nyuuzyou/wikis huggingface dataset:
Union of One-to-Many Relations with Disjoint Images is One-to-Many

== Theorem ==
Let $S_1, S_2, T_1, T_2$ be sets or classes.
Let $\RR_1$ be a one-to-many relation on $S_1 \times T_1$.
Let $\RR_2$ be a one-to-many relation on $S_2 \times T_2$.
Suppose that the images of $\RR_1$ and $\RR_2$ are disjoint.
Then $\RR_1 \cup \RR_2$ is a one-to-many relation on $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
== Proof ==
Let $\QQ = \RR_1 \cup \RR_2$.
Then $\QQ \subseteq \paren {S_1 \times T_1} \cup \paren {S_2 \times T_2} \subseteq \paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
Thus $\QQ$ is a relation on $\paren {S_1 \cup S_2} \times \paren {T_1 \cup T_2}$.
Let $T'_1$ and $T'_2$ be the images of $\RR_1$ and $\RR_2$, respectively.
Let $\tuple {x_1, y}, \tuple {x_2, y} \in \QQ$.
Then $y \in T'_1$ or $y \in T'_2$.
If $y \in T'_1$ then $y \notin T'_2$, so neither $\tuple {x_1, y}$ nor $\tuple {x_2, y}$ is in $\RR_2$, so these pairs are both in $\RR_1$.
As $\RR_1$ is one-to-many, $x_1 = x_2$.
A similar argument leads to the same result for $y \in T'_2$.
As this holds for all such $x_1, x_2, y$: $\QQ$ is a one-to-many relation.
Category:Relation Theory
Union of Open Intervals of Positive Reals is Set of Strictly Positive Reals

== Theorem ==
Let $\R_{>0}$ be the set of strictly positive real numbers.
For all $x \in \R_{>0}$, let $A_x$ be the open real interval $\openint 0 x$.
Then:
:$\ds \bigcup_{x \mathop \in \R_{>0} } A_x = \R_{>0}$
== Proof ==
Let $\ds A = \bigcup_{x \mathop \in \R_{>0} } A_x$.
Let $y \in A$.
Then by definition of union of family:
:$\exists x \in \R_{>0}: y \in A_x$
As $A_x \subseteq \R_{>0}$ it follows by definition of subset that:
:$y \in \R_{>0}$
So:
:$\ds \bigcup_{x \mathop \in \R_{>0} } A_x \subseteq \R_{>0}$
Let $y \in \R_{>0}$.
By the Axiom of Archimedes: 
:$\exists z \in \N: z > y$
and so:
:$y \in A_z$
That is by definition of union of family:
:$y \in \ds \bigcup_{x \mathop \in \R_{>0} } A_x$
So by definition of subset:
:$\ds \R_{>0} \subseteq \bigcup_{x \mathop \in \R_{>0} } A_x$
By definition of set equality:
:$\ds \bigcup_{x \mathop \in \R_{>0} } A_x = \R_{>0}$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $5$
Category:Real Intervals
Union of Open Irreducible Non-Disjoint Subspaces is Irreducible

