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I am self studying basic topology and have trouble proving the following question.

If $A$ and $B$ are compact, and if $W$ is a neighborhood of $A \times B$ in $X \times Y$, find a neighborhood $U$ of $A$ in $X$ and a neighborhood $V$ of $B$ in $Y$ such that $U \times V \subseteq W$.

Intuitively, in Euclidean space, I can see that such $U$ and $V$ must exist since one can trace around the "edge" of the compact space $A \times B$ and create an open product space that is only a little bit "larger" than $A \times B$ but smaller than $W$. Then the projection $U$ and $V$ would satisfy the requirement. But I could not think of a way to construct it formally. Any hint would be greatly appreciated!

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  • $\begingroup$ Start by observing that $W$ is a union of (possibly infinitely many) open rectangles $U_i\times V_i$. Use compactness to reduce the problem to the simpler case in which $W$ is a union of finitely many open rectangles. $\endgroup$
    – Algernon
    Commented Sep 25, 2020 at 18:23
  • $\begingroup$ Thanks! I am actually stuck right here. Let $I$ be the finite set indexing the finite rectangles $U_i \times V_i$. But how does one get the cover $U \times V$ from knowing $A \times B \subseteq \bigcup_{i \in I} U_i \times V_i \subseteq W$? $\endgroup$
    – tgeng
    Commented Sep 25, 2020 at 18:32
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    $\begingroup$ Trying this in the Euclidean plane would help. You have a closed rectangle $R=[a,b]\times[c,d]$ which is covered by finitely many open rectangles $S_1, S_2, ..., S_n$. How would you find an open rectangle $T$ which includes $R$ and is included in $S:=\bigcup_{i=1}^n S_i$? For each $x\in[a,b]$, the set $V_x:=\{y: (x,y)\in S\}$ is open. Furthermore, varying $x$ over $[a,b]$, there are only finitely many distinct sets $V_x$. Hence, $V:=\bigcap_{x\in[a,b]} V_x$ is open. Similarly, you can construct $U$. $\endgroup$
    – Algernon
    Commented Sep 25, 2020 at 18:54
  • $\begingroup$ ... and this argument works in general. $\endgroup$
    – Algernon
    Commented Sep 25, 2020 at 18:56
  • $\begingroup$ Thanks! I see it now. $\endgroup$
    – tgeng
    Commented Sep 25, 2020 at 20:58

1 Answer 1

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Algernon's argument seems to need a special case of Tychonoff's theorem to get to the finiteness of the union. Here is an argument which avoids that.

Lemma: Let $A\subseteq X$ be compact and $B\subseteq Y$ arbitrary. Let $W\subseteq X\times Y$ be open such that for each $x\in A$ there are a neighborhoods $U_x$ of $x$ and $V_x$ of $B$ with $U_x\times V_x\subseteq W$, then there are neighborhoods $U$ of $A$ and $V$ of $B$ with $U\times V\subseteq W$.

For the proof of the lemma, one can restrict by the compactness to finitely many neighborhoods and apply Algermon's argument with $V$ being a finite intersection.

To see that the hypotheses of the lemma are satisfied if $B$ is compact with $A\times B\subseteq W$, the lemma itself can be applied (with exchanged roles of $X$ and $Y$, $A:=B$ and $B:=\{x\}$): The hypothesis of the lemma is then satisfied by the mere definition of the product topology.

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