nyuuzyou/wikis huggingface dataset:
Union of Countable Sets of Sets/Proof 2

== Theorem ==
== Proof ==
Both $\AA$ and $\BB$ are countable.
Enumerate their elements by the sequences $\sequence {A_n}_{n \mathop \in \N}$ and $\sequence {B_n}_{n \mathop \in \N}$ respectively.
When one of $\AA$ and $\BB$ would be finite, achieve such sequences by allowing elements to occur multiple times.
Define $\CC := \set {A \cup B: A \in \AA, B \in \BB}$.
For each $C \in \CC$, define the set $N_C := \set {\tuple {i, j} \in \N \times \N: A_i \cup B_j = C}$.
The definition of $\CC$ ensures that $N_C \ne \O$ for all $C \in \CC$.
Consider the lexicographic ordering $\preccurlyeq_l$ on $\N \times \N$.
Combining the Well-Ordering Principle with Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering, $\preccurlyeq_l$ is a well-ordering on $\N \times \N$.
As $N_C \subseteq \N \times \N$ is non-empty, it thus has a smallest element; call it $n_C$.
Define now the map $\phi: \CC \to \N \times \N$ by $\map \phi C = n_C$.
Well-Ordering Minimal Elements are Unique ensures that $\phi$ is well-defined.
Now suppose that $C_1, C_2 \in \CC$ satisfy:
:$\map \phi {C_1} = \map \phi {C_2}$
that is, $n_{C_1} = n_{C_2}$, which in turn equal some $\tuple {i, j} \in \N \times \N$.
As $\tuple {i, j} \in N_{C_1}$ and $\tuple {i, j} \in N_{C_2}$, $C_1 = A_i \cup B_j = C_2$.
It follows that $\phi$ is an injection.
Finally, $\N \times \N$ is countable by Cartesian Product of Countable Sets is Countable.
Let $\psi: \N \times \N \to \N$ be an injection.
Then $\psi \circ \phi: \CC \to \N$ is also an injection by Composite of Injections is Injection.
Hence $\CC$ is countable.
Category:Union of Countable Sets of Sets
Union of Countable Sets of Sets/Proof 3

== Theorem ==
== Proof ==
Since $\AA$ and $\BB$ are countable, $\AA \times \BB$ is countable by Cartesian Product of Countable Sets is Countable.
Thus by Surjection from Natural Numbers iff Countable there is a surjection $f: \N \to \AA \times \BB$.
Let $\CC = \set {A \cup B: A \in \AA, B \in \BB}$.
Let $g: \AA \times \BB \to \CC$ be defined by letting $\map g {A, B} = A \cup B$.
$g$ is a surjection:
Let $C \in \CC$.
Then there exist $A \in \AA$ and $B \in \BB$ such that $C = A \cup B$.
By the definition of Cartesian product:
: $\tuple {A, B} \in \AA \times \BB$
Then:
: $\map g {A, B} = A \cup B = C$
It follows that $g$ is a surjection.
By Composite of Surjections is Surjection, $g \circ f: \N \to \CC$ is surjective.
Hence $\CC$ is countable.
Category:Union of Countable Sets of Sets
Union of Covers is Cover of Union

== Theorem ==
Let $\sequence {S_i}_{i \in I}$ be an indexed family of sets.
For each $i \in I$, let $\CC_i$ be a cover of $S_i$.
Then, $\ds \bigcup_{i \mathop \in I} \CC_i$ is a cover of $\ds \bigcup_{i \mathop \in I} S_i$.
== Proof ==
Let $\ds x \in \bigcup_{i \mathop \in I} S_i$ be arbitrary.
By definition of union, there is some $i \in I$ such that:
:$x \in S_i$
Then, by definition of cover, there is some $C \in \CC_i$ such that:
:$x \in C$
But, by definition of union:
:$\ds C \in \bigcup_{i \mathop \in I} \CC_i$
Hence, there is some $\ds C \in \bigcup_{i \mathop \in I} \CC_i$ such that $x \in C$.
As $x$ was arbitrary, the result follows from the definition of cover.
Category:Covers
Union of Derivatives is Subset of Derivative of Union

