Every sequence of real bounded functions has convergent sub?

[RBG]

I figured it out... how do I remove this question?

 

[geoffrey159]

I don't know the answer, but I have an answer in the case $x$ belongs to finite set of rationals $S = \{ x_1,...,x_p\} \subset \mathbb{Q}$. It may or may not be useful for what you want to prove.

1. $\{f_n(x_1) \}$ is a real bounded sequence so there is a converging subsequence $\{f_{\sigma_1(n)}(x_1) \}$ that converges to $f(x_1)$
2. $\{f_{\sigma_1(n)}(x_2) \}$ is a real bounded sequence so there is a converging subsequence $\{f_{(\sigma_2 \circ \sigma_1)(n)}(x_2) \}$ that converges to $f(x_2)$. Furthermore $\{f_{(\sigma_2 \circ \sigma_1)(n)}(x_1) \}$ converges to $f(x_1)$ as a subsequence of $\{f_{\sigma_1(n)}(x_1) \}$.
3. Repeating this process, you have $\{ f_{(\sigma_p\circ ... \circ \sigma_1)(n)}(x_i)\}$ convergerges to $f(x_i)$ for all $i = 1... p$.

 

[Samy_A]

I figured it out... how do I remove this question?

Why remove the question?

Wouldn't it be more helpful to other members of the forum to leave the question, and post the answer (or at least an outline of the answer)?