:Limit Ordinal",
"No Membership Loops",
"Definition:Contradiction",
"Proof by Contradiction",
"Definition:Limit Ordinal"
] |
proofwiki-5754 | No Infinitely Descending Membership Chains | Let $\omega$ denote the minimally inductive set.
Let $F$ be a mapping whose domain is $\omega$.
Then:
:$\exists n \in \omega: \map F {n^+} \notin \map F n$ | Let $F$ be a mapping whose domain is $\omega$.
By the axiom of replacement, the image of $F$ is a set.
Let the image of $F$ be denoted $\map \WW F$.
Then:
{{begin-eqn}}
{{eqn | o =
| r = \exists x \in \map \WW F: \paren {\map \WW F \cap x} = \O
| c = Axiom of Foundation
}}
{{eqn | o = \leadsto
| r = ... | Let $\omega$ denote the [[Definition:Minimally Inductive Set|minimally inductive set]].
Let $F$ be a [[Definition:Mapping|mapping]] whose [[Definition:Domain of Mapping|domain]] is $\omega$.
Then:
:$\exists n \in \omega: \map F {n^+} \notin \map F n$ | Let $F$ be a [[Definition:Mapping|mapping]] whose [[Definition:Domain of Mapping|domain]] is $\omega$.
By the [[Axiom:Axiom of Replacement|axiom of replacement]], the [[Definition:Image of Relation|image]] of $F$ is a [[Definition:Set|set]].
Let the [[Definition:Image of Relation|image]] of $F$ be denoted $\map \WW ... | No Infinitely Descending Membership Chains | https://proofwiki.org/wiki/No_Infinitely_Descending_Membership_Chains | https://proofwiki.org/wiki/No_Infinitely_Descending_Membership_Chains | [
"Set Theory"
] | [
"Definition:Minimally Inductive Set",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping"
] | [
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Axiom:Axiom of Replacement",
"Definition:Image (Set Theory)/Relation/Relation",
"Definition:Set",
"Definition:Image (Set Theory)/Relation/Relation",
"Axiom:Axiom of Foundation"
] |
proofwiki-5755 | Ordinals Isomorphic to the Same Well-Ordered Set | Let $A$ and $B$ be ordinals.
Let $\left({\prec, S}\right)$ be a strict well-ordering.
Let $\left({\in, A}\right)$ and $\left({\prec, S}\right)$ be order isomorphic.
Let $\left({\in, B}\right)$ and $\left({\prec, S}\right)$ be order isomorphic.
Then:
: $A = B$ | Let $\phi_1$ denote the mapping creating the order isomorphism between $\left({\in, A}\right)$ and $\left({\prec, S}\right)$.
Let $\phi_2$ denote the mapping creating the order isomorphism between $\left({\in, B}\right)$ and $\left({\prec, S}\right)$.
Then $\phi_2^{-1}: S \to B$ is an order isomorphism by Inverse of Or... | Let $A$ and $B$ be [[Definition:Ordinal|ordinals]].
Let $\left({\prec, S}\right)$ be a [[Definition:Strict Well-Ordering|strict well-ordering]].
Let $\left({\in, A}\right)$ and $\left({\prec, S}\right)$ be [[Definition:Order Isomorphism|order isomorphic]].
Let $\left({\in, B}\right)$ and $\left({\prec, S}\right)$ be... | Let $\phi_1$ denote the [[Definition:Mapping|mapping]] creating the [[Definition:Order Isomorphism|order isomorphism]] between $\left({\in, A}\right)$ and $\left({\prec, S}\right)$.
Let $\phi_2$ denote the [[Definition:Mapping|mapping]] creating the [[Definition:Order Isomorphism|order isomorphism]] between $\left({\i... | Ordinals Isomorphic to the Same Well-Ordered Set | https://proofwiki.org/wiki/Ordinals_Isomorphic_to_the_Same_Well-Ordered_Set | https://proofwiki.org/wiki/Ordinals_Isomorphic_to_the_Same_Well-Ordered_Set | [
"Ordinals",
"Order Isomorphisms"
] | [
"Definition:Ordinal",
"Definition:Strict Well-Ordering",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism"
] | [
"Definition:Mapping",
"Definition:Order Isomorphism",
"Definition:Mapping",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism",
"Inverse of Order Isomorphism is Order Isomorphism",
"Definition:Order Isomorphism",
"Composite of Order Isomorphisms is Order Isomorphism",
"Definition:Order I... |
proofwiki-5756 | Transfinite Recursion Theorem/Uniqueness | Let $f$ be a mapping with a domain $y$ where $y$ is an ordinal.
Let $f$ satisfy the condition that:
:$\forall x \in y: \map f x = \map G {f \restriction x}$
where $f \restriction x$ denotes the restriction of $f$ to $x$.
Let $g$ be a mapping with a domain $z$ where $z$ is an ordinal.
Let $g$ satisfy the condition that:... | Proof by transfinite induction:
Suppose that:
:$\forall x \in \alpha: \map f x = \map g x$
for some arbitrary ordinal $\alpha < y$.
Then $\alpha < z$.
{{explain|Find the link to the result proving this.}}
Hence:
{{begin-eqn}}
{{eqn | q = \forall x \in \alpha
| l = \map f x
| r = \map g x
}}
{{eqn | ll= \lea... | Let $f$ be a [[Definition:Mapping|mapping]] with a [[Definition:Domain of Mapping|domain]] $y$ where $y$ is an [[Definition:Ordinal|ordinal]].
Let $f$ satisfy the condition that:
:$\forall x \in y: \map f x = \map G {f \restriction x}$
where $f \restriction x$ denotes the [[Definition:Restriction of Mapping|restrictio... | Proof by [[Transfinite Induction/Schema 1|transfinite induction]]:
Suppose that:
:$\forall x \in \alpha: \map f x = \map g x$
for some arbitrary ordinal $\alpha < y$.
Then $\alpha < z$.
{{explain|Find the link to the result proving this.}}
Hence:
{{begin-eqn}}
{{eqn | q = \forall x \in \alpha
| l = \map f x
... | Transfinite Recursion Theorem/Uniqueness | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Uniqueness | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Uniqueness | [
"Transfinite Recursion Theorem"
] | [
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Ordinal",
"Definition:Restriction/Mapping",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 1",
"Equality of Restrictions",
"Transfinite Induction/Schema 1"
] |
proofwiki-5757 | Transfinite Induction/Schema 1 | Let $\map P x$ be a property
Suppose that:
:If $\map P x$ holds for all ordinals $x$ less than $y$, then $\map P y$ also holds.
Then $\map P x$ holds for all ordinals $x$. | It should be noted that for any two ordinals, $x < y \iff x \in y$.
Let $\map P x$ be a property that satisfies the above conditions.
{{AimForCont}} $y$ is an ordinal such that $\neg \map P y$.
By the Rule of Transposition, it follows that if $y$ is an ordinal such that $\neg \map P y$, then there exists an ordinal $x ... | Let $\map P x$ be a [[Definition:Property|property]]
Suppose that:
:If $\map P x$ holds for all [[Definition:Ordinal|ordinals]] $x$ less than $y$, then $\map P y$ also holds.
Then $\map P x$ holds for all [[Definition:Ordinal|ordinals]] $x$. | It should be noted that for any two [[Definition:Ordinal|ordinals]], $x < y \iff x \in y$.
Let $\map P x$ be a [[Definition:Propositional Function|property]] that satisfies the above conditions.
{{AimForCont}} $y$ is an [[Definition:Ordinal|ordinal]] such that $\neg \map P y$.
By the [[Rule of Transposition]], it f... | Transfinite Induction/Schema 1/Proof 1 | https://proofwiki.org/wiki/Transfinite_Induction/Schema_1 | https://proofwiki.org/wiki/Transfinite_Induction/Schema_1/Proof_1 | [
"Transfinite Induction"
] | [
"Definition:Property",
"Definition:Ordinal",
"Definition:Ordinal"
] | [
"Definition:Ordinal",
"Definition:Propositional Function",
"Definition:Ordinal",
"Rule of Transposition",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Smallest Element",
"Definition:Ordinal",
"Definition:Element",
"Definition:Ordinal",
"Definition:Strict Well-Ordering",
"Definition:... |
proofwiki-5758 | Transfinite Induction/Schema 1 | Let $\map P x$ be a property
Suppose that:
:If $\map P x$ holds for all ordinals $x$ less than $y$, then $\map P y$ also holds.
Then $\map P x$ holds for all ordinals $x$. | It should be noted that for any two ordinals, $x < y \iff x \in y$.
Let $\map P x$ be a property that satisfies the above conditions.
{{AimForCont}} $y$ is an ordinal such that $\neg \map P y$.
Let $S$ be the set defined as:
:$S = \set {x \in y^+: \neg \map P x}$
It is seen that this set is non-empty because by Set is ... | Let $\map P x$ be a [[Definition:Property|property]]
Suppose that:
:If $\map P x$ holds for all [[Definition:Ordinal|ordinals]] $x$ less than $y$, then $\map P y$ also holds.
Then $\map P x$ holds for all [[Definition:Ordinal|ordinals]] $x$. | It should be noted that for any two [[Definition:Ordinal|ordinals]], $x < y \iff x \in y$.
