nyuuzyou/wikis huggingface dataset:
Union Mapping/Examples/Absolute Value

== Examples of Union Mappings ==
Let $f_1: \R_{\ge 0} \to \R$ be the real function defined on the set of positive real numbers $\R_{\ge 0}$ as:
:$\forall x \in \R_{\ge 0}: \map {f_1} x = x$
Let $f_2: \R_{\le 0} \to \R$ be the real function defined on the set of negative real numbers $\R_{\le 0}$ as:
:$\forall x \in \R_{\le 0}: \map {f_2} x = -x$
Then:
:$f_1$ and $f_2$ are combinable mappings
and:
:the union mapping $f = f_1 \cup f_2$ is:
::$\forall x \in \R: \map f x = \size x$
:where $\size x$ denotes the absolute value of $x$.
== Proof ==
We have that:
:$\Dom {f_1} \cap \Dom {f_2} = \set 0$
and:
:$\map {f_1} 0 = 0 = \map {f_2} 0$
Hence by definition $f_1$ and $f_2$ are combinable.
Then we have that:
:$\forall x \in \R: \map f x = \begin {cases} x & : x \ge 0 \\ -x & : x \le 0 \end {cases}$
and the result follows by definition of the absolute value of $x$.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Restrictions and Extensions
Category:Union Mappings
Category:Absolute Value Function
Union Operation on Supersets of Subset is Closed

== Theorem ==
Let $S$ be a set.
Let $T \subseteq S$ be a given subset of $S$.
Let $\powerset S$ denote the power set of $S$
Let $\mathscr S$ be the subset of $\powerset S$ defined as:
:$\mathscr S = \set {Y \in \powerset S: T \subseteq Y}$
Then the algebraic structure $\struct {\mathscr S, \cup}$ is closed.
== Proof ==
Let $A, B \in \mathscr S$.
We have that:
and:
Thus we have:
Hence the result by definition of closed algebraic structure.
== Also see ==
* Intersection Operation on Supersets of Subset is Closed
== Sources ==
* : Chapter $5$: Subgroups: Exercise $1$
Category:Set Union
Union as Symmetric Difference with Intersection

== Theorem ==
Let $A$ and $B$ be sets.
Then:
:$A \cup B = \paren {A \symdif B} \symdif \paren {A \cap B}$
where:
:$A \cup B$ denotes set union
:$A \cap B$ denotes set intersection
:$A \symdif B$ denotes set symmetric difference
== Proof ==
From the definition of symmetric difference:
:$\paren {A \symdif B} \symdif \paren {A \cap B} = \paren {\paren {A \symdif B} \cup \paren {A \cap B} } \setminus \paren {\paren {A \symdif B} \cap \paren {A \cap B} }$
Also from the definition of symmetric difference:
:$\paren {A \symdif B} \cap \paren {A \cap B} = \paren {\paren {A \cup B} \setminus \paren {A \cap B} } \cap \paren {A \cup B}$
From Set Difference Intersection with Second Set is Empty Set:
:$\paren {S \setminus T} \cap T = \O$
Hence:
:$\paren {\paren {A \cup B} \setminus \paren {A \cup B} } \cap \paren {A \cup B} = \O$
This leaves:
Then:
Hence the result.
Category:Set Union
Category:Set Intersection
Category:Symmetric Difference
Union equals Intersection iff Sets are Equal

== Theorem ==
Let $S$ and $T$ be sets.
Then:
:$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S = T$
where:
:$S \cup T$ denotes set union
:$S \cap T$ denotes set intersection.
== Proof ==
From Intersection with Subset is Subset:
:$S \subseteq T \iff S \cap T = S$
From Union with Superset is Superset:
:$S \subseteq T \iff S \cup T = T$
That is:
:$T \subseteq S \iff S \cup T = S$
Thus:
:$\paren {S \cup T = S} \land \paren {S \cap T = S} \iff S \subseteq T \subseteq S$
By definition of set equality:
:$S = T \iff S \subseteq T \subseteq S$
Hence the result.
== Sources ==
* : $\S 1$: Sets and Functions: Problem $1$
Category:Set Union
Category:Set Intersection
Union from Synthetic Basis is Topology

