nyuuzyou/wikis huggingface dataset:
Compact Subset is Bounded Below Join Semilattice

== Theorem ==
Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Let $P = \left({K\left({L}\right), \precsim}\right)$ be an ordered subset of $L$,
where $K\left({L}\right)$ denotes the compact subset of $L$.
Then $P$ is a bounded below join semilattice.
== Proof ==
By Bottom is Compact:
:$\bot_L$ is a compact element,
where $\bot_L$ denotes the smallest element in $L$.
By definition of compact subset:
:$\bot_L \in K \left({L} \right)$
By definition of the smallest element:
:$\forall x \in K\left({L}\right): \bot_L \preceq x$
By definition of ordered subset:
:$\forall x \in K\left({L}\right): \bot_L \precsim x$
Thus by definition:
:$P$ is bounded below.
Thus by Compact Subset is Join Subsemilattice:
:$P$ is join semilattice.
== Sources ==
* 
* 
Category:Join and Meet Semilattices
Compact Subset is Join Subsemilattice

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\map K L$ be a compact subset of $L$.
Then $\map K L$ is join subsemilattice:
:$\forall x, y \in \map K L: x \vee y \in \map K L$
== Proof ==
Let $x, y \in \map K L$.
By definition of compact subset:
:$x$ and $y$ are compact.
By definition of compact:
:$x \ll x$ and $y \ll y$
By Way Below is Congruent for Join:
:$x \vee y \ll x \vee y$
By definition:
:$x \vee y$ is compact.
Thus by definition compact subset:
:$x \vee y \in \map K L$
== Sources ==
* 
* 
Category:Way Below Relation
Compact Subset of Compact Space is not necessarily Closed

== Theorem ==
A compact subset of a compact space is not necessarily closed.
== Proof ==
Let $S$ be a set containing more than one element.
Let $\tau = \set {S, \O}$ be the indiscrete topology on $S$.
Let $x \in S$.
Let $H = S \setminus \set x$.
Then $H$ is a proper subset of $S$.
Then from Subset of Indiscrete Space is Compact and Sequentially Compact, the subspace induced by $\tau$ on $H$ is a compact subspace of $\struct {S, \tau}$.
But from Closed Sets in Indiscrete Topology, the only closed subsets of $\struct {S, \tau}$ are $S$ and $\O$.
Thus $H$ is a compact subspace of $\struct {S, \tau}$ which is not closed in $S$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:Closed Sets
Category:Compact Spaces
Compact Subset of Hilbert Sequence Space is Closed

== Theorem ==
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Let $H$ be a compact subset of $\ell^2$
Then $H$ is closed in $\ell^2$.
== Proof ==
From Metric Space fulfils all Separation Axioms, $\ell^2$ is a Hausdorff space.
The result follows from Compact Subspace of Hausdorff Space is Closed.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $4$
Category:Hilbert Sequence Space
Compact Subset of Hilbert Sequence Space is Nowhere Dense

== Theorem ==
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Let $H$ be a compact subset of $\ell^2$.
Then $H$ is nowhere dense in $\ell^2$.
== Proof ==
By Compact Subset of Hilbert Sequence Space is Closed, $H$ is a closed set of $\ell^2$.
From Set is Closed iff Equals Topological Closure:
:$H^- = H$
where $H^-$ denotes the closure of $H$.
From Subset of Hilbert Sequence Space with Non-Empty Interior is not Compact:
:$H^\circ = \O$
where $H^\circ$ denotes the interior of $H$.
Thus:
:$\paren {H^-}^\circ = \O$
and the result follows by definition of nowhere dense.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $4$
Category:Hilbert Sequence Space
Category:Nowhere Denseness
Compact Subset of Normed Vector Space is Closed and Bounded

== Theorem ==
Let $\struct {X, \norm {\,\cdot\,}}$ be a normed vector space.
Let $K \subset X$ be a compact subset.
Then $K$ is closed and bounded.
== Proof ==
=== Closedness ===
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $K$.
Suppose, $\sequence {x_n}_{n \mathop \in \N}$ converges to $L \in K$. 
Then there is a subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ convergent to $L' \in K$. 
But $\sequence {x_{n_k}}_{k \mathop \in \N}$ is a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
Hence, $\sequence {x_{n_k}}_{k \mathop \in \N}$ converges to $L$.
By uniqueness of limits, $L = L'$.
By definition, $K$ is closed.
=== Boundedness ===
Suppose, $K$ is not bounded.
Then:
:$\forall n \in \N : \exists x_n \in K: \norm {x_n} > n$
Therefore, no subsequence of $\sequence {x_n}_{n \mathop \in \N}$ is convergent.
Hence, $K$ is not compact.
This is in contradiction with the assumption.
Thus, $K$ is bounded.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Boundedness
Category:Normed Vector Spaces
Compact Subsets of T3 Spaces

