Let \(\ts{A_\alpha}\) be a collection of subsets of \(X\); let \(X = \bigcup_\alpha A_\alpha\). Let \(f: X \to Y\); suppose that \(f \mid A_\alpha\) is continuous for each \(\alpha\).
Show that if the collection \(\ts{A_\alpha}\) is finite and each set \(A_\alpha\) is closed, then \(f\) is continuous.
Find an example where the collection \(\ts{A_\alpha}\) is countable and each \(A_\alpha\) is closed, but \(f\) is not continuous.
An indexed family of sets \(\ts{A_\alpha}\) is said to be locally finite if each point \(x\) of \(X\) has a neighborhood that intersects \(A_\alpha\) for only finitely many values of \(\alpha\). Show that if the family \(\ts{A_\alpha}\) is locally finite and each \(A_\alpha\) is closed, then \(f\) is continuous.
Solution
For a closed set \(C\subseteq Y\) put \(F_\alpha=(f|_{A_\alpha})^{-1}(C)=f^{-1}(C)\cap A_\alpha\), so that \(f^{-1}(C)=\bigcup_\alpha F_\alpha\) because the \(A_\alpha\) cover \(X\).
1. If \(A_\alpha\) is closed in \(X\), then \(F_\alpha\) is closed in \(X\) and \(F_\alpha\subseteq A_\alpha\).
Since \(f|_{A_\alpha}\) is continuous, \(F_\alpha\) is closed in the subspace \(A_\alpha\), and a closed subset of a closed subspace is closed in \(X\).
2. (a) If the \(A_\alpha\) are finitely many closed sets, \(f\) is continuous.
By step 1, \(f^{-1}(C)\) is a finite union of closed sets for every closed \(C\subseteq Y\).
3. (b) For the closed sets \(A_0=(-\infty,0]\) and \(A_n=[\frac1n,\infty)\), \(n\in\mathbb Z_+\), covering \(\mathbb R\), the function \(f\colon\mathbb R\to\mathbb R\) with \(f(x)=0\) for \(x\le0\) and \(f(x)=1\) for \(x>0\) is continuous on each \(A_n\) but not continuous.
The sets cover \(\mathbb R\), since every \(x>0\) exceeds some \(\frac1n\). The restriction of \(f\) to \(A_0\) is constant \(0\) and to each \(A_n\), \(n\ge1\), is constant \(1\). The set \(f^{-1}(\{0\})=(-\infty,0]\) is closed, but \(f^{-1}(\{1\})=(0,\infty)\) is not closed, although \(\{1\}\) is closed.
4. The union of a locally finite family \(\{F_\alpha\}\) of closed sets is closed.
Let \(F=\bigcup_\alpha F_\alpha\) and \(x\in\overline F\). Choose an open \(U\ni x\) meeting only \(F_{\alpha_1},\ldots,F_{\alpha_k}\). Then \(U\cap F\subseteq F_{\alpha_1}\cup\cdots\cup F_{\alpha_k}\), and \(x\in\overline{U\cap F}\) because \(U\) is an open neighborhood of \(x\). Hence \(x\in\overline{F_{\alpha_1}\cup\cdots\cup F_{\alpha_k}}=F_{\alpha_1}\cup\cdots\cup F_{\alpha_k}\subseteq F\).
5. (c) If \(\{A_\alpha\}\) is locally finite and each \(A_\alpha\) is closed, \(f\) is continuous.
By step 1, each \(F_\alpha\) is closed and \(F_\alpha\subseteq A_\alpha\), so \(\{F_\alpha\}\) is locally finite. By step 4, \(f^{-1}(C)=\bigcup_\alpha F_\alpha\) is closed for every closed \(C\subseteq Y\).
6. Q.E.D.
Steps 2, 3, and 5 answer (a), (b), and (c).