nyuuzyou/wikis huggingface dataset:
Finite Subset of Metric Space has no Limit Points

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $X \subseteq A$ such that $X$ is finite.
Then $X$ has no limit points.
== Proof ==
Let $x \in X$.
From Point in Finite Metric Space is Isolated, $x$ is an isolated point.
The result follows by definition of isolated point:
:$x$ is an isolated point $x$ is not a limit point.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 22$
Category:Metric Spaces
Finite Subset of Metric Space is Closed

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $S \subseteq A$ be finite.
Then $S$ is closed in $M$.
== Proof ==
From Metric Space is Hausdorff, $M$ is Hausdorff.
From Finite Subspace of Hausdorff Space is Closed, $S$ is closed.
Category:Closed Sets
Category:Metric Spaces
Finite Subset of Normed Vector Space is Closed

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $F \subseteq X$ be finite.
Then $F$ is closed in $M$.
== Proof ==
Suppose $F$ is empty.
By Empty Set is Closed in Normed Vector Space, $F$ is closed.
Suppose, for some $n \in \N$, that:
:$\ds F = \bigcup_{i \mathop = 1}^n \set {x_i}$
We have that Singleton in Normed Vector Space is Closed.
Hence $F$ is a finite union of closed sets.
By Finite Union of Closed Sets is Closed in Normed Vector Space, $F$ is closed.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Closed Sets (Normed Vector Spaces)
Finite Subset of Topological Vector Space is von Neumann-Bounded

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space over $\GF$. 
Let $F$ be a finite subset of $X$. 
Then $F$ is von Neumann-bounded.
== Proof ==
Let $F = \set {v_1, \ldots, v_m}$.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Multiple of Vector in Topological Vector Space Converges, we have: 
:$\ds \frac {v_i} n \to {\mathbf 0}_X$ as $n \to \infty$
for each $1 \le i \le n$.
Hence for each $1 \le i \le n$, there exists $N_i \in \N$ such that:
:$\ds \frac {v_i} n \in U$ for $n \ge N_i$
Let:
:$N = \max \set {N_1, \ldots, N_m}$
Then for $n \ge N$ we have:
:$v_i \in n U$ for each $1 \le i \le n$.
That is:
:$F \subseteq n U$ for each $1 \le i \le n$.
Since $U$ was an arbitrary open neighborhood of ${\mathbf 0}_X$, we have that $F$ is von Neumann-bounded.
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Finite Subsets form Directed Ordering

== Theorem ==
Let $I$ be a set.
Denote with $\FF$ the set of finite subsets of $I$.
Let $\subseteq$ be the subset relation on $\FF$.
Then $\subseteq$ is a directed ordering on $\FF$.
== Proof ==
From Subset Relation is Ordering, we know that $\subseteq$ is an ordering.
Now let $F, G \in \FF$.
From Set Union Preserves Subsets, conclude that $F \cup G \subseteq I$ as $F, G \subseteq I$.
From Union of Finite Sets is Finite, $F \cup G$ is a finite set.
Hence $F \cup G \in \FF$.
Furthermore, $F \subseteq F \cup G$ and $G \subseteq F \cup G$.
It follows that $\subseteq$ is a directed ordering.
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 2.$ Topology
Category:Directed Orderings
Category:Subset Relation
Finite Subsets form Directed Set

== Theorem ==
Let $I$ be a set.
Denote with $\FF$ the set of finite subsets of $I$.
Let $\subseteq$ be the subset relation on $\FF$.
Then $\struct {\FF, \subseteq}$ is a directed set.
== Proof ==
A fortiori from Finite Subsets form Directed Ordering.
Category:Preorder Theory
Category:Subset Relation
Finite Subsets form Ideal

== Theorem ==
Let $X$ be a set.
Let $\map {\operatorname {Fin} } X$ be the set of all finite subsets of $X$.
Then $\map {\operatorname {Fin} } X$ is ideal in $\struct {\powerset X, \subseteq}$
where $\powerset X$ denotes the power set of $X$.
== Proof ==
=== Non-Empty ===
By Empty Set is Subset of All Sets:
:$\O \subseteq X$ and $\O$ is finite.
By definition of $\operatorname {Fin}$:
:$\O \in \map {\operatorname {Fin} } X$
Thus by definition:
:$\map {\operatorname {Fin} } X$ is non-empty.
=== Directed ===
This follows from Finite Subsets form Directed Set.
=== Lower ===
Let $x \in \map {\operatorname {Fin} } X$, $y \in \powerset X$ such that
:$y \subseteq x$
By definition of $\operatorname {Fin}$:
:$x$ is a finite set.
By Subset of Finite Set is Finite:
:$y$ is a finite set.
By definition of power set:
:$y \subseteq X$
Thus by definition of $\operatorname {Fin}$:
:$y \in \map {\operatorname {Fin} } X$
Hence $\map {\operatorname {Fin} } X$ is ideal in $\struct {\powerset X, \subseteq}$
== Sources ==
* 
* 
Category:Preorder Theory
Category:Power Set
Finite Subspace of Dense-in-itself Metric Space is not Open

