nyuuzyou/wikis huggingface dataset:
Compact Hausdorff Space with no Isolated Points is Uncountable/Lemma

== Lemma for Compact Hausdorff Space with no Isolated Points is Uncountable ==
Let $T = \left({S, \tau}\right)$ be a Hausdorff space with no isolated points.
Let $U \subseteq S$ be a non-empty open set in $T$.
Let $x \in S$.
Then there is a non-empty open subset $V \subseteq U$ such that $x \notin V^-$, where $V^-$ is the closure of $V$.
== Proof ==
First we show that there is a $y \in U$ such that $y \ne x$:
By Law of Excluded Middle, either $x \in U$ or $x \notin U$.
Let $x \in U$.
We have that $x$ is not an isolated point.
Therefore:
:$\exists y \in U: y \ne x$
Let $x \notin U$.
As $U$ is non-empty it has an element $y$, and $y \ne x$.
Thus in either case, there is a $y \in U$ such that $y \ne x$.
By Closure Condition for Hausdorff Space, there is an open set $V$ such that $y \in V$ and $x \notin V^-$.
Category:Compact Spaces
Category:Hausdorff Spaces
Compact Hausdorff Topology is Maximally Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space which is compact.
Then $\tau$ is maximally compact.
== Proof ==
Let $\tau'$ be a topology on $S$ such that $\tau \subseteq \tau'$ but that $\tau \ne \tau'$.
Consider the identity mapping $I_S: \struct {S, \tau'} \to \struct {S, \tau}$.
From Separation Properties Preserved in Subspace, $I_S$ is a continuous bijection from a Hausdorff space to a compact Hausdorff space.
 $\struct {S, \tau'}$ is compact.
From Continuous Bijection from Compact to Hausdorff is Homeomorphism, $I_S$ is also open.
Hence:
:$\tau' \subseteq \tau$
Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$.
Hence by definition of set equality:
:$\tau' = \tau$
But this contradicts our hypothesis that $\tau' \ne \tau$.
By Proof by Contradiction, it follows that no topology which is strictly finer than $\tau$ can be compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:Hausdorff Spaces
Category:Compact Spaces
Compact Hausdorff Topology is Minimal Hausdorff

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space which is compact.
Then $\tau$ is the minimal subset of the power set $\powerset S$ such that $T$ is a Hausdorff space.
== Proof ==
 there exists a topology $\tau'$ on $S$ such that:
:$\tau' \subseteq \tau$ but $\tau' \ne \tau$
:$\tau'$ is a Hausdorff space.
From Equivalence of Definitions of Finer Topology:
:the identity mapping $I_S: \struct {S, \tau} \to \struct {S, \tau'}$ is continuous.
Let $A \in \tau$.
Then $S \setminus A \subseteq S$ is closed in $\struct {S, \tau}$.
By Closed Subspace of Compact Space is Compact, $S \setminus A$ is compact in $\struct {S, \tau}$.
From Continuous Image of Compact Space is Compact:
:$I_S \sqbrk {S \setminus A}$ is also compact.
By hypothesis, $\struct {S, \tau'}$ is a Hausdorff space.
From Compact Subspace of Hausdorff Space is Closed:
:$I_S \sqbrk {S \setminus A}$ is closed in $\struct {S, \tau'}$.
Hence:
:$A = S \setminus I_S \sqbrk {S \setminus A} \in \tau'$
By definition of subset:
:$\tau \subseteq \tau'$
Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$.
Hence by definition of set equality:
:$\tau' = \tau$
But this contradicts our hypothesis that $\tau' \ne \tau$.
By Proof by Contradiction, it follows that no topology which is strictly coarser than $\tau$ can be Hausdorff.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:Hausdorff Spaces
Category:Compact Spaces
Compact Hermitian Operator has Countable Point Spectrum

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $T \in \map {B_0} \HH$ be a compact Hermitian operator.
Then its point spectrum $\map {\sigma_p} T$ is countable.
== Proof ==
== Sources ==
* $\text {II}.5.1$
Category:Linear Transformations on Hilbert Spaces
Compact Idempotent is of Finite Rank