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_3$ space.
Let $A \subseteq S$ be compact in $T$.
Then for each $U \in \tau$ such that $A \subseteq U$:
:$\exists V \in \tau: A \subseteq V \subseteq V^- \subseteq U$
where $V^-$ denotes the closure of $V$.
== Proof ==
Let $A \subseteq S$ be compact in $T$.
Let $U \in \tau$ such that $A \subseteq U$.
Since $T$ is $T_3$:
:Each open set contains a closed neighborhood around each of its points:
::$\forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U$
Note that $\set {V_x: x \in A}$ forms an open cover for $A$.
By compactness of $A$, $\set {V_x: x \in A}$ has a finite subcover.
Therefore there is a finite subset $I \subseteq A$ such that: 
:$\set {V_x: x \in I}$ is an open cover for $A$.
Let $V = \ds \bigcup_{i \mathop \in I} V_x$, $N = \ds \bigcup_{i \mathop \in I} N_x$.
By definition of a topology, the set $\ds \bigcup_{i \mathop \in I} V_x$ is open since it is a union of open sets.
By Finite Union of Closed Sets is Closed in Topological Space, the set $\ds \bigcup_{i \mathop \in I} N_x$ is closed since it is a finite union of closed sets.
By Set Union Preserves Subsets, since $V_x \subseteq N_x \; \forall x \in I \subseteq U$:
:$V \subseteq N$
By Set Closure is Smallest Closed Set in Topological Space:
:$V \subseteq V^- \subseteq N$
Thus:
satisfies our claim.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
 
Category:T3 Spaces
Category:Compact Spaces
Compact Subspace of Hausdorff Space is Closed

== Theorem ==
Let $H = \struct {A, \tau}$ be a Hausdorff space.
Let $C$ be a compact subspace of $H$.
Then $C$ is closed in $H$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Hausdorff Spaces
Category:Compact Spaces
Category:Closed Sets
Category:Topological Subspaces
Category:Compact Subspace of Hausdorff Space is Closed
Compact Subspace of Hausdorff Space is Closed/Corollary

== Corollary to Compact Subspace of Hausdorff Space is Closed ==
A finite subspace of a Hausdorff space is closed.
== Proof ==
Follows directly from:
:Finite Topological Space is Compact
and:
:Compact Subspace of Hausdorff Space is Closed.
== Sources ==
* : $5$: Compact spaces: $5.4$: Properties of compact spaces: Proposition $5.4.2$
Category:Finite Topological Spaces‎
Category:Compact Subspace of Hausdorff Space is Closed
Compact Subspace of Hausdorff Space is Closed/Proof 1

== Theorem ==
== Proof ==
From Subspace of Hausdorff Space is Hausdorff, a subspace of a Hausdorff space is itself Hausdorff.
Let $a \in A \setminus C$.
We are going to prove that there exists an open set $U_a$ such that $a \in U_a \subseteq A \setminus C$.
For any single point $x \in C$, the Hausdorff condition ensures the existence of disjoint open set $\map U x$ and $\map V x$ containing $a$ and $x$ respectively.
Suppose there were only a finite number of points $x_1, x_2, \ldots, x_r$ in $C$.
Then we could take $\ds U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$ and get $a \in U_a \subseteq A \setminus C$.
Now suppose $C$ is not finite.
The set $\set {\map V x: x \in C}$ is an open cover for $C$.
As $C$ is compact, it has a finite subcover, say $\set {\map V {x_1}, \map V {x_2}, \dotsc, \map V {x_r} }$.
Let $\ds U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$.
Then $U_a$ is open because it is a finite intersection of open sets.
Also, $a \in U_a$ because $a \in \map U {x_i}$ for each $i = 1, 2, \ldots, r$.
Finally, if $b \in U_a$ then for any $i = 1, 2, \ldots, r$ we have $b \in \map U {x_i}$.
Because $\ds C \subseteq \bigcup_{i \mathop = 1}^r \map V {x_i}$:
:$b \notin \map V {x_i}$, so $b \notin C$
Thus:
:$U_a \subseteq A \setminus C$.
Then:
:$\ds A \setminus C = \bigcup_{a \mathop \in A \mathop \setminus C} U_a$
So $A \setminus C$ is open.
It follows that $C$ is closed.
== Sources ==
* : $5$: Compact spaces: $5.4$: Properties of compact spaces: Proposition $5.4.2$
Category:Compact Subspace of Hausdorff Space is Closed
Compact Subspace of Hausdorff Space is Closed/Proof 2

