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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 $$\delta=\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)\text{ and }\gamma=\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$$ exit, 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.

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.

In Bourbaki's TVS Chapter IV Section 5, the last part of the proof 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 simple(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 $$\delta=\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)\text{ and }\gamma=\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$$ exit, 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).

In Bourbaki's TVS IV page??, the proof 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 in 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 $\{u(x_m)|m\in N \}$ is bounded in $K$, but still I don't know a proof).

  Many Thanks for the advices. :)

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 $$\delta=\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)\text{ and }\gamma=\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$$ exit, 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.

[Sorry, editing in progress...]

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 simple(point-wise) convergence.\ $A$ - A nonempty subset of $C_s(X)$;

(a) $sup_{f\in A}|f(x)|<+\infty$ for each $x\in X$;  (b) for each infinite sequence if the iterated limits $$\delta=\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)$$ and $$\gamma=\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$$ exit, then $\delta=\gamma$.  (c) $\overbar{A}\subset C_s(X)$, i.e. each $u\in\overbar{A}$;

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

In Bourbaki's TVS Chapter IV Section 5, the last part of the proof of Proposition 2(in the 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 to show the set $\{u(x_m)|m\in N \}$ is bounded for the proof to get through). Thanks in advance.

In Bourbaki's TVS Chapter IV Section 5, the last part of the proof 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 simple(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 $$\delta=\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)\text{ and }\gamma=\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$$ exit, 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).

In Bourbaki's TVS IV page??, the proof 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 in 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 $\{u(x_m)|m\in N \}$ is bounded in $K$, but still I don't know a proof).

Many Thanks for the advices. :)