== Theorem ==
Let $T = \struct {S, \tau}$ be an irreducible topological space.
Let $U$ and $V$ be open irreducible subspaces of $T$.
Let their intersection $U \cap V$ be non-empty.
Then their union $U \cup V$ is an irreducible subspace of $T$.
== Proof ==
In view of the definition of the subspace topology $\tau_{U \cup V}$, it suffices to show that for all closed sets $A_1$ and $A_2$:
:$U \cup V \subseteq A_1 \cup A_2 \implies \exists i_0 \in \set {1, 2} : U \cup V \subseteq A_{i_0}$
To this end, let $A_1$ and $A_2$ closed sets such that:
:$U \cup V \subseteq A_1 \cup A_2$.
Then, in particular:
:$U \subseteq A_1 \cup A_2$.
Since $U$ is irreducible, we have:
:$(1):\quad \exists i_0 \in \set {1,2} : U \subseteq A_{i_0}$
In particular:
Thus:
:$V = \paren {V \setminus U} \cup \paren {V \cap A_{i_0} }$
Here, both $V \setminus U$ and $V \cap A_{i_0}$ are closed in $V$ with respect to the subspace topology $\tau_V$.
Since $U \cap V$ is non-empty, we have:
:$V \setminus U \subsetneq V$
That is, $V \setminus U$ is a proper subset of $V$.
Since $V$ is irreducible, $V \cap A_{i_0}$ is not a proper subset of $V$.
That is:
:$V = V \cap A_{i_0}$
Hence:
:$V \subseteq A_{i_0}$
Together with $(1)$, we conclude:
:$U \cup V \subseteq A_{i_0}$
Category:Irreducible Spaces
Union of Open Sets Not in System of Open Neighborhoods is Complement of Singleton Closure

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $x \in S$.
Let $\map \UU x$ denote the system of open neighborhoods of $x$.
Let $\set x^-$ denote the topological closure of $\set x$.
Then:
:$\bigcup \set{U \in \tau : U \notin \map \UU x} = S \setminus \set x^-$
== Proof ==
From Open Set Not in System of Open Neighborhoods Iff Subset of Complement of Singleton Closure:
:$\forall U \in \tau : \paren{U \notin \map \UU x \iff U \subseteq S \setminus \set x^-}$
From Union of Subsets is Subset:
:$\bigcup \set{U \in \tau : U \notin \map \UU x} \subseteq S \setminus \set x^-$
From Topological Closure is Closed:
:$\set x^-$ is a closed set
From Complement of Closed Set is Open Set:
:$S \setminus \set x^- \in \tau$
Since $x \notin S \setminus \set x^-$ then:
:$S \setminus \set x^- \in \set{U \in \tau : U \notin \map \UU x}$
From Set is Subset of Union:
:$S \setminus \set x^- \subseteq \bigcup \set{U \in \tau : U \notin \map \UU x}$
By definition of set equality:
:$\bigcup \set{U \in \tau : U \notin \map \UU x} = S \setminus \set x^-$
Category:Systems of Open Neighborhoods
Union of Open Sets of Metric Space is Open

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
The union of a set of open sets of $M$ is open in $M$.
== Proof ==
Let $I$ be any indexing set.
Let $U_i$ be open in $M$ for all $i \in I$.
Let $\ds x \in \bigcup_{i \mathop \in I} U_i$.
Then by definition of set union, $x \in U_k$ for some $k \in I$.
Since $U_k$ is open in $M$:
:$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
By Set is Subset of Union:
:$\ds U_k \subseteq \bigcup_{i \mathop \in I} U_i$
Hence by Subset Relation is Transitive:
:$\ds \map {B_\epsilon} x \subseteq \bigcup_{i \mathop \in I} U_i$
and the result follows by definition of open set in metric space.
== Sources ==
* : $2.24 a$
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.4$
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Proposition $2.3.19$
Category:Open Sets (Metric Spaces)
Category:Set Union
Union of Open Sets of Neighborhood Space is Open

== Theorem ==
Let $S$ be a neighborhood space.
Let $I$ be an indexing set.
Let $\family {U_\alpha}_{\alpha \mathop \in I}$ be a family of open sets of $\struct {S, \NN}$ indexed by $I$.
Then their union $\ds \bigcup_{\alpha \mathop \in I} U_i$ is an open set of $\struct {S, \NN}$.
== Proof ==
Let $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
Then $x \in U_\beta$ for some $\beta \in I$.
By definition of open set, $U_\beta$ is a neighborhood of $x$.
But from Set is Subset of Union:
:$\ds U_\beta \subseteq x \in \bigcup_{\alpha \mathop \in I} U_\alpha$
By neighborhood space axiom $N 3$ it follows that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is a neighborhood of $x$.
As $x$ is arbitrary, it follows that the above is true for all $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
It follows by definition that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is an open set of $\struct {S, \NN}$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$
Category:Neighborhood Spaces
Union of Open Sets of Normed Vector Space is Open

