Pointwise convergence imples uniform convergence in an infinite subset I came upon this statement in a stack answer.
Statement :
If $f_n$ is a sequence of real valued functions (not necessarily continuous or measurable) on $[0,1]$ such that $f_n$ converges point-wise to $0$, then there exists an infinite subset of $[0,1]$ where the convergence is uniform.
I couldn't prove it. I believe the claim is true because $[0,1]$ is uncountable but the set of sequences is countable only.
Any help would be appreciated in assistance to how to prove it.
 A: For every sequence $(F_n)_{n \in \omega}$ with $F_n:[0,1] \rightarrow \mathbb{R}$ converging pointwise to $0$, we can associate to every $x \in [0,1]$ an $f_x \in \mathbb{\omega}^\omega$ in the following way: Set $f_x(m):= \min\{n \in \mathbb{\omega}\,\, \colon \,\, \forall n' \geq n \,\,\,\,  \vert F_{n'}(x) \vert < \frac{1}{m}\}$.
Now the proof works in two steps.
In the first step we show that there exists $f^* \in \omega^\omega$ such that for every $k \in \omega$ the set $\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$ is uncountable.
We will construct such an $f^* \in \omega^\omega$ by induction on $k \in \omega$: Assume that $f^* \restriction k$ has already been constructed, and we have that the set $\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$ is uncountable. Since $$\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}= \bigcup_{l \in \omega}\,\, \{x \in [0,1] \,\, \colon \,\, f_x \restriction (k+1) \leq (f^* \restriction k)^\frown l\}$$ we find $l \in \omega$ such that $\{x \in [0,1] \,\, \colon \,\, f_x \restriction (k+1) \leq (f^* \restriction k)^\frown l\}$ is uncountable. Now set $f^*(k)=l$ and we see that $f^*\restriction (k+1)$ has the required properties.
In the second step we inductively construct $(x_k)_{k \in \omega} \subseteq [0,1]$ injective and $(f_k)_{k \in \omega} \subseteq \omega^\omega$ increasing such that for every $k \in \omega$ we have $f^* \leq f_k$ ,  $f_{x_k} \leq f_k$ and $f_k \restriction (k+1) = f_{k+1} \restriction (k+1)$. Once we have shown this, we can define $g \in \omega^\omega$ such that $g(k):=f_k(k)$ and see that $f_{x_k} \leq g$ for every $k \in \omega$. But this proves that $(F_n)_{n \in \omega}$ converges uniformly on $(x_k)_{k \in \omega}$.
To this end assume that $x_0,...,x_{k-1}$ and $f_0,...f_{k-1}$ with the required properties have already been constructed. Since $\{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$ is uncountable, we can find $x_k \in \{x \in [0,1] \,\, \colon \,\, f_x \restriction k \leq f^* \restriction k\}$ different from $x_0,...,x_{k-1}$. Set $f_k(m):=\max\{f_{k-1}(m), f_{x_k}(m)\}$, and note that since $f_{x_k} \restriction k \leq f^* \restriction k \leq f_{k-1} \restriction k$, we have $f_{k-1}\restriction k =f_k \restriction k$. This finishes the proof.
