nyuuzyou/wikis huggingface dataset:
Union of Set of Ordinals is Ordinal/Corollary/Proof 2

== Corollary to Union of Set of Ordinals is Ordinal ==
== Proof ==
So:
:$\ds \bigcup_{z \mathop \in y} \map F z \subseteq \On$
So $\ds \bigcup_{z \mathop \in y} \map F z$ is a transitive set.
Therefore $\ds \bigcup_{z \mathop \in y} \map F z$ is an ordinal.
:$\ds \bigcup_{z \mathop \in y} \map F z = \bigcup \Img y$
Let $U$ denote the universal class.
Therefore $\ds \bigcup_{z \mathop \in y} \map F z$ is a set, so it is a member of the class of all ordinals.
Category:Union of Set of Ordinals is Ordinal
Union of Set of Ordinals is Ordinal/Proof 1

== Theorem ==
== Proof ==
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.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Corollary $1.4$
Category:Union of Set of Ordinals is Ordinal
Union of Set of Ordinals is Ordinal/Proof 2

== Theorem ==
== Proof ==
From this, we conclude that $\bigcup A$ is a transitive set.
From Class is Transitive iff Union is Subclass, it follows that:
:$\bigcup A \subseteq A \subseteq \On$
Note that the above applies as well to sets as it does to classes.
By Subset of Well-Ordered Set is Well-Ordered, $A$ is also well-ordered by $\Epsilon$.
Thus by Alternative Definition of Ordinal, $\bigcup A$ is an ordinal.
== Sources ==
* : $\S 7.19$
Category:Union of Set of Ordinals is Ordinal
Union of Set of Sets is Non-empty iff some Set is Non-empty

== Theorem ==
Let $\SS$ be a set of sets.
Then:
:$\ds \bigcup \SS \ne \O \iff \exists S \in \SS: S \ne \O$
== Proof ==
Follows immediately from Union of Set of Sets is Empty iff Sets are Empty.
Category:Set Union
Category:Empty Set
Union of Set of Sets when a Set Intersects All

== Theorem ==
Let $F$ be a set of sets.
Let $S$ be a set or class.
Suppose that:
:$\forall A \in F: A \cap S \ne \O$
Then:
:$\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
== Proof ==
Suppose that $B \in F$.
Then $B \cap S$ has an element $x_B$.
Thus:
:$B \in \set {A \in F: x_B \in A}$
By the definition of union:
:$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
Suppose instead that:
:$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
Then by the definition of union, there exists an $x_B \in S$ such that:
:$B \in \set {A \in F: x_B \in A} \subseteq F$
Thus $B \in F$.
We have shown that:
:$\ds \forall B: \paren {B \in F \iff B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A} }$
Therefore by definition of set equality:
:$\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$
Category:Set Union
Union of Set of Singletons

== Theorem ==
Let $S$ be a set.
Let $T = \set {\set x: x \in S}$ be the set of all singletons of elements of $S$.
Then:
:$\ds \bigcup T = S$
where $\ds \bigcup T$ denotes the union of $T$.
== Proof ==
=== Union of $T$ Subset $S$ ===
Let $\ds x \in \bigcup T$.
By definition of union:
:$\exists A \in T: x \in A$
By definition of $T$:
:$\exists y \in S: A = \set y$
Then by definition of singleton:
:$x = y$
Thus $x \in S$.
=== $S$ Subset Union of $T$ ===
Let $x \in S$.
By definition of $T$:
:$\set x \in T$
By Set is Subset of Union/Set of Sets:
:$\ds \set x \subseteq \bigcup T$
By definition of singleton:
:$x \in \set x$
Thus by definition of subset:
:$\ds x \in \bigcup T$
Thus by definition of set equality:
:$\ds \bigcup T = S$
== Sources ==
* 
Category:Set Union
Category:Singletons
Union of Sigma-Algebras may not be Sigma-Algebra