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
The union of a set of open sets of $M$ is open in $M$.
== Proof ==
Let $I$ be any indexing set.
Let $U_i$ be open in $M$ for all $i \in I$.
Let $\ds x \in \bigcup_{i \mathop \in I} U_i$.
Then $x \in U_k$ for some $k \in I$.
Since $U_k$ is open in $M$:
:$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k \subseteq \bigcup_{i \mathop \in I} U_i$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
The result follows.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Open Sets (Normed Vector Spaces)
Category:Set Union
Union of Orderings is not necessarily Ordering

== Theorem ==
Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$.
Then $\RR \cup \SS$ is not necessarily an ordering on $A$.
== Proof ==
Let $\RR$ be an ordering as asserted.
Let $\SS$ be the dual ordering of $\RR$.
From Dual Ordering is Ordering, $\SS$ is an ordering.
By definition, the dual of $\RR$ is the inverse of $\RR$:
:$\SS = \RR^{-1}$
From Union of Antisymmetric Relation with Inverse is Antisymmetric iff Diagonal, it is generally not the case that $\RR \cup \RR^{-1}$ is antisymmetric.
Hence, in general, $\RR \cup \RR^{-1}$ is not an ordering.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.27 \ \text {(c)}$
Category:Orderings
Category:Set Union
Union of Ordinal is Subset of Itself

== Theorem ==
Let $\alpha$ be an ordinal.
Then:
:$\bigcup \alpha \subseteq \alpha$
where $\bigcup \alpha$ denotes the union of $\alpha$.
== Proof ==
Let $x \in \bigcup \alpha$.
Then:
:$\exists \beta \in \alpha: x \in \beta$
By Element of Ordinal is Ordinal, $\beta$ is an ordinal.
Thus:
:$x \in \beta$ and $\beta \in \alpha$
Hence by Ordinal Membership is Transitive:
:$x \in \alpha$
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.20 \ (1)$
Category:Ordinals
Category:Set Union
Category:Subsets
Union of Ordinals is Least Upper Bound

== Theorem ==
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 $A$:
:$\ds \forall x \in A: x \le A$
:$\ds \forall y \in A: y \le x \implies \bigcup A \le x$
== Proof ==
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 of $A$.
We shall denote $<$ for ordering on the ordinal numbers.
By Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, $<$ is the same as both $\in$ and $\subsetneq$.
Then:
Thus, by definition of subset:
:$\ds \bigcup A \subseteq x$
Therefore:
:$\ds \forall y \in A: y \le x \implies \bigcup A \le x$
== Sources ==
* : $\S 7.20$, $\S 7.21$
Category:Ordinals
Union of Overlapping Convex Sets in Toset is Convex

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $U$ and $V$ be convex sets in $S$.
Let $U \cap V \ne \O$.
Then $U \cup V$ is also convex.
=== Infinite Union ===
== Proof ==
Let $a,b,c \in S$
Let $a,c \in U \cup V$.
Let $a \prec b \prec c$.
If $a, c \in U$ then $b \in U$ because $U$ is convex.
Thus $b \in U \cup V$ by the definition of union.
Similarly, if $a,c \in V$ then $b \in U \cup V$.
Otherwise, , suppose that $a \in U$ and $c \in V$.
Since $U \cap V$ is nonempty by the premise, it has an element $p$.
Since $\preceq$ is a total ordering:
:$b \preceq p$ or $p \preceq b$.
If $b \preceq p$, then since $a \prec b$, $a,p \in U$, and $U$ is convex, we can conclude that
:$b \in U$
so $b \in U \cup V$.
A similar argument shows that it $p \preceq b$ then $b \in V$, so $b \in U \cup V$.
Thus in all cases we can conclude that $b \in U \cup V$, so $U \cup V$ is convex.
Category:Total Orderings
Union of Overlapping Convex Sets in Toset is Convex/Infinite Union