== Theorem ==
Let $\BB$ be a synthetic basis on a set $X$.
Let $\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB}$.
Then $\tau$ is a topology on $X$.
$\tau$ is called the topology arising from, or generated by, the basis $\BB$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Definition:Topology Generated by Synthetic Basis
* Definition:Topology Generated by Synthetic Sub-Basis
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:Topological Bases
Category:Union from Synthetic Basis is Topology
Union from Synthetic Basis is Topology/Proof 1

== Theorem ==
== Proof ==
We proceed to verify the open set axioms for $\tau$ to be a topology on $S$.
=== ===
Let $\AA \subseteq \tau$.
It is to be shown that:
:$\ds \bigcup \AA \in \tau$
Define:
:$\ds \AA' = \bigcup_{U \mathop \in \AA} \set {B \in \BB: B \subseteq U}$
By Union is Smallest Superset: Family of Sets, it follows that $\AA' \subseteq \BB$.
Hence, by Equivalence of Definitions of Topology Generated by Synthetic Basis and General Self-Distributivity of Set Union:
:$\ds \bigcup \AA = \bigcup_{U \mathop \in \AA} \bigcup \set {B \in \BB: B \subseteq U} = \bigcup \AA' \in \tau$
=== ===
Let $U, V \in \tau$.
It is to be shown that:
:$U \cap V \in \tau$
Define:
:$\OO = \set {A \cap B: A, B \in \BB, \, A \subseteq U, \, B \subseteq V}$
By the definition of a synthetic basis:
:$\forall A, B \in \BB: A \cap B \in \tau$
Hence, by the definition of a subset, it follows that $\OO \subseteq \tau$.
By , which has been verified above for $\tau$, we have:
:$\ds \bigcup \OO \in \tau$
Since set intersection preserves subsets, we have:
:$\ds \forall W \in \OO: \exists A, B \in \BB: W = A \cap B \subseteq U \cap V$
By Union is Smallest Superset: General Result, it follows that:
:$\ds \bigcup \OO \subseteq U \cap V$
By the definition of $\tau$, it follows from Set is Subset of Union: General Result that:
:$\ds \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: x \in A \cap B \subseteq \bigcup \OO$
That is, by the definition of a subset:
:$\ds U \cap V \subseteq \bigcup \OO$
By definition of set equality:
:$\ds U \cap V = \bigcup \OO \in \tau$
=== ===
By the definition of a synthetic basis, $\BB$ is a cover for $S$.
By Equivalent Conditions for Cover by Collection of Subsets, it follows that:
:$\ds S = \bigcup \BB \in \tau$
All the open set axioms are fulfilled, and the result follows.
== Also see ==
* Definition:Topology Generated by Synthetic Basis
* Definition:Topology Generated by Synthetic Sub-Basis
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.2$: Bases: Proposition $3.2.3$
Category:Union from Synthetic Basis is Topology
Union from Synthetic Basis is Topology/Proof 2

== Theorem ==
== Proof ==
We use Equivalence of Definitions of Topology Generated by Synthetic Basis.
We proceed to verify the open set axioms for $\tau$ to be a topology on $X$.
=== ===
Let $\AA \subseteq \tau$.
It is to be shown that:
:$\ds \bigcup \AA \in \tau$
By the definition of $\tau$, it follows from Set is Subset of Union: General Result that:
:$\ds \forall x \in \bigcup \AA: \exists U \in \AA: \exists B \in \BB: x \in B \subseteq U \subseteq \bigcup \AA$
By the transitivity of $\subseteq$, the result follows.
=== ===
Let $U, V \in \tau$.
It is to be shown that:
:$U \cap V \in \tau$
By the definition of a synthetic basis, we have that:
:$\forall A, B \in \BB: A \cap B \in \tau$
Therefore, it follows from Set is Subset of Union: General Result and Set Intersection Preserves Subsets that:
:$\ds \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: \exists C \in \BB: x \in C \subseteq A \cap B \subseteq U \cap V$
By the transitivity of $\subseteq$, the result follows.
=== ===
By the definition of a synthetic basis, we have that:
:$\forall x \in X: \exists B \in \BB: x \in B \subseteq X$
All the open set axioms are fulfilled, and the result follows.
== Also see ==
* Definition:Topology Generated by Synthetic Basis
* Definition:Topology Generated by Synthetic Sub-Basis
Category:Union from Synthetic Basis is Topology
Union is Associative

