nyuuzyou/wikis huggingface dataset:
Union of Blocks is Set of Points

== Theorem ==
Let $\struct {X, \BB}$ be a pairwise balanced design.
That is, let $\struct {X, \BB}$ be a design, with $\size X \ge 2$, and the number of occurrences of each pair of distinct points in $\BB$ be $\lambda$ for some $\lambda > 0$ constant.
Then the set union of all the subset elements in $\BB$ is precisely $X$.
=== Corollary ===
== Proof ==
Let $X = \set {x_1, x_2, \ldots, x_v}$.
Let $\BB = \multiset {y_1, y_2,\ldots, y_b}$, where the notation denotes a multiset.
Let $Y = \ds \bigcup_{i \mathop = 1}^b y_i$.
We shall show that $Y \subseteq X$ and $X \subseteq Y$.
=== $Y \subseteq X$: ===
By definition, $\BB$ is a multiset of subsets of $X$.
This means that each $y_i \in Y$ is a subset of $X$ for $i \in \closedint 1 b$.
By Union of Family of Subsets is Subset, $Y$ itself is a subset of $\BB$.
=== $X \subseteq Y$: ===
Let $x_i \in X$ be arbitrary.
Choose any $x_j \in X$ such that $i \ne j$.
Such an $x_j$ necessarily exists because $\card X \ge 2$.
Then $\set {x_i, x_j} \subseteq Y$ by the definition of blocks, as $\lambda > 0$ .
But by the definition of subset, this implies that $x_i$ is an element in $Y$.
Hence the result, as $x_i$ was arbitrary.
Category:Union of Blocks is Set of Points
Category:Design Theory
Union of Blocks is Set of Points/Corollary

== Corollary to Union of Blocks is Set of Points ==
Let $\struct {X, \BB}$ be a balanced incomplete block design.
Then the set union of all the subset elements in $\BB$ is precisely $X$.
== Proof ==
By the definition of balanced incomplete block design, $\struct {X, \BB}$ is itself a type of pairwise balanced design.
Thus the corollary follows from Union of Blocks is Set of Points.
Category:Union of Blocks is Set of Points
Union of Bounded Above Real Subsets is Bounded Above

== Theorem ==
Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ be bounded above.
Then $A \cup B$ is also bounded above.
== Proof ==
Let $A$ and $B$ both be bounded above.
Then by definition $A$ and $B$ both have an upper bound $U_A$ and $U_B$ respectively.
Suppose $U_A \le U_B$.
Then:
:$\forall a \in A: a \le U_B$
and also, by definition:
:$\forall b \in B: b \le U_B$
and so $U_B$ is an upper bound for $A$.
Otherwise, suppose $U_A > U_B$.
Then:
:$\forall b \in B: b \le U_A$
and also, by definition:
:$\forall a \in A: a \le U_A$
Let $x \in A \cup B$.
Then from the above, either $x \le U_A$ or $x \le U_B$.
So either $U_A$ or $U_B$ is an upper bound for $A \cup B$.
Hence, by definition, $A \cup B$ is bounded above by either $U_A$ or $U_B$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 2$
Category:Bounded Above Sets of Real Numbers
Category:Set Union
Union of Bounded Below Real Subsets is Bounded Below

== Theorem ==
Let $A$ and $B$ be sets of real numbers.
Let $A$ and $B$ be bounded below.
Then $A \cup B$ is also bounded below.
== Proof ==
Let $A$ and $B$ both be bounded below.
Then by definition $A$ and $B$ both have a lower bound $L_A$ and $L_B$ respectively.
Suppose $L_A \ge L_B$.
Then:
:$\forall a \in A: a \ge L_B$
and also, by definition:
:$\forall b \in B: b \ge L_B$
and so $L_B$ is a lower bound for $A$.
Otherwise, suppose $L_A < L_B$.
Then:
:$\forall b \in B: b \ge L_A$
and also, by definition:
:$\forall a \in A: a \ge L_A$
Let $x \in A \cup B$.
Then from the above, either $x \ge L_A$ or $x \ge L_B$.
So either $L_A$ or $L_B$ is a lower bound for $A \cup B$.
Hence, by definition, $A \cup B$ is bounded below by either $L_A$ or $L_B$.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 3$
Category:Bounded Below Sets of Real Numbers
Category:Set Union
Union of Chain in Set of Finite Character with Countable Union is Maximal Element

