-1
$\begingroup$

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 $(f_n) \to f$ $\square$

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.