# Sequence of function on $\mathbb{R}$ Cauchy iff convergent

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

• 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. – Mark Dec 21 '18 at 17:04
• @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. – BigbearZzz Dec 21 '18 at 17:11
• Yes, that's pretty much everything what can be said about this question. – Mark Dec 21 '18 at 17:14
• @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/… – Sei Sakata Dec 21 '18 at 17:35
• 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. – Mark Dec 21 '18 at 18:07

## 1 Answer

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.