== Theorem ==
== Proof ==
For a subset $S \subseteq A$, let $S^{\complement}$ denote the relative complement of $S$ in $A$.
Consider an arbitrary point $x \in C^{\complement}$.
Define the set:
:$\ds \OO = \set {V \in \tau: \exists U \in \tau: x \in U \subseteq V^\complement}$
By Empty Intersection iff Subset of Complement:
:$U \subseteq V^\complement \iff U \cap V = \O$
Hence, by the definition of a Hausdorff space, it follows that $\OO$ is an open cover for $C$.
By the definition of a compact subspace, there exists a finite subcover $\FF$ of $\OO$ for $C$.
By the Principle of Finite Choice, there exists an $\FF$-indexed family $\family {U_V}_{V \mathop \in \FF}$ of elements of $\tau$ such that:
:$\forall V \in \FF: x \in U_V \subseteq V^\complement$
Define:
:$\ds U = \bigcap_{V \mathop \in \FF} U_V$
By General Intersection Property of Topological Space:
:$U \in \tau$
Clearly, $x \in U$.
We have that:
From Subset Relation is Transitive, we have that $U \subseteq C^\complement$.
Hence $C^\complement$ is a neighborhood of $x$.
From Set is Open iff Neighborhood of all its Points:
:$C^\complement \in \tau$
That is, $C$ is closed in $H$.
Category:Compact Subspace of Hausdorff Space is Closed
Compact Subspace of Linearly Ordered Space

== Theorem ==
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Then $Y$ is a compact subspace of $\struct {X, \tau}$ both of the following hold:
:$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
:$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.
== Proof ==
=== Forward Implication ===
Let $S$ be a non-empty subset of $Y$.
By Compact Subspace of Linearly Ordered Space: Lemma 1, $S$ has a supremum $k$ in $Y$.
 that $S$ has an upper bound $b$ in $X$ such that $b \prec k$.
Let:
:$\AA = \set {s^\preceq: s \in S} \cup \set {b^\succeq}$
where:
:$s^\preceq$ denotes the lower closure of $s$ in $S$
:$b^\succeq$ denotes the upper closure of $s$ in $S$.
Then $\AA$ is an open cover of $Y$.
But $\AA$ has no finite subcover, contradicting the fact that $Y$ is compact.
A similar argument proves the corresponding statement for infima, so $(2)$ holds.
=== Reverse Implication ===
== Also see ==
* Heine–Borel Theorem: Dedekind-Complete Space
* Connected Subspace of Linearly Ordered Space
== Sources ==
* : $\S 5$: Problem $\text C$
Category:Linearly Ordered Spaces
Category:Compact Spaces
Category:Compact Subspace of Linearly Ordered Space
Compact Subspace of Linearly Ordered Space/Lemma 1