== Theorem ==
Let $S$ be a set of sets of finite character.
Let its union $\ds \bigcup S$ be countable.
Then $\ds \bigcup S$ is a maximal element of $S$ under the subset relation.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles
Category:Union of Chain in Set of Finite Character with Countable Union is Maximal Element
Category:Finite Character
Category:Maximal Elements
Union of Chain in Set of Finite Character with Countable Union is Maximal Element/Proof

== Theorem ==
== Proof ==
Let $S$ be as .
 $\ds \bigcup S$ is not a maximal element of $S$ under the subset relation.
Then:
:$\exists T \subseteq S: \ds \bigcup S \subsetneqq T$
Thus:
:$\exists x \in S: x \ne \ds \bigcup S$
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles: Exercise $6.1$
Category:Union of Chain in Set of Finite Character with Countable Union is Maximal Element
Union of Chain of Convex Sets in Vector Space is Convex

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $\Gamma$ be a chain of convex sets. 
Let:
:$\ds C = \bigcup \Gamma$
Then $C$ is convex.
== Proof ==
Let $t \in \closedint 0 1$ and $x, y \in C$. 
Then there exists $C_1, C_2 \in \Gamma$ such that $x \in C_1$ and $y \in C_2$.
Since $\Gamma$ is a chain, we have $C_1 \subseteq C_2$ or $C_2 \subseteq C_1$.
 suppose that $C_1 \subseteq C_2$.
Then $x, y \in C_2$.
Since $C_2$ is convex, we have $t x + \paren {1 - t} y \in C_2$.
So $t x + \paren {1 - t} y \in \bigcup \Gamma = C$.
So $C$ is convex.
Category:Convex Sets (Vector Spaces)
Union of Chain of Orderings is Ordering

== Theorem ==
Let $S$ be a set.
Let $C$ be a non-empty chain of orderings on $S$.
Then $\bigcup C$ is an ordering on $S$.
== Proof ==
Let $\preceq$ be an arbitrary element of $C$.
Let ${\sim} = \bigcup C$.
Checking in turn each of the criteria for an ordering:
Let $a, b \in S$.
Let $a \sim b$ and $b \sim a$.
Since $a \sim b$ and $b \sim a$, there exist $\preceq_1, \preceq_2 \, \in \, C$ such that $a \preceq_1 b$ and $b \preceq_2 a$.
Since $C$ is a chain:
:${\preceq_1} \subset {\preceq_2}$
or:
:${\preceq_2} \subset {\preceq_1}$
, suppose ${\preceq_1} \subset {\preceq_2}$.
Then it must hold that $a \preceq_2 b$.
Since:
:$a \preceq_2 b$
:$b \preceq_2 a$
:${\preceq_2} \in T$ is an ordering
it follows that:
:$a = b$
and so $\sim$ is antisymmetric.
=== Reflexivity ===
Let $a \in S$.
Because $\preceq$ is an ordering:
:$a \preceq a$
As ${\preceq} \subseteq {\sim}$:
:$a \sim a$
and so $\sim$ is reflexive.
=== Transitivity ===
Let $a, b, c \in S$.
Let $a \sim b$ and $b \sim c$.
Thus there are elements $\preceq_1$ and $\preceq_2$ of $C$ such that:
:$a \preceq_1 b$ and $b \preceq_2 c$
Since $C$ is a chain, it must hold (as above) that:
:$a \preceq_2 b$ or $b \preceq_1 c$
, suppose $a \preceq_2 b$
Since:
:$a \preceq_2 b$
:$b \preceq_2 c$
:${\preceq_2} \in T$ is an ordering
it follows that:
:$a \preceq_2 c$
Since ${\preceq_2} \subseteq {\sim}$:
:$a \sim c$
So $\sim$ has been shown to be transitive.
$\sim$ has been shown to be reflexive, transitive and antisymmetric.
Hence by definition it is an ordering.
Category:Set Theory
Category:Order Theory
Union of Chain of Ordinal Sequences is Ordinal Sequence

== Theorem ==
Let $C$ be a chain of ordinal sequences.
Then $\bigcup C$ is an ordinal sequence.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Lemma $5.2 \ (1)$
Category:Ordinal Sequences
Category:Chains (Order Theory)
Union of Chain of Proper Ideals is Proper Ideal

