Here you can see my attempt at the proof. I am sure I did something wrong because my prof asked me to show it for rationals and I "somehow" showed it for all reals. I would appreciate it if someone can spot my mistake! Thanks

The reference to the theorem 3.6 is for Rudin "Principles of Mathematical Analysis":

Theorem 3.6: If $\{p_n\}$ is a sequence in a compact metric space X, then some subsequence of $\{p_n\}$ converges to a point of X

my attempt at the proof

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    $\begingroup$ 1. Theorem 3.6 guarantees the existence of "some" subsequence, but when you apply it in your solution you say "each" subsequence. 2. More importantly, you need to pick a subsequence $n_k$ such that the claim holds simultaneously for any rational $x$. if you [correctly] apply Theorem 3.6, the subsequence you get for one $x$ might be different from the subsequence you get for a different $x$. $\endgroup$ – angryavian Jan 6 at 22:31
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    $\begingroup$ Consider carefully the difference between the statements "for every $x$ there exists a subsequence..." and "there exists a subsequence such that for every $x$..." You have proved the first one, but you were supposed to prove the second, which is harder. $\endgroup$ – Nate Eldredge Jan 6 at 22:32
  • $\begingroup$ @NateEldredge my bad. makes a lot of sense!! $\endgroup$ – Kaan Yolsever Jan 6 at 22:33
  • $\begingroup$ @angryavian thanks!! $\endgroup$ – Kaan Yolsever Jan 6 at 22:35

You chose a subsequence working for only one $x$. There is zero guarantee that this same subsequence will work for another.

More precisely, your compactness argument is as follows:

Let $x$ be a real number. The sequence $(f_n(x))_n$ is bounded, thus there exists an increasing $\varphi_x : \mathbb{N} \rightarrow \mathbb{N}$ such that $(f_{\varphi_x(n)}(x))_n$ is convergent.

But if $y$ is another real number, there is no reason why $(f_{\varphi_x(n)}(y))_n$ is convergent.

  • $\begingroup$ isn't the argument valid for any x in the domain [a,b]? $\endgroup$ – Kaan Yolsever Jan 6 at 22:29
  • $\begingroup$ Yes, but: « for every $x$ there exists a subsequence that works » does not imply « there exists a subsequence that works for every $x$ ». It is the same reason why, say: « for each human being, there exists a scalar that is their age » holds, but « there exists a scalar such that, for each human being, said scalar is their age » does not. $\endgroup$ – Mindlack Jan 6 at 22:34
  • $\begingroup$ thanks. understood! i'll accept your answer when i can $\endgroup$ – Kaan Yolsever Jan 6 at 22:35

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