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$\begingroup$

May I refer you to: alt text

taken from page 92 of Infinite Dimensional Analysis by Charalambos D. Aliprantis.

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?2011-01-04
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    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 $f^{-1}(y) \subseteq G \subseteq f^{-1}(W)$.2011-01-04
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    Here is a screenshot of Corollary 3.43: http://i.imgur.com/ZKZaI.jpg2011-01-04
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    @Nuno: Thanks: I've uploaded the image into the question in the place of the ugly long link.2011-01-04
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    @Arturo Magidin: Thank you too for adding it to the question. It's much better now.2011-01-05

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