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Problem

Let $\{A_\alpha\}$ be a collection of subsets of $X$; let $X=\bigcup_{\alpha}A_\alpha$. Let $f:X\rightarrow Y$;suppose that $f\vert_{A_\alpha}$ is continuous for each $\alpha$.

An index family of sets $\{A_\alpha\}$ is defined to be locally finite if each point $x$ has a neighborhood that intersects $A_\alpha$ for only finitely many values of $\alpha$. Show that if the family $\{A_\alpha\}$ is locally finite and each $A_\alpha$ closed, then $f$ is continuous.

Attempted Solution

It suffices to show that $f^{-1}\left(V\right)$ is closed in $X$ for any closed set $V$ in $Y$. Pick an arbitrary $x\in\overline{f^{-1}\left(V\right)}$ , then we have $U\cap f^{-1}\left(V\right)\neq\emptyset$ for any open set $U$ containing $x$. By local finiteness, $\exists$ an open neighborhood $N$ such that $x\in N$ and $N\cap A_{\alpha}\neq\emptyset$ for finitely $\alpha,$ namely $\left\{ A_{i}\right\} _{i=1}^{k}$. Since $x\in U\cap N,$ we have \begin{align} U\cap f^{-1}\left(V\right)\cap\left(\cup_{i=1}^{k}A_{i}\right) & =U\cap\left[\cup_{i=1}^{k}\left(f^{-1}\left(V\right)\cap A_{i}\right)\right]\\ & =U\cap\left(\cup_{i=1}^{k}f\vert_{A_{i}}^{-1}\left(V\right)\right)\\ & \supset U\cap N\cap\left(\cup_{i=1}^{k}f\vert_{A_{i}}^{-1}\left(V\right)\right)\neq\emptyset \end{align} from which we have that $x\in\overline{\cup_{i=1}^{k}f\vert_{A_{i}}\left(V\right)}$. Note that $f\vert_{A_{i}}$ is continuous, so $f\vert_{A_{i}}^{-1}\left(V\right)=f^{-1}\left(V\right)\cap A_{i}$ is closed in the subspace topology of $A_{i}.$ So $f\vert_{A_{i}}^{-1}\left(V\right)=F_{i}\cap A_{i}$ for some closed set $F_{i}\subset X$. Since $A_{i}$ is closed, $f\vert_{A_{i}}^{-1}\left(V\right)$ is closed in $X$ as well. Thus, $x\in\cup_{i=1}^{k}f\vert_{A_{i}}^{-1}\left(V\right)=f^{-1}\left(V\right)\cap\left(\cup_{i=1}^{k}A_{i}\right)\subset f^{-1}\left(V\right)$, from which we conclude $\overline{f^{-1}\left(V\right)}\subset f^{-1}\left(V\right)$ and as a result, $f^{-1}\left(V\right)$ is closed because it contains all the limit points.

Question

(1).This is is problem in Munkrs topology. I tried to solve it and I think I had it. I really appreciate is anyone can take a look at my solution.

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  • $\begingroup$ @hermes why is it so small to read though. I can see a full size pic here. $\endgroup$ – Mathemagician Nov 6 '15 at 1:54
  • $\begingroup$ Sorry I mean on cell phone, it looks smaller than other text. $\endgroup$ – Math Wizard Nov 6 '15 at 2:13
  • $\begingroup$ @hermes corrected! $\endgroup$ – Mathemagician Nov 6 '15 at 2:44
  • $\begingroup$ For future reference, detexify.kirelabs.org/classify.html is amazing for finding the $\LaTeX$ command for a given symbol. $\endgroup$ – Math1000 Nov 6 '15 at 3:17
  • $\begingroup$ @Math1000 thanks $\endgroup$ – Mathemagician Nov 6 '15 at 3:25
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It's a bit clearer to build it up:

Lemma: if $A_i, i \in I$ is a locally finite family, then $$\overline{\bigcup_{i \in I} A_i} = \bigcup_{i \in I} \overline{A_i}\text{.}$$

Proof: clearly, for any $j \in I$: $A_j \subseteq \cup_{i \in I} A_i$, so $\overline{A_j} \subseteq \overline{\cup_{i \in I} A_i}$. As $j \in I$ is arbitrary, $\cup_{i \in I} \overline{A_i} \subseteq \overline{\cup_{i \in I} A_i}$ as well, which covers one inclusion, which always holds.

If $x \in \overline{\cup_{i \in I} A_i}$, let $V$ be any neighbourhood such that $J = \{ i \in I: A_i \cap V \neq \emptyset\}$ is finite. Suppose (for a contradiction) that $x \notin \overline{A_j}$ for all $j \in J$. Then there are open neighbourhoods $U_j$ of $x$, for each $j \in J$, such that $U_j \cap A_j = \emptyset$. But then $O = V \cap \cap_{j \in J} U_j$ is also an open neighbourhood of $x$ that $O$ misses all $A_i$: those with index outside $J$ because of the $V$, and we also miss all members $A_j, j \in J$ due to the $U_j$. But this contradicts that $x \in \overline{\cup_{i \in I} A_i}$. So for some $j \in J$, $x \in \overline{A_j}$, which shows the other inclusion.

As a corollary: the locally finite union of closed sets is closed (which is more general than the finite case, known from the axioms).

Lemma 2: if $A$ is a closed subset of $X$ and $F \subseteq A$ is closed in $A$, it is also closed in $X$.

Proof: $F = C \cap A$, for some closed subset of $X$ (definition of subspace topology). But then $F$ is a finite intersection of two closed sets of $X$ and so itself closed in $X$.

Proof of the result:

Let $F$ be closed in $Y$. Then

$$f^{-1}[F] = X \cap f^{-1}[F] = (\bigcup_{i \in I} A_i) \cap f^{-1}[F] = \bigcup_{i \in I} (f^{-1}[F] \cap A_i) = \bigcup_{i \in I} (f|_{A_i})^{-1}[F])$$

The right hand side is a locally finite union of closed sets of $X$ (closed in $X$ because lemma 2 applies and the restricted maps are continuous on their domain). So lemma 1 implies that $f^{-1}[F]$ is closed.

Your proof also tries to prove the lemma 1, essentially (and lemma 2 is mentioned too in passing), but I think it's a bit less clear. The lemmas are also useful in their own right.

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  • $\begingroup$ Hi I am sorry for down voting your answer. I hope i can undo it. I misclicked it on my phone app.. $\endgroup$ – Mathemagician Nov 6 '15 at 21:00
  • $\begingroup$ you can click up vote to remove down vote. $\endgroup$ – Math Wizard Nov 7 '15 at 1:19
  • $\begingroup$ @hermes it says that i voted already and my vote the locked... $\endgroup$ – Mathemagician Nov 7 '15 at 6:59
  • $\begingroup$ Then you need to edit the post and vote again. $\endgroup$ – Math Wizard Nov 7 '15 at 7:11
  • $\begingroup$ @hermes I have edited it and waiting for peer review now. Thanks for tip! $\endgroup$ – Mathemagician Nov 7 '15 at 23:10

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