== Theorem ==
Let $H$ be a Hilbert space.
Let $T \in \map {B_0} H$ be a compact linear operator.
Let $T$ be idempotent.
Then:
:$T \in \map {B_{00} } H$
that is, $T$ is a bounded finite rank operator.
== Proof ==
Let $\sequence {y_n}$ be a bounded sequence in the range of $T$. 
Since $T$ is idempotent:
:$\forall n \in \N: T y_n = y_n$
Since $T$ is compact, $\sequence {T y_n}$ and hence $\sequence {y_n}$, contains a convergent subsequence.
Thus the range of $T$ has the Bolzano-Weierstrass property.
By Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Sufficient Condition, the range has finite dimension.
== Sources ==
* : $\text {II}.4$: Exercise $4$
Category:Linear Transformations on Hilbert Spaces
Compact Linear Transformation is Bounded

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $\map {B_0} {\HH, \KK}$ be the space of compact linear transformations $\HH \to \KK$. 
Let $\map B {\HH, \KK}$ be the space of bounded linear transformations $\HH \to \KK$. 
Let $T \in \map {B_0} {\HH, \KK}$.
Then $T \in \map B {\HH, \KK}$.
That is:
:$\map {B_0} {\HH, \KK} \subseteq \map B {\HH, \KK}$
== Proof ==
Let $T : \HH \to \KK$ be a linear transformation.
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$. 
Let $\norm \cdot_\KK$ be the inner product norm on $\KK$.
We show that:
:if $T$ is not bounded, then it is not compact.
That is: 
:if $T \not \in \map B {\HH, \KK}$, then $T \not \in \map {B_0} {\HH, \KK}$.
Then, we will obtain:
:$\map {B_0} {\HH, \KK} \subseteq \map B {\HH, \KK}$
Suppose that $T$ is not bounded.
That is: 
:there does not exist a real number $c > 0$ such that $\norm {T h}_\KK \le c \norm h_\HH$ for all $h \in \HH$. 
Then:
:for each $n \in \N$, there exists some $x_n \in \HH$ such that $\norm {T x_n}_\KK > n \norm {x_n}_\HH$.
Now define sequence $\sequence {h_n}$ by: 
:$\ds h_n = \frac {x_n} {\norm {x_n} }$
for each $n \in \N$. 
We have: 
:$\norm {h_n} = 1$
for each $n \in \N$, so $\sequence {h_n}$ is bounded. 
We also have: 
:$\norm {T h_n} > n$
for each $n \in \N$. 
So, for any increasing sequence of positive integers $\sequence {n_j}$ we have:
:$\norm {T h_{n_j} } > n_j$
Note that $n_j \to \infty$. 
So, for every subsequence $\sequence {h_{n_j} }$ of $\sequence {h_n}$ we have that: 
:$\sequence {T h_{n_j} }$ is unbounded. 
So from Convergent Sequence in Normed Vector Space is Bounded, we have that: 
:$\sequence {T h_{n_j} }$ does not converge. 
So:
:the bounded sequence $\sequence {h_n}$ does not have a subsequence $\sequence {h_{n_j} }$ such that $\sequence {T h_{n_j} }$ converges.
So:
:$T$ is not compact.
Hence the demand.
== Sources ==
* $\text {II}.4.2 \ \text {(a)}$
Category:Linear Transformations on Hilbert Spaces
Category:Compact Linear Transformations
Compact Linear Transformations Composed with Bounded Linear Operator

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $T \in \map {B_0} {H, K}$ be a compact linear transformation.
Let $A \in \map B H, B \in \map B K$ be bounded linear operators.
Then the compositions $T A$ and $B T$ are also compact linear transformations.
== Proof ==
Let $\sequence {h_n}_{n \in \N}$ is a bounded sequence in $H$.
That is, there exists a $M > 0$ such that:
:$\forall n \in \N : \norm {h_n}_H \le M$
Then $\sequence {A h_n}_{n \in \N}$ is also bounded, since:
:$\norm {A h_n}_H \le \norm {A}_{\map B H} \norm {h_n}_H \le \norm {A}_{\map B H} M$
As $T$ is compact,
the sequence $\sequence {T A h_n}_{n \in \N}$ has a subsequence convergent in $K$.
Hence $T A$ is compact.
Let $\sequence {k_n}_{n \in \N}$ is a bounded sequence in $K$.
As $T$ is compact,
the sequence $\sequence {T k_n}_{n \in \N}$ has a subsequence convergent in $H$.
Say $\sequence {T k_{n_r} }_{r \in \N}$ is convergent.
Recall that $B$ is continuous due to Continuity of Linear Transformation between Normed Vector Spaces.
Thus $\sequence {B T k_{n_r} }_{r \in \N}$ is convergent.
Hence $B T$ is compact.
== Sources ==
* $\text {II}.4.2 \text {(c)}$
Category:Linear Transformations on Hilbert Spaces
Compact Metric Space is Complete

