Point-Set

Take \(\left\{{B_{{\varepsilon}\over 2}(y) {~\mathrel{\Big\vert}~}y\in Y}\right\}\rightrightarrows Y\), pull back to an open cover of \(X\), has Lebesgue number \(\delta_L > 0\), then \(x' \in B_{\delta_L}(x) \implies f(x), f(x') \in B_{{\varepsilon}\over 2}(y)\) for some \(y\).

• Let \(\left\{{A_i}\right\} \rightrightarrows A\) be a covering of \(A\) by sets open in \(A\).
• Each \(A_i = B_i \cap A\) for some \(B_i\) open in \(B\) (definition of subspace topology)
• Define \(V = \left\{{B_i}\right\}\), then \(V \rightrightarrows A\) is an open cover.
• Since \(A\) is closed, \(W\coloneqq B\setminus A\) is open
• Then \(V\cup W\) is an open cover of \(B\), and has a finite subcover \(\left\{{V_i}\right\}\)
• Then \(\left\{{V_i \cap A}\right\}\) is a finite open cover of \(A\).

Let \(f:X\to f(X)\) be continuous. Take an open covering \(\mathcal{U} \rightrightarrows f(X)\), then \(f^{-1}(\mathcal{U}) \rightrightarrows X\), which is cover by opens since \(f\) is continuous. Take a finite subcover by compactness of \(X\), then they push forward to a finite subcover of \(f(X)\).

Show that \(f^{-1}\) is continuous by showing \(f\) is a closed map. If \(A\subseteq X\) is closed in a compact space, \(A\) is compact. The continuous image of a compact set is compact, so \(f(A)\) is compact. A compact set in a Hausdorff space is closed, so \(f(A)\) is closed in \(Y\).

See Munkres page 104.

• For each \(y\in Y\) choose neighborhoods \(A_y, B_y \subseteq Y\) such that \begin{align*} (x, y) \in A_y \times B_y \subseteq U .\end{align*}
• By compactness of \(Y\), reduce this to finitely many \(B_y \rightrightarrows Y\) so \(Y = \bigcup_{j=1}^n B_{y_j}\)
• Set \(O\coloneqq\cap_{j=1}^n B_{y_j}\); this works.

\todo[inline]{Check this proof!}

• Let \(\left\{{U_j\times V_j {~\mathrel{\Big\vert}~}j\in J}\right\} \rightrightarrows X\times Y\).
• Fix a point \(x_0\in X\), then \(\left\{{x_0}\right\}\times Y \subset N\) for some open set \(N\).
• By the tube lemma, there is a \(U^x \subset X\) such that the tube \(U^x \times Y \subset N\).
• Since \(\left\{{x_0}\right\}\times Y \cong Y\) which is compact, there is a finite subcover \(\left\{{U_j \times V_j {~\mathrel{\Big\vert}~}j\leq n}\right\} \rightrightarrows\left\{{x_0}\right\}\times Y\).
• “Integrate the \(X\)”: write \begin{align*}W = \cap_{j=1}^n U_j,\end{align*} then \(x_0 \in W\) and \(W\) is a finite intersection of open sets and thus open.
• Claim: \(\left\{{U_j \times V_j {~\mathrel{\Big\vert}~}j\leq n}\right\}\rightrightarrows W\times Y\)
  • Let \((x, y) \in W\times Y\); want to show \((x, y)\in U_j \times V_j\) for some \(j\leq n\).
  • Then \((x_0, y) \in \left\{{x_0}\right\}\times Y\) is on the same horizontal line
  • \((x_0, y)\in U_j \times V_j\) for some \(j\) by construction
  • So \(y\in V_j\) for this \(j\)
  • Since \(x\in W\), \(x\in U_j\) for every \(j\), thus \(x\in U_j\).
  • So \((x, y) \in U_j \times V_j\)