== Theorem ==
Let $X$ be a set.
Let $\struct {X, \Sigma_1}$ and $\struct {X, \Sigma_2}$ be $\sigma$-algebras.
Then $\Sigma_1 \cup \Sigma_2$ may not be a $\sigma$-algebra on $X$. 
== Proof ==
Let $X = \set {1, 2, 3}$. 
Let: 
:$\Sigma_1 = \set {\O, \set 1, \set {2, 3}, \set {1, 2, 3} }$
and:
:$\Sigma_2 = \set {\O, \set 2, \set {1, 3}, \set {1, 2, 3} }$
Then $\Sigma_1$ and $\Sigma_2$ are $\sigma$-algebras on $X$.
We have: 
:$\Sigma_1 \cup \Sigma_2 = \set {\O, \set 1, \set 2, \set {1, 3}, \set {2, 3}, \set {1, 2, 3} }$
However, this does not contain: 
:$\set 1 \cup \set 2 = \set {1, 2}$
Since $\sigma$-algebras are closed under countable union, $\Sigma_1 \cup \Sigma_2$ therefore cannot be a $\sigma$-algebra. 
Category:Sigma-Algebras
Union of Simply Connected Sets with Path-Connected Intersection is Simply Connected

== Theorem ==
Let $\struct {T, \tau}$ be a topological space.
Consider all subsets of $T$ as subspaces, equipped with the subspace topology induced by $\tau$.
Let $U$ and $V$ be open subsets of $T$ that are simply connected.
Let $U \cap V$ be non-empty and path-connected.
Let $U \cup V = T$.
Then $\struct {T, \tau}$ is simply connected.
== Proof ==
== Sources ==
* : $9$: The Fundamental Group: $\S 59$: The Fundamental Group of $S^n$
Category:Simply Connected Spaces
Union of Singleton

== Theorem ==
Consider the set of sets $A$ such that $A$ consists of just one set $x$:
:$A = \set x$
Then the union of $A$ is $x$:
:$\bigcup A = x$
== Proof ==
Let $A = \set x$.
From the definition of set union:
:$\bigcup \set x = \set {y: \exists z \in \set x: y \in z}$
from which it follows directly that:
:$\bigcup \set x = \set {y: y \in x}$
as $x$ is the only set in $\set x$.
That is:
:$\bigcup A = x$
== Also see ==
* Intersection of Singleton
== Sources ==
* : $\S 4$: Unions and Intersections
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.2: \ \text{(i)}$
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.5. \ \text {(a)}$
Category:Set Union
Category:Singletons
Union of Slow g-Tower is Well-Orderable

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a slowly progressing mapping on $M$.
Let $M$ be a slow $g$-tower.
Then $\ds \bigcup M$ is well-orderable.
== Proof ==
Let $M$ be a slow $g$-tower.
From Slow $g$-Tower is Slowly Well-Ordered under Subset Relation we have that $M$ is slowly well-ordered under $\subseteq$.
The result follows from Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Theorem $4.5$
Category:G-Towers
Category:Well-Orderings
Category:Class Union
Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable