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\AA$ be a set of convex subsets of $S$.
For any $P, Q \in \AA$, let there be elements $C_0, \dotsc, C_n \in \AA$ such that:
:$C_0 = P$
:$C_n = Q$
:For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
Then $\bigcup \AA$ is convex in $S$.
== Proof ==
Let $a, c \in \bigcup \AA$.
Let $b \in S$.
Let $a \prec b \prec c$.
Since $a, c \in \bigcup \AA$, there are $P, Q \in \AA$ such that $a \in P$ and $c \in Q$.
By the premise, there are elements $C_0, \dots, C_n \in \AA$ such that:
:$C_0 = P$
:$C_n = Q$
:For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
Applying Union of Overlapping Intervals is Interval inductively:
:$\ds \bigcup_{k \mathop = 0}^n C_k$ is convex.
Since $\ds a, c \in \bigcup_{k \mathop = 0}^n C_k$, by the definition of convexity:
:$\ds b \in \bigcup_{k \mathop = 0}^n C_k$
Thus:
:$\ds b \in \bigcup \AA$
Since this holds for all such triples $a, b, c$, it follows that $\AA$ is convex.
Category:Total Orderings
Union of Path-Connected Sets with Common Point is Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {B_\alpha}_{\alpha \mathop \in A}$ be a family of path-connected sets of $T$.
Let $\exists x \in \ds \bigcap \family {B_\alpha}_{\alpha \mathop \in A}$.
Then
:$\ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$ is a path-connected set of $T$.
== Proof ==
Let $B = \ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$.
Let $a, b \in B$.
Then 
:$\exists \alpha, \beta \in A: a \in B_\alpha \land b \in B_\beta$.
As $B_\alpha$ is a path-connected set in $T$ then $a$ and $x$ are path-connected points.
Similarly, $x$ and $b$ are path-connected points.
From Joining Paths makes Another Path, $a$ and $b$ are path-connected points.
Since $a$ and $b$ were arbitrary points then $B$ is a path-connected set of $T$.
== Sources ==
* : $\S 4$ Connectedness and Compactness, Proposition $4.9 \ \text {(d)}$
Category:Connected Spaces
Union of Power Sets

== Theorem ==
The union of the power sets of two sets $S$ and $T$ is a subset of the power set of their union:
:$\powerset S \cup \powerset T \subseteq \powerset {S \cup T}$
=== Union of Power Sets not always Equal to Powerset of Union ===
Equality does not hold in general:
== Proof ==
== Also see ==
* Intersection of Power Sets
== Sources ==
* : $\S 5$: Complements and Powers
* : $\S 2$. Sets of sets: Exercise $5 \ \text{(b)}$
* : Chapter $1$: Sets and Logic: Exercise $7 \ \text{(i)}$
Category:Power Set
Category:Set Union
Category:Subsets
Union of Power Sets not always Equal to Powerset of Union

== Theorem ==
The union of the power sets of two sets $S$ and $T$ is not necessarily equal to the power set of their union.
== Proof ==
Proof by Counterexample:
Let $S = \set {1, 2, 3}, T = \set {2, 3, 4}, X = \set {1, 2, 3, 4}$.
But note that $X \nsubseteq S \land X \nsubseteq T$.
Thus:
So:
:$\powerset {S \cup T} \ne \powerset S \cup \powerset T$
== Sources ==
* : $\S 2$. Sets of sets: Exercise $5 \ \text{(b)}$
* : Chapter $1$: Sets and Logic: Exercise $7$
Category:Power Set
Category:Set Union
Category:Subsets
Union of Primitive Recursive Sets