== Theorem ==
Let $R$ be a ring with unity.
Let $\struct {P, \subseteq}$ be the ordered set consisting of all ideals of $R$, ordered by inclusion.
Let $\sequence {I_\alpha}_{\alpha \mathop \in A}$ be a non-empty chain of proper ideals in $P$.
Let $\ds I = \bigcup_{\alpha \mathop \in A} I_\alpha$ be their union.
Then $I$ is a proper ideal of $R$.
== Proof ==
By Union of Chain of Ideals is Ideal, $I$ is an ideal.
It remains to show that $I \subsetneq R$.
The ideals $I_\alpha$ are all proper, so none of them contain the unity.
Thus $I$ does not contain $1$, which means $I \subsetneq R$.
== Also see ==
* Union of Chain of Ideals is Ideal
Category:Ring Theory
Category:Ideal Theory
Union of Chain of Subalgebras is Subalgebra

== Theorem ==
Let $K$ be a field. 
Let $A$ be an algebra over $K$.
Let $\BB$ be a $\subseteq$-chain of subalgebras of $A$. 
Then $\bigcup \BB$ is a subalgebra of $A$. 
Further, if each $A \in \BB$ is commutative, then so is $\bigcup \BB$. 
== Proof ==
From Union of Chain of Submodules is Submodule, $\bigcup \BB$ is a linear subspace of $A$.
Now let $x, y \in \bigcup \BB$.
Then there exists $B_1, B_2 \in \BB$ such that $x \in B_1$ and $y \in B_2$.
Since $\BB$ is a chain, we either have $B_1 \subseteq B_2$ or $B_2 \subseteq B_1$. 
Hence we can pick $i \in \set {1, 2}$ such that $x, y \in B_i$. 
Since $B_i$ is a subalgebra, we have $x y \in B_i$ and hence $x y \in \bigcup \BB$.
Hence $\bigcup \BB$ is a subalgebra of $A$. 
Further, if each $A \in \BB$ is commutative, then in particular so is $B_i$.
So $x y = y x$.
Category:Algebras
Union of Chain of Submodules is Submodule

== Theorem ==
Let $R$ be a ring.
Let $M$ be an $R$-module.
Let $\NN$ be a chain of submodules with respect to inclusion.
Let:
:$N = \bigcup \NN$
Then $N$ is a submodule of $M$. 
== Proof ==
We use the Submodule Test.
=== Proof of $(\text {SM} 1)$ ===
Let $x, y \in N$. 
Then there exists $N_1, N_2 \in \NN$ with $x \in N_1$ and $y \in N_2$.
Since $\NN$ is a chain, we have $N_1 \subseteq N_2$ or $N_2 \subseteq N_1$. 
Hence both $x$ and $y$ are contained in $N_i$ for some $i \in \set {1, 2}$. 
Fix such an $i$. 
Since $N_i$ is a submodule, we have $x + y \in N_i$. 
Hence $x + y \in \NN$. 
=== Proof of $(\text {SM} 2)$ ===
Let $x \in N$ and $\lambda \in R$. 
Then there exists $N' \in \NN$ such that $x \in N'$. 
Since $N'$ is a submodule, we have $\lambda x \in N'$.
Hence $\lambda x \in \NN$.
We conclude that $N$ is a submodule of $M$ by the Submodule Test.
Category:Submodules
Union of Class is Subclass implies Class is Transitive

== Theorem ==
Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let:
:$\bigcup A \subseteq A$
Then $A$ is transitive.
== Proof ==
Let $\bigcup A \subseteq A$.
Let $x \in \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of subclass:
:$x \in A$
Thus we have that:
:$x \in y \land y \in A \implies x \in A$
It follows by definition that $A$ is a transitive class.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.3. \ \text {(b)}$
Category:Class is Transitive iff Union is Subclass
Union of Class is Transitive if Every Element is Transitive

== Theorem ==
Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let $A$ be such that every element of $A$ is transitive.
Then $\bigcup A$ is also transitive.
== Proof ==
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.3. \ \text {(d)}$
Category:Class Union
Category:Transitive Classes
Category:Union of Class is Transitive if Every Element is Transitive
Union of Class is Transitive if Every Element is Transitive/Proof