Let $\map P x$ be a [[Definition:Propositional Function|property]] that satisfies the above conditions.
{{AimForCont}} $y$ is an [[Definition:Ordinal|ordinal]] such that $\neg \map P y$.
Let $S$ be the [[Definition:Set|set]] ... | Transfinite Induction/Schema 1/Proof 2 | https://proofwiki.org/wiki/Transfinite_Induction/Schema_1 | https://proofwiki.org/wiki/Transfinite_Induction/Schema_1/Proof_2 | [
"Transfinite Induction"
] | [
"Definition:Property",
"Definition:Ordinal",
"Definition:Ordinal"
] | [
"Definition:Ordinal",
"Definition:Propositional Function",
"Definition:Ordinal",
"Definition:Set",
"Definition:Non-Empty Set",
"Set is Element of Successor",
"Definition:Ordinal",
"Successor Set of Ordinal is Ordinal",
"Element of Ordinal is Ordinal",
"Definition:Ordinal",
"Rule of Transposition... |
proofwiki-5759 | Transfinite Induction/Principle 2 | Let $A$ be a class satisfying the following conditions:
* $\O \in A$
* $\forall x \in A: x^+ \in A$
* If $y$ is a limit ordinal, then $\paren {\forall x < y: x \in A} \implies y \in A$
where $x^+$ denotes the successor of $x$.
Then $\On \subseteq A$. | {{NotZFC}}
We shall prove this using the first principle of transfinite induction.
Assume $y$ is an ordinal such that $y \subseteq A$.
This implies $\forall x < y: x \in A$.
By definition, $y$ is a limit ordinal, $y = \O$, or $\exists x: y = x^+$.
If $y$ is a limit ordinal, then, since:
:$\forall x < y: x \in A$
it fol... | Let $A$ be a [[Definition:Class (Class Theory)|class]] satisfying the following conditions:
* $\O \in A$
* $\forall x \in A: x^+ \in A$
* If $y$ is a [[Definition:Limit Ordinal|limit ordinal]], then $\paren {\forall x < y: x \in A} \implies y \in A$
where $x^+$ denotes the [[Definition:Successor Set|successor]] of $x... | {{NotZFC}}
We shall prove this using [[Transfinite Induction/Principle 1|the first principle of transfinite induction]].
Assume $y$ is an [[Definition:Ordinal|ordinal]] such that $y \subseteq A$.
This implies $\forall x < y: x \in A$.
By definition, $y$ is a [[Definition:Limit Ordinal|limit ordinal]], $y = \O$, or ... | Transfinite Induction/Principle 2 | https://proofwiki.org/wiki/Transfinite_Induction/Principle_2 | https://proofwiki.org/wiki/Transfinite_Induction/Principle_2 | [
"Transfinite Induction"
] | [
"Definition:Class (Class Theory)",
"Definition:Limit Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Transfinite Induction/Principle 1",
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Limit Ordinal",
"Definition:Ordinal",
"Definition:Ordinal",
"Transfinite Induction/Principle 1",
"Category:Transfinite Induction"
] |
proofwiki-5760 | Transfinite Induction/Schema 2 | Let $\map \phi x$ be a property satisfying the following conditions:
:$(1): \quad \map \phi \O$ is true
:$(2): \quad$ If $x$ is an ordinal, then $\map \phi x \implies \map \phi {x^+}$
:$(3): \quad$ If $y$ is a limit ordinal, then $\paren {\forall x < y: \map \phi x} \implies \map \phi y$
where $x^+$ denotes the success... | It should be noted that for any two ordinals, $x \lt y \iff x \in y$.
Let $\map \phi x$ be a property that satisfies the above conditions.
{{AimForCont}} $y$ be an ordinal such that $\neg \map \phi y$.
It is noted that $y \ne \O$.
Therefore $y$ must be either a successor ordinal or a limit ordinal.
If $y$ is a successo... | Let $\map \phi x$ be a [[Definition:Property|property]] satisfying the following conditions:
:$(1): \quad \map \phi \O$ is [[Definition:True|true]]
:$(2): \quad$ If $x$ is an [[Definition:Ordinal|ordinal]], then $\map \phi x \implies \map \phi {x^+}$
:$(3): \quad$ If $y$ is a [[Definition:Limit Ordinal|limit ordinal]]... | It should be noted that for any two [[Definition:Ordinal|ordinals]], $x \lt y \iff x \in y$.
Let $\map \phi x$ be a [[Definition:Property|property]] that satisfies the above conditions.
{{AimForCont}} $y$ be an [[Definition:Ordinal|ordinal]] such that $\neg \map \phi y$.
It is noted that $y \ne \O$.
Therefore $y$ ... | Transfinite Induction/Schema 2/Proof 1 | https://proofwiki.org/wiki/Transfinite_Induction/Schema_2 | https://proofwiki.org/wiki/Transfinite_Induction/Schema_2/Proof_1 | [
"Transfinite Induction"
] | [
"Definition:Property",
"Definition:True",
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Successor Ordinal",
"Definition:Ordinal"
] | [
"Definition:Ordinal",
"Definition:Property",
"Definition:Ordinal",
"Definition:Successor Ordinal",
"Definition:Limit Ordinal",
"Definition:Successor Ordinal",
"Definition:Ordinal",
"Rule of Transposition",
"Set is Element of Successor",
"Definition:Limit Ordinal",
"Rule of Transposition",
"Def... |
proofwiki-5761 | Transfinite Induction/Schema 2 | Let $\map \phi x$ be a property satisfying the following conditions:
:$(1): \quad \map \phi \O$ is true
:$(2): \quad$ If $x$ is an ordinal, then $\map \phi x \implies \map \phi {x^+}$
:$(3): \quad$ If $y$ is a limit ordinal, then $\paren {\forall x < y: \map \phi x} \implies \map \phi y$
where $x^+$ denotes the success... | Define the class:
:$A := \set {x \in \On: \map \phi x = \T}$.
Then $\map \phi x = \T$ is equivalent to the statement:
:that $x \in A$
The three conditions in the hypothesis become:
:$(1a): \quad \O \in A$
:$(2a): \quad x \in A \implies x^+ \in A$
:$(3a): \quad \paren {\forall x < y: x \in A} \implies y \in A$
These are... | Let $\map \phi x$ be a [[Definition:Property|property]] satisfying the following conditions:
:$(1): \quad \map \phi \O$ is [[Definition:True|true]]
:$(2): \quad$ If $x$ is an [[Definition:Ordinal|ordinal]], then $\map \phi x \implies \map \phi {x^+}$
:$(3): \quad$ If $y$ is a [[Definition:Limit Ordinal|limit ordinal]]... | Define the [[Definition:Class (Class Theory)|class]]:
:$A := \set {x \in \On: \map \phi x = \T}$.
Then $\map \phi x = \T$ is [[Definition:Logical Equivalence|equivalent]] to the [[Definition:Statement|statement]]:
:that $x \in A$
The three conditions in the hypothesis become:
:$(1a): \quad \O \in A$
:$(2a): \quad x ... | Transfinite Induction/Schema 2/Proof 2 | https://proofwiki.org/wiki/Transfinite_Induction/Schema_2 | https://proofwiki.org/wiki/Transfinite_Induction/Schema_2/Proof_2 | [
"Transfinite Induction"
] | [
"Definition:Property",
"Definition:True",
"Definition:Ordinal",
"Definition:Limit Ordinal",
"Definition:Successor Ordinal",
"Definition:Ordinal"
] | [
"Definition:Class (Class Theory)",
"Definition:Logical Equivalence",
"Definition:Statement",
"Transfinite Induction/Principle 2"
] |
proofwiki-5762 | Transfinite Recursion Theorem/Theorem 2 | Let $\Dom x$ denote the domain of $x$.
Let $\Img x$ denote the image of the mapping $x$.
{{explain|We infer that $x$ is a mapping, but what is its context?}}
Let $G$ be a class of ordered pairs $\tuple {x, y}$ satisfying at least one of the following conditions:
:$(1): \quad x = \O$ and $y = a$
{{explain|What is $a$?}}... | {{begin-eqn}}
{{eqn | l = \map F \O
| r = \map G {F \restriction \O}
| c = {{Hypothesis}}
}}
{{eqn | r = \map G \O
| c = Restriction of $\O$
}}
{{eqn | r = a
| c = Definition of $G$
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = \map F {\beta^+}
| r = \map G {F \restriction \beta^+... | Let $\Dom x$ denote the [[Definition:Domain of Mapping|domain]] of $x$.
Let $\Img x$ denote the [[Definition:Image of Mapping|image of the mapping]] $x$.
