# Inclusion in a proof (cts image of a compact metrizable space is metrizable)

May I refer you to:

Why $G \subset f^{-1}(W)$ ? I don't see this inclusion. Can you please help?

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Not everyone will have access to the page (I get a message saying I have no access or have used up my access for this book); is it really that much work for you to summarize the point in addition to providing a discreet link to the page for the benefit of those who can? –  Arturo Magidin Jan 4 '11 at 4:45
You will need to take appropriate basis elements to guarantee that $G \subseteq f^{-1}(W)$. Since $W$ is open in $Y$ and $y \in W$, $f^{-1}(W)$ is open in $X$ and contains $f^{-1}(y)$. For each $x \in f^{-1}(y)$, let $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq f^{-1}(W)$. The collection $C = \{B_x : x \in f^{-1}(y)\}$ is an open cover of $f^{-1}(y)$. Now use the fact that $f^{-1}(y)$ is compact to obtain a finite subcover of $C$ containing $f^{-1}(y)$ and let $G$ be their union. Thus $f^{-1}(y) \subseteq G \subseteq f^{-1}(W)$. –  Nuno Jan 4 '11 at 5:10
Here is a screenshot of Corollary 3.43: i.imgur.com/ZKZaI.jpg –  Nuno Jan 4 '11 at 5:42
@Nuno: Thanks: I've uploaded the image into the question in the place of the ugly long link. –  Arturo Magidin Jan 4 '11 at 17:31
@Arturo Magidin: Thank you too for adding it to the question. It's much better now. –  Nuno Jan 5 '11 at 3:44

An alternative proof: Of course $Y = f[X]$ is compact and Hausdorff. It has a countable network (which is like a base of a space but without the requirement that its members are open sets), as a continuous image of a space with a countable network. A compact Hausdorff space with a countable network is second-countable (Arhangel'skij's theorem) and we are done by Urysohn. For a proof of the stuff on networks, I have a post here that explains it (at the end it goes into networks).