Proof (not given in class). 1. \(\Rightarrow \) 2.: suppose the subspace topology on \(K\) is compact, and let \(\mathscr F\) be a collection of open subsets of \(X\) such that \(K\subseteq \bigcup \mathscr F.\) The collection \(\mathscr F_K = \{U\cap K: U\in \mathscr F\}\) 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 \(\mathscr F\) which still covers \(K.\)

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

By condition 2., the collection \(\{U_\alpha :\alpha \in I\}\) has a finite subcollection, say \(U_{\alpha _1},\dots ,U_{\alpha _n},\) which still covers \(K.\) Hence \(\mathscr C\) has finite subcover \(U_{\alpha _1}\cap K,\dots ,U_{\alpha _n}\cap K,\) 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 \(\mathscr G\) of open sets in \(Y\) covers \(f(X):\) that is, \(f(X) \subseteq \bigcup \mathscr G.\)

Consider the collection \(\mathscr C = \{f^{-1}(V): V \in \mathscr G\}\) of subsets of \(X.\) We claim that \(\mathscr C\) 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 \(\mathscr C\) are open in \(X.\) Also,

\[ X = f^{-1}(f(X)) \subseteq f^{-1}\Bigl (\bigcup \mathscr G\Bigr ) = \bigcup \{ f^{-1}(V): V\in \mathscr G\} \]

which shows that \(\mathscr C\) covers \(X.\)

Since \(X\) is compact, \(\mathscr C\) has finite subcover, say \(f^{-1}(V_1),\dots ,f^{-1}(V_n).\) Then \(V_1,\dots ,V_n\) is a finite subcollection of \(\mathscr G\) 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 \(X\xrightarrow [f]{\sim } Y,\) 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 \(\mathscr C\) of open subsets of \(X.\) Then

\[ \mathscr C \cup \{X\setminus F\} \]

is an open cover for the whole of \(X.\)

Since \(X\) is compact, \(\mathscr C \cup \{X\setminus F\}\) 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 \(\mathscr C\) 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:

(†)  \(z\in X\setminus K\)  \(\Rightarrow \)  \(\exists \) 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

\[ U(x) \ni x, \quad V_x(z)\ni z: \quad U(x)\cap V_x(z)=\emptyset . \]

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

\[ V(z) = V_{x_1}(z) \cap \dots \cap V_{x_n}(z). \]

(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 \(V(z) \cap U_{x_i}(z) \subseteq V_{x_i}(z) \cap U_{x_i}(z) = \emptyset \) and so

\[ V(z) \cap (U(x_1)\cup \dots \cup U(x_n)) = \bigcup _{i=1}^n V(z) \cap U(x_i) = \emptyset . \]

Since \(K\) is contained in the union \(U(x_1)\cup \dots \cup U(x_n),\) 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 \(f^{-1}\colon Y \to K\) 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. □