{{explain|We infer that $x$ is a mapping, but what is its context?}}
Let $G$ be a [[Definition:Class (Class Theory)|class]] of [[Definition:Ordered Pair|ordered pa... | {{begin-eqn}}
{{eqn | l = \map F \O
| r = \map G {F \restriction \O}
| c = {{Hypothesis}}
}}
{{eqn | r = \map G \O
| c = Restriction of $\O$
}}
{{eqn | r = a
| c = Definition of $G$
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = \map F {\beta^+}
| r = \map G {F \restriction \beta... | Transfinite Recursion Theorem/Theorem 2 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Theorem_2 | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Theorem_2 | [
"Transfinite Recursion Theorem"
] | [
"Definition:Domain (Set Theory)/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Class (Class Theory)",
"Definition:Ordered Pair",
"Definition:Limit Ordinal",
"Definition:Class of All Ordinals"
] | [
"Union of Ordinals is Least Upper Bound",
"Transfinite Induction"
] |
proofwiki-5763 | Transfinite Recursion Theorem/Corollary | Let $x$ be an ordinal.
Let $G$ be a mapping
There exists a unique mapping $f$ that satisfies the following properties:
:The domain of $f$ is $x$
:$\forall y \in x: \map f y = \map G {f \restriction y}$ | Construct $K$ and $F$ as in the First Principle of Transfinite Recursion.
Set $f = \paren {F \restriction x}$.
Then since $x \subseteq \Dom F$, the domain of $f$ is $x$.
{{begin-eqn}}
{{eqn | q = \forall y \in x
| l = \map f y
| r = \map F y
| c = from the fact that $f$ is a restriction
}}
{{eqn | r =... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $G$ be a [[Definition:Mapping|mapping]]
There exists a [[Definition:Unique|unique]] mapping $f$ that satisfies the following properties:
:The [[Definition:Domain of Mapping|domain]] of $f$ is $x$
:$\forall y \in x: \map f y = \map G {f \restriction y}$ | Construct $K$ and $F$ as in the [[First Principle of Transfinite Recursion]].
Set $f = \paren {F \restriction x}$.
Then since $x \subseteq \Dom F$, the domain of $f$ is $x$.
{{begin-eqn}}
{{eqn | q = \forall y \in x
| l = \map f y
| r = \map F y
| c = from the fact that $f$ is a restriction
}}
{{eq... | Transfinite Recursion Theorem/Corollary | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Corollary | https://proofwiki.org/wiki/Transfinite_Recursion_Theorem/Corollary | [
"Transfinite Recursion Theorem"
] | [
"Definition:Ordinal",
"Definition:Mapping",
"Definition:Unique",
"Definition:Domain (Set Theory)/Mapping"
] | [
"Transfinite Recursion Theorem/Theorem 1",
"Transfinite Recursion Theorem/Theorem 1",
"Definition:Mapping",
"Definition:Mapping",
"Transfinite Induction/Principle 1",
"Transfinite Induction",
"Definition:Mapping",
"Definition:Unique"
] |
proofwiki-5764 | Inverse of Product/Monoid/General Result | Let $\struct {S, \circ}$ be a monoid whose identity is $e$.
Let $a_1, a_2, \ldots, a_n \in S$ be invertible for $\circ$, with inverses ${a_1}^{-1}, {a_2}^{-1}, \ldots, {a_n}^{-1}$.
Then $a_1 \circ a_2 \circ \cdots \circ a_n$ is invertible for $\circ$, and:
:$\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \c... | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$
$\map P 1$ is (trivially) true, as this just says:
:$\paren {a_1}^{-1} = {a_1}^{-1}$ | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a_1, a_2, \ldots, a_n \in S$ be [[Definition:Invertible Element|invertible]] for $\circ$, with [[Definition:Inverse Element|inverses]] ${a_1}^{-1}, {a_2}^{-1}, \ldots, {a_n}^{-1}$.
Then $a_1 \circ a... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$
$\map P 1$ is (trivially) true, as this just says:
:$\p... | Inverse of Product/Monoid/General Result | https://proofwiki.org/wiki/Inverse_of_Product/Monoid/General_Result | https://proofwiki.org/wiki/Inverse_of_Product/Monoid/General_Result | [
"Monoids",
"Inverse Elements",
"Inverse of Product",
"Proofs by Induction"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Invertible Element",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Invertible Element"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-5765 | Inverse of Group Product | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | We have that a group is a monoid, all of whose elements are invertible.
The result follows from Inverse of Product/Monoid/General Result.
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a, b \in G$, with [[Definition:Inverse Element|inverses]] $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | We have that a [[Definition:Group|group]] is a [[Definition:Monoid|monoid]], all of whose [[Definition:Element|elements]] are [[Definition:Invertible Element|invertible]].
The result follows from [[Inverse of Product/Monoid/General Result]].
{{Qed}} | Inverse of Group Product/General Result/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Group_Product | https://proofwiki.org/wiki/Inverse_of_Group_Product/General_Result/Proof_1 | [
"Group Theory",
"Inverse Elements",
"Inverse of Product",
"Inverse of Group Product"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Group",
"Definition:Monoid",
"Definition:Element",
"Definition:Invertible Element",
"Inverse of Product/Monoid/General Result"
] |
proofwiki-5766 | Inverse of Group Product | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
$\map P 1$ is (trivially) true, as this just says:
:$\paren {a_1}^{-1} = a_1^{-1}$
=== Basis for the Induction ===
$\map P 2$ is th... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a, b \in G$, with [[Definition:Inverse Element|inverses]] $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
$\map P 1$ is (trivially) true, as this just says:
:$\paren {... | Inverse of Group Product/General Result/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Group_Product | https://proofwiki.org/wiki/Inverse_of_Group_Product/General_Result/Proof_2 | [
"Group Theory",
"Inverse Elements",
"Inverse of Product",
"Inverse of Group Product"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Inverse of Group Product",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"General Associativity Theorem",
"Inverse of Group Product/General Result/Proof 2",
"Inverse of Group Pro... |
proofwiki-5767 | Inverse of Group Product | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | {{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }
| r = \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}
| c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}
| c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {a \circ e} \circ a... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a, b \in G$, with [[Definition:Inverse Element|inverses]] $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | {{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }
| r = \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}
| c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}
| c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {a \circ e} \circ a... | Inverse of Group Product/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Group_Product | https://proofwiki.org/wiki/Inverse_of_Group_Product/Proof_1 | [
"Group Theory",
"Inverse Elements",
"Inverse of Product",
"Inverse of Group Product"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Group Product Identity therefore Inverses"
] |
proofwiki-5768 | Inverse of Group Product | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | We have that a group is a monoid, all of whose elements are invertible.
The result follows from Inverse of Product in Monoid.
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a, b \in G$, with [[Definition:Inverse Element|inverses]] $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | We have that a [[Definition:Group|group]] is a [[Definition:Monoid|monoid]], all of whose [[Definition:Element|elements]] are [[Definition:Invertible Element|invertible]].
The result follows from [[Inverse of Product in Monoid]].
{{Qed}} | Inverse of Group Product/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Group_Product | https://proofwiki.org/wiki/Inverse_of_Group_Product/Proof_2 | [
"Group Theory",
"Inverse Elements",
"Inverse of Product",
"Inverse of Group Product"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Group",
"Definition:Monoid",
"Definition:Element",
"Definition:Invertible Element",
"Inverse of Product/Monoid"
] |
proofwiki-5769 | Inverse of Group Product | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | {{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {a \circ b}^{-1}
| r = e
| c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
| l = a \circ \paren {b \circ \paren {a \circ b}^{-1} }
| r = e
| c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
| l = b \circ \paren {a \circ b}^... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a, b \in G$, with [[Definition:Inverse Element|inverses]] $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$ | {{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {a \circ b}^{-1}
| r = e
| c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
| l = a \circ \paren {b \circ \paren {a \circ b}^{-1} }
| r = e
| c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
| l = b \circ \paren {a \circ b}^... | Inverse of Group Product/Proof 3 | https://proofwiki.org/wiki/Inverse_of_Group_Product | https://proofwiki.org/wiki/Inverse_of_Group_Product/Proof_3 | [
"Group Theory",
"Inverse Elements",
"Inverse of Product",
"Inverse of Group Product"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Group Product Identity therefore Inverses"
] |
proofwiki-5770 | Inverse of Group Product/General Result | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a_1, a_2, \ldots, a_n \in G$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.
Then:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ | We have that a group is a monoid, all of whose elements are invertible.
The result follows from Inverse of Product/Monoid/General Result.
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a_1, a_2, \ldots, a_n \in G$, with [[Definition:Inverse Element|inverses]] $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.
Then:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \c... | We have that a [[Definition:Group|group]] is a [[Definition:Monoid|monoid]], all of whose [[Definition:Element|elements]] are [[Definition:Invertible Element|invertible]].
The result follows from [[Inverse of Product/Monoid/General Result]].
{{Qed}} | Inverse of Group Product/General Result/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Group_Product/General_Result | https://proofwiki.org/wiki/Inverse_of_Group_Product/General_Result/Proof_1 | [
"Inverse of Group Product"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Group",
"Definition:Monoid",
"Definition:Element",
"Definition:Invertible Element",
"Inverse of Product/Monoid/General Result"
] |
proofwiki-5771 | Inverse of Group Product/General Result | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a_1, a_2, \ldots, a_n \in G$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.
Then:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$ | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
$\map P 1$ is (trivially) true, as this just says:
:$\paren {a_1}^{-1} = a_1^{-1}$
=== Basis for the Induction ===
$\map P 2$ is th... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a_1, a_2, \ldots, a_n \in G$, with [[Definition:Inverse Element|inverses]] $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.