== Theorem ==
Set union is associative:
:$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$
=== Family of Sets ===
== Proof ==
Therefore:
:$x \in A \cup \paren {B \cup C}$ $x \in \paren {A \cup B} \cup C$
Thus it has been shown that:
: $A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$
== Also see ==
* Intersection is Associative
* Set Difference is not Associative
* Symmetric Difference is Associative
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(c)}$
* : $\S 4$: Unions and Intersections
* : A Note on Set-Theoretic Concepts: $(2)$
* : Exercise $1.1: \ 8 \ \text{(d)}$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 2.1$: Sets: Exercise $\text{B} \ 4$
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : $\S 1.8$. Sets of sets: Example $27$
* : $\S 4.2$. Commutative and associative operations: Example $66$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.1$
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections: $1$
* : $1$ Set Theory: $1$. Sets and Functions: $1.2$: Operations on sets
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 6 \ \text{(a)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(e)}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(c)}$
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $122 \ \text{(c)}$
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $1 \ \text{(vi)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets
* : $\S 2$
* : Appendix $\text{A}.2$: Theorem $\text{A}.10$
* : $\text {(i)}$
* : $\text {(iii)}$
Category:Set Union
Category:Direct Proofs
Category:Examples of Associative Operations‎
Category:Union is Associative
Union is Associative/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets.
Let $\ds I = \bigcup_{\lambda \mathop \in \Lambda} I_\lambda$ denote the union of $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda \mathop \in \Lambda} \paren {\bigcup_{i \mathop \in I_\lambda} S_i}$
== Proof ==
For every $\lambda \in \Lambda$, let $\ds T_\lambda = \bigcup_{i \mathop \in I_\lambda} S_i$.
Then:
Thus:
:$\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda \mathop \in \Lambda} T_\lambda = \bigcup_{\lambda \mathop \in \Lambda} \paren {\bigcup_{i \mathop \in I_\lambda} S_i}$
== Also see ==
* General Associativity of Set Intersection
== Sources ==
* : $\S 9$: Families
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.2$
Category:Union is Associative
Union is Commutative

== Theorem ==
Set union is commutative:
:$S \cup T = T \cup S$
=== Family of Sets ===
== Proof ==
== Also see ==
* Intersection is Commutative
* Set Difference is Anticommutative
* Symmetric Difference is Commutative
== Sources ==
* : Chapter $\text {I}$: Sets of Points: $1 \cdot 1$. The algebra of sets
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $2 \ \text{(b)}$
* : $\S 4$: Unions and Intersections
* : A Note on Set-Theoretic Concepts: $(1)$
* : Exercise $1.1: \ 8 \ \text{(c)}$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 2.1$: Sets: Exercise $\text{B} \ 6$
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : $\S 1.4$. Union: Example $15$
* : $\S 4.2$. Commutative and associative operations: Example $63$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.1$
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections: $1$
* : $1$ Set Theory: $1$. Sets and Functions: $1.2$: Operations on sets
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 6 \ \text{(b)}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(b)}$
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $122 \ \text{(a)}$
* : Chapter $1$: Sets and mappings: $\S 1.2$: Sets: Exercise $1 \ \text{(iv)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets
* : $\S 2$
* : Appendix $\text{A}.2$: Theorem $\text{A}.10$
* : $\text {(ii)}$
* : $\text {(ii)}$
Category:Set Union
Category:Examples of Commutative Operations
Category:Union is Commutative
Union is Commutative/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.
Let $\ds I = \bigcup_{i \mathop \in I} S_i$ denote the union of $\family {S_i}_{i \mathop \in I}$.
Let $J \subseteq I$ be a subset of $I$.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k = \bigcup_{k \mathop \in \relcomp I J} S_k \cup \bigcup_{j \mathop \in J} S_j$
where $\relcomp I J$ denotes the complement of $J$ relative to $I$.
== Proof ==
We have that both $\ds \bigcup_{j \mathop \in J} S_j$ and $\ds \bigcup_{k \mathop \in \relcomp I J} S_k$ are sets.
Hence by Union is Commutative we have:
:$\bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k = \bigcup_{k \mathop \in \relcomp I J} S_k \cup \bigcup_{j \mathop \in J} S_j$
It remains to be demonstrated that $\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$.
So:
That is:
:$\ds x \in \bigcup_{i \mathop \in I} S_i \iff x \in \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$
The result follows by definition of set equality.
== Also see ==
* General Commutativity of Set Intersection
== Sources ==
* : $\S 9$: Families
Category:Union is Commutative
Union is Dominated by Disjoint Union