== Theorem ==
A compact metric space is complete.
== Proof ==
Follows directly from:
:Compact Space is Countably Compact
:Countably Compact Metric Space is Sequentially Compact
:Sequentially Compact Metric Space is Complete
== Sources ==
* : $3.11 b$
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Complete Metric Spaces
Category:Compact Spaces
Compact Metric Space is Separable

== Theorem ==
Let $\struct {X, d}$ be a compact metric space.
Then $\struct {X, d}$ is separable.
== Proof ==
From Compactness and Sequential Compactness are Equivalent in Metric Spaces, $\struct {X, d}$ is sequentially compact. 
From Sequentially Compact Metric Space is Separable, $\struct {X, d}$ is separable.
Category:Separable Spaces
Category:Compact Spaces
Category:Metric Spaces
Compact Metric Space is Totally Bounded

== Theorem ==
Let $M = \struct {A, d}$ be a metric space which is compact.
Then $M$ is totally bounded.
== Proof ==
Let $M = \struct {A, d}$ be compact.
Let $\epsilon > 0$.
Then the family $\set {\map {B_\epsilon} x: x \in A}$ of open $\epsilon$-balls forms an open cover of $A$.
By the definition of compact, every open cover for $A$ has a finite subcover.
That is, there are points $x_0, \ldots, x_n$ such that:
:$\ds A = \bigcup_{0 \mathop \le i \mathop \le n} \map {B_\epsilon} {x_i}$
as required. 
== Also see==
* Heine-Borel Theorem/Metric Space
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Totally Bounded Metric Spaces
Category:Compact Spaces
Compact Set of Irrational Numbers is Nowhere Dense

== Theorem ==
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the Euclidean topology $\tau_d$.
Let $S \subseteq \R \setminus \Q$ be a compact set of $\R \setminus \Q$.
Then $S$ is nowhere dense in $\R \setminus \Q$.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $8$
Category:Irrational Number Space
Category:Compact Spaces
Category:Nowhere Denseness
Compact Set of Rational Numbers is Nowhere Dense