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let:
:$\FF \subseteq \powerset S$ be a set of subsets of $S$
where $\powerset S$ denotes the power set of $S$.
Then:
:$\ds \bigcup_{A \mathop \in \FF} A' \subseteq \paren {\bigcup_{A \mathop \in \FF} A}'$
where $A'$ denotes the derivative of $A$.
== Proof ==
Let $\ds x \in \bigcup_{A \mathop \in \FF} A'$.
Then by definition of union there exists $A \in \FF$ such that:
:$(1): \quad x \in A'$
By Set is Subset of Union:
:$\ds A \subseteq \bigcup_{A \mathop \in \FF} A$
Then by Derivative of Subset is Subset of Derivative:
:$\ds A' \subseteq \paren {\bigcup_{A \mathop \in \FF} A}'$
Hence by $(1)$ the result:
:$\ds x \in \paren {\bigcup_{A \mathop \in \FF} A}'$
follows by definition of subset.
== Sources ==
* 
* 
Category:Set Derivatives
Union of Disjoint Singletons is Doubleton

== Theorem ==
Let $\set a$ and $\set b$ be singletons such that $a \ne b$.
Then:
:$\set a \cup \set b = \set {a, b}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4$: Unions and Intersections
Category:Set Union
Category:Singletons
Category:Doubletons
Category:Union of Disjoint Singletons is Doubleton
Union of Disjoint Singletons is Doubleton/Proof 1

== Theorem ==
== Proof ==
Let $x \in \set a \cup \set b$.
Then by the Axiom of Unions:
:$x = a \lor x = b$
It follows from the Axiom of Pairing that:
:$x \in \set {a, b}$
Thus by definition of subset:
:$\set a \cup \set b \subseteq \set {a, b}$
Let $x \in \set {a, b}$.
Then by the Axiom of Pairing:
:$x = a \lor x = b$
So by the Axiom of Unions:
:$x \in \set a \cup \set b$
Thus by definition of subset:
:$\set {a, b} \subseteq \set a \cup \set b$
The result follows by definition of set equality.
Category:Union of Disjoint Singletons is Doubleton
Union of Doubleton

== Theorem ==
Let $x$ and $y$ be sets.
Let $\set {x, y}$ be a doubleton.
Then $\bigcup \set {x, y}$ is a set such that:
:$\bigcup \set {x, y} = x \cup y$
== Proof ==
Then, from Axiom of Unions, it follows that $x \cup y$ is a set.
== Sources ==
* : $\S 0.2$. Sets
* : $\S 5.7$
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.4. \ \text {(a)}$
Category:Class Union
Category:Doubleton Classes
Union of Elements of Power Set

== Theorem ==
Let $S$ be a set.
Then:
:$\ds S = \bigcup_{X \mathop \in \powerset S} X$
where $\powerset S$ denotes the power set of $S$.
== Proof ==
By Subset of Union:
:$\ds \forall X \in \powerset S: X \subseteq \bigcup_{X \mathop \in \powerset S} X$
From Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$.
So:
:$\ds S \subseteq \bigcup_{X \mathop \in \powerset S} X$
From Union is Smallest Superset:
:$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
where $\mathbb S \subseteq \powerset S$.
But as $\powerset S \subseteq \powerset S$ from Set is Subset of Itself:
:$\ds \paren {\forall X \in \powerset S: X \subseteq S} \iff \bigcup \powerset S \subseteq S$
The is no more than the definition of the power set, making it a tautological statement, and so:
:$\ds \bigcup \powerset S \subseteq S$
The result follows by definition of set equality.
== Sources ==
* : $\S 5$: Complements and Powers
Category:Power Set
Category:Set Union
Union of Empty Set

== Theorem ==
Consider the set of sets $\mathbb S$ such that $\mathbb S$ is the empty set $\O$.
Then the union of $\mathbb S$ is $\O$:
:$\mathbb S = \O \implies \bigcup \mathbb S = \O$
== Proof ==
Let $\mathbb S = \O$.
Then from the definition:
:$\bigcup \mathbb S = \set {x: \exists X \in \mathbb S: x \in X}$
from which it follows directly:
:$\bigcup \mathbb S = \O$
as there are no sets in $\mathbb S$.
== Also presented as ==
Using the terminology of indexed families, this can also be written as:
:$\ds \bigcup_{i \mathop \in \O} S_i = \set {x: \exists i \in \O: x \in S_i} = \O$
== Also see ==
* Intersection of Empty Set
== Sources ==
* : $\S 4$: Unions and Intersections
* : $\S 9$: Families
* : $1$. Preliminary ideas; sets, vector spaces: $1.1$. Sets
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.7$
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.2: \ \text{(ii)}$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
* : Appendix $\text{A}$: Set Theory: Families
Category:Set Union
Category:Empty Set
Union of Equivalence Classes is Whole Set