== Theorem ==
Let $M = \struct {A, d}$ be a metric space that is dense-in-itself.
Let $U$ be a finite subset of $A$.
Then $U$ is not an open set of $M$.
== Proof ==
Let $U = \set {x_0, x_1, \ldots, x_n}$.
 $U$ is open.
Let:
:$\ds D = \min_{i \mathop \ne j} \map d {x_i, x_j}$
Let $x_j \in U$.
Consider the open ball $\map {B_{D/2} } {x_j}$ of $x_j$.
From Open Ball of Metric Space is Open Set, $\map {B_{D/2} } {x_j}$ is an open set.
Then from Finite Intersection of Open Sets of Metric Space is Open, we can see that $U \cap \map {B_{D/2} } {x_j}$ is also an open set.
However, we have that:
:$\dfrac D 2 < \ds \min_{i \mathop \ne j} \map d {x_i, x_j}$
Thus the open ball $\map {B_{D / 2} } {x_j}$ is simply the singleton $\set {x_j}$.
But then $x_j$ is an isolated point of $M$.
This contradicts the fact that $M$ is dense-in-itself.
So in fact, $U$ cannot be open.
The result follows.
Category:Metric Spaces
Category:Denseness
Finite Summation does not Change under Permutation

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ be a finite set.
Let $f: S \to \mathbb A$ be a mapping.
Let $\sigma: S \to S$ be a permutation.
Then we have the equality of summations over finite sets:
:$\ds \sum_{s \mathop \in S} \map f s = \sum_{s \mathop \in S} \map f {\map \sigma s}$
== Proof ==
This is a special case of Change of Variables in Summation over Finite Set.
Category:Summations
Finite Suprema Set and Lower Closure is Ideal

== Theorem ==
Let $P = \struct {S, \vee, \preceq}$ be a join semilattice.
Let $X$ be a non-empty subset of $S$.
Then
:$\map {\operatorname{finsups} } X^\preceq$ is ideal in $P$.
where
:$\map {\operatorname{finsups} } X$ denotes the finite suprema set of $X$,
:$X^\preceq$ denotes the lower closure of $X$.
== Proof ==
By Finite Suprema Set and Lower Closure is Smallest Ideal:
:$X \subseteq \map {\operatorname{finsups} } X^\preceq$
By definition of non-empty set:
:$\map {\operatorname{finsups} } X^\preceq$ is a non-empty set.
We will prove that
:$\map {\operatorname{finsups} } X$ is directed.
Let $x, y \in \map {\operatorname{fininfs} } X$
By definition of finite suprema set:
:$\exists A \in \map {\operatorname {Fin} } X: x = \sup A \land A$ admits a supremum
and
:$\exists B \in \map {\operatorname {Fin} } X: y = \sup B \land B$ admits an supremum
where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $X$.
Define $C = A \cup B$.
By Union of Subsets is Subset:
:$C \subseteq X$
By Union of Finite Sets is Finite:
:$C$ is finite.
Then
:$C \in \map {\operatorname {Fin} } X$
By Existence of Non-Empty Finite Suprema in Join Semilattice:
:$C \ne \O \implies C$ admits a supremum.
By Union is Empty iff Sets are Empty:
:$C = \O \implies A = \O$
So
:$C$ admits a supremum.
By definition of finite suprema set:
:$\sup C \in \map {\operatorname{finsups} } X$
By Set is Subset of Union:
:$A \subseteq C$ and $B \subseteq C$
Thus by Supremum of Subset:
:$x \preceq \sup C$ and $y \preceq \sup C$
Hence $\map {\operatorname{finsups} } X$ is directed.
By Directed iff Lower Closure Directed:
:$\map {\operatorname{finsups} } X^\preceq$ is directed.
By Lower Closure is Lower Section:
:$\map {\operatorname{finsups} } X^\preceq$ is lower.
Hence $\map {\operatorname{finsups}} X^\preceq$ is ideal in $P$.
== Sources ==
* 
* 
* 
Category:Order Theory
Category:Suprema
Finite Suprema Set and Lower Closure is Smallest Ideal