== Theorem ==
Let $N$ be a class.
Let $N$ be slowly well-ordered under the subset relation.
Then $\bigcup N$ is well-orderable.
== Proof ==
Let $N$ be slowly well-ordered under $\subseteq$.
Let $A = \bigcup N$.
For $a \in A$, let $\map F a$ denote the smallest element of $N$ that contains $a$.
Then for $a, b \in A$, we define $a \preccurlyeq b \iff \map F a \subseteq \map F b$.
It will be shown that $\preccurlyeq$ is a well-ordering on $A$.
By Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable, it is sufficient to show that $F$ is an injection.
So, let us assume that $\map F a = \map F b$ for arbitrary $a, b \in A$.
By condition $S_1$ of the definition of slowly well-ordered under $\subseteq$, $\O \in N$ is the smallest element of $N$.
We have that $a \in \map F a$ by definition.
Hence $\map F a$ cannot be the smallest element of $N$.
Suppose $\map F a$ were a limit element of $N$.
Then by condition $S_3$ of the definition of slowly well-ordered under $\subseteq$, $\map F a$ would be the union of its lower section $\map \bigcup {\paren {\map F a}^\subset}$
But no element of the lower section of $\map F a$ can contain $a$.
Hence $\map \bigcup {\paren {\map F a}^\subset}$ cannot contain $a$.
Hence $\map F a$ cannot be a limit element of $N$.
By Categories of Elements under Well-Ordering, $\map F a$ must therefore be the immediate successor element of some $x \in N$.
As $a$ is the immediate predecessor of $x$, it follows by the definition of $\map F a$ that $a \notin x$.
However, by condition $S_2$ of the definition of slowly well-ordered under $\subseteq$, $\map F a$ contains exactly $1$ more element than $x$.
Hence as $a \in \map F a$ but $a \notin x$ it must be the case that $a$ must be that $1$ more element.
Hence:
:$\map F a = x \cup \set a$
As $\map F a = \map F b$ :
:$\map F b = x \cup \set a$
As $\map F b$ is the smallest element of $N$ that contains $b$, we have:
:$b \ne x$
Hence:
:$b \in \set a$
which means:
:$b = a$
We have shown that:
:$\map F a = \map F b \implies a = b$
and so by definition $F$ is an injection.
Thus, by Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable, the result follows.
The following results also hold:
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
=== Corollary 4 ===
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Theorem $4.4$
Category:Well-Orderings
Category:Subset Relation
Category:Class Union
Category:Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable
Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable/Corollary 1

== Corollary to Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable ==
Let $N$ be a class.
Let $N$ be slowly well-ordered under the subset relation.
For $a \in \ds \bigcup N$, let $\map F a$ denote the smallest element of $N$ that contains $a$.
For $a, b \in \ds \bigcup N$, we define $a \preccurlyeq b \iff \map F a \subseteq \map F b$.
We have that:
:$a \preccurlyeq b \iff \forall x \in N: b \subseteq x \implies a \subseteq x$
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Exercise $4.1 \ \text{(a)}$
Category:Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable
Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable/Corollary 2

== Corollary to Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable ==
Let $N$ be a class.
Let $N$ be slowly well-ordered under the subset relation.
For $a \in \bigcup N$, let $\map F a$ denote the smallest element of $N$ that contains $a$.
For $a, b \in \bigcup N$, we define $a \preccurlyeq b \iff \map F a \subseteq \map F b$.
We have that:
:$\forall x \in N, a \in \bigcup N: x \in \map F a \implies x \preccurlyeq a$
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Exercise $4.1 \ \text{(b)}$
Category:Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable
Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable/Corollary 3

== Corollary to Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable ==
Let $N$ be a class.
Let $N$ be slowly well-ordered under the subset relation.
For $a \in \bigcup N$, let $\map F a$ denote the smallest element of $N$ that contains $a$.
For $a, b \in \bigcup N$, we define $a \preccurlyeq b \iff \map F a \subseteq \map F b$.
In the above, $\bigcup N$ means the union of $N$.
We have that:
:$\forall a \in \bigcup N:$ if $a$ is not the greatest element of $\bigcup N$, then the immediate successor of $a$ is the smallest element of $\bigcup N \setminus \map F a$.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Exercise $4.1 \ \text{(c)}$
Category:Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable
Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable/Corollary 4

== Corollary to Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable ==
Let $N$ be a class.
Let $N$ be slowly well-ordered under the subset relation.
For $a \in \ds \bigcup N$, let $\map F a$ denote the smallest element of $N$ that contains $a$.
For $a, b \in \ds \bigcup N$, we define $a \preccurlyeq b \iff \map F a \subseteq \map F b$.
We have that:
:Every immediate successor element of $N$ is $\map F a$ for some $a \in N$.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Exercise $4.1 \ \text{(d)}$
Category:Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable
Union of Subclass is Subclass of Union of Class