== Theorem ==
Let $I$ be an indexing set.
For all $i \in I$, let $S_i$ be a set.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i \preccurlyeq \bigsqcup_{i \mathop \in I} S_i$
where $\preccurlyeq$ denotes domination, $\bigcup$ denotes union, and $\bigsqcup$ denotes disjoint union.
== Proof ==
For all $\ds x \in \bigcup_{i \mathop \in I} S_i$, there exists a $\map i x \in I$ such that $x \in S_{\map i x}$.
Thus the mapping $\ds \iota : \bigcup_{i \mathop \in I} S_i \to \bigsqcup_{i \mathop \in I} S_i$ defined by:
:$\map \iota x = \tuple {x, \map i x}$
is an injection.
Category:Set Theory
Union is Increasing

== Theorem ==
Let $U$ be a set.
Let $\FF$ and $\GG$ be sets of subsets of $U$.
Then $\FF \subseteq \GG \implies \bigcup \FF \subseteq \bigcup \GG$.
That is, $\bigcup$ is an increasing mapping from $\struct {\powerset {\powerset U}, \subseteq}$ to $\struct {\powerset U, \subseteq}$, where $\powerset U$ is the power set of $U$.
== Proof ==
Let $\FF \subseteq \GG$.
Let $x \in \bigcup \FF$.
Then by the definition of union:
:$\exists S \in \FF: x \in S$
By the definition of subset:
:$S \in \GG$
Thus by the definition of union:
:$x \in \bigcup \GG$
Since this holds for all $x \in \bigcup \FF$:
:$\bigcup \FF \subseteq \bigcup \GG$
Category:Set Union
Union is Increasing Sequence of Sets

== Theorem ==
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of sets.
Then: 
:the sequence $\ds \sequence {\bigcup_{k \mathop = 1}^n D_k}_{n \mathop \in \N}$ is increasing. 
== Proof ==
We have: 
:$\ds \bigcup_{k \mathop = 1}^{n + 1} D_k = D_{n + 1} \cup \bigcup_{k \mathop = 1}^n D_k$
From Set is Subset of Union, we have: 
:$\ds \bigcup_{k \mathop = 1}^n D_k \subseteq D_{n + 1} \cup \bigcup_{k \mathop = 1}^n D_k$
so:
:$\ds \bigcup_{k \mathop = 1}^n D_k \subseteq \bigcup_{k \mathop = 1}^{n + 1} D_k$
So:
:$\ds \sequence {\bigcup_{k \mathop = 1}^n D_k}_{n \mathop \in \N}$ is increasing.
Category:Set Union
Category:Increasing Sequences of Sets
Union is Smallest Superset