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $S \subseteq \Q$ be a compact set of $\Q$.
Then $S$ is nowhere dense in $\Q$.
== Proof ==
By Compact Subspace of Hausdorff Space is Closed, $S$ is closed in $\Q$.
By Set is Closed iff Equals Topological Closure, $S = S^-$.
 $S$ is not nowhere dense in $\Q$.
Then $S^-$ contains some non-empty open set.
From Basis for Euclidean Topology on Real Number Line, the set of all open real intervals of $\R$ form a basis for $\struct {\R, \tau_d}$.
By Basis for Topological Subspace, the set of all intersections of $\Q$ and open real intervals of $\R$ form a basis for $\struct {\Q, \tau_d}$.
So there exists $V = \Q \cap \openint {a'} {b'} \subseteq S^- = S$ for some $a' < b'$.
By Between two Real Numbers exists Irrational Number we have:
:$\exists r \in \R \setminus \Q: a' < r < b'$
By Between two Real Numbers exists Rational Number we have:
:$\exists a, b \in \Q: a' < a < r < b < b'$
Thus $\Q \cap \closedint a b$ is a closed subset of $S$ in $\Q$.
By Closed Subspace of Compact Space is Compact, $\Q \cap \closedint a b$ is compact.
Now we construct an open cover for $\Q \cap \closedint a b$:
Let $C = \openint r {b + 1} \cap \Q$.
Let $C_q = \openint {a - 1} q \cap \Q$ for each $q \in \openint a r \cap \Q$.
Then $C$ and $C_q$ are open sets.
Let $\CC = \set C \cup \set {C_q: q \in \openint a r \cap \Q}$.
To show that $\CC$ is an open cover for $\Q \cap \closedint a b$:
:Pick any $x \in \Q \cap \closedint a b$.
:Suppose $x > r$. 
:Then $r < x \le b < b + 1$.
:Therefore $x \in \openint r {b + 1} = C \in \CC$.
:Suppose $x < r$.
:By Between two Real Numbers exists Rational Number we have:
::$\exists q \in \Q: a - 1 < a \le x < q < r$
:Then $x \in \openint {a - 1} q$ and $q \in \openint a r$.
:Thus $x \in C_q \in \CC$.
:Therefore $x \in \bigcup \CC$, showing that $\CC$ is an open cover for $\Q \cap \closedint a b$. 
Since $\Q \cap \closedint a b$ is compact, $\CC$ has a finite subcover, $\CC'$.
This finite subcover can only contain a finite number of sets in $\set {C_q: q \in \openint a r \cap \Q}$.
Therefore $\set {q: C_q \in \CC'}$ has a greatest element, $q'$.
We have that $q' \in \openint a r \cap \Q$.
By Between two Real Numbers exists Rational Number we have:
:$\exists q^* \in \Q: a < q' < q^* < r < b$
So $q^* \in \Q \cap \closedint a b$.
But for all $q \in \hointl a {q'} \cap \Q$:
:$q^* \notin C_q = \openint {a - 1} q \cap \Q$
We also have $q^* \notin C = \openint r {b + 1} \cap \Q$.
Hence $q^* \notin \bigcup \CC'$.
So $\CC'$ is not a cover for $\Q \cap \closedint a b$.
This is a contradiction.
Thus $S$ is nowhere dense in $\Q$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $8$
Category:Rational Number Space
Category:Compact Spaces
Category:Nowhere Denseness
Compact Sets in Countable Complement Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then the compact sets of $T$ are exactly the finite subsets of $S$.
== Proof ==
From Finite Topological Space is Compact, any finite subset of $S$ is compact.
 $H \subseteq S$ is an infinite compact set.
Take a (countably) infinite sequence $\sequence {a_n}_{n \mathop \ge 0}$ of distinct elements of $H$.
Consider the open sets:
:$V_m := S \setminus \set {a_{m + n} }_{n \mathop \ge 0}$
for $m \ge 0$, which satisfy $V_{m_1} \subset V_{m_2}$ if $m_1 < m_2$.
Then:
:$\ds H \subseteq \bigcup_{m \mathop \ge 0} V_m$
is an open cover of $H$.
Since $H$ is compact, it has a finite open subcover, say:
:$\ds H \subseteq \bigcup_{i \mathop = 0}^N V_{m_i}$
with $m_0 < m_1 < \cdots < m_N$.
But then:
:$\ds \bigcup_{i \mathop = 0}^N V_{m_i} = V_{m_N}$ 
and so:
:$\ds a_{m_N + 1}, a_{m_N + 2}, \ldots \notin V_{m_N} = \bigcup_{i \mathop = 0}^N V_{m_i}$
which implies $a_{m_N + 1}, a_{m_N + 2}, \ldots \notin H$, a contradiction.
Hence the result by Proof by Contradiction.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $2$
Category:Compact Spaces
Category:Countable Complement Topology
Compact Sets in Fortissimo Space

== Theorem ==
A subset of a Fortissimo space is compact it is finite.
== Proof ==
Let $T = \struct {S, \tau}$ be a Fortissimo space.
=== Necessary Condition ===
By Finite Topological Space is Compact, all finite sets are compact in $T$.
=== Sufficient Condition ===
We prove the contrapositive.
Let $A$ be an infinite subset of $S$.
Let $C$ be a countably infinite subset of $A$ that does not contain $p$.
For each $x \in A$, $C \setminus \set x$ is a countably infinite set.
Hence $\relcomp S {C \setminus \set x} = \relcomp S C \cup \set x$ is open in $T$.
Thus $\CC = \set {\relcomp S C \cup \set x: x \in A}$ is an open cover of $A$.
However, each subset of the $\CC$ contains exactly $1$ element in $C$, so a finite subset of $\CC$ can only contain a finite number of elements in $C$.
Therefore $\CC$ has no finite subcover.
Hence $A$ is not compact.
This shows that if a subset of a Fortissimo space is compact, it must be finite.
Category:Fortissimo Space
Category:Compact Spaces
Compact Sets of Double Pointed Topology

