# Compact and open subspaces using products

I'm preparing for a topology prelim, and there's one question related to compactness that I'm trying to work on. Here it is:

Let $C$ be a compact subspace of $X$, and $K$ be a compact subspace of $Y$. Now let $U$ be an open set in $X \times Y$ that contains $C \times K$. Show that there exist open subspaces $V$ of $X$ and $W$ of $Y$ such that $$C \times K \subset V \times W \subset U.$$ Well, certainly $C \times K$ is a compact subspace of $X \times Y$, right? Then this would mean that every covering of $C \times K$ by open sets in $X \times Y$ contains a finite subcollection covering $C \times K$. Can't you use $U$ as the open covering, and then somehow get a finite subcollection which might have something to do with the space $V \times W$? What am I missing here? I would appreciate some helpful input, thanks.

My second attempt:
Alright, I might as well let $U = A \times B$, $A,B$ open such that $C \subset A, K \subset B$. Now we could let $\mathcal{C}$ be an open covering of $C$; since $C$ is compact, there is a finite subcollection $\{V_1, V_2, … ,V_n\}$ of $\mathcal{C}$ covering $C$. Next, let $V= A \cap (\bigcup_{i=1}^n V_i)$. Then, $V \subset A$, $V$ open and $C \subset V$. Similarly, we can find $W$ such that $W \subset B$, $W$ open and $K \subset W$ so that $$C \times K \subset V \times W \subset A \times B= U.$$ I hope this was the right approach to take.

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It's true that $C\times K$ is a compact subspace of $X\times Y$, but taking $\{U\}$ as its open cover, the only finite subcover would be $\{U\}$ again. –  Thomas E. Dec 24 '12 at 22:26
About your second attempt: Let $X = Y = \Bbb R$ and set $C = K = [-1, 1]$. Now what happens if $U$ is the open disc of radius $2$ around the origo? What would $A$ and $B$ be? (You might not as well assume $U$ to be a basis element.) –  Arthur Dec 24 '12 at 22:57

Let $C$ and $K$ be compact subsets of $X$ and $Y$ respectively, and $U$ an open set in $X\times Y$ s.t. $C\times K\subset U$. Here's how you can find those $V$ and $W$:

Fix $y\in K$. Now for each $x\in C$ we have $(x,y)\in U$, so there exists basis elements of the product topology $B_{x}\times G_{x}\subset U$ so that $x\in B_{x}$ and $y\in G_{x}$. Since $C$ is compact and $C\subset \bigcup_{x\in C}B_{x}$, there exists a finite $I\subset C$ so that $C\subset \bigcup_{x\in I}B_{x}$. Denote $V_{y}=\bigcup_{x\in I}B_{x}$ and $W_{y}=\bigcap_{x\in I}G_{x}$. Now $C\subset V_{y}$ and $y\in W_{y}$, where $V_{y}\subset X$ and $W_{y}\subset Y$ are open sets with $V_{y}\times W_{y}\subset U$. This could be done for each $y\in K$. Since $K$ is compact and $K\subset \bigcup_{y\in K}W_{y}$, there exists a finite $J\subset K$ so that $K\subset\bigcup_{y\in J}W_{y}$. Denote $V=\bigcap_{y\in J}V_{y}$ and $W=\bigcup_{y\in J}W_{y}$. Now $C\subset V$ and $K\subset W$, where $V\subset X$ and $W\subset Y$ are open sets with $V\times W\subset U$.

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This is a variation on the tube lemma.

First, cover $C\times K$ with finitely many basis elements $V\times W_i\subseteq U$, with $1\leq i\leq n$. We construct these by observing that any point $(c, k) \in C\times K$ can be covered by an open set $V'\times W'\subseteq U$, with $C\subseteq V'$ (this is possible by compactness of $\{c\}\times K \subseteq U$). Now chose a finite subcover $\{V_i\times W_i\}$, and set $V = \bigcap V_i$.

Now, set $W= \bigcup W_i$, and we're done.

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