# Bounded sequence of functions implies convergent subsequence

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

• 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 $x$. – angryavian Jan 6 at 22:31
• 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. – Nate Eldredge Jan 6 at 22:32
• @NateEldredge my bad. makes a lot of sense!! – Kaan Yolsever Jan 6 at 22:33
• @angryavian thanks!! – 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.
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.
• 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. – Mindlack Jan 6 at 22:34