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a join semilattice.
Let $X$ be a subset of $S$.
Then $X \subseteq \map {\operatorname {finsups} } X^\preceq$ and
:for every ideal $I$ in $L$: $X \subseteq I \implies \map {\operatorname {finsups} } X^\preceq \subseteq I$
where
:$\map {\operatorname {finsups} } X$ denotes the finite suprema set of $X$,
:$X^\preceq$ denotes the lower closure of $X$.
== Proof ==
By Set is Subset of Finite Suprema Set:
:$X \subseteq \map {\operatorname {finsups} } X$
By Lower Closure of Subset is Subset of Lower Closure:
:$X^\preceq \subseteq \map {\operatorname {finsups} } X^\preceq$
By Set is Subset of Lower Closure:
:$X \subseteq X^\preceq$
Thus by Subset Relation is Transitive:
:$X \subseteq \map {\operatorname {finsups} } X^\preceq$
Let $I$ be a ideal in $L$ such that
:$X \subseteq I$
Let $x \in \map {\operatorname {finsups} } X^\preceq$
By definition of lower closure of subset:
:$\exists y \in \map {\operatorname {finsups} } X: x \preceq y$
By definition of finite suprema set:
:$\exists A \in \map {\operatorname {Fin} } X: y = \sup A \land A$ admits a supremum
where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $S$.
By Subset Relation is Transitive:
:$A \subseteq I$
By Directed in Join Semilattice with Finite Suprema:
:$A \ne \O \implies \sup A \in I$
By Supremum of Empty Set is Smallest Element and Bottom in Ideal:
:$A = \O \implies \sup A = \bot \in I$
where $\bot$ denotes the smallest element of $S$.
So
:$y \in I$
Thus by definition of lower section:
:$x \in I$
== Sources ==
* 
Category:Order Theory
Category:Suprema
Finite Symmetric Group is Ambivalent

== Theorem ==
Let $n$ be a natural number.
Let $S_n$ be a symmetric group of order $n$.
Then $S_n$ is ambivalent.
== Proof ==
The Conjugacy Classes of Symmetric Group are determined uniquely by the cycle type of the elements.
Since any element in $S_n$ is of the same cycle type with its inverse, they are in the same conjugacy class.
Hence they are conjugates of each other.
This implies that $S_n$ is ambivalent.
Category:Ambivalent Groups
Finite T1 Space is Discrete

== Theorem ==
Let $S$ be a finite set.
Let $T = \left({S, \tau}\right)$ be a $T_1$ (Fréchet) space.
Then $\tau$ is the discrete topology on $S$.
== Proof 1 ==
== Proof 2 ==
Category:Finite Topological Spaces
Category:T1 Spaces
Category:Discrete Topology
Category:Finite T1 Space is Discrete
Finite T1 Space is Discrete/Proof 1

== Theorem ==
== Proof ==
Let $T = \struct {S, \tau}$ be a $T_1$ space on a finite set $S$.
Let $U \subseteq S$ be any subset of $S$. 
Let $H = \relcomp S U$ be the complement of $U$ relative to $S$.
Then by Relative Complement of Relative Complement we have that $U = \relcomp S H$.
We can write $H$ as:
:$\ds H = \bigcup_{x \mathop \in H} \set x$
From Equivalence of Definitions of $T_1$ Space, $\forall x \in H: \set x$ is closed in $T$.
As $S$ is a finite set, it follows that $H$ is a finite union of closed sets of $T$.
By Topology Defined by Closed Sets, $H$ is therefore closed in $T$.
By definition of closed set, $U = \relcomp S H$ is open in $T$.
As $U$ is arbitrary, it follows that:
:$\forall U \subseteq S: S \in \tau$
by definition of a topology.
Hence the result, by definition of the discrete topology.
Category:Finite T1 Space is Discrete
Finite T1 Space is Discrete/Proof 2

