# Every sequence of real bounded functions has convergent sub?

1. Dec 10, 2015

### RBG

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

Last edited: Dec 10, 2015
2. Dec 10, 2015

### 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 $i = 1... p$.

3. Dec 11, 2015

### Samy_A

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)?

Last edited: Dec 11, 2015