== Theorem ==
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Let $Y$ be a compact subspace of $\struct {X, \tau}$.
Then $\struct {Y, \preceq \restriction_Y}$ is a complete lattice, where $\restriction$ denotes restriction.
== Proof ==
 $Y$ is not a complete lattice.
Then there is an $S \subseteq Y$ with no least upper bound in $Y$.
Let $U = \set {b \in Y: b \text { is an upper bound of } S}$.
Note that $U$ may be empty.
Let:
:$\UU = \set {u^\succeq: u \in U}$
:$\LL = \set {s^\preceq: s \in S}$.
where:
:$u^\succeq$ is the upper closure of $u$ in $U$
:$s^\preceq$ is the lower closure of $s$ in $S$.
Then $\UU \cup \LL$ covers $Y$:
Let $y \in Y$.
Suppose $y$ is an upper bound for $S$.
Then because $S$ has no least upper bound in $Y$, there exists a $y' \in U$ such that $y' \prec y$.
Thus $y \in y'^\succeq$, so, in the case where $y$ is an upper bound for $S$:
:$\ds y \in \bigcup \UU$
Otherwise, that is if $y$ is not an upper bound for $S$, there exists an $s \in S$ such that $y \prec s$.
Hence in this case:
:$\ds y \in \bigcup \LL$
$\UU \cup \LL$ has no finite subcover:
Let $F_U$ and $F_L$ be finite subsets of $U$ and $S$, respectively.
Define:
:$\FF_U = \set {u^\succeq: u \in F_S}$
:$\FF_L = \set {s^\preceq: s \in F_L}$
We will show that $\FF_U \cup \FF_L$ does not cover $Y$.
Suppose to the contrary that $\FF_U \cup \FF_L$ is a cover for $Y$.
Since $Y$ is non-empty, $F_L$ or $F_U$ is non-empty.
If $F_L$ is non-empty, it has a maximum $m$.
Since $S$ has no least upper bound, it has no maximum
Therefore there is an $s \in S$ such that $m \prec s$.
Thus $\ds s \notin \bigcup \FF_L$.
By the definition of upper bound, no upper bound of $S$ precedes $m$.
So $\ds m \notin \bigcup \FF_U$.
Thus $\FF_L \cup \FF_U$ does not cover $Y$.
This is a contradiction of our supposition that it does.
Hence it cannot be the case that $\FF_U \cup \FF_L$ is a cover for $Y$.
If $F_U$ is non-empty, it has a minimum $m$.
Then since $S$ has no least upper bound in $Y$, there is a $u \in U$ such that $u \prec m$.
Thus $\ds u \notin \bigcup \FF_U$.
But since $u$ is an upper bound of $S$, it is not succeeded by any element of $S$.
So $\ds u \notin \bigcup \FF_L$.
Thus $\FF_L \cup \FF_U$ does not cover $Y$, a contradiction.
So $\UU \cup \LL$ is an open cover of $Y$ with no finite subcover
So $Y$ is not compact, which contradicts the supposition.
By Proof by Contradiction, it follows that $\struct {Y, \preceq \restriction_Y}$ is a complete lattice.
Category:Compact Subspace of Linearly Ordered Space
Compact Subspace of Linearly Ordered Space/Lemma 2

== Lemma ==
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Let the following hold:
:$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
:$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.
Then:
:$\struct {Y, \preceq', \tau'}$ is a linearly ordered space.
== Proof ==
By definition of a generalized ordered space:
:$\tau'$ has a sub-basis consisting of upper and lower sections in $Y$.
To prove that $\struct {Y, \preceq', \tau'}$ is a linearly ordered space, we need to show that each element of this sub-basis is open in the $\preceq'$ order topology.
Let $U$ be a non-empty, $\tau'$-open upper section in $Y$.
If $U = \O$ or $U = Y$, then $U$ is open in the $\preceq'$-order topology by the definition of topology.
Suppose then that $\O \subsetneqq U \subsetneqq Y$.
Let $C = Y \setminus U$.
Then $C$ is non-empty and $\tau'$-closed.
$Y$ is $\tau'$-closed by Closed Set in Linearly Ordered Space.
Thus $C$ is also $\tau'$-closed.
$C$ has a supremum in $X$ by the premise.
Let $c = \sup_X C$
Since $C$ is $\tau'$-closed, Closed Set in Linearly Ordered Space shows that $c \in C$.
Since $c \in C = Y \setminus U$ and $U$ is an upper section in $Y$:
:$c \prec u$ for all $u \in U$.
Furthermore, if $y \in Y$ and $c \prec y$, then by the definition of supremum, $y \notin C$, so $y \in U$.
Thus:
:$U = {c^\succeq}_Y$
where ${c^\succeq}_Y$ denotes the upper closure of $c$ in $Y$.
So $U$ is open in the $\preceq'$-order topology for $Y$.
The same argument shows that $\tau'$-open lower sections in $Y$ are open in the $\preceq'$-order topology.
Thus $\tau'$ is the $\preceq'$-order topology.
Category:Compact Subspace of Linearly Ordered Space
Compact Subspace of Linearly Ordered Space/Reverse Implication

== Theorem ==
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Let the following hold:
:$(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
:$(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.
Then $Y$ is a compact subspace of $\struct {X, \tau}$.
== Proof 1 ==
== Proof 2 ==
Category:Compact Subspace of Linearly Ordered Space
Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 1