Then:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \c... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
$\map P 1$ is (trivially) true, as this just says:
:$\paren {... | Inverse of Group Product/General Result/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Group_Product/General_Result | https://proofwiki.org/wiki/Inverse_of_Group_Product/General_Result/Proof_2 | [
"Inverse of Group Product"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Inverse of Group Product",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"General Associativity Theorem",
"Inverse of Group Product/General Result/Proof 2",
"Inverse of Group Pro... |
proofwiki-5772 | Well-Ordered Transitive Subset is Equal or Equal to Initial Segment | Let $\struct {\prec, A}$ be a well-ordered set.
For every $x \in A$, let every $\prec$-initial segment $A_x$ be a set.
Let $B$ be a subclass of $A$ such that
:$\forall x \in A: \forall y \in B: \paren {x \prec y \implies x \in B}$.
That is, $B$ must be $\prec$-transitive.
Then:
:$A = B$
or:
:$\exists x \in A: B = A_x$ | Let $A \ne B$.
Then $B \subsetneq A$.
Therefore:
{{begin-eqn}}
{{eqn | l = A \setminus B
| o = \ne
| r = \O
| c = Set Difference with Proper Subset
}}
{{eqn | ll= \leadsto
| q = \exists x \in A \setminus B
| l = \paren {A \setminus B} \cap A_x
| r = \O
| c = Proper Well-Orderin... | Let $\struct {\prec, A}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
For every $x \in A$, let every $\prec$-[[Definition:Initial Segment|initial segment]] $A_x$ be a [[Definition:Set|set]].
Let $B$ be a [[Definition:Subset|subclass]] of $A$ such that
:$\forall x \in A: \forall y \in B: \paren {x \prec y \... | Let $A \ne B$.
Then $B \subsetneq A$.
Therefore:
{{begin-eqn}}
{{eqn | l = A \setminus B
| o = \ne
| r = \O
| c = [[Set Difference with Proper Subset]]
}}
{{eqn | ll= \leadsto
| q = \exists x \in A \setminus B
| l = \paren {A \setminus B} \cap A_x
| r = \O
| c = [[Proper Wel... | Well-Ordered Transitive Subset is Equal or Equal to Initial Segment | https://proofwiki.org/wiki/Well-Ordered_Transitive_Subset_is_Equal_or_Equal_to_Initial_Segment | https://proofwiki.org/wiki/Well-Ordered_Transitive_Subset_is_Equal_or_Equal_to_Initial_Segment | [
"Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Initial Segment",
"Definition:Set",
"Definition:Subset"
] | [
"Set Difference with Proper Subset",
"Proper Well-Ordering determines Smallest Elements",
"Set Difference with Superset is Empty Set",
"Modus Tollendo Tollens",
"Definition:Strict Total Ordering"
] |
proofwiki-5773 | Inverse of Inverse/General Algebraic Structure | Let $\struct {S, \circ}$ be an algebraic structure with an identity element $e$.
Let $x \in S$ be invertible, and let $y$ be an inverse of $x$.
Then $x$ is also an inverse of $y$. | Let $x \in S$ be invertible, where $y$ is an inverse of $x$.
Then:
:$x \circ y = e = y \circ x$
by definition.
{{qed}} | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] with an [[Definition:Identity Element|identity element]] $e$.
Let $x \in S$ be [[Definition:Invertible Element|invertible]], and let $y$ be an [[Definition:Inverse Element|inverse]] of $x$.
Then $x$ is also an [[... | Let $x \in S$ be [[Definition:Invertible Element|invertible]], where $y$ is an [[Definition:Inverse Element|inverse]] of $x$.
Then:
:$x \circ y = e = y \circ x$
by definition.
{{qed}} | Inverse of Inverse/General Algebraic Structure | https://proofwiki.org/wiki/Inverse_of_Inverse/General_Algebraic_Structure | https://proofwiki.org/wiki/Inverse_of_Inverse/General_Algebraic_Structure | [
"Inverse Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Invertible Element",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Invertible Element",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-5774 | Inverse of Group Inverse | Let $\struct {G, \circ}$ be a group.
Let $g \in G$, with inverse $g^{-1}$.
Then:
:$\paren {g^{-1} }^{-1} = g$ | Let $g \in G$.
Then:
{{begin-eqn}}
{{eqn | l = g
| o = \in
| r = G
| c =
}}
{{eqn | ll= \leadsto
| l = e
| r = g^{-1} \circ g
| c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
| l = \paren {g^{-1} }^{-1} \circ e
| r = \paren {g^{-1} }^{-1} \circ \paren {g^{-1} \cir... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $g \in G$, with [[Definition:Inverse Element|inverse]] $g^{-1}$.
Then:
:$\paren {g^{-1} }^{-1} = g$ | Let $g \in G$.
Then:
{{begin-eqn}}
{{eqn | l = g
| o = \in
| r = G
| c =
}}
{{eqn | ll= \leadsto
| l = e
| r = g^{-1} \circ g
| c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
| l = \paren {g^{-1} }^{-1} \circ e
| r = \paren {g^{-1} }^{-1} \circ \paren {g^{-1} \ci... | Inverse of Group Inverse/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Group_Inverse | https://proofwiki.org/wiki/Inverse_of_Group_Inverse/Proof_1 | [
"Group Theory",
"Inverse Elements",
"Inverse of Group Inverse"
] | [
"Definition:Group",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [] |
proofwiki-5775 | Inverse of Group Inverse | Let $\struct {G, \circ}$ be a group.
Let $g \in G$, with inverse $g^{-1}$.
Then:
:$\paren {g^{-1} }^{-1} = g$ | Let $g \in G$.
Then:
{{begin-eqn}}
{{eqn | l = g g^{-1}
| r = e
| c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
| l = g
| r = \paren {g^{-1} }^{-1}
| c = Group Product Identity therefore Inverses
}}
{{end-eqn}}
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $g \in G$, with [[Definition:Inverse Element|inverse]] $g^{-1}$.
Then:
:$\paren {g^{-1} }^{-1} = g$ | Let $g \in G$.
Then:
{{begin-eqn}}
{{eqn | l = g g^{-1}
| r = e
| c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
| l = g
| r = \paren {g^{-1} }^{-1}
| c = [[Group Product Identity therefore Inverses]]
}}
{{end-eqn}}
{{qed}} | Inverse of Group Inverse/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Group_Inverse | https://proofwiki.org/wiki/Inverse_of_Group_Inverse/Proof_2 | [
"Group Theory",
"Inverse Elements",
"Inverse of Group Inverse"
] | [
"Definition:Group",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Group Product Identity therefore Inverses"
] |
proofwiki-5776 | Condition for Injective Mapping on Ordinals | Let $F$ be a mapping satisfying the following properties:
:$(1): \quad$ The domain of $F$ is $\On$, the class of all ordinals
:$(2): \quad$ For all ordinals $x$, $\map F x = \map G {F \restriction x}$
:$(3): \quad$ For all ordinals $x$, $\map G {F \restriction x} \in \paren {A \setminus \Img x}$ where $\Img x$ is the i... | {{NotZFC}}
Let $x$ be an ordinal.
Then $\map F x = \map G {F \restriction x}$ and $\map G {F \restriction x} \in A$ by hypothesis.
Therefore, $\map F x \in A$.
This satisfies the first statement.
Take two distinct ordinals $x$ and $y$.
{{WLOG}}, assume $x \in y$ (we are justified in this by Ordinal Membership is Tricho... | Let $F$ be a [[Definition:Mapping|mapping]] satisfying the following properties:
:$(1): \quad$ The [[Definition:Domain of Mapping|domain]] of $F$ is $\On$, the [[Definition:Class of All Ordinals|class of all ordinals]]
:$(2): \quad$ For all [[Definition:Ordinal|ordinals]] $x$, $\map F x = \map G {F \restriction x}$
:$... | {{NotZFC}}
Let $x$ be an [[Definition:Ordinal|ordinal]].
Then $\map F x = \map G {F \restriction x}$ and $\map G {F \restriction x} \in A$ by hypothesis.
Therefore, $\map F x \in A$.
This satisfies the first statement.
Take two [[Definition:Distinct Elements|distinct]] [[Definition:Ordinal|ordinals]] $x$ and $y$.... | Condition for Injective Mapping on Ordinals | https://proofwiki.org/wiki/Condition_for_Injective_Mapping_on_Ordinals | https://proofwiki.org/wiki/Condition_for_Injective_Mapping_on_Ordinals | [
"Class of All Ordinals",
"Class Mappings"
] | [
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Class of All Ordinals",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Image (Set Theory)/Mapping/Element",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Injective",
"Definition:Class (Class Theory)/P... | [
"Definition:Ordinal",
"Definition:Distinct/Plural",
"Definition:Ordinal",
"Ordinal Membership is Trichotomy",
"Definition:Distinct/Plural",
"Definition:Ordinal",
"Definition:Injection",
"Definition:Injection",
"Definition:Set",
"Definition:Set",
"Burali-Forti Paradox",
"Definition:Set",
"Def... |
proofwiki-5777 | Maximal Injective Mapping from Ordinals to a Set | Let $F$ be a mapping satisfying the following properties:
{{explain|What is $G$?}}
:The domain of $F$ is $\On$, the class of all ordinals
:For all ordinals $x$, $\map F x = \map G {F \restriction x}$.