== Theorem ==
Let $A, B \subseteq \N$ be subsets of the set of natural numbers $\N$.
Let $A$ and $B$ both be primitive recursive.
Then $A \cup B$, the union of $A$ and $B$, is primitive recursive.
== Proof ==
$A$ and $B$ are primitive recursive, therefore so are their characteristic functions $\chi_A$ and $\chi_B$.
Let $n \in \N$ be a natural number.
Then $n \in A \cup B \iff \chi_A \left({n}\right) + \chi_B \left({n}\right) > 0$.
So:
Thus $A \cup B$ is defined by substitution from the primitive recursive functions $\operatorname{sgn}$, $\operatorname{add}$, $\chi_A$ and $\chi_B$.
Hence the result.
Category:Primitive Recursive Functions
Category:Set Union
Union of Real Intervals is not necessarily Real Interval

== Theorem ==
Let $I_1$ and $I_2$ be real intervals.
Then $I_1 \cup I_2$ is not necessarily a real interval.
== Proof ==
Proof by Counterexample:
Consider the real intervals:
:$I_1 = \left({0 \,.\,.\, 2}\right)$
:$I_2 = \left({4 \,.\,.\, 6}\right)$
Then we have that:
:$1 < 3 < 5$
where:
:$1 \in I_1 \cup I_2$
:$5 \in I_1 \cup I_2$
but:
:$3 \notin I_1 \cup I_2$
Thus $I_1 \cup I_2$ is not a real interval.
== Sources ==
* : $\S 2.1$: Sets: Exercise $\text{B} \ 7$
Category:Real Intervals
Union of Recursively Enumerable Sets

== Theorem ==
Let $S, T \subseteq \N$ be recursively enumerable sets.
Then $S \cup T$ is recursively enumerable.
== Proof ==
Suppose $S = \empty$.
Then by Union with Empty Set:
:$S \cup T = T$
which is recursively enumerable by assumption.
Suppose $T = \empty$.
For the same reason:
:$S \cup T = S$
which is also recursively enumerable by assumption.
Now, suppose neither $S$ nor $T$ is empty.
By Recursively Enumerable Set is Image of Primitive Recursive Function/Corollary, there exist primitive recursive $f, g : \N \to \N$ such that:
:$S = \Img f$
:$T = \Img g$
Define:
:$\map h {x, c} = \begin{cases}
\map f x & : c = 0 \\
\map g x & : \text{otherwise}
\end{cases}$
By:
* Equality Relation is Primitive Recursive
* Definition by Cases is Primitive Recursive/Corollary
$h$ is a primitive recursive function.
By Primitive Recursive Function is Total Recursive Function, it is thus a recursive function.
It remains to be shown that $S \cup T = \Img h$.
Let $x \in S \cup T$.
Then $x \in \Img f$ or $x \in \Img g$.
If $x \in \Img f$, then there is some $z \in \N$ such that $\map f z = x$.
Therefore, $\map h {z, 0} = x$.
Likewise, if $x \in \Img g$, there is some $z \in \N$ such that $\map g z = x$.
Therefore, $\map h {z, 1} = x$.
Thus, $S \cup T \subseteq \Img h$.
Now, $x \in \Img h$.
Then, there are $z, c \in \N$ such that:
:$\map h {z, c} = x$
Depending on $c$, either:
:$x = \map f z$
or:
:$x = \map g z$
In either case, $x \in \Img f \cup \Img g$.
Therefore, $\Img h \subseteq S \cup T$.
Hence, by definition of set equality:
:$S \cup T = \Img h$
Therefore, by definition, $S \cup T$ is recursively enumerable.
Category:Recursion Theory
Union of Reflexive Relations is Reflexive

== Theorem ==
The union of two reflexive relations is also a reflexive relation.
== Proof ==
Let $\RR_1$ and $\RR_2$ be reflexive relations on a set $S$.
From Relation Contains Diagonal Relation iff Reflexive, we have that:
: $\Delta_S \subseteq \RR_1$
: $\Delta_S \subseteq \RR_2$
Hence from Subset Relation is Transitive:
: $\Delta_S \subseteq \RR_1 \cup \RR_2$
Hence the result, by definition of reflexive relation.
Category:Reflexive Relations
Category:Set Union
Union of Regular Open Sets is not necessarily Regular Open

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ and $V$ be regular open sets of $T$.
Then $U \cup V$ is not also necessarily a regular open set of $T$.
== Proof ==
Proof by Counterexample:
By Open Real Interval is Regular Open, the open real intervals:
:$\openint 0 {\dfrac 1 2}, \openint {\dfrac 1 2} 1$
are both regular open sets of $\R$.
Consider $A$, the union of the adjacent open intervals:
:$A := \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1$
From Interior of Closure of Interior of Union of Adjacent Open Intervals:
:$A^{- \circ} = \openint 0 1$
Thus $A$ is not a regular open set of $\R$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(c)}$
Category:Regular Open Sets
Category:Open Sets
Category:Set Union
Union of Relation with Inverse is Symmetric Relation

== Theorem ==
Let $\RR$ be a relation on a set $S$.
Let $\RR^{-1}$ denote the inverse of $\RR$.
Then $\RR \cup \RR^{-1}$, the union of $\RR$ with $\RR^{-1}$, is symmetric.
== Proof ==
Let $\tuple {a, b} \in \RR \cup \RR^{-1}$.
By definition of union, either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {a, b} \in \RR^{-1}$
;Case 1
If $\tuple {a, b} \in \RR$, then by definition of inverse relation:
:$\tuple {b, a} \in \RR^{-1}$
But from Set is Subset of Union:
:$\tuple {b, a} \in \RR \cup \RR^{-1}$
;Case 2
If $\tuple {a, b} \in \RR^{-1}$, then by definition of inverse relation:
:$\tuple {b, a} \in \paren {\RR^{-1} }^{-1}$
From Inverse of Inverse Relation:
:$\tuple {b, a} \in \RR$
But from Set is Subset of Union:
:$\tuple {b, a} \in \RR \cup \RR^{-1}$
Hence by Proof by Cases:
:$\tuple {a, b} \in \RR \cup \RR^{-1} \implies \tuple {b, a} \in \RR \cup \RR^{-1}$
and $\RR \cup \RR^{-1}$ is symmetric by definition.
Category:Set Union
Category:Inverse Relations
Category:Symmetric Relations
Union of Relations Compatible with Operation is Compatible

== Theorem ==
Let $\struct {S, \circ}$ be a closed algebraic structure.
Let $\FF$ be a family of relations on $S$.
Let each element of $\FF$ be compatible with $\circ$.
Let $\QQ = \bigcup \FF$.
Then $Q$ is a relation compatible with $\circ$.
== Proof ==
Let $x, y, z \in S$.
Let $x \mathrel \QQ y$.
Then for some $\RR \in \FF$:
:$x \mathrel \RR y$.
Since $\RR$ is a relation compatible with $\circ$:
:$\tuple {x \circ z} \mathrel \RR \tuple {y \circ z}$
Since $\RR \in \FF$:
:$\RR \subseteq \bigcup \FF = \QQ$
Thus 
:$\tuple {x \circ z} \mathrel \QQ \tuple {y \circ z}$
We have shown that:
:$\forall x, y, z \in S: x \mathrel \QQ y \implies \tuple {x \circ z} \mathrel \QQ \tuple {y \circ z}$
A similar argument shows that:
:$\forall x, y, z \in S: x \mathrel \QQ y \implies \tuple {z \circ x} \mathrel \QQ \tuple {z \circ y}$
So $Q$ is a relation compatible with $\circ$.
Category:Compatible Relations
Union of Relations is Relation

== Theorem ==
Let $S$ and $T$ be sets.
Let $\FF$ be a family of relations from $S$ to $T$.
Let $\ds \RR = \bigcup \FF$, the union of all the elements of $\FF$.
 