== Theorem ==
Let $\RR \subseteq S \times S$ be an equivalence on a set $S$.
Then the set of $\RR$-classes constitutes the whole of $S$.
== Proof ==
We have that:
and:
Then:
== Also see ==
* Fundamental Theorem on Equivalence Relations
* Equivalence Classes are Disjoint
* Equivalence Class is not Empty
== Sources ==
* : Appendix $\text{A}.3$: Equivalence Relations: Theorem $\text{A}.2$
* : $\S 17.5 \ \text{(i)}$: Equivalence classes
* : Chapter $1$: Preliminaries: Exercises: $1.8$
* : $1$: Set Theory and Logic: $\S 3$: Relations
* 
Category:Equivalence Relations
Category:Quotient Sets
Category:Fundamental Theorem on Equivalence Relations
Union of Equivalences/Proof 1

== Theorem ==
== Proof ==
This can be shown by giving an example.
Let $S = \set {a, b, c}$, and let $\RR_1$ and $\RR_2$ be equivalences on $S$ such that:
: $\eqclass a {\RR_1} = \eqclass b {\RR_1} = \set {a, b}$
: $\eqclass c {\RR_1} = \set c$
: $\eqclass a {\RR_2} = \set a$
: $\eqclass b {\RR_2} = \eqclass c {\RR_2} = \set {b, c}$
Let $\RR_3 = \RR_1 \cup \RR_2$.
Then:
However:
So $\RR_3$ is not transitive, and therefore $\RR_3 = \RR_1 \cup \RR_2$ is not an equivalence relation.
Category:Union of Equivalences
Union of Event with Complement is Certainty

== Theorem ==
Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
Let $A \in \Sigma$ be an events of $\EE$, so that $A \subseteq \Omega$.
Then:
:$A \cup \overline A = \Omega$
where $\overline A$ is the complementary event to $A$.
That is, $A \cup \overline A$ is a certainty.
== Proof ==
By definition:
:$A \subseteq \Omega$
and:
:$\overline A = \relcomp \Omega A$
From Union with Relative Complement:
:$A \cup \overline A = \Omega$
We then have from Kolmogorov axiom $(2)$ that:
:$\map \Pr \Omega = 1$
The result follows by definition of certainty.
Category:Complementary Events
Category:Set Union
Union of Events/Examples/Both Prime and Even

== Examples of Unions of Events ==
Consider the experiment $\EE$ such that $2$ (positive) integers are drawn at random from a table of random numbers.
Let $A$ be the event that at least $1$ of these integers is prime.
Let $B$ be the event that at least $1$ of these integers is even.
Then their union $A \cup B$ means:
:either of the $2$ integers is even or either of the $2$ integers is prime.
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Example $1.2$
Category:Examples of Unions of Events
Union of Events/Examples/Defective Devices

== Examples of Unions of Events ==
Consider the experiment $\EE$ such that $3$ devices are checked as to whether they are operational.
Let $A$ be the event that at least $1$ of these $3$ devices is defective.
Let $B$ be the event that all $3$ devices are sound.
Then their union $A \cup B$ is a certainty.
== Proof ==
Let $A$ occur.
Then it is not the case that all $3$ devices are sound.
Therefore $B$ does not occur.
Let $B$ occur.
Then it is not the case that at least $1$ device is defective.
Therefore $A$ does not occur.
Thus it is seen that:
:$B = \overline A$
that is, $B$ is the complementary event to $A$.
Thus:
:$A \cup B$
is the same thing as:
:$A \cup \overline A$
and the result follows from Union of Event with Complement is Certainty.
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Problem $4 \ \text {(a)}$
Category:Examples of Unions of Events
Union of Exteriors contains Exterior of Intersection/Mistake

