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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.

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  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Mar 6 '16 at 23:01
  • $\begingroup$ @EricWofsey Is the right method to consider sequential continuity and sequential compactness? $\endgroup$ – Anthony Peter Mar 6 '16 at 23:02
  • $\begingroup$ Yes, sequences seem like a great way to attack this. $\endgroup$ – Eric Wofsey Mar 6 '16 at 23:02
  • $\begingroup$ @EricWofsey Is the idea to somehow bring the convergence of the subsequence back up to the entire sequence? or is it something else entirely? $\endgroup$ – Anthony Peter Mar 6 '16 at 23:04
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    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Mar 6 '16 at 23:21
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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 $B$.

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  • $\begingroup$ 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? $\endgroup$ – Anthony Peter Mar 6 '16 at 23:33
  • $\begingroup$ Yes, and that their common limit has to be equal to $f(x,y)$. $\endgroup$ – Eric Wofsey Mar 6 '16 at 23:41
  • $\begingroup$ If I may ask just briefly before I accept your answer, would you have considered yourself capable of solving this type of problem after your first course in analysis? Or is it only after spending the undergrad years tackling other areas that when you return to problems like this (even without necessarily practising analysis much more) that the solution sketches become much clearer? $\endgroup$ – Anthony Peter Mar 6 '16 at 23:45
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    $\begingroup$ I certainly wouldn't worry if you have trouble solving this. Unless there's a much easier solution than the one I see, I would consider this a not unsolvable but definitely challenging problem for someone who has just finished a first course in analysis (it could be a starred exercise in a textbook). I won't venture a guess as to whether I could have solved it after my first analysis course, but at least it probably would have taken me a lot longer than it did now. $\endgroup$ – Eric Wofsey Mar 6 '16 at 23:49
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    $\begingroup$ Yep, looks good! $\endgroup$ – Eric Wofsey Mar 7 '16 at 1:25

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