:For all ordinals $x$, if $A \setminus \Img x \ne \O$, then $\map G {F \restriction x} \in A \setminus \Img x$ where $\... | Set $B$ equal to the class of all ordinals $x$ such that $A \setminus \Img x \ne \O$.
Assume $B = \On$.
Then:
{{begin-eqn}}
{{eqn | l = B
| r = \On
}}
{{eqn | ll= \leadsto
| q = \forall x
| l = \map F x
| r = \map G {F \restriction x}
| c = Definition of $B$
}}
{{eqn | ll= \leadsto
... | Let $F$ be a [[Definition:Mapping|mapping]] satisfying the following properties:
{{explain|What is $G$?}}
:The [[Definition:Domain of Mapping|domain]] of $F$ is $\On$, the [[Definition:Class of All Ordinals|class of all ordinals]]
:For all [[Definition:Ordinal|ordinals]] $x$, $\map F x = \map G {F \restriction x}$.
:F... | Set $B$ equal to the [[Definition:Class (Class Theory)|class]] of all ordinals $x$ such that $A \setminus \Img x \ne \O$.
Assume $B = \On$.
Then:
{{begin-eqn}}
{{eqn | l = B
| r = \On
}}
{{eqn | ll= \leadsto
| q = \forall x
| l = \map F x
| r = \map G {F \restriction x}
| c = Definitio... | Maximal Injective Mapping from Ordinals to a Set | https://proofwiki.org/wiki/Maximal_Injective_Mapping_from_Ordinals_to_a_Set | https://proofwiki.org/wiki/Maximal_Injective_Mapping_from_Ordinals_to_a_Set | [
"Class of All Ordinals",
"Class Mappings"
] | [
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Class of All Ordinals",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Set",
"Definition:Injection",
"Definition:Mapping",
"Definition:Mapping",
"Transfinite Recur... | [
"Definition:Class (Class Theory)",
"Condition for Injective Mapping on Ordinals",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Set",
"Union of Ordinals is Least Upper Bound",
"Definition:Mapping",
"Definition:Mapping",
"Ordinal Membership Trichotomy",
"Definition:Injection"
] |
proofwiki-5778 | Power of Element/Semigroup | Let $\struct {S, \oplus}$ be a magma.
Let $a \in S$.
Let $n \in \N_{>0}$.
Let $\tuple {a_1, a_2, \ldots, a_n}$ be the ordered $n$-tuple defined by $a_k = a$ for each $k \in \N_n$.
Then:
:$\ds \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a$
where:
:$\ds \bigoplus_{k \mathop = 1}^n a_k$ is the composite of $\tuple {a_1, a_... | The proof will proceed by the Principle of Mathematical Induction on $\N$.
Let $T$ be the set defined as:
:$\ds T := \set {n \in \N: \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a}$
First, recall the definition of the composite of $\tuple {a_1, a_2, \ldots, a_n}$ for $\oplus$:
:$\ds \bigoplus_{k \mathop = 1}^n a_k = \beg... | Let $\struct {S, \oplus}$ be a [[Definition:Magma|magma]].
Let $a \in S$.
Let $n \in \N_{>0}$.
Let $\tuple {a_1, a_2, \ldots, a_n}$ be the [[Definition:Ordered Tuple|ordered $n$-tuple]] defined by $a_k = a$ for each $k \in \N_n$.
Then:
:$\ds \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a$
where:
:$\ds \bigoplus_{k ... | The proof will proceed by the [[Principle of Mathematical Induction]] on $\N$.
Let $T$ be the [[Definition:Set|set]] defined as:
:$\ds T := \set {n \in \N: \bigoplus_{k \mathop = 1}^n a_k = \oplus^n a}$
First, recall the definition of the [[Definition:Composite (Abstract Algebra)|composite]] of $\tuple {a_1, a_2, \l... | Power of Element/Semigroup | https://proofwiki.org/wiki/Power_of_Element/Semigroup | https://proofwiki.org/wiki/Power_of_Element/Semigroup | [
"Semigroups",
"Powers (Abstract Algebra)"
] | [
"Definition:Magma",
"Definition:Ordered Tuple",
"Definition:Iterated Binary Operation",
"Definition:Power of Element/Semigroup"
] | [
"Principle of Mathematical Induction",
"Definition:Set",
"Definition:Iterated Binary Operation",
"Definition:Power of Element/Semigroup",
"Principle of Mathematical Induction"
] |
proofwiki-5779 | Order Isomorphism between Ordinals and Proper Class | Let $\struct {A, \prec}$ be a strict well-ordering.
Let $A$ be a proper class.
Let the initial segment of $x$ be a set for every $x \in A$.
Then we may make the following definitions:
Set $G$ equal to the collection of ordered pairs $\tuple {x, y}$ such that:
:$y \in A \setminus \Img x$
:$\paren {A \setminus \Img x} \c... | {{NotZFC}} | Let $\struct {A, \prec}$ be a [[Definition:Strict Well-Ordering|strict well-ordering]].
Let $A$ be a [[Definition:Proper Class|proper class]].
Let the [[Definition:Initial Segment|initial segment]] of $x$ be a [[Definition:Set|set]] for every $x \in A$.
Then we may make the following definitions:
Set $G$ equal to ... | {{NotZFC}} | Order Isomorphism between Ordinals and Proper Class | https://proofwiki.org/wiki/Order_Isomorphism_between_Ordinals_and_Proper_Class | https://proofwiki.org/wiki/Order_Isomorphism_between_Ordinals_and_Proper_Class | [
"Class of All Ordinals",
"Class Mappings"
] | [
"Definition:Strict Well-Ordering",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Initial Segment",
"Definition:Set",
"Transfinite Recursion Theorem/Theorem 1",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Class of All Ordinals",
"Definition:Ordinal",
"D... | [] |
proofwiki-5780 | Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping | Let $S$ be a set.
Let $\struct {S, \prec}$ be a strict well-ordering.
Then there exists a unique ordinal $x$ and unique mapping $f$ such that $f: x \to S$ is an order isomorphism. | The existence of $x$ and $f$ follows from the Counting Theorem.
The uniqueness of $x$ follows from the Counting Theorem.
The uniqueness of $f$ follows from Order Isomorphism between Wosets is Unique.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $\struct {S, \prec}$ be a [[Definition:Strict Well-Ordering|strict well-ordering]].
Then there exists a [[Definition:Unique|unique]] [[Definition:Ordinal|ordinal]] $x$ and [[Definition:Unique|unique]] [[Definition:Mapping|mapping]] $f$ such that $f: x \to S$ is an [[Definition... | The existence of $x$ and $f$ follows from the [[Counting Theorem]].
The uniqueness of $x$ follows from the [[Counting Theorem]].
The uniqueness of $f$ follows from [[Order Isomorphism between Wosets is Unique]].
{{qed}} | Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping | https://proofwiki.org/wiki/Strict_Well-Ordering_Isomorphic_to_Unique_Ordinal_under_Unique_Mapping | https://proofwiki.org/wiki/Strict_Well-Ordering_Isomorphic_to_Unique_Ordinal_under_Unique_Mapping | [
"Ordinals",
"Well-Orderings"
] | [
"Definition:Set",
"Definition:Strict Well-Ordering",
"Definition:Unique",
"Definition:Ordinal",
"Definition:Unique",
"Definition:Mapping",
"Definition:Order Isomorphism"
] | [
"Counting Theorem",
"Counting Theorem",
"Order Isomorphism between Wosets is Unique"
] |
proofwiki-5781 | Unique Isomorphism between Ordinal Subset and Unique Ordinal | Let $\On$ be the class of all ordinals.
Let $S \subset \On$ where $S$ is a set.
Then there exists a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi : x \to S$ is an order isomorphism. | Since $S \subset \On$, $\struct {S, \in}$ is a strict well-ordering.
{{explain|Link to result justifying this statement.}}
The result follows directly from Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping.
{{qed}} | Let $\On$ be the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $S \subset \On$ where $S$ is a [[Definition:Set|set]].
Then there exists a [[Definition:Unique|unique]] [[Definition:Mapping|mapping]] $\phi$ and a unique [[Definition:Ordinal|ordinal]] $x$ such that $\phi : x \to S$ is an [[Definition:O... | Since $S \subset \On$, $\struct {S, \in}$ is a [[Definition:Strict Well-Ordering|strict well-ordering]].
{{explain|Link to result justifying this statement.}}
The result follows directly from [[Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping]].
{{qed}} | Unique Isomorphism between Ordinal Subset and Unique Ordinal | https://proofwiki.org/wiki/Unique_Isomorphism_between_Ordinal_Subset_and_Unique_Ordinal | https://proofwiki.org/wiki/Unique_Isomorphism_between_Ordinal_Subset_and_Unique_Ordinal | [
"Ordinals",
"Order Isomorphisms"
] | [
"Definition:Class of All Ordinals",
"Definition:Set",
"Definition:Unique",
"Definition:Mapping",
"Definition:Ordinal",
"Definition:Order Isomorphism"
] | [
"Definition:Strict Well-Ordering",
"Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping"
] |
proofwiki-5782 | Initial Segment of Ordinals under Lexicographic Order | Let $\preccurlyeq_l$ denote the lexicographic order for the set $\paren {\On \times \On}$.