== Theorem ==
Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
Then $X \subseteq S \times D$ is compact in $\tau$ for some compact set $C$ of $\tau_S$:
:$\map {\pr_1} X = C$
where $\pr_1$ denotes the first projection on $S \times D$.
=== Corollary ===
== Proof ==
=== Necessary Condition ===
Suppose that $X \subseteq S \times D$ is a compact set in $\vartheta$.
It is to be shown that $C = \map {\pr_1} X$ is compact in $\tau$.
This follows from Compactness Properties Preserved under Projection Mapping.
=== Sufficient Condition ===
Suppose $\map {\pr_1} X = C$ for some compact $C \subseteq S$.
Let $\UU$ be an open cover for $X$, that is:
:$\ds \bigcup \UU \supseteq X$
By Open Sets of Double Pointed Topology, each $U \in \UU$ is of the form:
:$U' \times D$
with $U'$ open in $\vartheta$; note that $U' = \map {\pr_1} U$.
By Cartesian Product Distributes over Union, it follows that:
:$\ds \bigcup \UU = \paren {\bigcup_{U \mathop \in \UU} \map {\pr_1} U} \times D$
Since $X \subseteq \ds \bigcup \UU$, it follows by Image Preserves Subsets and Image of Union under Relation that:
:$C = \map {\pr_1} X \subseteq \bigcup_{U \mathop \in \UU} \map {\pr_1} U$
showing that:
:$\VV := \set {\map {\pr_1} U: U \in \UU}$
is an open cover for $C$.
Since $C$ is compact, we find a finite subcover $\VV'$ of $\VV$ such that:
:$C \subseteq \ds \bigcup \VV'$
Finally, we will show that:
:$\UU' := \set {V \times D: V \in \VV'} \subseteq \UU$
forms a finite subcover of $\UU$.
That is, that:
:$X \subseteq \ds \bigcup \UU'$
So let $\tuple {s, d} \in X$.
By assumption on $X$, we have that $s \in C$, and since $\VV'$ is a open cover for $C$:
:$\exists V \in \VV': s \in V$
It is immediate that $\tuple {s, d} \in V \times D$ by definition of Cartesian product.
Since $V \times D \in \UU'$, it follows by definition of set union that:
:$\tuple {s, d} \in \ds \bigcup \UU'$
That is, $\UU'$ is a finite subcover for $X$, and hence $X$ is compact.
Category:Double Pointed Topologies
Compact Sets of Double Pointed Topology/Corollary

== Corollary to Compact Sets of Double Pointed Topology ==
Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
$\struct {S \times D, \tau}$ is compact $\struct {S, \tau_S}$ is compact.
== Proof ==
By Projection is Surjection, it follows that:
:$\map {\pr_1} {S \times D} = S$
where $\pr_1$ is the first projection on $S \times D$.
The result follows by Compact Sets of Double Pointed Topology.
Category:Double Pointed Topologies
Compact Space in Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $\set p$ is compact in $T$.
== Proof ==
Any open cover of $\set p$ has a finite subcover: any single set that contains $p$ is a cover for $\set p$.
So $\set p$ is compact in $T$.
== Also see ==
* Finite Topological Space is Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $5$
Category:Compact Spaces
Category:Particular Point Topology
Compact Space is Countably Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is countably compact.
== Proof ==
Let $T = \struct {S, \tau}$ be a compact space.
Then by definition every open cover of $S$ has a finite subcover.
So every countable open cover of $S$ has a finite subcover.
Hence by definition $T$ is countably compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Compact Spaces
Category:Countably Compact Spaces
Compact Space is Paracompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is paracompact.
== Proof ==
From the definition, $T$ is compact every open cover of $S$ has a finite subcover.
From Subcover is Refinement of Cover, it follows that every open cover of $S$ has an open refinement which is locally finite.
This is precisely the definition of paracompact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Compact Spaces
Category:Paracompact Spaces
Compact Space is Pseudocompact

