[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.