Definition

Cover and compact

1. Cover

   A collection $\mathcal{O}$ of open subsets of the topological space $X$ is called an open cover if $\cup \mathcal{O}=X$.

2. Compact space

   A topological space $X$ is called compact if every open cover of $X$ has a finite subcollection that is also an open cover of $X$.

Examples

1. Examples
   • The space $\mathbb{R}$ is not compact
   • The subspace $A=\{0\}\cup\{\frac{1}{n}\mid n\in \mathbb{N}\}$ of $\mathbb{R}$ is compact.
   • Any finite topological space is compact.
   • The subspace $(0,1]\subseteq \mathbb{R}$ is not compact.

Compact subspaces

Cover of subspace

1. Cover of subspace

   Let $Y\subseteq X$. A collection $\mathcal{O}$ of subsets of $X$ is said to cover $Y$ if $Y\subseteq \cup \mathcal{O}$.

2. :B_lemma:

   Let $Y\subseteq X$ be a subspace. Then $Y$ is compact if an only if every covering of $Y$ by open sets (in $X$) contains a finite subcollection covering $Y$.

Proof

• Assume that $Y$ is compact. Let $\mathcal{O}=\{U_{\alpha}\}$ be a covering of $Y$, where each $U_{\alpha}$ is open in $X$.

• Then $\{U_{\alpha}\cap Y\}$ is a cover of $Y$ by open sets in $Y$. By compactness of $Y$, we have that there is a finite subset $\{U_{1},U_{2},\ldots,U_{r}\}\subseteq \mathcal{O}$ such that

  $$Y= (U_{1}\cap Y)\cup\cdots\cup (U_{r}\cap Y).$$
• It follows then that $\{U_{i}\}\subseteq \mathcal{O}$ is a finite open cover of $Y$.
• Now, let $\mathcal{O}'$ be a cover of $Y$ by open sets in $Y$. For each $U_{\alpha}'\in \mathcal{O}'$, let $U_{\alpha}$ open in $X$ such that $U_{\alpha}'=U_{\alpha}\cap Y$. Then $\{U_{\alpha}\}$ covers $Y$.
• Now, if $\{U_{\alpha_{i}}\}_{i=1}^{r}$ covers $Y$, we have that $\{U'_{\alpha_{i}}\}_{i=1}^{r}\subseteq \mathcal{O}'$ covers $Y$ as well.

Theorems

1. Closed in compact

   Every closed subset of a compact space is compact.

2. Proof
   • Let $Y\subseteq X$ with $Y$ closed and $X$ compact. Let $\mathcal{O}$ be a cover of $Y$ by open sets in $X$.
   • Then $\mathcal{O}'=\mathcal{O}\cup\{X-Y\}$ is an open cover of $X$. Since $X$ is compact, a finite subcollection $\mathcal{F}\subseteq \mathcal{O}'$ covers $X$.
   • Then $\mathcal{F}-\{X-Y\}$ is a subcollection of $\mathcal{O}$ and is a finite subcover of $Y$. \qed

3. Compact in Hausdorff

   Every compact subset of a Hausdorff space is closed.

4. Proof
   • Let $Y\subseteq X$ with $Y$ compact and $X$ Hausdorff. We will prove that $X-Y$ is open.
   • Let $x_{0}\in X-Y$. For each $y\in Y$, let $U_{y},V_{y}$ be disjoint neighborhoods of $x_{0},y$ respectively.
   • Then $\{V_{y}\}_{y\in Y}$ is an open cover of $Y$. Choose $y_{1},y_{2},\ldots,y_{r}\in Y$ such that $Y\subseteq V_{y_{1}}\cup\cdots\cup V_{y_{r}}$.
   • It follows that $U_{y_{1}}\cap\cdots\cap U_{y_{r}}$ is a neigborhood of $x_{0}$ that is disjoint from $V_{y_{1}}\cup\cdots\cup V_{y_{r}}$ and so is contained in $X-Y$. \qed

A corollary

This statement follows from the proof of the previous theorem:

1. Corollary

   If $Y\subseteq X$ is compact, where $X$ is Hausdorff, and $x\not\in Y$, there are disjoint open sets $U,V$ such that $x\in U$ and $Y\subseteq V$.

Image of compact

1. Teorema

   Let $f\colon X\to Y$ be continuous, with $X$ compact. Then $f(X)$ is compact.

2. :B_{ignoreheading}:
   • Let $\{V_{\alpha}\}$ be an open cover of $f(X)$.
   • Then $\{f^{-1}(V_{\alpha})\}$ is an open cover of $X$. Since $X$ is compact, there are $\alpha_{1},\ldots,\alpha_{n}$ such that $X\subseteq f^{-1}(V_{\alpha_{1}})\cup\cdots\cup f^{-1}(V_{\alpha_{n}})$.
   • It follows that $f(X)\subseteq V_{\alpha_{1}}\cup\cdots\cup V_{\alpha_{n}}$. $\square$

3. Teorema

   Let $f\colon X\to Y$ be a continuous and biyective function, where $X$ is compact and $Y$ is Hausdorff. Then $f$ is a homeomorphism.

4. :B_{ignoreheading}:
   • Since $f$ is continuous and biyective, in order to show that $f$ is a homeomorphism it is enough to show it is closed.
   • Let $C\subseteq X$ be closed. Since $X$ is compact, we have that $C$ is compact.
   • Since $f$ is continuous, we have that $f(C)$ is compact.
   • Finally, since $Y$ is Hausdorff, we have that $f(C)$ is closed. $\square$

5. Teorema

   If $X$ and $Y$ are compact spaces, then $X\times Y$ is compact.