== Source Work ==
:
:Part $\text I$: Basic Definitions
::Section $1$. General Introduction
:::Closures and Interiors
== Mistake ==
:The exterior of the union of sets is always contained in the intersection of the exteriors, and similarly, the exterior of the intersection is contained in the union of the exteriors; equality holds only for finite unions and intersections.
== Correction ==
Part of the above statement is true. See:
* Intersection of Exteriors contains Exterior of Union
* Exterior of Finite Union equals Intersection of Exteriors
However, the statement:
:" ... the exterior of the intersection is contained in the union of the exteriors ..."
is shown to be false by Exterior of Intersection contains Union of Exteriors.
The statement:
:"... equality holds only for finite unions and intersections ..." is also shown to be untrue, as follows.
Consider where $T = \R$, the real number line.
Let $H = \openint 0 2 \cup \openint 3 4, K = \openint 1 3$.
For these sets:
We see that $3 \in \paren {H \cap K}^e$, but $3 \notin H^e \cup K^e$.
So:
:$H^e \cup K^e \subset \paren {H \cap K}^e$
but:
:$H^e \cup K^e \ne \paren {H \cap K}^e$
== Acknowledgements ==
This error was confirmed by 
HappyJoe on mathhelpforum on 17th April 2011.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Mistakes/Counterexamples in Topology
Union of Family/Examples/Size of y-1 lt n and Size of y+1 gt 1 over n

== Example of Union of Family ==
Let $I$ be the indexing set $I = \set {1, 2, 3, \ldots}$
Let $\family {T_n}$ be the indexed family of subsets of the set of real numbers $\R$, defined as:
:$T_n = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}$
Then:
:$\ds \bigcup_{n \mathop \in I} T_n = \R \setminus \set {-1}$
== Proof ==
From Size of y-1 lt n and Size of y+1 gt 1 over n:
:$T_n = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}$
We have that:
:$\paren {\openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n} } \subseteq \paren {\openint {1 - \paren {n + 1} } {-1 - \dfrac 1 {n + 1} } \cup \openint {-1 + \dfrac 1 {n + 1} } {1 + \paren {n + 1} } }$
That is:
:$T_n \subseteq T_{n + 1}$
and so:
:$\ds \bigcup_{n \mathop \in I} T_n = \lim_{n \mathop \to \infty} \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}$
As $n \to \infty$, we have:
and it follows that:
:$\lim_{n \mathop \to \infty} \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n} = \openint \gets {-1} \cup \openint {-1} \to$
whence the result.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text {A}$
Category:Unions of Families
Union of Filtered Sets is Filtered

== Theorem ==
Let $\struct {S, \preceq}$ be a preordered set.
Let $A$ be a set of subsets of $S$ such that
:$\forall X \in A: X$ is filtered
and
:$\forall X, Y \in A: \exists Z \in A: X \cup Y \subseteq Z$
Then:
:$\bigcup A$ is also filtered.
== Proof ==
Let $x, y \in \bigcup A$.
By definition of union:
:$\exists X \in A: x \in X$
and
:$\exists Y \in A: y \in Y$
By assumption:
:$\exists Z \in A: X \cup Y \subseteq Z$
By definition of union:
:$x, y \in X \cup Y$
By definition of subset:
:$x, y \in Z$
By assumption:
:$Z$ is filtered.
By definition of filtered:
:$\exists z \in Z: z \preceq x \land z \preceq y$
Thus by definition of union:
:$z \in \bigcup A$
Hence
:$\bigcup A$ is filtered.
== Sources ==
* 
* 
Category:Preorder Theory
Union of Finite Sets is Finite

== Theorem ==
Let $S$ and $T$ be finite sets.
Then $S \cup T$ is a finite set.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 6$ Finite Sets: Exercise $6.2$
Category:Set Union
Category:Finite Sets
Category:Union of Finite Sets is Finite
Union of Finite Sets is Finite/Proof 1

== Theorem ==
== Proof ==
If $S$ or $T$ is empty, the result is trivial.
Otherwise, let $f: \N_{<n} \to S$ and $g: \N_{<m} \to T$ be bijections, where $\N_{<n}$ is an initial segment of $\N$.
Now define $h: \N_{< n + m} \to S \cup T$ by:
:$\map h i = \begin{cases} \map f i : & \text {if $i < n$} \\ \map g {i - n} : & \text{if $i \ge n$} \end{cases}$
By Set Finite iff Surjection from Initial Segment of Natural Numbers, it suffices to show that $h$ is surjective.
Let $s \in S$.
Then:
Next let $t \in T$. 
We have that:
:$n + \map {g^{-1} } t < n + m$
so that:
:$n + \map {g^{-1} } t \in \N_{< n + m}$
Then:
Thus it has been shown that for all $s \in S$ and $t \in T$, there exists an element of $\N_{< n + m}$ which maps to it.
So by definition of set union, for all $x \in S \cup T$:
:$\exists y \in \N_{< n + m}: \map h y = x$
Hence, by definition, $h$ is a surjection.
Therefore $S \cup T$ is finite.
== Sources ==
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.8$
Category:Union of Finite Sets is Finite
Union of Finite Sets is Finite/Proof 2