== Theorem ==
== Proof ==
We have from Finite Complement Topology is Minimal $T_1$ Topology that $\tau$ space is an expansion of a finite complement space.
But from the definition, any such finite complement space on a finite set is discrete.
But as the Discrete Topology is Finest Topology, $\tau$ must itself be the discrete topology on $S$.
Category:Finite T1 Space is Discrete
Finite Topological Space is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.
Then $T$ is compact.
== Proof by selecting finite subcover ==
== Proof by finiteness of the topology ==
== Sources ==
* : $5$: Compact spaces: $5.2$: Definition of compactness
* : $5$: Compact spaces: Exercise $5.10: 1$
Category:Finite Topological Spaces
Category:Compact Spaces
Category:Finite Topological Space is Compact
Finite Topological Space is Compact/Proof 1

== Theorem ==
== Proof ==
From Cardinality of Empty Set, the empty set is finite.
Suppose $S$ is empty.
Then by Empty Set is Compact Space, $T$ is compact.
Suppose $S$ is non-empty.
Let $\VV$ be an open cover of $T$.
For each $x \in S$, define $\VV_x$ to be $\set {V \in \VV : x \in V}$
Since $S$ is finite, and since by definition of a cover, each $x\in S$ is contained in at least one $V$ in $\VV$, we have that $\set {\VV_x : x \in S}$ is a finite collection of nonempty sets.
From Principle of Finite Choice, there is a choice function which selects one $V_x$ from $\VV_x$ for each $x \in S$. By definition of $\VV_x$, such a $V_x$ contains $x$.
Since there were only finitely many $\VV_x$, this provides finitely many open sets $V_x \in \VV$ such that $\ds S \subseteq \bigcup_{x \mathop \in S} V_x$.
Thus $\set {\VV_x : x \in S}$ is a finite subcover of $\VV$.
The result follows by definition of compact.
Category:Finite Topological Space is Compact
Finite Topological Space is Compact/Proof 2

== Theorem ==
== Proof ==
From Power Set of Finite Set is Finite, the power set of $S$ is finite.
Since the topology $\tau$ is by definition a set of subsets of $S$, it follows that $\tau$ is finite as well.
Let $\VV$ be an open cover of $S$.
By definition $\VV \subseteq \tau$ and so is also a finite set.
From Set is Subset of Itself, $\VV \subseteq \VV$.
Thus by definition $\VV$ is a finite subcover of $\VV$.
The result follows by definition of compact.
Category:Finite Topological Space is Compact
Finite Totally Ordered Set is Well-Ordered

== Theorem ==
Every finite totally ordered set is well-ordered.
== Proof ==
Let $\struct {S, \preceq}$ be a finite totally ordered set.
 $\preceq$ is not a well-founded relation.
From Infinite Sequence Property of Well-Founded Relation, $\struct {S, \preceq}$ is well-founded there is no infinite sequence $\sequence {a_n}$ of elements of $S$ such that $\forall n \in \N: a_{n + 1} \prec a_n$.
Let us construct such an infinite sequence.
Then at least some of the terms would be repeated in that sequence.
So there would be, for example:
:$s_i \preceq s_j \preceq s_k \preceq s_i$
But then we have:
:$s_i \preceq s_j$ and $s_j \preceq s_k$
and as $\preceq$ is transitive, we have:
:$s_i \preceq s_k$
But as we also have $s_k \preceq s_i$
we have $s_i \preceq s_k$ and $s_k \preceq s_i$ where $s_i \ne s_k$.
That is, $\preceq$ is not antisymmetric.
That is, $\preceq$ is therefore not a totally ordered set.
From this contradiction it follows that $\preceq$ is well-founded.
The result follows from the definition of well-ordered set.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
* : $\S 3.5$: Well-ordered sets. Ordinal Numbers: Example $1$
Category:Total Orderings
Category:Well-Orderings
Finite Tree has Leaf Nodes/Proof 2