Let the ordinal number $1$ denote the successor of $\O$.
Then the initial segment of $\tuple {1, \O}$ with respect to the lexicographic order $\preccurlyeq_l$ is a proper class.
This initial segment shall be denoted $\paren {\On ... | Define the mapping $F: \On \to \On \times \On$ as:
:$\forall x \in \On: \map F x = \tuple {\O, x}$
Then, $F: \On \to \paren {\On \times \On}_{\tuple {1, \O} }$, since $\O < 1$.
{{explain|The above statement does not make linguistic sense. The entity $F: \On \to \paren {\On \times \On}_{\tuple {1, \O} }$ is a noun, and ... | Let $\preccurlyeq_l$ denote the [[Definition:Lexicographic Order|lexicographic order]] for the set $\paren {\On \times \On}$.
Let the ordinal number $1$ denote the [[Definition:Successor Set|successor]] of $\O$.
Then the [[Definition:Initial Segment|initial segment]] of $\tuple {1, \O}$ with respect to the lexicograp... | Define the [[Definition:Mapping|mapping]] $F: \On \to \On \times \On$ as:
:$\forall x \in \On: \map F x = \tuple {\O, x}$
Then, $F: \On \to \paren {\On \times \On}_{\tuple {1, \O} }$, since $\O < 1$.
{{explain|The above statement does not make linguistic sense. The entity $F: \On \to \paren {\On \times \On}_{\tuple {... | Initial Segment of Ordinals under Lexicographic Order | https://proofwiki.org/wiki/Initial_Segment_of_Ordinals_under_Lexicographic_Order | https://proofwiki.org/wiki/Initial_Segment_of_Ordinals_under_Lexicographic_Order | [
"Ordinals",
"Lexicographic Order"
] | [
"Definition:Lexicographic Order",
"Definition:Successor Mapping/Successor Set",
"Definition:Initial Segment",
"Definition:Class (Class Theory)/Proper Class"
] | [
"Definition:Mapping",
"Definition:Class of All Ordinals",
"Equality of Ordered Pairs",
"Definition:Injection",
"Burali-Forti Paradox"
] |
proofwiki-5783 | Canonical Order Well-Orders Ordered Pairs of Ordinals | The canonical order, $R_0$ strictly well-orders the ordered pairs of ordinal numbers. | === Strict Ordering ===
Let $\tuple {x, y} \mathrel {R_0} \tuple {x, y}$.
Then:
:$\map \max {x, y} < \map \max {x, y} \lor \tuple {x, y} \mathrel {\operatorname {Le} } \tuple {x, y}$
{{explain|Where is the construct $\operatorname{Le}$ defined, and why can $\le$ not be used instead?}}
Both lead to contradictions, so:
:... | The [[Definition:Canonical Order|canonical order]], $R_0$ [[Definition:Strict Well-Ordering|strictly well-orders]] the [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Ordinal|ordinal numbers]]. | === Strict Ordering ===
Let $\tuple {x, y} \mathrel {R_0} \tuple {x, y}$.
Then:
:$\map \max {x, y} < \map \max {x, y} \lor \tuple {x, y} \mathrel {\operatorname {Le} } \tuple {x, y}$
{{explain|Where is the construct $\operatorname{Le}$ defined, and why can $\le$ not be used instead?}}
Both lead to contradictions, s... | Canonical Order Well-Orders Ordered Pairs of Ordinals | https://proofwiki.org/wiki/Canonical_Order_Well-Orders_Ordered_Pairs_of_Ordinals | https://proofwiki.org/wiki/Canonical_Order_Well-Orders_Ordered_Pairs_of_Ordinals | [
"Ordinals",
"Well-Orderings"
] | [
"Definition:Canonical Order",
"Definition:Strict Well-Ordering",
"Definition:Ordered Pair",
"Definition:Ordinal"
] | [
"Definition:Antireflexive Relation",
"Definition:Transitive Relation"
] |
proofwiki-5784 | Initial Segment of Canonical Order is Set | Let $R_0$ denote the canonical ordering of $\paren {\On \times \On}$.
Then, for all $\tuple {x, y} \in \paren {\On \times \On}$, the $R_0$-initial segment is a set.
{{explain|The $R_0$-initial segment of what?}} | Let $z = \map \max {x, y}^+$.
Let $\tuple {v, w} \mathrel R_0 \tuple {x, y}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \max {v, w}
| o = \le
| r = \map \max {x, y}
}}
{{eqn | o = \lt
| r = z
}}
{{end-eqn}}
Thus, the initial segment:
: $\paren {\On \times \On}_{\tuple {x, y} } \subseteq \paren {z \times z}... | Let $R_0$ denote the [[Definition:Canonical Order|canonical ordering]] of $\paren {\On \times \On}$.
Then, for all $\tuple {x, y} \in \paren {\On \times \On}$, the $R_0$-[[Definition:Initial Segment|initial segment]] is a [[Definition:Set|set]].
{{explain|The $R_0$-[[Definition:Initial Segment|initial segment]] of wh... | Let $z = \map \max {x, y}^+$.
Let $\tuple {v, w} \mathrel R_0 \tuple {x, y}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \max {v, w}
| o = \le
| r = \map \max {x, y}
}}
{{eqn | o = \lt
| r = z
}}
{{end-eqn}}
Thus, the [[Definition:Initial Segment|initial segment]]:
: $\paren {\On \times \On}_{\tuple {x,... | Initial Segment of Canonical Order is Set | https://proofwiki.org/wiki/Initial_Segment_of_Canonical_Order_is_Set | https://proofwiki.org/wiki/Initial_Segment_of_Canonical_Order_is_Set | [
"Ordinals",
"Order Theory"
] | [
"Definition:Canonical Order",
"Definition:Initial Segment",
"Definition:Set",
"Definition:Initial Segment"
] | [
"Definition:Initial Segment",
"Axiom of Subsets Equivalents",
"Definition:Initial Segment"
] |
proofwiki-5785 | Ordinal Addition is Closed | Let $\On$ be the class of all ordinals.
Then:
:$\forall x, y \in \On: x + y \in \On$
That is: the sum $x+y$ is an ordinal. | Using Transfinite Induction on $y$: | Let $\On$ be the [[Definition:Class of All Ordinals|class of all ordinals]].
Then:
:$\forall x, y \in \On: x + y \in \On$
That is: the [[Definition:Ordinal Addition|sum]] $x+y$ is an [[Definition:Ordinal|ordinal]]. | Using [[Transfinite Induction/Schema 2|Transfinite Induction]] on $y$: | Ordinal Addition is Closed | https://proofwiki.org/wiki/Ordinal_Addition_is_Closed | https://proofwiki.org/wiki/Ordinal_Addition_is_Closed | [
"Ordinal Arithmetic"
] | [
"Definition:Class of All Ordinals",
"Definition:Ordinal Addition",
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2"
] |
proofwiki-5786 | Ordinal Addition by Zero | Let $x$ be an ordinal.
Let $\O$ be the zero ordinal.
Then:
:$x + \O = x = \O + x$
where $+$ denotes ordinal addition. | By definition of ordinal addition, it is immediate that:
:$x + \O = x$
{{qed|lemma}}
We shall use Transfinite Induction on $x$ to prove $\O + x = x$ | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $\O$ be the [[Definition:Zero (Ordinal)|zero ordinal]].
Then:
:$x + \O = x = \O + x$
where $+$ denotes [[Definition:Ordinal Addition|ordinal addition]]. | By definition of [[Definition:Ordinal Addition|ordinal addition]], it is immediate that:
:$x + \O = x$
{{qed|lemma}}
We shall use [[Transfinite Induction/Schema 2|Transfinite Induction]] on $x$ to prove $\O + x = x$ | Ordinal Addition by Zero | https://proofwiki.org/wiki/Ordinal_Addition_by_Zero | https://proofwiki.org/wiki/Ordinal_Addition_by_Zero | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Zero (Ordinal)",
"Definition:Ordinal Addition"
] | [
"Definition:Ordinal Addition",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5787 | Limit Ordinal Equals its Union | Let $\lambda$ be a limit ordinal.
Then:
:$\lambda = \bigcup \lambda$
where $\bigcup \lambda$ denotes the union of $\lambda$. | From Union of Ordinal is Subset of Itself:
:$\bigcup \lambda \subseteq \lambda$
Suppose $x \in \lambda$.
By Successor of Ordinal Smaller than Limit Ordinal is also Smaller:
:$x^+ < \lambda$
and so:
:$x \in x^+$ and $x^+ \in \lambda$
from which:
:$x \in \bigcup \lambda$
That is:
:$\lambda \subseteq \bigcup \lambda$
Henc... | Let $\lambda$ be a [[Definition:Limit Ordinal|limit ordinal]].
Then:
:$\lambda = \bigcup \lambda$
where $\bigcup \lambda$ denotes the [[Definition:Union of Set of Sets|union]] of $\lambda$. | From [[Union of Ordinal is Subset of Itself]]:
:$\bigcup \lambda \subseteq \lambda$
Suppose $x \in \lambda$.