== Theorem ==
Let $A$ and $B$ be classes.
Let $\bigcup A$ and $\bigcup B$ denote the union of $A$ and union of $B$ respectively.
Let $A$ be a subclass of $B$:
:$A \subseteq B$
Then $\bigcup A$ is a subclass of $\bigcup B$:
:$\bigcup A \subseteq \bigcup B$
== Proof ==
Let $x \in \bigcup A$.
Then:
:$\exists y \in A: x \in y$
But as $A \subseteq B$ it follows that $y \in B$.
That is:
:$\exists y \in B: x \in y$
That is:
:$x \in \bigcup B$
Hence the result by definition of subclass.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.2. \ \text {(a)}$
Category:Class Union
Union of Subgroups

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H, K \le G$ be subgroups of $G$.
Let neither $H \subseteq K$ nor $K \subseteq H$.
Then $H \cup K$ is not a subgroup of $G$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
As neither $H \subseteq K$ nor $K \subseteq H$, it follows from Set Difference with Superset is Empty Set‎ that neither $H \setminus K = \O$ nor $K \setminus H = \O$.
So, let $h \in H \setminus K, k \in K \setminus H$.
Thus, $h \notin K, k \notin H$.
If $\struct {H \cup K, \circ}$ is a group, then it must be closed.
If $\struct {H \cup K, \circ}$ is closed, then $h \circ k \in H \cup K \implies h \circ k \in H \lor h \circ k \in K$.
If $h \circ k \in H$ then $h^{-1} \circ h \circ k \in H \implies k \in H$.
If $h \circ k \in K$ then $h \circ k \circ k^{-1} \in K \implies h \in K$.
So $h \circ k$ can be in neither $H$ nor $K$.
Therefore $\struct {H \cup K, \circ}$ is not closed.
Therefore $H \cup K$ is not a subgroup of $G$.
== Examples ==
== Sources ==
* : $\S 5.2$. Subgroups: Example $94$
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.13$
* : $\S 1.9$
* : Chapter $6$: An Introduction to Groups: Exercise $7$
Category:Subgroups
Category:Set Union
Category:Union of Subgroups
Union of Subgroups/Corollary 1

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H, K \le G$.
Let $H \cup K$ be a subgroup of $G$.
Then either $H \subseteq K$ or $K \subseteq H$.
== Proof ==
 neither $H \subseteq K$ nor $K \subseteq H$.
Then from Union of Subgroups it follows that $H \cup K$ is not a subgroup of $G$.
The result follows by Proof by Contradiction.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $7$
Category:Union of Subgroups
Union of Subgroups/Corollary 2

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H, K \le G$.
Let $H \vee K$ be the join of $H$ and $K$.
Then $H \vee K = H \cup K$ $H \subseteq K$ or $K \subseteq H$.
== Proof ==
From the definition of join, $H \vee K$ is the smallest subgroup of $G$ containing $H \cup K$.
The result follows from Union of Subgroups.
== Sources ==
* : $\S 1.9$: Exercise $5.8$
Category:Union of Subgroups
Union of Subgroups/Examples/Subgroups of S3

== Examples of Union of Subgroups ==
Let $S_3$ denote the Symmetric Group on $3$ Letters, whose Cayley table is given as:
Consider the subgroups $H, K \le G$:
:$H = \set {e, \tuple {12} }$
:$K = \set {e, \tuple {13} }$
We have that:
:$H \cup K = \set {e, \tuple {12}, \tuple {13} }$
and:
:$\tuple {12} \circ \tuple {13} = \tuple {123}$
But $\tuple {123} \notin H \cup K$.
Hence $H \cup K$ is not closed and so is not a group.
The result follows by definition of subgroup.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{J}$
* : Chapter $4$: Subgroups: Exercise $2$
Category:Union of Subgroups
Union of Subset of Family is Subset of Union of Family