== Theorem ==
== Proof ==
Let the proposition we are proving in the case of a tree of order $k$ be denoted $\map P k$.
=== Basis for the Induction ===
That $\map P 2$ is true follows as the Unique Tree of Order 2 contains exactly two leaves (viz. its two nodes).
This is our basis for the induction.
=== Induction Hypothesis ===
Suppose $\map P i$ is true for all $i$ such that $2 \le i \le k$.
Then we need to demonstrated that $\map P i$ is true for all $i$ such that $2 \le i \le k + 1$.
This is our induction hypothesis.
=== Induction Step ===
This is our induction step:
Let $T_{k + 1}$ be a tree of order $k + 1$.
Pick an edge $e = \set {v_1, v_2}$ of $T_{k + 1}$.
Consider the subgraph $T_0$ obtained by removing $e$.
By Edge of Tree is Bridge, $e$ is a bridge and removing it results in a subgraph with two components.
By Connected Subgraph of Tree is Tree each such component is a tree.
Let $T_i$ be the component containing $v_i$.
Denote its order by $k_i$, where $i \in \set{1, 2}$
Since no nodes have been removed from $T_{k + 1}$, $k_1 + k_2 = k + 1$.
There are two cases:
==== Case 1 ====
Neither $T_1$ nor $T_2$ is of order $1$.
Then $2 \le k_i \le k - 1, i = 1, 2$.
By the induction hypothesis, both $T_1$ and $T_2$ have at least two leaves.
Either $v_1$ and $v_2$ are both leaves of the respective trees, or one of them is not.
Suppose that $v_1$ and $v_2$ are both leaves of the respective trees.
Then by adding $e$ back to the graph and thus recreating $T_{k + 1}$, $v_1$ and $v_2$ cease to be leaves.
However, at least one other leaf from each subgraph $T_i, i = 1, 2$ survives the addition, becoming a leaf of $T_{k + 1}$.
Therefore $T_{k + 1}$ has at least two leaves.
Suppose that either $v_1$ or $v_2$ is not a leaf of its respective tree.
Then $T_{k + 1}$ has at least three leaves.
==== Case 2 ====
$T_1$ or $T_2$ is of order $1$.
, assume that $T_1$ is of order $1$, that is $k_1 = 1$.
This implies that $k_2 = k$.
Then, by the induction hypothesis, $T_2$ has at least two leaf nodes.
Also $k_1 = 1$ implies that $v_1$ is a leaf of $T_{k + 1}$.
Suppose that $v_2$ is one of the leaves of $T_2$.
Then by adding $e$ back and thus recreating $T_{k + 1}$, $v_2$ is no longer a leaf, though at least one other leaf of $T_2$ survives the addition, becoming a leaf of $T_{k + 1}$.
Therefore $T_{k + 1}$ has at least two leaves.
Suppose that $v_2$ is not one of the leaves of $T_2$.
Then $T_{k + 1}$ has at least three leaves.
In both cases, the thesis follows by the Second Principle of Mathematical Induction.
Category:Finite Tree has Leaf Nodes
Finite Union of Bounded Subsets

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then the union of any finite number of bounded subsets of $M$ is itself bounded.
== Proof ==
It is sufficient to prove this for two subsets, as the general result follows by induction.
Suppose $S_1$ and $S_2$ are bounded subsets of $M = \struct {A, d}$.
Let $a_1, a_2 \in A$.
Let $K_1, K_2 \in \R$ such that:
:$(1): \quad \forall x \in S_1: \map d {x, a_1} \le K_1$
:$(2): \quad \forall x \in S_2: \map d {x, a_2} \le K_2$
, let $a = a_1$ and $K = \max \set {K_1, K_2 + \map d {a_1, a_2} }$.
Then $\forall x \in S_1 \cup S_2$:
Either:
:$x \in S_1$ and so $\map d {x, a} \le K_1 \le K$
or:
:$x \in S_2$ and so $\map d {x, a} = \map d {x, a_1} \le \map d {x, a_2} + \map d {a_2, a_1} \le K_2 + \map d {a_2, a_1} \le K$
Hence the result.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Proposition $2.2.15$
Category:Boundedness
Finite Union of Closed Sets is Closed/Normed Vector Space