6. :B_{ignoreheading}:
   • First, we assume that we have spaces $X,Y$, with $Y$ compact. Let $x_{0}\in X$ and $N\subseteq X\times Y$ open such that $\{x_{0}\}\times Y\subseteq N$. We will prove then that there is a neighborhood $W$ of $x_{0}$ such that $W\times Y\subseteq N$.
   • For each $y\in Y$, since $(x_{0},y)\in N$, which is open in $X\times Y$, there is a basis element $U_{y}\times V_{y}\subseteq N$ containing $(x_{0},y)$. The collection $\{U_{y}\times V_{y}\}$ covers the compact set $\{x_{0}\}\times Y$, and so we can cover it with finitely many such basis elements: $U_{1}\times V_{1},\ldots,U_{n}\times V_{n}$.
   • Let $W=U_{1}\cap\cdots\cap U_{n}$. Then $W$ is a neighborhood of $x_{0}$. We prove the finite subcover also cover the tube $W\times Y$.
   • Let $(x,y)\in W\times Y$. Then $(x_{0},y)\in U_{j}\times V_{j}$ for some $j$. Since $x\in W$ implies $x\in U_{j}$, it follows that $(x,y)\in U_{j}\times V_{j}$, and so $W\times Y\subseteq N$.

• We now assume $X,Y$ are compact spaces, and let $\mathcal{O}$ be an open cover of $X\times Y$.
• Given $x\in X$, since $\{x\}\times Y$ is compact and is covered by $\mathcal{O}$, it can also be covered by finitely many $U_{1},\ldots, U_{n}$ elements of $\mathcal{O}$. Let $N=U_{1}\cup\cdots\cup U_{n}$.
• By the tube lemma, there is a neighborhood $W_{x}$ of $x$ such that $W\times Y\subseteq N$. Then the collection $\{W_{x}\}$ covers $X$. Since $X$ is compact, there is a finite subcollection $W_{1},\ldots,W_{m}$ covering $X$.
• Finally, since the finite collection of tubes $W_{1}\times Y,\ldots,W_{m}\times Y$ covers $X\times Y$, and each tube can be covered by a finite subcollection of $\mathcal{O}$, we are done. $\square$

Property of finite intersection

1. Finite intersection property

   Let $\mathcal{C}$ be a collection of subsets of $X$. We say that $\mathcal{C}$ has the finite intersection property if for every finite subcolection $\{F_{1},\ldots,F_{r}\}\subseteq \mathcal{C}$ we have that $F_{1}\cap\cdots\cap F_{r}\ne\emptyset$.

2. Teorema

   A topological space $X$ is compact if an only if for every collection of closed sets $\mathcal{C}$ that has the finite intersection property, we have that $\cap\mathcal{C}\ne\emptyset$.

3. :B_{ignoreheading}:

   • Given a collection $\mathcal{O}$ of subsets of $X$, define the collection $\mathcal{C}=\{X-U\mid U\in \mathcal{O}\}$. Then we have:
     • $\mathcal{O}$ is a collection of open sets if and only if $\mathcal{C}$ is a collection of closed sets.
     • $\mathcal{O}$ is a cover of $X$ if and only if $\cap \mathcal{C}=\emptyset$.
     • The finite subcolection $\{U_{1},\ldots,U_{n}\}\subseteq \mathcal{O}$ covers $X$ if and only if the intersection of the corresponding $X-U_{i}$ is empty.

Compact ordered sets

Closed intervals on ordered sets

1. Teorema

   Let $X$ be a totally ordered set with the least upper bound property (every non-empty subset that has an upper bound has a least upper bound). Then, in the ordered topology, every closed interval in $X$ is compact.

2. :B_{ignoreheading}:
   • Let $a,b\in X$, $a<b$, and let $\mathcal{O}$ be a covering of $[a,b]$ by sets open in $[a,b]$ with respect to the subspace topology (which is the same as the order topology on $[a,b]$). We wish to prove there is a finite subcover of $\mathcal{O}$.
   • We prove first that if $x\in [a,b]$, $x\ne b$, then there is a point $y\in [a,b]$, $y>x$, such that the inverval $[x,y]$ can be covered by at most two elements of $\mathcal{O}$.
   • If $x$ has an immediate successor $y\in X$, then $y\in [a,b]$ and $[x,y]=\{x,y\}$

3. Corollary

   Every closed interval in $\mathbb{R}$ is compact.

4. Teorema

   A subset $A\subseteq \mathbb{R}^{n}$ is compact if and only if is closed and bounded in the euclidian metric $d$, or in the metric $\rho$. ($\rho(x,y)=\max \vert x_{i}-y_{i}\vert$)

Maximum value theorem

1. Maximum value theorem

   Let $f\colon X\to Y$ be continuous, where $Y$ is a totally ordered set with the order topology. If $X$ is compact, there are points $c,d\in X$ such that $f(c)\leq f(x)\leq f(d)$ for every $x\in X$.

The reals are uncountable

1. Teorema

   Let $X$ be a non-empty compact Hausdorff space. If every point of $X$ is a limit point of $X$, then $X$ is uncountable.

2. Corollary

   Every closed interval in $\mathbb{R}$ is uncountable.