By [[Successor of Ordinal Smaller than Limit Ordinal is also Smaller]]:
:$x^+ < \lambda$
and so:
:$x \in x^+$ and $x^+ \in \lambda$
from which:
:$x \in \bigcup \lambda$
That is:
:$\lambda \subseteq \bigcup \l... | Limit Ordinal Equals its Union | https://proofwiki.org/wiki/Limit_Ordinal_Equals_its_Union | https://proofwiki.org/wiki/Limit_Ordinal_Equals_its_Union | [
"Set Union",
"Limit Ordinals"
] | [
"Definition:Limit Ordinal",
"Definition:Set Union/Set of Sets"
] | [
"Union of Ordinal is Subset of Itself",
"Successor of Ordinal Smaller than Limit Ordinal is also Smaller",
"Definition:Set Equality"
] |
proofwiki-5788 | Substitutivity of Equality | Let $x$ and $y$ be sets.
Let $\map P x$ be a well-formed formula of the language of set theory.
Let $\map P y$ be the same proposition $\map P x$ with some (not necessarily all) free instances of $x$ replaced with free instances of $y$.
Let $=$ denote set equality.
:$x = y \implies \paren {\map P x \iff \map P y}$ | By induction on the well-formed parts of $\map P x$.
The proof shall use $\implies$ and $\neg$ as the primitive connectives. | Let $x$ and $y$ be [[Definition:Set|sets]].
Let $\map P x$ be a [[Definition:Well-Formed Formula|well-formed formula]] of the [[Definition:Language of Set Theory|language of set theory]].
Let $\map P y$ be the same proposition $\map P x$ with some (not necessarily all) [[Definition:Free Variable|free]] instances of $... | By [[Induction on Well-Formed Formulas|induction on the well-formed parts]] of $\map P x$.
The proof shall use $\implies$ and $\neg$ as the [[Definition:Language of Set Theory|primitive connectives]]. | Substitutivity of Equality | https://proofwiki.org/wiki/Substitutivity_of_Equality | https://proofwiki.org/wiki/Substitutivity_of_Equality | [
"Set Theory",
"Equality"
] | [
"Definition:Set",
"Definition:Well-Formed Formula",
"Definition:Language of Set Theory",
"Definition:Free Variable",
"Definition:Set Equality"
] | [
"Induction on Well-Formed Formulas",
"Definition:Language of Set Theory"
] |
proofwiki-5789 | Membership is Left Compatible with Ordinal Addition | Let $x$, $y$, and $z$ be ordinals.
Let $<$ denote membership relation $\in$, since $\in$ is a strict well-ordering on the ordinals.
Then:
:$x < y \implies \paren {z + x} < \paren {z + y}$ | The proof proceeds by transfinite induction on $y$.
In the proof, we shall use $\in$, $\subsetneq$, and $<$ interchangeably.
We are justified in this by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. | Let $x$, $y$, and $z$ be [[Definition:Ordinal|ordinals]].
Let $<$ denote [[Definition:Membership Relation|membership relation]] $\in$, since $\in$ is a [[Definition:Strict Well-Ordering|strict well-ordering]] on the [[Definition:Ordinal|ordinals]].
Then:
:$x < y \implies \paren {z + x} < \paren {z + y}$ | The proof proceeds by [[Transfinite Induction/Schema 2|transfinite induction]] on $y$.
In the proof, we shall use $\in$, $\subsetneq$, and $<$ interchangeably.
We are justified in this by [[Transitive Set is Proper Subset of Ordinal iff Element of Ordinal]]. | Membership is Left Compatible with Ordinal Addition | https://proofwiki.org/wiki/Membership_is_Left_Compatible_with_Ordinal_Addition | https://proofwiki.org/wiki/Membership_is_Left_Compatible_with_Ordinal_Addition | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Membership Relation",
"Definition:Strict Well-Ordering",
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal"
] |
proofwiki-5790 | Ordinal Addition is Left Cancellable | Let $x, y, z$ be ordinals.
Then:
:$\paren {z + x} = \paren {z + y} \implies x = y$
That is, ordinal addition is left cancellable. | For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably.
This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
Note that:
{{begin-eqn}}
{{eqn | l = x < y
| o = \implies
| r = \paren {z + x} < \paren {z + y}
| c = Membership is Left Compatible with Ord... | Let $x, y, z$ be [[Definition:Ordinal|ordinals]].
Then:
:$\paren {z + x} = \paren {z + y} \implies x = y$
That is, [[Definition:Ordinal Addition|ordinal addition]] is [[Definition:Left Cancellable Operation|left cancellable]]. | For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably.
This is justified by [[Transitive Set is Proper Subset of Ordinal iff Element of Ordinal]].
Note that:
{{begin-eqn}}
{{eqn | l = x < y
| o = \implies
| r = \paren {z + x} < \paren {z + y}
| c = [[Membership is Left Compatibl... | Ordinal Addition is Left Cancellable | https://proofwiki.org/wiki/Ordinal_Addition_is_Left_Cancellable | https://proofwiki.org/wiki/Ordinal_Addition_is_Left_Cancellable | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Ordinal Addition",
"Definition:Left Cancellable Operation"
] | [
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal",
"Membership is Left Compatible with Ordinal Addition",
"Membership is Left Compatible with Ordinal Addition",
"No Membership Loops",
"Definition:Conditional/Consequent",
"Rule of Transposition",
"Ordinal Membership is Trichotomy"
] |
proofwiki-5791 | Supremum Inequality for Ordinals | Let $A \subseteq \On$ and $B \subseteq \On$ be ordinals.
Then:
:$\ds \forall x \in A: \exists y \in B: x \le y \implies \bigcup A \le \bigcup B$ | {{begin-eqn}}
{{eqn | l = x
| o = <
| r = \bigcup A
}}
{{eqn | ll= \leadsto
| q = \exists z
| l = x
| o = <
| r = z
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \land
| l = z < A
| o = <
| r = A
| c =
}}
{{eqn | ll= \leadsto
| q = \exists y \in B: \ex... | Let $A \subseteq \On$ and $B \subseteq \On$ be [[Definition:Ordinal|ordinals]].
Then:
:$\ds \forall x \in A: \exists y \in B: x \le y \implies \bigcup A \le \bigcup B$ | {{begin-eqn}}
{{eqn | l = x
| o = <
| r = \bigcup A
}}
{{eqn | ll= \leadsto
| q = \exists z
| l = x
| o = <
| r = z
| c = {{Defof|Set Union}}
}}
{{eqn | lo= \land
| l = z < A
| o = <
| r = A
| c =
}}
{{eqn | ll= \leadsto
| q = \exists y \in B: \ex... | Supremum Inequality for Ordinals | https://proofwiki.org/wiki/Supremum_Inequality_for_Ordinals | https://proofwiki.org/wiki/Supremum_Inequality_for_Ordinals | [
"Ordinals"
] | [
"Definition:Ordinal"
] | [
"Union of Ordinals is Least Upper Bound"
] |
proofwiki-5792 | Subset is Right Compatible with Ordinal Addition | Let $x, y, z$ be ordinals.
Then:
:$x \le y \implies \paren {x + z} \le \paren {y + z}$ | The proof proceeds by transfinite induction on $z$. | Let $x, y, z$ be [[Definition:Ordinal|ordinals]].
Then:
:$x \le y \implies \paren {x + z} \le \paren {y + z}$ | The proof proceeds by [[Transfinite Induction/Schema 2|transfinite induction]] on $z$. | Subset is Right Compatible with Ordinal Addition | https://proofwiki.org/wiki/Subset_is_Right_Compatible_with_Ordinal_Addition | https://proofwiki.org/wiki/Subset_is_Right_Compatible_with_Ordinal_Addition | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-5793 | Ordinal Subtraction when Possible is Unique | Let $x$ and $y$ be ordinals such that $x \le y$.
Then there exists a unique ordinal $z$ such that $x + z = y$.
That is:
:$x \le y \implies \exists! z \in \On: x + z = y$ | By transfinite induction on $y$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] such that $x \le y$.
Then there exists a [[Definition:Unique|unique]] [[Definition:Ordinal|ordinal]] $z$ such that $x + z = y$.
That is:
:$x \le y \implies \exists! z \in \On: x + z = y$ | By [[Transfinite Induction/Schema 2|transfinite induction]] on $y$. | Ordinal Subtraction when Possible is Unique | https://proofwiki.org/wiki/Ordinal_Subtraction_when_Possible_is_Unique | https://proofwiki.org/wiki/Ordinal_Subtraction_when_Possible_is_Unique | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Unique",
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2"
] |
proofwiki-5794 | Equality of Successors | Let $x$ and $y$ be ordinals.
Let $x^+$ denote the successor set of $x$.
