Given $(f_k)$ a sequence of bounded functions, find subsequence $(x_{n_j})$ of $(x_n)$ such that $(f_k(x_{n_j}))$ is convergent for all $k$. Let $(x_n)$ be a sequence in $\mathbb{R}$.
Let $(f_k:\mathbb{R}\to\mathbb{R})$ be a sequence of bounded functions in $\mathbb{R}$.
Prove that there is a subsequence $(x_{n_j})$ of $(x_n)$ such that $(f_k(x_{n_j}))$ is convergent for all $k\in\mathbb{N}$.
I am able to prove this by induction when $(f_k)$ is a finite sequence.
Basically, if we have $f_{k}\circ (x\circ \alpha)$ being convergent for all $k\le m$, then $f_{m+1}\circ(x\circ\alpha)$ is a bounded sequence,
so there is an index sequence $\beta$ such that $f_{m+1}\circ(x\circ\alpha\circ\beta)$ is convergent.
Then $f_k\circ(x\circ\alpha\circ\beta)$ is also convergent for all $k<m+1$ because it is a subsequence of $f_k\circ(x\circ\alpha)$.
However, I don't see a way to carry this argument over to a countable sequence $(f_k)$.
 A: First we note that if we fix $k$ and let $y_n$ be any sequence then $f_k(y_n)$ admits a convergent subsequence. Indeed a bounded sequence always admits a convergent subsequence.
Now we construct an infinite matrix
$$
\left(
\begin{matrix}\ 
y_{11} & y_{21} & y_{31} & \dots \\
\ y_{12} & y_{22} & y_{32} & \dots \\
\ y_{13} & y_{23} & y_{33} & \dots \\
 \ y_{14} & y_{24} & y_{34} & \dots \\
\ \dots 
\end{matrix} \right)\ . $$
To form the first row we extract a convergent subsequence $(x_{n_k})$ so that $f_1(x_{n_k})$ converges as $k\to \infty$. Set $y_{k1}=x_{n_k}$. For the second row we proceed similarly looking for a convergent subsequence of
$$f_2(y_{11}),\  f_2(y_{21}),\  f_2(y_{31}), f_2(y_{41}) \  \dots $$
Proceeding this way we can make sure that:

*

*on any given row  there is displayed a subsequence of the previous row ,

*the rows of the matrix
$$
\left(
\begin{matrix}\ 
f_1(y_{11}) & f_1(y_{21}) & f_1(y_{31}) & \dots \\
\ f_2(y_{12}) & f_2(y_{22}) & f_2(y_{32}) & \dots \\
\ f_3(y_{13}) & f_3(y_{23}) & f_3(y_{33}) & \dots \\
 \ f_4(y_{14}) & f_4(y_{24}) & f_4(y_{34}) & \dots \\
\ \dots 
\end{matrix} \right)$$
all converge.

Then the sequence $z_i=y_{ii}$ ($i = 1, 2, 3, \dots$) displayed on the diagonal is a subsequence of the original sequence $x_n$ having the desired property. This is because for each $k$, up to disregarding the first $k-1$ initial terms, $z_i$ is a subsequence of the $k^\text{th}$ row of the matrix we constructed.
A: While Antonio Alfieri gave a great illustration of the diagonal method (and I accepted their answer), I would also like to give a more detailed answer here, following Vercingetorix's comment.

I have proved that
there is a sequence of indexes {$\alpha_n:\mathbb{N}\to\mathbb{N}$}
such that for any $n\in\mathbb{N}$,

*

*$f_k\circ x\circ\alpha_n$ converges for any $0\le k\le n$.

*$\alpha_n$ is a subsequence of $\alpha_k$ for any $0\le k\le n$, i.e. there is strictly increasing $\gamma_{kn}:\mathbb{N}\to\mathbb{N}$ such that $\alpha_n=\alpha_k\circ\gamma_{kn}$.

Define $\beta:\mathbb{N}\to\mathbb{N}$ by $\beta(n)=\alpha_n(n)$ for all $n\in\mathbb{N}$.
For any $n<m$,
$$
\begin{align*}
\beta(n)&=\alpha_n(n) \\
&< \alpha_n(m) \\
&\le\alpha_n(\gamma_{nm}(m)) \\
&=\alpha_m(m) \\
&=\beta(m).
\end{align*}
$$
Then $\beta$ is strictly increasing,
so $x\circ\beta$ is a subsequence of $x$.
Let $k\in\mathbb{N}$.
Define $\gamma_k:\mathbb{N}\to\mathbb{N}$ by
$\gamma_k(n)=\gamma_{kn}(n)$ for all $n\in\mathbb{N}$.
For all $n<m$,
we have
$$
\begin{align*}
\alpha_k(\gamma_{kn}(n))
&= \alpha_n(n) \\ 
&< \alpha_m(m) \\
&=\alpha_k(\gamma_{km}(m)),
\end{align*}
$$
so $\gamma_{kn}(n)<\gamma_{km}(m)$ because $\alpha_k$ is strictly increasing.
Then $\gamma_k$ is strictly increasing,
so $\alpha_k\circ\gamma_k$ is a subsequence of $\alpha_k$.
For any $n\ge k$, we have
$\beta(n) = \alpha_n(n) = \alpha_k(\gamma_{kn}(n)) = \alpha_k(\gamma_k(n))$,
so $\beta$ is eventually a subsequence of $\alpha_k$.
Thus, $f_k\circ x\circ \beta$ is eventually a subsequence of $f_k\circ x\circ \alpha_k$,
which is convergent,
so $f_k\circ x\circ \beta$ is convergent.