== Theorem ==
Let $\struct {K, \tau}$ be a compact space.
Then:
:$\struct {K, \tau}$ is pseudocompact
== Proof ==
Follows immediately from:
* Compact Space is Countably Compact
* Countably Compact Space is Pseudocompact
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Pseudocompact Spaces
Category:Compact Spaces
Compact Space is Sigma-Compact

== Theorem ==
Every compact space is $\sigma$-compact.
== Proof ==
By definition, a $\sigma$-compact space is the union of countably many compact sets.
A compact space is the union of exactly one compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Compact Spaces
Category:Sigma-Compact Spaces
Compact Space is Strongly Locally Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is a strongly locally compact space.
== Proof==
From Underlying Set of Topological Space is Clopen, $S$ is clopen in $T$.
From Closed Set Equals its Closure, $S = S^-$.
So every point of $S$ is contained in an open set (that is, $S$) whose closure (that is, $S$ again) is compact (as $T = \struct {S, \tau}$ itself is compact).
That is precisely the definition of a strongly locally compact space.
== Also see ==
* Compact Space is Weakly Locally Compact
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Compact Spaces
Category:Strongly Locally Compact Spaces
Compact Space is Weakly Locally Compact

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a compact space.
Then $T$ is a weakly locally compact.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Compact Spaces
Category:Weakly Locally Compact Spaces
Category:Compact Space is Weakly Locally Compact
Compact Space is Weakly Locally Compact/Proof 1

== Theorem ==
== Proof ==
Let $T$ be a compact space.
By definition, $T$ is weakly locally compact each point of $S$ has a compact neighborhood in $T$.
Let $x \in S$ be an arbitrary point of $S$.
By Space is Neighborhood of all its Points, $S$ is a neighborhood of $x$.
Thus $x$ has a compact neighborhood in $T$.
Hence the result by definition of weakly locally compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Compact Space is Weakly Locally Compact
Compact Space is Weakly Sigma-Locally Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is a weakly $\sigma$-locally compact space.
== Proof==
Let $T = \struct {S, \tau}$ be a compact space.
We have that:
:Compact Space is Sigma-Compact
:Compact Space is Weakly Locally Compact
Hence by definition $T$ is weakly $\sigma$-locally compact space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Weakly Sigma-Locally Compact Spaces
Category:Compact Spaces
Compact Space satisfies Finite Intersection Axiom

== Theorem ==
=== Definition by Open Covers ===
=== Definition by Finite Intersection Axiom ===
== Proof ==
Let every open cover of $S$ have a finite subcover.
Let $\AA$ be any set of closed subsets of $S$ satisfying $\ds \bigcap \AA = \O$.
We define the set:
:$\VV := \set {S \setminus A : A \in \AA}$
which is trivially an open cover of $S$.
From De Morgan's Laws: Difference with Union:
:$\ds S \setminus \bigcup \VV = \bigcap \set {S \setminus V : V \in \VV} = \bigcap \set {A : A \in \AA} = \O$
and therefore:
:$S = \ds \bigcup \VV$
By definition, there exists a finite subcover $\tilde \VV \subseteq \VV$.
We define:
:$\tilde \AA := \set {S \setminus V : V \in \tilde \VV}$
then $\tilde \AA \subseteq \AA$ by definition of $\VV$.
Because $\tilde \VV$ covers $S$, it follows directly that:
:$\ds \bigcap \tilde \AA = \bigcap \set {S \setminus V : V \in \tilde \VV} = S \setminus \bigcup \tilde \VV = \O$
Thus, in every set $\AA$ of closed subsets of $S$ satisfying $\ds \bigcap \AA = \O$, there exists a finite subset $\tilde \AA$ such that $\ds \bigcap \tilde \AA = \O$.
That is, $S$ satisfies the Finite Intersection Axiom.
The converse works exactly as the previous, but with the roles of the open cover and $\AA$ reversed.
== Also see ==
* Equivalence of Definitions of Compact Topological Space
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Compact Spaces