== Theorem ==
== Proof ==
Let $A$ be such that every $y \in A$ is transitive.
Let $x \in \bigcup A$.
Then $x$ is an element of some element $y$ of $A$.
We have that $y$ is transitive.
Hence, by definition of transitive class:
:$x \subseteq y$
Because $y \in A$, by definition of union of class:
:$y \subseteq \bigcup A$
So:
:$x \subseteq \bigcup A$
As this is true for all $x \in A$, it follows by definition that $\bigcup A$ is transitive.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity: Theorem $10.3$
Category:Union of Class is Transitive if Every Element is Transitive
Union of Closed Intervals of Positive Reals is Set of Positive Reals

== Theorem ==
Let $\R_{>0}$ be the set of strictly positive real numbers.
For all $x \in \R_{> 0}$, let $B_x$ be the closed real interval $\closedint 0 x$.
Then:
:$\ds \bigcup_{x \mathop \in \R_{>0} } B_x = \R_{\ge 0}$
== Proof ==
Let $\ds B = \bigcap_{x \mathop \in \R_{>0} } B_x$.
Let $y \in B$.
Then by definition of union of family:
:$\exists x \in \R_{>0}: y \in B_x$
As $B_x \subseteq \R_{>0}$ it follows by definition of subset that:
:$y = 0$
or
:$y \in \R_{>0}$
In either case, $y \in \R_{\ge 0}$
So:
:$\ds \bigcap_{x \mathop \in \R_{>0} } B_x \subseteq \R_{\ge 0}$
Let $y \in \R_{\ge 0}$.
If $y = 0$ then $y \in \R_{\ge 0}$ by definition.
Otherwise, by the Axiom of Archimedes: 
:$\exists z \in \N: z > y$
and so:
:$y \in B_z$
That is by definition of union of family:
:$\ds y \in \bigcap_{x \mathop \in \R_{\ge 0} } B_x$
So by definition of subset:
:$\ds \R_{\ge 0} \subseteq \bigcap_{x \mathop \in \R_{>0} } B_x$
By definition of set equality:
:$\ds \bigcup_{x \mathop \in \R_{>0} } B_x = \R_{\ge 0}$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $5$
Category:Real Intervals
Union of Closed Locally Finite Set of Subsets is Closed

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $\FF$ be a closed locally finite set of subsets of $T$.
Let $E = \ds \bigcup \FF$.
Then:
:$E$ is closed in $T$.
== Proof ==
Let:
:$\UU = \ds \leftset{U \in \tau : \set{F \in \FF : U \cap F \ne \O}}$ is finite $\ds \rightset{}$
By definition of closed locally finite set of subsets:
:$\forall x \in S : \exists U \in \tau : x \in U : \set{F \in \FF : U \cap F \ne \O}$ is finite
That is:
:$\forall x \in S : \exists U \in \UU : x \in U$
Hence
:$\UU$ is a open cover of $T$ by definition.
Let $U \in \UU$.
Let $\FF_U= \set{F \in \FF : F \cap U \ne \O}$ 
We have:
From Closed Set in Topological Subspace:
:$\forall F \in \FF_U : U \cap F$ is closed in the subspace topology on $U$
From :
:$U \cap E$ is closed in the subspace topology on $U$
Since $U$ was an arbitrary element of $\UU$:
:$\forall U \in \UU : U \cap E$ is closed in the subspace topology on $U$
From Characterization of Closed Set by Open Cover:
:$E$ is closed in $T$ 
Category:Locally Finite Sets of Subsets
Union of Closure with Closure of Complement is Whole Space

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
Let $H^-$ denote the closure of $H$ in $T$.
Let $S \setminus H$ denote the complement of $H$ relative to $S$.
Then:
:$H^- \cup \left({S \setminus H}\right)^- = S$
== Proof ==
We have that:
:$H^- \cup \left({S \setminus H}\right)^- \subseteq S$
by definition of $S$.
From Union with Relative Complement:
:$H \cup \left({S \setminus H}\right) = S$
From Set is Subset of its Topological Closure:
From Set Union Preserves Subsets:
:$H \subseteq H^-, \left({S \setminus H}\right) \subseteq \left({S \setminus H}\right)^- \implies H \cup \left({S \setminus H}\right) \subseteq H^- \cup \left({S \setminus H}\right)^-$
which means:
:$S \subseteq H^- \cup \left({S \setminus H}\right)^-$
The result follows by definition of equality of sets.
Category:Set Closures
Category:Set Union
Category:Relative Complement
Union of Closures of Elements of Locally Finite Set is Closed

