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.



Pointwise convergence imples uniform convergence in an infinite subset
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.


