# Continuity in Slices + Mapping Compact Sets to Compact Sets = Continuous

The following is a problem from Berkeley's Grad Prelims:

Let $f:\mathbb R^2 \to \mathbb R$ satisfy

(i) Given any $x_0,y_0 \in \mathbb R$, $y \mapsto f(x_0,y)$ and $x \mapsto f(x,y_0)$ are continuous.

(ii) For each compact $K$, $f(K)$ is compact.

Prove $f$ is continuous.

Attempt:

Let $(x_n,y_n) \to (x,y)$. We wish to show $f((x_n,y_n)) \to f((x,y))$.

Sequential compactness gives us a subsequence $$f((x_{n_k},y_{n_k})) \to f((\tilde x, \tilde y)) \in f(K).$$

I'm un sure how to proceed to show (i) $f((\tilde x, \tilde y)) = f((x,y))$, and (ii) how the existence of such a subsequence and the slice continuity give convergence of the entire sequence.

Originally I had hoped that the slice continuity would make $\{f((x_n,y_n))\}$ cauchy, but that doesn't seem to be the case.

I'm looking for a small hint as to how to proceed. Please, no answers.

• It might be hard to give a small hint that will get you very far; unless I'm missing something, this problem is rather complicated to solve. – Eric Wofsey Mar 6 '16 at 23:01
• @EricWofsey Is the right method to consider sequential continuity and sequential compactness? – Anthony Peter Mar 6 '16 at 23:02
• Yes, sequences seem like a great way to attack this. – Eric Wofsey Mar 6 '16 at 23:02
• @EricWofsey Is the idea to somehow bring the convergence of the subsequence back up to the entire sequence? or is it something else entirely? – Anthony Peter Mar 6 '16 at 23:04
• Ultimately you're going to use something like the fact that if every subsequence of a sequence has a subsequence that converges to a point $p$, then the original sequence converges to $p$. But there's a lot more work to be done to get to the point that you can use that fact; I attempted to give a hint of the ideas involved in my answer. – Eric Wofsey Mar 6 '16 at 23:21

You're on the right track, but instead of only taking $K=\{(x_n,y_n)\}\cup\{(x,y)\}$ and using compactness of $f(K)$, you need to use the fact that $K'=A\cup\{(x,y)\}$ is compact for every subset $A\subseteq\{(x_n,y_n)\}$, and hence so is $f(K')$. You are also going to have to split into cases according to whether $f(K)$ is infinite or finite (or really, according to whether there is any infinite subset of $K$ whose image is finite); in the finite case, you will need to use the fact that $f$ is continuous on slices. In the infinite case, you might find the following fact (which you should prove if you haven't seen it before) helpful: if $B\subseteq\mathbb{R}$ is infinite and $p\in\mathbb{R}$ is such that $C\cup\{p\}$ is compact for every subset $C\subseteq B$, then $p$ is the unique limit point of the set $B$.
• Is the second half (the cases) to ensure that the subsequences given by the sequential compactness of $f(K')$ all converge to the same limit? – Anthony Peter Mar 6 '16 at 23:33
• Yes, and that their common limit has to be equal to $f(x,y)$. – Eric Wofsey Mar 6 '16 at 23:41