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $\AA$ be a locally finite set of subsets of $T$.
Then:
:$\ds \paren {\bigcup \AA}^- = \bigcup \set{A^- : A \in \AA}$
where $A^-$ denotes the closure of $A$ in $T$.
== Proof ==
From Closures of Elements of Locally Finite Set is Locally Finite:
:$\set{A^- : A \in \AA}$ is also locally finite
From Union of Closed Locally Finite Set of Subsets is Closed:
:$\bigcup \set{A^- : A \in \AA}$ is closed in $T$
We have:
From Closure of Union contains Union of Closures
:$\ds \bigcup \set{A^- : A \in \AA} \subseteq \paren {\bigcup \AA}^-$
By definition of set equality:
:$\ds \paren {\bigcup \AA}^- = \bigcup \set{A^- : A \in \AA}$
Category:Locally Finite Sets of Subsets
Union of Closures of Singleton Rationals is Rational Space

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the usual (Euclidean) topology $\tau_d$.
Let $B_\alpha$ denote the singleton containing the rational number $\alpha$.
Then the union of the closures in the set of real numbers $\R$ of all $B_\alpha$ is $\Q$:
:$\ds \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \Q$
== Proof ==
Let $\alpha \in \Q$.
By Real Number is Closed in Real Number Line, $B_\alpha = \set \alpha$ is closed in $\R$.
From Closed Set Equals its Closure, it follows that:
:$B_\alpha = \map \cl {B_\alpha}$
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $1$
Category:Rational Number Space
Category:Singletons
Union of Conjugacy Classes is Normal

== Theorem ==
Let $G$ be a group.
Let $H \le G$.
Then $H$ is normal in $G$ $H$ is a union of conjugacy classes of $G$.
== Proof ==
Hence the result.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $21$
Category:Normal Subgroups
Union of Connected Sets with Common Point is Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {B_\alpha}_{\alpha \mathop \in A}$ be a family of 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 connected set of $T$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4$ Connectedness and Compactness, Proposition $4.9 \ \text {(d)}$
Category:Connected Spaces
Category:Union of Connected Sets with Common Point is Connected
Union of Connected Sets with Common Point is Connected/Proof 1

== Theorem ==
== Proof ==
Let $B = \ds \bigcup_{\alpha \mathop \in A} B_\alpha$.
From definition 3 of a connected set, $B$ is connected in $T$ the subspace $\struct {B, \tau_B}$ is a connected space.
From definition 4 of a connected space, $\struct {B, \tau_B}$ is connected the only clopen sets in $\struct {B, \tau_B}$ are $B$ and $\O$.
Let $U$ be any clopen set of the subspace of $\struct {B, \tau_B}$.
Let $V = B \setminus U$.
From Complement of Clopen Set is Clopen, $V$ is also clopen.
Hence $U, V$ are disjoint clopen sets such that $B = U \cup V$. 
 assume $x \in U$.
From Set is Subset of Union:
:$\ds \forall \alpha \in A : B_\alpha \subseteq B = U \cup V$
From Connected Subset of Union of Disjoint Open Sets:
:$\forall \alpha \in A : B_\alpha \subseteq U$
From Union is Smallest Superset:
:$B \subseteq U$
Since $U \subseteq B$ then $U = B$.
From Set Difference with Self is Empty Set,
:$V = B \setminus U = B \setminus B = \O$
Hence the only clopen sets in $\struct {B, \tau_B}$ are $B$ and $\O$.
The result follows. 
== Sources ==
* : $\S 4$ Connectedness and Compactness, Proposition $4.9 \ \text {(d)}$
Category:Union of Connected Sets with Common Point is Connected
Union of Connected Sets with Non-Empty Intersections is Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $I$ be an indexing set.
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.
Let $\AA$ be such that no two of its elements are disjoint:
:$\forall B, C \in \AA: B \cap C \ne \O$
Then $\ds \bigcup \AA$ is itself connected.
=== Corollary ===
== Proof ==
Let $A := \ds \bigcup \AA$.
Let $D = \set {0, 1}$, with the discrete topology.
Let $f: A \to D$ be a continuous mapping.
To show that $A$ is connected, we need to show that $f$ is not a surjection.
From Connected Set in Subspace, each $C \in \AA$ is connected in $A$.
From Restriction of Continuous Mapping is Continuous, the restriction $f \restriction_C$ is continuous for all $C \in \AA$.
From Continuous Image of Connected Space is Connected, $f \sqbrk C$ is connected in $D$ for all $C \in \AA$.
From Discrete Space is Totally Disconnected, $f \sqbrk C$ is a singleton for all $C \in \AA$:
:$f \sqbrk C = \set {\map \epsilon C}$
where $\map \epsilon C = 0$ or $1$.
By hypothesis:
:$\forall B, C \in \AA: B \cap C \ne \O$
Hence:
:$\forall B, C \in \AA: \map \epsilon B = \map \epsilon C$
Thus $f$ is constant on $A$ as required.
== Sources ==
* : $6.2$: Connectedness: Proposition $6.2.15$
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Connected Spaces
Category:Union of Connected Sets with Non-Empty Intersections is Connected
Union of Connected Sets with Non-Empty Intersections is Connected/Corollary

