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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.

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    $\begingroup$ If $\mathfrak{c}$ is a real-valued measurable cardinal then this follows from Egorov's theorem... $\endgroup$ Jul 12, 2021 at 19:25
  • $\begingroup$ mathoverflow.net/questions/45784/… says this is "not hard" but doesn't give a proof. (It mainly discusses whether "infinite" can be strengthened to "uncountable".) $\endgroup$ Jul 12, 2021 at 19:27
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    $\begingroup$ Crossposted at math.stackexchange.com/questions/4196015/…: when you do this, please make sure each question contains a link to the other. $\endgroup$ Jul 12, 2021 at 19:37
  • $\begingroup$ @NateEldredge apologies. I'll remember that in the future. $\endgroup$
    – Kr Dpk
    Jul 12, 2021 at 19:39
  • $\begingroup$ @NateEldredge I saw that answer. I have commented on the question seeking an explanation, but the question is 10yrs old, so don't know how helpful that would be. $\endgroup$
    – Kr Dpk
    Jul 12, 2021 at 19:43

1 Answer 1

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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.

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  • $\begingroup$ Here $\omega$ denotes the set of natural numbers, and for every $k \in \omega$ we have $k=\{0,...,k-1\}$. Also note that the proof cannot be much simpler, since your statement holds iff for every $(f_x)_{x \in [0,1]} \subseteq \omega^\omega$ there exists $X \subseteq [0,1]$ infinite and $g \in \omega^\omega$ such that $f_x \leq g$ for every $x \in X$. $\endgroup$ Jul 12, 2021 at 23:38
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    $\begingroup$ Thanks for the answer. $\endgroup$
    – Kr Dpk
    Jul 13, 2021 at 5:05

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