Then, $x = y \iff x^+ = y^+$ | {{begin-eqn}}
{{eqn | l = x = y
| o = \leadsto
| r = x^+ = y^+
| c = Substitutivity of Equality
}}
{{end-eqn}}
Conversely,
{{begin-eqn}}
{{eqn | l = x^+ = y^+
| o = \leadsto
| r = \bigcup x^+ = \bigcup y^+
| c = Substitutivity of Equality
}}
{{eqn | o = \leadsto
| r = x = y
... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Then, $x = y \iff x^+ = y^+$ | {{begin-eqn}}
{{eqn | l = x = y
| o = \leadsto
| r = x^+ = y^+
| c = [[Substitutivity of Equality]]
}}
{{end-eqn}}
Conversely,
{{begin-eqn}}
{{eqn | l = x^+ = y^+
| o = \leadsto
| r = \bigcup x^+ = \bigcup y^+
| c = [[Substitutivity of Equality]]
}}
{{eqn | o = \leadsto
| r... | Equality of Successors | https://proofwiki.org/wiki/Equality_of_Successors | https://proofwiki.org/wiki/Equality_of_Successors | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Substitutivity of Equality",
"Substitutivity of Equality",
"Union of Successor Ordinal"
] |
proofwiki-5795 | Infinite Ordinal can be expressed Uniquely as Sum of Limit Ordinal plus Finite Ordinal | Let $x$ be an ordinal.
Suppose $x$ satisfies $\omega \subseteq x$.
Then $x$ has a unique representation as $\paren {y + z}$ where $y$ is a limit ordinal and $z$ is a finite ordinal. | Take $K_{II}$ to be the set of all limit ordinals.
Then set $y = \bigcup \set {w \in K_{II}: w \le x}$
The set $\set {w \in K_{II}: w \le x}$ is non-empty because $\omega \subseteq x$.
By Union of Ordinals is Least Upper Bound, $y \in K_{II}$ and $y \le x$.
By Ordinal Subtraction when Possible is Unique, there is a uni... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Suppose $x$ satisfies $\omega \subseteq x$.
Then $x$ has a [[Definition:Unique|unique]] representation as $\paren {y + z}$ where $y$ is a [[Definition:Limit Ordinal|limit ordinal]] and $z$ is a [[Definition:Finite Ordinal|finite ordinal]]. | Take $K_{II}$ to be the set of all [[Definition:Limit Ordinal|limit ordinals]].
Then set $y = \bigcup \set {w \in K_{II}: w \le x}$
The set $\set {w \in K_{II}: w \le x}$ is [[Definition:Non-Empty Set|non-empty]] because $\omega \subseteq x$.
By [[Union of Ordinals is Least Upper Bound]], $y \in K_{II}$ and $y \le x... | Infinite Ordinal can be expressed Uniquely as Sum of Limit Ordinal plus Finite Ordinal | https://proofwiki.org/wiki/Infinite_Ordinal_can_be_expressed_Uniquely_as_Sum_of_Limit_Ordinal_plus_Finite_Ordinal | https://proofwiki.org/wiki/Infinite_Ordinal_can_be_expressed_Uniquely_as_Sum_of_Limit_Ordinal_plus_Finite_Ordinal | [
"Ordinal Arithmetic"
] | [
"Definition:Ordinal",
"Definition:Unique",
"Definition:Limit Ordinal",
"Definition:Finite Ordinal"
] | [
"Definition:Limit Ordinal",
"Definition:Non-Empty Set",
"Union of Ordinals is Least Upper Bound",
"Ordinal Subtraction when Possible is Unique",
"Ordinal Subtraction when Possible is Unique",
"Equality is Transitive",
"Ordinal Addition is Associative",
"Limit Ordinals Preserved Under Ordinal Addition"... |
proofwiki-5796 | Finite Ordinal Plus Transfinite Ordinal | Let $n$ be a finite ordinal.
Let $x$ be a transfinite ordinal.
Then:
: $n + x = x$ | By Transfinite Induction on $x$.
The proof will use $<$, $\in$, and $\subset$ interchangeably. This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. | Let $n$ be a [[Definition:Finite Ordinal|finite ordinal]].
Let $x$ be a [[Definition:Transfinite Ordinal|transfinite ordinal]].
Then:
: $n + x = x$ | By [[Transfinite Induction/Schema 2|Transfinite Induction]] on $x$.
The proof will use $<$, $\in$, and $\subset$ interchangeably. This is justified by [[Transitive Set is Proper Subset of Ordinal iff Element of Ordinal]]. | Finite Ordinal Plus Transfinite Ordinal | https://proofwiki.org/wiki/Finite_Ordinal_Plus_Transfinite_Ordinal | https://proofwiki.org/wiki/Finite_Ordinal_Plus_Transfinite_Ordinal | [
"Ordinal Arithmetic",
"Transfinite Arithmetic",
"Finite Ordinals"
] | [
"Definition:Finite Ordinal",
"Definition:Transfinite Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal"
] |
proofwiki-5797 | Class is Transitive iff Union is Subclass | A class $A$ is transitive {{iff}}:
:$\bigcup A \subseteq A$ | === Necessary Condition ===
{{:Class is Transitive iff Union is Subclass/Necessary Condition}}{{qed|lemma}} | A [[Definition:Class (Class Theory)|class]] $A$ is [[Definition:Transitive Set|transitive]] {{iff}}:
:$\bigcup A \subseteq A$ | === [[Class is Transitive iff Union is Subclass/Necessary Condition|Necessary Condition]] ===
{{:Class is Transitive iff Union is Subclass/Necessary Condition}}{{qed|lemma}} | Class is Transitive iff Union is Subclass | https://proofwiki.org/wiki/Class_is_Transitive_iff_Union_is_Subclass | https://proofwiki.org/wiki/Class_is_Transitive_iff_Union_is_Subclass | [
"Class Theory",
"Transitive Classes",
"Class is Transitive iff Union is Subclass"
] | [
"Definition:Class (Class Theory)",
"Definition:Transitive Class"
] | [
"Union of Transitive Class is Subclass"
] |
proofwiki-5798 | Union of Successor Ordinal | Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Then:
:$\map \bigcup {x^+} = x$ | {{begin-eqn}}
{{eqn | l = \map \bigcup {x^+}
| r = \map \bigcup {x \cup \set x}
| c = {{Defof|Successor Set}}
}}
{{eqn | r = \paren {\bigcup x \cup \bigcup \set x}
| c = Set Union is Self-Distributive/Sets of Sets
}}
{{eqn | r = \paren {\bigcup x \cup x}
| c = Union of Singleton
}}
{{eqn | r = x... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $x^+$ denote the [[Definition:Successor Set|successor]] of $x$.
Then:
:$\map \bigcup {x^+} = x$ | {{begin-eqn}}
{{eqn | l = \map \bigcup {x^+}
| r = \map \bigcup {x \cup \set x}
| c = {{Defof|Successor Set}}
}}
{{eqn | r = \paren {\bigcup x \cup \bigcup \set x}
| c = [[Set Union is Self-Distributive/Sets of Sets]]
}}
{{eqn | r = \paren {\bigcup x \cup x}
| c = [[Union of Singleton]]
}}
{{eqn... | Union of Successor Ordinal | https://proofwiki.org/wiki/Union_of_Successor_Ordinal | https://proofwiki.org/wiki/Union_of_Successor_Ordinal | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Set Union is Self-Distributive/Sets of Sets",
"Union of Singleton",
"Class is Transitive iff Union is Subclass",
"Category:Ordinals"
] |
proofwiki-5799 | Set Union is Self-Distributive | Set union is self-distributive:
:$\forall A, B, C: \paren {A \cup B} \cup \paren {A \cup C} = A \cup B \cup C = \paren {A \cup C} \cup \paren {B \cup C}$
where $A, B, C$ are sets. | We have:
* Union is Associative
* Union is Commutative
* Set Union is Idempotent
The result follows from Associative Commutative Idempotent Operation is Self-Distributive.
{{qed}}
Category:Set Union
Category:Examples of Self-Distributive Operations
Category:Set Union is Self-Distributive
lmqzsopzfpajvmeihd58p06yoombkpd | [[Definition:Set Union|Set union]] is [[Definition:Self-Distributive Operation|self-distributive]]:
:$\forall A, B, C: \paren {A \cup B} \cup \paren {A \cup C} = A \cup B \cup C = \paren {A \cup C} \cup \paren {B \cup C}$
where $A, B, C$ are [[Definition:Set|sets]]. | We have:
* [[Union is Associative]]
* [[Union is Commutative]]
* [[Set Union is Idempotent]]
The result follows from [[Associative Commutative Idempotent Operation is Self-Distributive]].
{{qed}}
[[Category:Set Union]]
[[Category:Examples of Self-Distributive Operations]]
[[Category:Set Union is Self-Distributive]]
l... | Set Union is Self-Distributive | https://proofwiki.org/wiki/Set_Union_is_Self-Distributive | https://proofwiki.org/wiki/Set_Union_is_Self-Distributive | [
"Set Union",
"Examples of Self-Distributive Operations",
"Set Union is Self-Distributive"
] | [
"Definition:Set Union",
"Definition:Self-Distributive Operation",
"Definition:Set"
] | [
"Union is Associative",
"Union is Commutative",
"Set Union is Idempotent",
"Associative Commutative Idempotent Operation is Self-Distributive",
"Category:Set Union",
"Category:Examples of Self-Distributive Operations",
"Category:Set Union is Self-Distributive"
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.