== Corollary to Union of Connected Sets with Non-Empty Intersections is Connected ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $I$ be an indexing set.
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.
Let $B$ be a connected set of $T$ such that:
:$\forall C \in \AA: B \cap C \ne \O$
Then $\ds B \cup \bigcup \AA$ is connected.
== Proof 1 ==
== Proof 2 ==
Category:Union of Connected Sets with Non-Empty Intersections is Connected
Category:Connected Spaces
Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 1

== Corollary to Union of Connected Sets with Non-Empty Intersections is Connected ==
== Proof ==
Let $C \in \AA$.
From Union of Connected Sets with Non-Empty Intersections is Connected applied to $B$ and $C$, the union $B \cup C$ is connected.
Thus the set $\tilde \AA = \set {B \cup C: C \in \AA}$ satisfies the conditions of the theorem.
Hence the result.
== Sources ==
* : $6.2$: Connectedness: Corollary $6.2.16$
Category:Union of Connected Sets with Non-Empty Intersections is Connected
Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 2

== Corollary to Union of Connected Sets with Non-Empty Intersections is Connected ==
== Proof ==
Let $\ds H = B \cup \bigcup \AA$
 that $H$ is not connected.
That is, that $H$ is disconnected.
We have that:
:$\forall C \in \AA: B \cap C \ne \O$
Thus:
:$\exists x \in H: x \in B \cap C$
From the definition of disconnected, there exist separated sets $U, V$ whose union is $H$.
So:
:$\forall s \in S: s \in U \lor s \in V$
 $s \in V$.
Then $H, V$ serve as separated sets whose union is a cover for $\family {A_\alpha}_{\alpha \mathop \in I}$.
Note that the disconnection is valid as:
:$\exists x \in U \cap \family {A_\alpha}_{\alpha \mathop \in I}$
As $s \in V$, it follows that the $\family {A_\alpha}_{\alpha \mathop \in I}$ are disconnected.
From this contradiction it follows that $s \notin V$.
Hence $s \in U$.
This holds for arbitrary $s \in H$.
Hence:
:$H \subseteq U$
But as $U$ and $V$ are separated, $U \cap H$ is empty, as required per the definition of a disconnected set.
This contradicts our deduction that $H \subseteq U$.
Hence the sets $U, V$ are not separated sets in $H$.
Thus $H$ is connected in $T$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Functions and Products
Category:Union of Connected Sets with Non-Empty Intersections is Connected
Union of Countable Sets of Sets

== Theorem ==
Let $\AA$ and $\BB$ be countable sets of sets.
Then:
:$\set {A \cup B: A \in \AA, B \in \BB}$
is also countable.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Set Union
Category:Countable Sets
Category:Union of Countable Sets of Sets
Union of Countable Sets of Sets/Proof 1

== Theorem ==
== Proof ==
Since $\AA$ is countable, its contents can be arranged in a sequence:
:$\AA = \set {A_1, A_2, \ldots}$
Let $B \in \BB$.
Consider the sequence of sets:
:$\sequence {A_1 \cup B, A_2 \cup B, \ldots}$
We may leave out any possible repetitions, and obtain a countable set:
:$\set {A \cup B: A \in \AA}$
for every $B \in \BB$.
Thus as $B$ varies over all the elements of $\BB$, we obtain the countable family:
:$\sequence {A \cup B: A \in \AA}_{\paren {B \mathop \in \BB} }$
From Countable Union of Countable Sets is Countable, their union is countable.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
Category:Union of Countable Sets of Sets