== Theorem ==
== Proof ==
Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.
Let $\preceq'$ be the restriction of $\preceq$ to $Y$.
=== Lemma ===
The premises immediately show that $\struct {Y, \preceq'}$ is a complete lattice.
By Complete Linearly Ordered Space is Compact, $Y$ is a compact subspace of $X$.
Category:Compact Subspace of Linearly Ordered Space
Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 2

== Theorem ==
== Proof ==
Let $\FF$ be an ultrafilter on $Y$.
For $S \in \FF$, let $\map f S = \inf S$.
Let $p = \sup \map f \FF$.
Then $\FF$ converges to $p$:
=== Upward rays ===
Let $a \in X$ with $a \prec p$.
Since $\FF$ is an ultrafilter, either $Y \cap {\uparrow} a \in \FF$ or $Y \cap {\bar \downarrow} a \in \FF$.
 that $Y \cap {\bar \downarrow} a \in \FF$.
For each $S \in \FF$:
:$S \cap {\bar \downarrow} a \in \FF$ because an ultrafilter is a filter.
:$S \cap {\bar \downarrow} a \ne \O$ because a filter on a set is proper.
By applying the definition of supremum to $p$, it follows that there exists an $S \in \FF$ such that $a \prec \inf S$.
By the definition of infimum:
:$S \cap {\bar \downarrow} a = \O$
which is a contradiction.
Thus $Y \cap {\uparrow} a \in \FF$.
=== Downward rays ===
Let $b \in X$ with $p \prec b$.
Either $Y \cap {\downarrow} b \in \FF$ or $Y \cap {\bar \uparrow} b \in \FF$.
 that $Y \cap {\bar \uparrow} b \in \FF$.
Let $b' = \map \inf {Y \cap {\bar \uparrow} b}$.
We have that $b$ is a lower bound of $Y \cap {\bar \uparrow} b$
So by the definition of infimum:
:$b \preceq b'$
Since $p \prec b$ and $b \preceq b'$, $p \prec b'$ by Extended Transitivity.
By the definition of $b'$ and the definition of $f$:
:$b' \in \map f \FF$
But this contradicts the fact that $p$ is the supremum, and hence an upper bound, of $\map f \FF$.
By the definition of the order topology, the upward and downward rays containing each point form a neighborhood sub-basis for that point.
Thus by the Neighborhood Sub-Basis Criterion for Filter Convergence, $\FF$ converges.
Since every ultrafilter on $Y$ converges, $Y$ is compact by Equivalence of Definitions of Compact Topological Space.
Category:Compact Subspace of Linearly Ordered Space
Compact Subspace of Metric Space is Bounded

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $C$ be a subspace of $M$.
If $C$ is compact, then it is bounded.
== Proof 1==
Let $a \in M$.
Let $n \in \N_{>0}$.
Let $\map {B_n} a$ be the open $n$-ball of $a$.
By definition of open $n$-ball, $\forall x \in C: \map d {x, a} < n$, there exists $n \in \N$ such that:
:$\ds C \subseteq \bigcup_{n \mathop = 1}^\infty \map {B_n} a$
Thus the collection:
:$\set {\map {B_n} a: n \in \N}$ 
forms an open cover of $C$.
Because $C$ is compact, it has a finite subcover, say: 
:$\set {\map {B_{n_1} } a, \map {B_{n_2} } a, \ldots, \map {B_{n_r} } a}$
Let $n = \max \set {n_1, n_2, \ldots, n_r}$.
Then:
:$\ds C \subseteq \bigcup_{n \mathop = 1}^r \map {B_{n_r} } a = \map {B_n} a$
The result follows by definition of bounded.
== Proof 2==
We are given that $C$ is compact.
By Compact Metric Space is Totally Bounded, $C$ is totally bounded.
So $C$ is bounded.
Hence the result.
== Also see==
* Heine-Borel Theorem/Metric Space
== Sources ==
* : $5$: Compact spaces: $5.4$: Properties of compact spaces: Proposition $5.4.1$
Category:Metric Subspaces
Category:Compact Spaces
Compact Subspace of Metric Space is Sequentially Compact in Itself