== Theorem ==
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $J \subseteq I$
Then:
:$\ds \bigcup_{\alpha \mathop \in J} A_\alpha \subseteq \bigcup_{\alpha \mathop \in I} A_\alpha$
where $\ds \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.
== Proof ==
== Also see ==
* Intersection of Family is Subset of Intersection of Subset of Family
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $3 \ \text{(a)}$
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $3 \ \text{(b)}$
Category:Subsets
Category:Set Union
Category:Indexed Families
Union of Subsets is Subset

== Theorem ==
Let $S_1$, $S_2$, and $T$ be sets.
Let $S_1$ and $S_2$ both be subsets of $T$.
Then:
:$S_1 \cup S_2 \subseteq T$
That is:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$
=== Set of Sets ===
=== Subset of Power Set ===
=== Family of Sets ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B vii}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $3$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(iii)}$
Category:Set Union
Category:Subsets
Category:Union of Subsets is Subset
Union of Subsets is Subset/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$.
== Proof ==
Suppose that $\forall i \in I: S_i \subseteq X$.
Consider any $\ds x \in \bigcup_{i \mathop \in I} S_i$.
By definition of set union:
:$\exists i \in I: x \in S_i$
But as $S_i \subseteq X$ it follows that $x \in X$.
Thus it follows that:
:$\ds \bigcup_{i \mathop \in I} S_i \subseteq X$
So:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.4 \ \text{(i)}$
Category:Union of Subsets is Subset
Union of Subsets is Subset/Proof 1

== Theorem ==
== Proof ==
Let:
:$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T}$
Then:
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(g)}$
Category:Union of Subsets is Subset
Union of Subsets is Subset/Proof 2

== Theorem ==
== Proof ==
Let $x \in S_1 \cup S_2$.
By the definition of union, either $x \in S_1$ or $x \in S_2$.
By hypothesis, $S_1 \subseteq T$ and $S_2 \subseteq T$.
By definition of subset:
:$x \in S_1 \implies x \in T$
:$x \in S_2 \implies x \in T$
By Proof by Cases it follows that $x \in T$.
Hence the result by definition of subset.
== Sources ==
* : $\S 1.4$. Union: Example $15$
* : $\S 7.2 \ \text{(ii)}$: Unions and Intersections
Category:Union of Subsets is Subset
Union of Subsets is Subset/Set of Sets

== Theorem ==
Let $T$ be a set.
Let $\mathbb S$ be a set of sets.
Suppose that for each $S \in \mathbb S$, $S \subseteq T$.
Then:
:$\ds \bigcup \mathbb S \subseteq T$
== Proof ==
Let $x \in \ds \bigcup \mathbb S$.
By the definition of union, there exists an $S \in \mathbb S$ such that $x \in S$.
By premise, $S \subseteq T$.
By the definition of subset, $x \in T$.
Since this result holds for each $x \in \ds \bigcup \mathbb S$:
:$\ds \bigcup \mathbb S \subseteq T$
Category:Union of Subsets is Subset
Union of Subsets is Subset/Subset of Power Set

== Theorem ==
Let $S$ and $T$ be sets.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
== Proof ==
Let $\mathbb S \subseteq \powerset S$.
Suppose that $\forall X \in \mathbb S: X \subseteq T$.
Consider any $\ds x \in \bigcup \mathbb S$.
By definition of set union, it follows that:
:$\exists X \in \mathbb S: x \in X$
But as $X \subseteq T$ it follows that $x \in T$.
Thus it follows that:
:$\ds \bigcup \mathbb S \subseteq T$
So:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
Category:Set Union
Category:Subsets
Union of Symmetric Differences

== Theorem ==
Let $R, S, T$ be sets.
Then:
:$\paren {R \symdif S} \cup \paren {S \symdif T} = \paren {R \cup S \cup T} \setminus \paren {R \cap S \cap T}$
where $R \symdif S$ denotes the symmetric difference between $R$ and $S$.
== Proof ==
From the definition of symmetric difference, we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Thus, expanding:
== Sources ==
* $\S 2$: Problem $2$
Category:Symmetric Difference
Category:Set Union
Union of Topologies is not necessarily Topology