== Theorem ==
Let $S_1$ and $S_2$ be sets.
Then $S_1 \cup S_2$ is the smallest set containing both $S_1$ and $S_2$.
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$
=== Set of Sets ===
=== General Result ===
== Proof ==
=== Necessary Condition ===
From Union of Subsets is Subset:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$
=== Sufficient Condition ===
Let $\paren {S_1 \cup S_2} \subseteq T$:
Similarly for $S_2$:
That is:
:$\paren {S_1 \cup S_2} \subseteq T \implies \paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T}$
Thus from the definition of equivalence:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$
== Also see ==
* Intersection is Largest Subset
Category:Set Union
Category:Subsets
Union is Smallest Superset/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \iff \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$.
== Proof ==
=== Necessary Condition ===
From Union of Family of Subsets is Subset we have that:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
=== Sufficient Condition ===
Now suppose that $\ds \bigcup_{i \mathop \in I} S_i \subseteq X$.
Consider any $i \in I$ and take any $x \in S_i$.
From Set is Subset of Union of Family we have that:
:$\ds S_i \subseteq \bigcup_{i \mathop \in I} S_i$
Thus:
:$\ds x \in \bigcup_{i \mathop \in I} S_i$
But:
:$\ds \bigcup_{i \mathop \in I} S_i \subseteq X$
So it follows that $S_i \subseteq X$.
So:
:$\ds \bigcup_{i \mathop \in I} S_i \subseteq X \implies \paren {\forall i \in I: S_i \subseteq X}$
Hence:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \iff \bigcup_{i \mathop \in I} S_i \subseteq X$
== Also see ==
* Intersection is Largest Subset: Family of Sets
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.1 \ (2)$
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $4 \ \text{(b)}$
Category:Set Union
Category:Subsets
Category:Indexed Families
Union is Smallest Superset/General Result

== Theorem ==
Let $S$ and $T$ be sets.
Let $\powerset S$ denote the power set of $S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
=== Family of Sets ===
In the context of a family of sets, the result can be presented as follows:
== Proof ==
Let $\mathbb S \subseteq \powerset S$.
By Union of Subsets is Subset: Subset of Power Set:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
Now suppose that $\ds \bigcup \mathbb S \subseteq T$.
Consider any $X \in \mathbb S$ and take any $x \in X$.
From Set is Subset of Union: General Result we have that:
:$\ds X \subseteq \bigcup \mathbb S$
Thus:
:$\ds x \in \bigcup \mathbb S$
But:
:$\ds \bigcup \mathbb S \subseteq T$
So it follows that:
:$X \subseteq T$
So:
:$\ds \bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$
Hence:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
== Also see ==
* Intersection is Largest Subset
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.6 \ \text{(f)}$
* : $\S 1.4$: Exercise $1.4.4 \ \text{(i)}$
Category:Set Union
Category:Subsets
Union is Smallest Superset/Set of Sets

== Theorem ==
Let $T$ be a set.
Let $\mathbb S$ be a set of sets.
Then:
:$\paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
== Proof ==
By Union of Subsets is Subset: Set of Sets:
:$\paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
For the converse implication, suppose that $\bigcup \mathbb S \subseteq T$.
Consider any $X \in \mathbb S$ and take any $x \in X$.
From Set is Subset of Union: Set of Sets we have that $X \subseteq \bigcup \mathbb S$.
Thus $x \in \bigcup \mathbb S$.
But $\bigcup \mathbb S \subseteq T$.
So it follows that $X \subseteq T$.
So:
:$\bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$
Hence:
:$\paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
== Also see ==
* Intersection is Largest Subset
Category:Set Union
Category:Subsets
Union of Antisymmetric Relation with Inverse is Antisymmetric iff Diagonal