== Theorem ==
== Proof ==
Note that $\card {S \cup T} \le \card {S \times T}$ by Cardinal of Union Less than Cardinal of Cartesian Product.
The theorem follows from the fact that $S \times T$ is finite by Product of Finite Sets is Finite.
== Sources ==
* : $\S 10.29 \ (1)$
Category:Union of Finite Sets is Finite
Union of Functions Theorem

== Theorem ==
Let $X$ be a set.
Let $\sequence {X_i: i \in \N}$ be an exhausting sequence of sets on $X$.
For each $i \in \N$, let $g_i: X_i \to Y$ be a mapping such that:
:$g_{i + 1} \restriction X_i = g_i$
where $g_{i + 1} \restriction X_i$ denotes the restriction of $g_{i + 1}$ to $g_i$.
Then:
:$\ds \bigcup \set {g_i: i \in \N}$
is a mapping from $X$ to $Y$.
=== Corollary ===
== Proof ==
By definition, $\ds g = \bigcup \set {g_i: i \in \N}$ is a relation whose domain is $X$.
 $g$ is not a mapping.
Then for some $x \in X$ and $i, h \in \N$:
:$(1): \quad x \in X_i, \map {g_i} x \ne \map {g_{i + h} } x$
Let $k \in \N$ be the smallest such that:
:$\map {g_i} x \ne g_{i + k}$
where $x$ and $i$ are the same as in $(1)$.
Then:
:$\map {g_{i + k - 1} } x = g_i$
But then:
:$g_{i + k} \restriction X_{i + k - 1} = g_{i + k - 1}$
From this contradiction it follows that our supposition that $g$ is not a mapping must be false.
That is: 
:$\ds \bigcup \set {g_i: i \in \N}$
is a mapping from $X$ to $Y$.
== Sources ==
* : $\S 1.18$: Sequences Defined Inductively: Theorem $18.1$
Category:Mapping Theory
Category:Named Theorems
Union of Functions Theorem/Corollary

== Theorem ==
Let $X$ be a set.
Let $\sequence {X_i: i \in \N}$ be an exhausting sequence of sets on $X$.
For each $i \in \N$, let $g_i: X_i \to Y$ be a mapping such that:
:$g_{i + 1} \restriction X_i = g_i$
where $g_{i + 1} \restriction X_i$ denotes the restriction of $g_{i + 1}$ to $g_i$.
For each $i \in \N$, let $g_i : X_i \to Y$ be invertible.
Then $\ds \bigcup \set {g_i: i \in \N}$ is invertible and:
:$\ds \paren {\bigcup \set {g_i: i \in \N} }^{-1} = \bigcup \set {g_i^{-1}: i \in \N}$
== Proof ==
Category:Inverse Mappings
Union of Horizontal Sections is Horizontal Section of Union

== Theorem ==
Let $X$, $Y$ and $A$ be sets.
Let $\set {E_\alpha : \alpha \in A}$ be a set of subsets of $X \times Y$. 
Let $y \in Y$. 
Then: 
:$\ds \paren {\bigcup_{\alpha \mathop \in A} E_\alpha}^y = \bigcup_{\alpha \mathop \in A} \paren {E_\alpha}^y$
where:
:$\ds \paren {\bigcup_{\alpha \mathop \in A} E_\alpha}^y$ is the $y$-horizontal section of $\ds \bigcup_{\alpha \mathop \in A} E_\alpha$
:$\paren {E_\alpha}^y$ is the $y$-horizontal section of $E_\alpha$.
 
== Proof ==
Note that:
:$\ds x \in \bigcup_{\alpha \mathop \in A} \paren {E_\alpha}^y$
: 
:$x \in \paren {E_\alpha}^y$ for some $\alpha \in A$.
From the definition of the $y$-horizontal section, this is equivalent to:
:$\tuple {x, y} \in E_\alpha$ for some $\alpha \in A$. 
This in turn is equivalent to: 
:$\ds \tuple {x, y} \in \bigcup_{\alpha \mathop \in A} E_\alpha$
Again applying the definition of the $y$-horizontal section, this is the case :
:$\ds x \in \paren {\bigcup_{\alpha \mathop \in A} E_\alpha}^y$
So:
:$\ds x \in \bigcup_{\alpha \mathop \in A} \paren {E_\alpha}^y$ $\ds x \in \paren {\bigcup_{\alpha \mathop \in A} E_\alpha}^y$
giving:
:$\ds \paren {\bigcup_{\alpha \mathop \in A} E_\alpha}^y = \bigcup_{\alpha \mathop \in A} \paren {E_\alpha}^y$
Category:Set Union
Category:Horizontal Section of Sets
Union of Image of Convergent Sequence and Limit in Topological Space is Compact

