Proof (not given in class). 1. $\Rightarrow$ 2.: suppose the subspace topology on $K$ is compact, and let $\mathcal{F}$ be a collection of open subsets of $X$ such that $K\subseteq \bigcup \mathcal{F}.$ The collection ${\mathcal{F}}_K=\{U\cap K:U\in \mathcal{F}\}$ of subsets of $K$ is clearly an open cover of $K.$ By assumption, $K$ is compact so this open cover must have a finite subcover, say $\{U_1\cap K,\dots ,U_n\cap K\}.$ Then $\{U_1,\dots ,U_n\}$ is a finite subcollection of $\mathcal{F}$ which still covers $K.$

2. $\Rightarrow$ 1.: to show that $K$ is compact, we let $\mathcal{C}$ be an open cover of $K.$ By definition of subspace topology, $\mathcal{C}$ is of the form $\{U_{\text{𝛼}}\cap K:\text{𝛼}\in I\}$ where $U_{\text{𝛼}}$ are open in $X.$ Clearly, for $\mathcal{C}$ to cover $K,$ one must have $K\subseteq \bigcup _{\text{𝛼}\in I}{U}_{\text{𝛼}}.$

By condition 2., the collection $\{U_{\text{𝛼}}:\text{𝛼}\in I\}$ has a finite subcollection, say $U_{{\text{𝛼}}_1},\dots ,U_{{\text{𝛼}}_n},$ which still covers $K.$ Hence $\mathcal{C}$ has finite subcover $U_{{\text{𝛼}}_1}\cap K,\dots ,U_{{\text{𝛼}}_n}\cap K,$ verifying the definition of “compact” for $K.$ □

Proof. We will use Criterion 4.1 of compactness for a subset to show that $f(X)$ is a compact set. Suppose a collection $\mathcal{G}$ of open sets in $Y$ covers $f(X):$ that is, $f(X)\subseteq \bigcup \mathcal{G}.$

Consider the collection $\mathcal{C}=\{f^{-1}(V):V\in \mathcal{G}\}$ of subsets of $X.$ We claim that $\mathcal{C}$ is an open cover of $X.$ If $V$ is open in $Y,$ “$f$ is continuous” means that $f^{-1}(V)$ is open in $X,$ so all members of $\mathcal{C}$ are open in $X.$ Also,

$$X=f^{-1}(f(X))\subseteq f^{-1}(\bigcup \mathcal{G})=\bigcup \{f^{-1}(V):V\in \mathcal{G}\}$$

which shows that $\mathcal{C}$ covers $X.$

Since $X$ is compact, $\mathcal{C}$ has finite subcover, say $f^{-1}(V_1),\dots ,f^{-1}(V_n).$ Then $V_1,\dots ,V_n$ is a finite subcollection of $\mathcal{G}$ which covers $f(X).$ We have verified Criterion 4.1, hence $f(X)$ is a compact set in $Y.$ □

Proof. If $X$ is compact and $X\underset{f}{\overset{\sim }{\to }}Y,$ then $f$ is continuous, so $f(X)$ must be compact by Theorem 4.2. Yet $f(X)=Y$ because $f,$ being a homeomorphism, is surjective. □

Proof. Let $X$ be a compact topological space and let $F$ be a closed subset of $X.$ We want to use Criterion 4.1, so we let $F$ be covered by a family $\mathcal{C}$ of open subsets of $X.$ Then

$$\mathcal{C}\cup \{X\setminus F\}$$

is an open cover for the whole of $X.$

Since $X$ is compact, $\mathcal{C}\cup \{X\setminus F\}$ has a finite subcover of $X.$ This finite subcover of $X$ can be $U_1,\dots ,U_n$ or $U_1,\dots ,U_n,X\setminus F.$ In either case, the sets $U_1,\dots ,U_n$ form a finite subcollection of $\mathcal{C}$ which must cover $F.$ (This last step is illustrated by Figure 4.1.) □

Proof. Let $X$ be a Hausdorff topological space and let $K$ be a compact subset of $X.$ Letting $z$ be any point of $X\setminus K,$ it is enough to prove:

(†)  $z\in X\setminus K$  $\Rightarrow$  $\mathrm{\exists }$ open $V(z):$ $z\in V(z)$ and $V(z)\subseteq X\setminus K.$

Indeed, if (†) holds then, in the same way as in the proof of Proposition 3.4 $X\setminus K=\bigcup _{z\in X\setminus K}V(z)$ is an open set, so $K$ is closed.

For each $x\in K,$ the Hausdorff property gives us open neighbourhoods

$$U(x)\ni x,V_x(z)\ni z:U(x)\cap V_x(z)=\mathrm{\varnothing }.$$

The open sets $\{U(x):x\in K\}$ cover $K,$ so by Criterion 4.1 there is a finite subcollection $U(x_1),\dots ,U(x_n)$ which still covers $K.$ Put

$$V(z)=V_{x_1}(z)\cap \cdots \cap V_{x_n}(z).$$

(The construction of the open neighbourhood $V(z)$ is illustrated by Figure 4.2.)

As a finite intersection of open sets, $V(z)$ is open. Moreover, by construction $V(z)\cap U_{x_i}(z)\subseteq V_{x_i}(z)\cap U_{x_i}(z)=\mathrm{\varnothing }$ and so

$$V(z)\cap (U(x_1)\cup \cdots \cup U(x_n))=\bigcup _{i=1}^{n}V(z)\cap U(x_i)=\mathrm{\varnothing }.$$

Since $K$ is contained in the union $U(x_1)\cup \cdots \cup U(x_n),$ it follows that $V(z)$ does not intersect $K.$ We have therefore proved (†) and the Proposition. □

Proof. $f$ is already assumed to be bijective and continuous, hence to show that $f$ is a homeomorphism, we need to prove that the inverse function $f^{-1}:Y\to K$ is continuous. We will use the closed set criterion of continuity, Proposition 2.5. Let $F\subset K$ be closed in $K.$ The $f^{-1}$-preimage of $F$ is $(f^{-1}{)}^{-1}(F)=f(F):$

• • a closed subset of a compact is compact (Proposition 4.3) so $F$ is compact,

• • a continuous image of a compact is compact (Theorem 4.2), so $f(F)$ is compact,

• • a compact subset of the Hausdorff space $Y$ is closed (Proposition 4.4), so $f(F)$ is closed in $Y.$

We have shown that the function $f^{-1}$ is such that the preimage of a closed set is closed. Hence, by the closed set criterion of continuity, $f^{-1}$ is continuous. □