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Let $A$ and $B$ be two compact subsets of $\mathbb{R}$. Let $f:A \times B \to \mathbb{R}$ be a continuous function on $A \times B$.

For each $a\in A$, define $B_a=\{b\in B:f(a,b)=0\}$, and suppose each $B_a$ is a singleton, so we may define a $g:A\to\Bbb R$ such that $g(a)$ is the unique element of $B_a$ for each $a\in A$. Is $g$ a continuous function of $a$?

I tried the following. But I have the feeling it's not correct.

Assume $g$ is not continuous at $a \in A$, then there exists a sequence $\{a_n\} \subset A$ converging to $a$ in $A$, for which the sequence $\{b_n\}=\{g(a_n)\} \subset B$ does not converges to $b=g(a)$. Since $B$ is compact, by the Bolzano-Weierstrass theorem, $\{b_n\}$ has a subsequence $\{b'_n\}$ converging to some $b'\ne b$ as $\{b_n\}$ does not converges to $b$. Let $\{a'_n\} \subset A$ be the subsequence of $\{a_n\}$ induces by $\{b'_n\}$. Since $\{a_n\}$ converges to $a$ then every subsequence of $\{a_n\}$ converges to $a$ and $\{a'_n\}$ converges to $a$. Since $g(a)$ contains a unique element $b$, then $b'=b$ which is a contraction since $\{b'_n\}$ is a subsequence of $\{b_n\}$. (The previous sentence seems suspicious).
The claim follows.

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    It is equivalent. I was simply making explicit that it was a set of points that we're dealing with (though it happens to be a singleton), and not a point. One could indeed say that $B_a=\arg_{b\in B}\bigl(f(a,b)=0\bigr),$ instead.2012-06-27

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Let's assume that $g$ is not continuous at $a$. Thus, there is a sequence $\{a_n\}$ such that $a_n$ converges to $a$, but such that $b_n:=g(a_n)$ fails to converge to $b:=g(a)$.

By definition, this means that there exists some $\epsilon$ such that for all $N$, there is some $n\geq N$ with $|b_n-b|>\epsilon$--that is, there are infinitely many $n$ such that $|b_n-b|>\epsilon$. Let $\{b_n'\}$ be a subsequence of $\{b_n\}$ with $|b_n'-b|>\varepsilon$ for all $n$ (possible by "infinitely many"), and observe that neither $\{b_n'\}$ nor any of its subsequences can converge to $b$ (I leave it to you to justify this).

Since $\{b_n'\}$ is a sequence of points of the compact set $B$, then by B-W, there is a convergent subsequence $\{b_n''\}$ of $\{b_n'\}$. As noted above, $\{b_n''\}$ must converge to something other than $b$. Observing that $\{b_n''\}$ is a subsequence of $\{b_n\}$, too, then we may assume without loss of generality that $b_n'=b_n''$ for all $n$, meaning that in fact $b_n'$ converges, say to $b'\neq b$. (The only thing you hadn't done is justified why we could assume that your chosen subsequence $\{b_n'\}$ converges. So close!)

Then we induce $\{a_n'\}$ from $\{b_n'\}$ and continue as you described.

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    You're very welcome.2012-06-28
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I found the following proof by borrowing from Berge Maximum theorem proof strategy.
Do you think it's ok?
(The beginning is similar then my previous proof, I have indicated the new content)

I improved this paragraph per Cameron help
Assume $g$ is not continuous at $a \in A$, then there exists a sequence $\{a_n\} \subset A$ converging to $a$ and an $\epsilon>0$ such that for each $n$, $b_n=g(a_n)$ and $|b_n-b|>\epsilon$ where $b=g(a)$. Since $B$ is compact, by the Bolzano-Weierstrass theorem, $\{b_n\}$ has a subsequence $\{b'_n\}$ converging to some $b'\ne b$ as $\{b_n\}$ is never in the neighborhood $(b-\epsilon,b+\epsilon)$ of $b$.

Let $\{a_n'\} \subset A$ be the subsequence of $\{a_n\}$ induced by $\{b'_n\}$. Since $\{a_n\}$ converges to $a$ then every subsequence of $\{a_n\}$ converges to $a$ and $\{a'_n\}$ converges to $a$.
new content
Since $f$ is continuous then $f(a'_n,b'_n)$ converges to $f(a,b')$. Since $g(a)$ is the singleton $b$ then $b'$ is not a root of $f(a,\cdot)$. Whence, either $f(a,b')>f(a,b)$ or $f(a,b'). Since $B$ is compact there is sequence $\{\bar b_n\} \subset B$ converging to $b$. By continuity of $f$, $f(a'_n,\bar b_n)$ converges to $f(a,b)$.

If $f(a,b')>f(a,b)$, then there is an $n_0$ for which $f(a'_n,b'_n)>f(a'_n,\bar b_n)$ when $n \ge n_0$. Since $b'_n=g(a'_n)$, then $0>f(a'_n,\bar b_n)$ for $n \ge n_0$ and $0>f(a,b)$, which is a contradiction since $b$ is a root of $f(a,\cdot)$.
Similarly, we cannot have $f(a,b') and hence $f(a,b')=f(a,b)$.
Since, $g(a)$ is the singleton $b$, $b=b'$, which is a contradiction. Thus, $g$ is continuous at $a$.

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    Alright. I've taken$a$look at the new content, now, and it looks just fine! Still haven't quite got the first bit fixed, though. I want to point out, though, that everything you're saying is correct. It simply isn't fully justified. Since you're so close (nice work, by the way), I will go ahead and put$a$more detailed explanation in an answer so I don't run out of room. Hopefully, it will help you out.2012-06-28