== Theorem ==
Let $\struct {X, \tau}$ be a topological space. 
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence such that: 
:$x_n \to x$
Then: 
:$\ds \set {x_n : n \in \N} \cup \set x$ is compact. 
== Proof ==
Let $\family {U_\alpha}_{\alpha \in A}$ be an open cover for $\ds \set {x_n : n \in \N} \cup \set x$.
Take $\beta \in A$ be such that $x \in U_\beta$. 
Since $x_n \to x$, there exists $N \in \N$ such that $x_n \in U_\beta$ for $n \ge N$. 
For each $n < N$, we can select $U_{\alpha_1}, \ldots, U_{\alpha_{N - 1} }$ such that $x_j \in U_{\alpha_j}$ for each $1 \le j \le N - 1$.
Then we have: 
:$\ds \set {x_n : n \in \N} \cup \set x \subseteq U_\beta \cup \bigcup_{j \mathop = 1}^{N - 1} U_{\alpha_j}$ 
So every open cover has a finite subcover.
So $\ds \set {x_n : n \in \N} \cup \set x$ is compact.
Category:Compact Spaces
Union of Indexed Family of Sets Equal to Union of Disjoint Sets

== Theorem ==
Let $\family {E_n}_{n \mathop \in \N}$ be a countable indexed family of sets where at least two $E_n$ are distinct.
Then there exists a countable indexed family of disjoint sets $\family {F_n}_{n \mathop \in \N}$ defined by:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
satisfying:
:$\ds \bigsqcup_{n \mathop \in \N} F_n = \bigcup_{n \mathop \in \N} E_n$
where $\bigsqcup$ denotes disjoint union.
=== Corollary ===
=== General Result ===
== Proof ==
Denote:
where:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
We first show that $E = F$.
That $x \in E \implies x \in F$ follows from the construction of $F$ from subsets of $E$.
Thus $E \subseteq F$.
Then:
so $F \subseteq E$.
Thus $E = F$ by definition of set equality.
To show that the sets in $F$ are (pairwise) disjoint, consider an arbitrary $x \in F$. 
Then $x \in F_k$ for some $F_k$.
By the Well-Ordering Principle, there exists a smallest such $k$.
Then:
:$\forall j < k: x \notin F_j$
Choose any distinct $\ell, m \in \N$. 
We have:
If $m > \ell$, then:
If $m < \ell$, then:
So the sets $F_\ell, F_m$ are disjoint. 
Thus $F$ is the disjoint union of sets equal to $E$:
:$\ds \bigcup_{k \mathop \in \N} E_k = \bigsqcup_{k \mathop \in \N} F_k$
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $12$
Category:Indexed Families
Category:Union of Indexed Family of Sets Equal to Union of Disjoint Sets
Union of Indexed Family of Sets Equal to Union of Disjoint Sets/Corollary

== Corollary to Union of Indexed Family of Sets Equal to Union of Disjoint Sets ==
Let $\family {E_n}_{n \mathop \in \N}$ be a countable indexed family of sets where at least two $E_n$ are distinct.
Let the countable indexed family of disjoint sets $\family {F_n}_{n \mathop \in \N}$ defined by:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
satisfying:
:$\ds \bigsqcup_{n \mathop \in \N} F_n = \bigcup_{n \mathop \in \N} E_n$
where $\bigsqcup$ denotes disjoint union.
The countable family $\family {F_k}_{k \mathop \in \N}$ can be constructed by:
:$\ds F_k = \bigcap_{j \mathop = 0}^{k \mathop - 1} \paren {E_k \setminus E_j}$
== Proof ==
We have that:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
satisfies:
:$\ds \bigsqcup_{n \mathop \in \N} F_n = \bigcup_{n \mathop \in \N} E_n$
from Union of Indexed Family of Sets Equal to Union of Disjoint Sets.
Then by a direct application of De Morgan's laws:
:$\ds E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j} = \bigcap_{j \mathop = 0}^{k \mathop - 1} \paren {E_k \setminus E_j}$
Category:Union of Indexed Family of Sets Equal to Union of Disjoint Sets