# Question on the proof of the Stability Theorem in Guillemin & Pollack

I'm having trouble with the following part in the proof of the stability theorem on Pg 36 of Guillemin and Pollack's Differential Topology: They write that since $X$ is compact, it follows that any open neighbourhood of $X\times \lbrace 0 \rbrace$ in $X \times [0,1]$ contains $X \times [0,\epsilon]$, for a suitable $\epsilon$. I cannot see how $X$'s compactness has anything to do with this: this depends only on the local compactness of the closed interval.

But, compactness is an essential condition for the theorem and this is the only point where it enters the proof (except part(e) of the theorem). Any help will be greatly appreciated.

Added I have a feeling that they want $X\times[o,\epsilon]$ to be compact, then prove something for open sets indexed by all points in this set (i.e. the obvious open cover) and then pass onto a finite cover...

Added I think I was reading the proof wrong. I am just writing this in case someone makes a similar silly mistake. My line of thought was that any open nbd of the slice $X\times\lbrace 0 \rbrace$ in $X \times I$ is of the form $X \times G$ where $G$ is open in $I$, since $I$ is locally compact, we can find a $[0,\epsilon]$ inside $G$ and $X\times [0,\epsilon]$ will cover $X\times\lbrace 0 \rbrace$. But what they want to do in the proof is to prove something locally at a point on $X\times \lbrace 0 \rbrace$ and boost it to the whole thing.

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 You are referencing parts of a theorem and the readers do not know what that is. Recommend You can add the theorem or the relevant parts and also the book you are using. – Amzoti Jan 1 at 20:53

## 1 Answer

For every $x\in X$ there is an open neighbourhood1 of the form $U_x\times [0,\epsilon_x)$ around $(x,0)$ within the given open neighbourhood of $X\times\{0\}$. By compactness, finitely many of these suffice to cover all of $X\times \{0\}$. Letting $\epsilon$ smaller than the finitely many involved $\epsilon_x$ gives us the desired result.

1 Just remember that a set $S\subset X\times Y$ is open in the product topology by definition if for each $(x,y)\in S$ there is a set of the form $U\times V$ with $U,V$ open in $X, Y$, respectively, and $(x,y)\in U\times V\subseteq S$. And of course the essential open neighbourhoods of $0\in[0,1]$ are of the form $[0,\epsilon)$ with $\epsilon>0$

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 Yes, I think this is what they want to say. Thanks a lot for clearing this up. – Apoha Jan 1 at 20:55 How strange that I need more points than i have to vote up this answer... – Apoha Jan 1 at 20:56 @Apoha, if the answer helps you, then you can click the ✓ (check) mark next to the answer to accept it as the correct answer. (You need a certain amount of reputation to upvote posts, but you can always accept posts on your questions.) – George V. Williams Jan 1 at 21:09