== Theorem ==
Let $\RR$ be an antisymmetric relation on a set $S$.
Then:
:$\RR \cup \RR^{-1}$ is antisymmetric
:$\RR = \Delta_S$
where:
:$\RR^{-1}$ denotes the inverse of $\RR$
:$\Delta_S$ denotes the diagonal relation
:$\cup$ denotes set union.
== Proof ==
As asserted, let $\RR$ be an antisymmetric relation.
=== Necessary Condition ===
Let $\RR = \Delta_S$.
From Inverse of Diagonal Relation,
:$\RR^{-1} = \Delta_S$
Hence:
:$\RR \cup \RR^{-1} = \Delta_S$
From Relation is Reflexive Symmetric and Antisymmetric iff Diagonal Relation, we have that the diagonal relation $\Delta_S$ is antisymmetric.
Hence if $\RR = \Delta_S$, then $\RR \cup \RR^{-1}$ is antisymmetric.
=== Sufficient Condition ===
Let $\RR \cup \RR^{-1}$ be antisymmetric.
Let $\tuple {x, y} \in \RR$.
By definition of inverse relation:
:$\tuple {x, y} \in \RR^{-1}$
Thus by definition of set union:
:$\tuple {x, y} \in \RR \cup \RR^{-1}$
and
:$\tuple {y, x} \in \RR \cup \RR^{-1}$
As $\RR \cup \RR^{-1}$ is antisymmetric:
:$x = y$
Hence:
:$\forall \tuple {x, y} \in \RR: x = y$
and it follows that:
:$\RR = \Delta_S$
Category:Diagonal Relation
Category:Antisymmetric Relations
Category:Set Union
Union of Balanced Sets in Vector Space is Balanced

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$. 
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of balanced subsets of $X$.
Then:
:$\ds E = \bigcup_{\alpha \mathop \in I} E_\alpha$ is balanced.
== Proof ==
Let $x \in E$.
Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.
We aim to show that $\lambda x \in E$. 
Since $x \in E$, there exists $\alpha \in I$ with $x \in E_\alpha$. 
Since $E_\alpha$ is balanced, we have $\lambda x \in E_\alpha$. 
So:
:$\ds \lambda x \in \bigcup_{\alpha \mathop \in I} E_\alpha = E$
Since $x$ and $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$ were arbitrary, $E$ is balanced.
Category:Balanced Sets
Category:Set Union
Union of Bijections with Disjoint Domains and Codomains is Bijection

== Theorem ==
Let $A$, $B$, $C$, and $D$ be sets or classes.
Let $A \cap B = C \cap D = \O$.
Let $f: A \to C$ and $g: B \to D$ be bijections.
Then $f \cup g: A \cup B \to C \cup D$ is also a bijection.
=== Corollary ===
== Proof ==
By the definition of bijection, $f$ and $g$ are many-to-one and one-to-many relations.
By Union of Many-to-One Relations with Disjoint Domains is Many-to-One and Union of One-to-Many Relations with Disjoint Images is One-to-Many:
:$f \cup g$ is many-to-one and one-to-many.
Thus to show $f \cup g$ is a bijection requires us only to demonstrate that it is both left-total and right-total.
We will show that $f \cup g$ is left-total.
Let $x \in A \cup B$.
Then $x \in A$ or $x \in B$.
If $x \in A$ then since $f$ is left-total there is a $y \in C$ such that $\tuple {x, y} \in f$.
By the definition of union, $\tuple {x, y} \in f \cup g$.
If $x \in B$ then since $g$ is left-total there is a $y \in D$ such that $\tuple {x, y} \in g$.
Then by the definition of union, $\tuple {x, y} \in f \cup g$.
As this holds for all $x$, $f \cup g$ is left-total.
The proof that $f \cup g$ is right-total is similar.
Thus it has been demonstrated that:
:$f \cup g$ is many-to-one
:$f \cup g$ is one-to-many
:$f \cup g$ is left-total
:$f \cup g$ is right-total
and therefore, by definition, a bijection.
Category:Bijections
Category:Set Union
Category:Union of Bijections with Disjoint Domains and Codomains is Bijection
Union of Bijections with Disjoint Domains and Codomains is Bijection/Corollary

== Corollary to Union of Bijections with Disjoint Domains and Codomains is Bijection ==
Let $A$, $B$, $C$, and $D$ be sets or classes.
Let $A \cap B = C \cap D = \O$.
Let $f: A \to C$ and $g: D \to B$ be bijections.
Then $f \cup g^{-1}: A \cup B \to C \cup D$ is also a bijection.
== Proof ==
By definition of bijection, $g^{-1}: B \to D$ is a bijection.
Hence the result by Union of Bijections with Disjoint Domains and Codomains is Bijection.
Category:Union of Bijections with Disjoint Domains and Codomains is Bijection