Theorem: Let $(f_n)$ be a sequence of functions on $I \subseteq \mathbb{R}$. Then $(f_n)$ pointwise convergent iff pointwise cauchy.

Here, I only prove "$\Longleftarrow$" since the converse is very straightforward.

Proof attempt: Suppose $\left(f_n \right)$ cauchy, then $\forall \epsilon >0 \: \forall x \in I \: \exists N_o \in \mathbb{N}: \forall m,n \in \mathbb{N}$ $$n,m \geq N_0 \implies \mid f_n(x)-f_m(x) \mid < \frac{\epsilon}{3} $$

Let $M$ be an upper bound of $f_{N_0}(x)$. Then $$\mid f_n(x)-f_{N_0}(x) \mid < \frac{\epsilon}{3} \implies \mid f_n(x) \mid < \frac{\epsilon}{3}+M$$ and $n$ is arbitrary, so $(f_n)$ is bounded. Now by Bolzano-Weierstrass theorem, $(f_n)$ has a convergent subsequence. Let $(f_{n_{k}})$ be a such sequence and $(f_{n_{k}}) \to f$. Then $\exists N_1 \in \mathbb{N}:$ $$n_k \geq N_1 \implies \mid f_{n_{k}}(x)-f(x) \mid<\frac{\epsilon}{3} $$ Combining the terms yields $$\mid f_{n_{k}}(x)-f(x)+f_n(x)-f_{N_0}(x) \mid \leq \mid f_{n_{k}}(x)-f(x) \mid + \mid f_n(x)-f_{N_0}(x) \mid <\frac{2\epsilon }{3}$$ Now, let $n_k \geq N_0$. Then $$\mid f_n(x)-f(x)\mid <\frac{2\epsilon }{3}+\mid f_{N_0}(x)-f_{n_k}(x)\mid $$ But since $n_k \geq N_0$, $\mid f_{N_0}(x)-f_{n_k}(x)\mid < \frac{\epsilon}{3}$ so $\mid f_n(x)-f(x)\mid <\epsilon$. And we can conclude that $(f_n) \to f$ $\square$

  • $\begingroup$ You need to take $n_k$ which is also bigger than $N_1$. Other than that it is fine. Actually this result is absolutely trivial if you know that a sequence of real numbers converges in $\mathbb{R}$ iff it is Cauchy. $\endgroup$ – Mark Dec 21 '18 at 17:04
  • $\begingroup$ @Mark I just saw your comment after I wrote my answer. Surprisingly, it's almost what you wrote verbatim. It seems like there isn't much to say after all. $\endgroup$ – BigbearZzz Dec 21 '18 at 17:11
  • $\begingroup$ Yes, that's pretty much everything what can be said about this question. $\endgroup$ – Mark Dec 21 '18 at 17:14
  • $\begingroup$ @Mark yes i understand that this proof is trivial under that assumption, however, the proof for that theorem seemed very long and complicated if not proven in a geneal metric space setting as seen in this page. proofwiki.org/wiki/… $\endgroup$ – Sei Sakata Dec 21 '18 at 17:35
  • $\begingroup$ Well, there are different proofs but note that you just proved it in your question. When you took a specific $x$ you started to work with a sequence $f_n(x)$ which is a Cauchy sequence of numbers. And you showed it has a limit. $\endgroup$ – Mark Dec 21 '18 at 18:07

It seems fine to me except that you should take $n_k \geq N_1$ in the last part of the argument instead.

The result also follows immediately from the fact that a sequence of real number $a_n$ is convergent if and only if it is Cauchy.


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