== Theorem ==
Let $\tau_1$ and $\tau_2$ be topologies on a set $S$.
Then $\tau_1 \cup \tau_2$ is not necessarily also a topology on $S$.
== Proof ==
Let $S := \set {0, 1, 2}$ be a set.
Let:
:$\tau_1 := \set {\O, \set 0, \set 1, \set {0, 1}, S}$ 
:$\tau_2 := \set {\O, \set 0, \set 2, \set {0, 2}, S}$
be topologies on $S$. 
Then:
:$\tau := \tau_1 \cup \tau_2 = \set {\O, \set 0, \set 1, \set 2, \set {0, 1} \set {0, 2}, S}$
For $\tau$ to be a topology the union of any number of elements of $\tau$ should also be in $\tau$.
But:
:$\set 1 \cup \set 2 = \set {1, 2} \not \in \tau$ 
Therefore $\tau$ is not a topology on $S$.
Hence the result. 
== Also see ==
* Union of Topologies on Singleton or Doubleton is Topology where it is shown that this result does not hold if $\size S \le 2$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 4$
Category:Topology
Union of Topologies on Singleton or Doubleton is Topology

== Theorem ==
Let $S$ be a set which is either a singleton or a doubleton.
Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies on $S$.
Then $\tau := \ds \bigcup_{i \mathop \in I} {\tau_i}$ is also a topology on $S$.
== Proof ==
Let $S$ be a singleton.
Let $S = \set a$. 
From Topology on Singleton is Indiscrete Topology, the only possible topology on $S$ is the indiscrete topology. 
From Set Union is Idempotent, the union of any number of indiscrete topologies on $S$ is the indiscrete topology.
Thus the union of any number of topologies on a singleton is a topology on that singleton.
Let $S$ be a doubleton.
Let $S = \set {a, b}$.
By Topologies on Doubleton, there are $4$ possible different topologies on $S$:
It remains to show that every union of elements in $\set {\tau_1, \tau_2, \tau_3, \tau_4}$ is a topology on $S$.
By inspection, the following statements are seen to hold:
:$\tau_1 \cup \tau_2 = \tau_2$
:$\tau_1 \cup \tau_3 = \tau_3$
:$\tau_1 \cup \tau_4 = \tau_4$
:$\tau_2 \cup \tau_3 = \tau_4$
:$\tau_2 \cup \tau_4 = \tau_4$
:$\tau_3 \cup \tau_4 = \tau_4$
Now let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies for a set $S$.
By the above, if $\tau_4$ is one of the $\tau_i$:
:$\ds \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_4$
So assume that $\tau_4$ is not one of the $\tau_i$.
If both $\tau_2$ and $\tau_3$ are in $\family {\tau_i}_{i \mathop \in I}$, then by the above:
:$\ds \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_4$
Now suppose that $\tau_3$ is in $\family {\tau_i}_{i \mathop \in I}$ and $\tau_2$ is not.
Then by the above it follows that:
:$\tau = \tau_3$
Now suppose that $\tau_2$ is in $\family {\tau_i}_{i \mathop \in I}$ and $\tau_3$ is not.
Then by the above it follows that:
:$\tau = \tau_2$
Finally, assume that $\tau_2$ and $\tau_3$ are not in $\family {\tau_i}_{i \mathop \in I}$.
Then the only topology in $\family {\tau_i}_{i \mathop \in I}$ is $\tau_1$. 
In this, we find:
:$\tau = \tau_1$
Hence the result.
== Also see ==
* Union of Topologies is not necessarily Topology, which shows that when $\size S \ge 3$ this result is false.
Category:Topology
Category:Singletons
Category:Doubletons
Category:Set Union