Then $\RR$ is a relation from $S$ to $T$.
== Proof ==
By the definition of a relation from $S$ to $T$, each element of $\FF$ is a subset of $S \times T$.
By Union of Subsets is Subset: Set of Sets:
:$\RR \subseteq S \times T$
Therefore, by the definition of a relation from $S$ to $T$, $\RR$ is a relation from $S$ to $T$.
Category:Relation Theory
Union of Relative Complements of Nested Subsets

== Theorem ==
Let $R \subseteq S \subseteq T$ be sets with the indicated inclusions.
Then:
:$\relcomp T S \cup \relcomp S R = \relcomp T R$
where $\complement$ denotes relative complement.
Phrased via Set Difference as Intersection with Relative Complement:
:$\paren {T \setminus S} \cup \paren {S \setminus R} = T \setminus R$
where $\setminus$ denotes set difference.
== Proof ==
From Union with Set Difference:
:$T = T \setminus S \cup S$
and therefore by Set Difference is Right Distributive over Union:
:$T \setminus R = \paren {\paren {T \setminus S} \setminus R} \cup \paren {S \setminus R}$
Now, by Set Difference with Union and Union with Superset is Superset:
:$\paren {T \setminus S} \setminus R = T \setminus \paren {S \cup R} = T \setminus S$
Combining the above yields:
:$T \setminus R = \paren {T \setminus S} \cup \paren {S \setminus R}$
== Also see ==
* Set Difference is Subset of Union of Differences is the inclusion retained when the subset conditions are dropped
Category:Relative Complement
Category:Set Union
Union of Right-Total Relations is Right-Total

== Theorem ==
Let $S_1, S_2, T_1, T_2$ be sets or classes.
Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be right-total relations.
Then $\RR_1 \cup \RR_2$ is right-total.
== Proof ==
Define the predicates $L$ and $R$ by:
:$\map L X \iff \text {$X$ is left-total}$
:$\map R X \iff \text {$X$ is right-total}$
== Also see ==
* Union of Left-Total Relations is Left-Total
Category:Relation Theory
Union of Set of Dense-in-itself Sets is Dense-in-itself

== Theorem ==
Let $T$ be a topological space.
Let $\FF \subseteq \powerset T$ such that 
:every element of $\FF$ is dense-in-itself.
Then the union $\bigcup \FF$ is also dense-in-itself.
== Proof ==
By Dense-in-itself iff Subset of Derivative:
:$\forall A \in \FF: A \subseteq A'$
where $A'$ denotes the derivative of $A$.
Then by Set Union Preserves Subsets:
:$\ds \bigcup \FF \subseteq \bigcup_{A \mathop \in \FF} A'$
By Union of Derivatives is Subset of Derivative of Union:
:$\ds \bigcup_{A \mathop \in \FF} A' \subseteq \paren {\bigcup \FF}'$
Then by Subset Relation is Transitive:
:$\ds \bigcup \FF \subseteq \paren {\bigcup \FF}'$
The result follows by Dense-in-itself iff Subset of Derivative.
== Sources ==
* 
* 
Category:Denseness
Category:Set Derivatives
Union of Set of Ordinals is Ordinal/Corollary

== Corollary to Union of Set of Ordinals is Ordinal ==
Let $y$ be a set.
Let $\On$ be the class of all ordinals.
Let $F: y \to \On$ be a mapping.
Then:
:$\bigcup \map F y \in \On$
where $\map F y$ is the image of $y$ under $F$.
== Proof 1 ==
== Proof 2 ==
Category:Union of Set of Ordinals is Ordinal
Union of Set of Ordinals is Ordinal/Corollary/Proof 1

== Corollary to Union of Set of Ordinals is Ordinal ==
== Proof ==
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.
By the epsilon relation $\bigcup \map F y$ is well-ordered.
Thus $\bigcup \map F y$ is a member of $\On$, the class of all ordinals.
Category:Union of Set of Ordinals is Ordinal