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then the union of finitely many closed sets of $M$ is itself closed.
== Proof ==
Let $\ds \bigcup_{i \mathop = 1}^n F_i$ be the union of a finite number of closed sets of $M$.
By definition of closed set, each of the $X \setminus F_i$ is by definition open in $M$.
Then from De Morgan's laws:
:$\ds X \setminus \bigcup_{i \mathop = 1}^n F_i = \bigcap_{i \mathop = 1}^n \paren {X \setminus F_i}$
We have that $\ds \bigcap_{i \mathop = 1}^n \paren {X \setminus F_i}$ is the intersection of a finite number of open sets of $M$.
By Finite Intersection of Open Sets of Normed Vector Space is Open, $\ds \bigcap_{i \mathop = 1}^n \paren {X \setminus F_i} = X \setminus \bigcup_{i \mathop = 1}^n F_i$ is likewise open in $M$.
Then by definition of closed set, $\ds \bigcup_{i \mathop = 1}^n F_i$ is closed in $M$.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Finite Union of Closed Sets is Closed
Category:Closed Sets (Normed Vector Spaces)
Finite Union of Closed Sets is Closed/Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then the union of finitely many closed sets of $T$ is itself closed.
== Proof ==
Let $\ds \bigcup_{i \mathop = 1}^n V_i$ be the union of a finite number of closed sets of $T$.
Then from De Morgan's laws:
:$\ds S \setminus \bigcup_{i \mathop = 1}^n V_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus V_i}$
By definition of closed set, each of the $S \setminus V_i$ is by definition open in $T$.
We have that $\ds \bigcap_{i \mathop = 1}^n \paren {S \setminus V_i}$ is the intersection of a finite number of open sets of $T$.
Therefore, by definition of a topology, $\ds \bigcap_{i \mathop = 1}^n \paren {S \setminus V_i} = S \setminus \bigcup_{i \mathop = 1}^n V_i$ is likewise open in $T$.
Then by definition of closed set, $\ds \bigcup_{i \mathop = 1}^n V_i$ is closed in $T$.
== Sources ==
* : $2.24 d$
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets: Theorem $1$
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $4$
Category:Finite Union of Closed Sets is Closed
Finite Union of Compact Sets is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $n \in \N$ be a natural number.
Let $\sequence {U_i}_{1 \mathop \le i \mathop \le n}$ be a finite sequence of compact subsets of $T$.
Let $\UU_n := \ds \bigcup_{i \mathop = 1}^n U_i$ be the union of $\sequence {U_i}$.
Then $\UU_n$ is compact in $T$.
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\UU_n := \ds \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$.
$\map P 0$ is the case:
:$\UU_0 := \ds \bigcup_{i \mathop = 1}^0 U_i$
From Union of Empty Set:
:$\ds \bigcup_{i \mathop = 1}^0 U_i = \O$
From Empty Set is Compact Space it follows that:
:$\UU_0$ is compact in $T$.
$\map P 1$ is true, as this just says:
:$U_1$ is compact in $T$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$U_1 \cup U_2$ is compact in $T$.
This is proved in Union of Two Compact Sets is Compact.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\UU_k := \ds \bigcup_{i \mathop = 1}^k U_i$ is compact in $T$.
Then we need to show:
:$\UU_{k+1} := \ds \bigcup_{i \mathop = 1}^{k + 1} U_i$ is compact in $T$.
=== Induction Step ===
This is our induction step:
We have that:
:$\ds \bigcup_{i \mathop = 1}^{k + 1} U_i = \paren {\bigcup_{i \mathop = 1}^k U_i} \cup U_{k + 1}$
By the induction hypothesis:
:$\UU_k$ is compact in $T$.
By the basis for the induction:
:$\UU_k \cup U_{k + 1}$ is compact in $T$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$.
Category:Compact Spaces
Category:Set Union
Finite Union of Countable Sets is Countable

== Theorem ==
The union of a finite number of countable sets is countable.
== Proof ==
Let $S_0, \ldots, S_{n - 1}$ be countable sets.
For $i \in \set {0, \ldots, n - 1}$, let $f_i: \N \to S_i$ be a surjection.
These exist by Surjection from Natural Numbers iff Countable.
Now define $f: \N \to \ds \bigcup_{i \mathop = 0}^{n - 1} S_i$ by:
:$\map f m := \map {f_i} {\floor {\dfrac m n} }$
where:
:$i$ is the unique element of $\set {0, \ldots, n - 1}$ such that $m \equiv i \pmod n$
:$\floor x$ denotes the floor of $x$.
Now let $x \in \ds \bigcup_{i \mathop = 0}^{n - 1} S_i$ be arbitrary.
Let $k$ be the smallest natural number such that $x \in S_k$.
Let $l$ be the smallest natural number such that $\map {f_k} l = x$.
These are guaranteed to exist by definition of set union and surjectivity of $f_k$, respectively.
Then we have:
:$\map f {l n + k} = \map {f_k} {\floor {\dfrac {l n + k} n} } = \map {f_k} l = x$
since $k < n$.
Since $x$ was arbitrary, $f$ is a surjection.
Hence by Surjection from Natural Numbers iff Countable, $\ds \bigcup_{i \mathop = 0}^{n - 1} S_i$ is countable
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Exercise $18.17$
Category:Set Union
Category:Countable Sets