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $C \subseteq M$ be a subspace of $M$ such that $C$ is compact.
Then $C$ is sequentially compact in itself.
== Proof ==
Let $C \subseteq M$ be compact.
Let $\sequence {x_n}$ be a sequence in $C$.
Let $S$ be the range of $\sequence {x_n}$.
Thus $S \subseteq C$ and $S$ may be either finite or infinite.
=== Finite Range ===
Let $S$ be finite.
Then at least one $x \in S$ must be repeated infinitely often in $\sequence {x_n}$.
Its occurrences form a subsequence converging to $x$.
=== Infinite Range ===
Let $S$ be infinite.
From Subsequence of Sequence in Metric Space with Limit, it is enough to show that $S$ has a limit point in $C$.
 $S$ has no limit point in $C$.
Then for each $x \in C$, there exists $\epsilon \in \R_{>0}$ such that the open $\epsilon$-ball $\map {B_\epsilon} x$ contains no $y \in S: y \ne x$.
That is:
:$\map {B_\epsilon} x \cap S = \set x$
or
:$\map {B_\epsilon} x \cap S = \O$
Because $C$ is compact, the open cover $\BB := \set {\map {B_\epsilon} x: x \in C}$ has a finite subcover.
But each $\map {B_\epsilon} x \in \BB$ contains at most one point of $S$.
Therefore the union of any finite subset of $\BB$ contains only finitely many points of $S$.
So no finite subset of $\BB$ covers $S$, let alone $C$.
Thus $S$ must have a limit point in $C$.
== Sources ==
* : $7.2$: Sequential compactness: Proposition $7.2.6$
* 
Category:Compact Spaces
Category:Sequentially Compact Spaces
Category:Metric Spaces
Compact Subspace of Real Numbers is Closed and Bounded

== Theorem ==
Let $\R$ be the real number line considered as a Euclidean space.
Let $S \subseteq \R$ be compact subspace of $\R$.
Then $S$ is closed and bounded in $\R$.
== Proof 1 ==
== Proof 2 ==
Category:Compact Spaces
Category:Analysis
Category:Compact Subspace of Real Numbers is Closed and Bounded
Compact Subspace of Real Numbers is Closed and Bounded/Proof 1

== Theorem ==
== Proof ==
From:
: Non-Closed Set of Real Numbers is not Compact
: Unbounded Set of Real Numbers is not Compact
the result follows by the Rule of Transposition.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Compact Subspace of Real Numbers is Closed and Bounded
Compact Subspace of Real Numbers is Closed and Bounded/Proof 2

== Theorem ==
== Proof ==
From Real Number Line is Metric Space, $\left({\R, d}\right)$ is a metric space, where $d$ denotes the Euclidean metric on $\R$.
Therefore, the result follows from:
: Metric Space is Hausdorff
: Compact Subspace of Hausdorff Space is Closed
and:
: Compact Metric Space is Totally Bounded
: Totally Bounded Metric Space is Bounded
Category:Compact Subspace of Real Numbers is Closed and Bounded
Compact Subspace of Topological Vector Space is von Neumann-Bounded

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$. 
Let $K$ be a compact subspace (where subspace is meant in the topological sense) of $X$.
Then $K$ is von Neumann-bounded.
== Proof ==
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
We aim to find $s > 0$ such that $E \subseteq t V$ for $t > s$.
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists $W \subseteq V$ such that $W$ is a balanced open neighborhood of ${\mathbf 0}_X$.
From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, we have: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty n W$
In particular: 
:$\ds K \subseteq \bigcup_{n \mathop = 1}^\infty n W$
Since $K$ is compact, there exists $n_1, n_2, \ldots, n_k$ such
:$\ds K \subseteq \bigcup_{i \mathop = 1}^k n_i W$
Relabel these $n_i$ so that: 
:$n_1 < n_2 < \ldots < n_k$
From Finite Union of Dilations of Balanced Set, we then have: 
:$\ds \bigcup_{i \mathop = 1}^k n_i W = n_k W$
so that:
:$K \subseteq n_k W$
Since $W$ is balanced, for $t > n_k$ we have: 
:$\ds \frac {n_k} t W \subseteq W$
so that: 
:$n_k W \subseteq t W$
Then, we have: 
:$K \subseteq t W$
Since $W \subseteq V$, we obtain: 
:$K \subseteq t V$
for $t > n_k$.
Since $V$ was arbitrary, we have that $K$ is von Neumann-bounded.
== Sources ==
* : $1.15$: Theorem
Category:Balanced Sets
Category:Compact Spaces
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces