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On Bourbaki's TVS Chapter IV pages 33-34, the last part of Proposition 2 can be formulated as follows:

Notations:

$K$ - The underlying field which is the real or complex number field;

$X$ - A compact Hausdorff topological space;

$D$ - A dense subset of $X$;

$C_s(X)$ - The space of continuous $K$-valued functions, equipped with the topology of point-wise convergence.

$A$ - A nonempty subset of $C_s(X)$;

$\overline{A}$ - The closure of $A$ in the product space $K^X$;

Consider the following statements:

(a) $sup_{f\in A}|f(x)|<+\infty$ for each $x\in X$;

(b) for each infinite sequence $\{f_n\}$ in $A$ and each infinite sequence $\{x_n\}$ in $D$, if the iterated limits $$\delta=\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)\text{ and }\gamma=\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$$ exist, then $\delta=\gamma$;

(c) $\overline{A}\subset C_s(X)$, i.e. each $u\in\overline{A}$ is continuous on $X$;

Then (a) and (b) imply (C).

On Bourbaki's TVS chapter IV page 34, the proof given there is via reductio ad adsurdum. Assume $u\in \overline{A}$ is discontinuous at $a\in X$, then two sequences $\{f_n\}$ in $A$ and $\{x_n\}$ in $D$ are constructed, satisfying some inequalities, which can be used to contradict (b), provided there is at least a subsequence of $\{u(x_n)\}$ converges in $K$. Then on the same page, 4th line from the bottom, it states that "Since $u(X)$ is a compact subset in $K$..." Why this is true? (Actually one only needs the set $\{u(x_m)|m\in N \}$ to be bounded in $K$, but still I don't know a proof). Many Thanks for the advice.

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    $\begingroup$ Hi Chris, can I suggest you make this question as self-contained as possible? Not everyone has Bourbaki volumes at hand. For example, what is $K$? Is $X$ compact and $u$ continuous? Etc. Thanks! $\endgroup$
    – B R
    Commented Aug 16, 2012 at 14:38
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    $\begingroup$ In addition, if ever a question was "too localized"... $\endgroup$
    – Igor Rivin
    Commented Aug 16, 2012 at 17:27
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    $\begingroup$ Question seems logically complete and coherent now, so am voting to reopen. $\endgroup$
    – Tom Leinster
    Commented Aug 20, 2012 at 19:45
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    $\begingroup$ My impression is that the proof or the statement is incomplete. A small tweak in the proof certainly gives that either $u$ is continuous at $a$, or $|u(x)|\to\infty$ as $x\to a$ with $x\neq a$. Perhaps one can find a counterexample with that in mind. If one compactifies $K$ with $\infty$ in the usual way, and in condition (iv) of the original proposition one allows $\gamma=\infty$ and $\delta=\infty$, then the proof works. $\endgroup$
    – GH from MO
    Commented Aug 21, 2012 at 17:40
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    $\begingroup$ As suggested by GH, if one replaces (b) with the strengthened: (b)' for each infinite sequence $\\{f_n\\}$ in $A$ and each infinite sequence $\\{x_n\\}$ in $D$, if the iterated limits $$\delta=\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)\text{ and }\gamma=\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$$ exist in the one-point compactification $\overline{K}$ of $K$, then $\delta=\gamma$; Then one can show that (a) and (b)' imply (c), with some modifications of the original proof of Bourbaki. $\endgroup$
    – Chris
    Commented Aug 26, 2012 at 8:05

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