Need to prove $f$ continuous at $x_0$ iff for every monotonic sequence $(x_n)$ converging to $x_0$ we have $\lim f(x_n)=f(x_0)$

Question

This was a problem that the Professor went over in class, but I am having trouble understanding and finishing the proof. The full question is:

$f:I \rightarrow \mathbb R$ is continuous at $x_0 \in I$ if and only if for any monotonic sequence $x_n$, with $x_n \rightarrow x_0$ we have $f(x_n) \rightarrow f(x_0)$

This is his solution (I'll mention where I am confused):

We know that for every monotonic sequence $$(x_n) \rightarrow x_0, f(x_n) \rightarrow f(x_0)$$ Want to show that $$x_n \rightarrow x_0 \implies f(x_n) \rightarrow f(x_0) $$ for any $x_n$. With $x_n \rightarrow x_0$ we know there exists a monotonic convergent subsequence, $$x_{n_k} \rightarrow x_0 .$$ Now we want to show $\{f(x_n)\}_n$ is Cauchy.

(Where did this come from? Why does this need to be shown? From what I understand, $\{f(x_n)\}_n$ is a subsequence of the function $f$? Or would this technically be called a subfunction? I don't know if there is such a thing)

For any $\epsilon > 0, \exists N_{\epsilon}$ such that $$|f(x_m)-f(x_n)|<\epsilon \quad for \quad m,n>N_{\epsilon} $$ If not, $\exists \epsilon_0, \forall N\in \mathbb R$ such that $$|f(x_m)-f(x_n)|\geq \epsilon \quad for \quad some \quad m,n>N$$

I am supposed to finish the proof by showing that $\{f(x_n)\}_n$ is Cauchy, but I am not sure how to do this and don't know where to begin. Sorry if the question involves a lot of explaining, it isn't a homework problem to be turned in but I need to understand what is going on (if I can show it is Cauchy, however, I do get a bit of extra credit, so please don't give me the answer off the bat).

Thanks for any help! This function\continuity chapter really has me scratching my head.

Answer 1

I have a proof of the result here, not via Cauchy. It appeals to this lemma:

Lemma: $x_n$ converges to $x$, if and only if, for every subsequence $x_{n_k}$, there exists a sub-subsequence such that $x_{n_{k_l}}$ converges to $x$

Proof: see this thread, and recall that compactness is not needed, as shown by Ragib Zaman's answer.

We take any sequence $x_n$ converges to $x$, then take any subsequence of $x_n$, call it $x_{n_k}$, then $x_{n_k}$ has a monotone subsequence $x_{n_{k_l}}$, such that $f(x_{n_{k_l}})$ converges to $f(x)$

Then for consider $f(x_n)$ for any sequence $x_n$ converging to $x$, for any subsequence $f(x_{n_k})$, it has a sub-subsequence $f(x_{n_{k_l}})$ converges to $f(x)$. Now we use the lemma, we have shown $f(x_n)$ converges to $f(x)$, for any choice of sequence $x_n$ converging to $x$.

Answer 2

I'm not entirely sure if this is what you're asking, but here you go anyway. Note that $\{f(x_n)\}$ is just a sequence of real numbers. It is taking all of the $x_n$ of a given sequence and plugging them into the function $f$ to get a new sequence of a real numbers. For example, if $f(x) = x^2 + x$, and $\{x_n\}$ is the sequence defined by $x_n = 1/n$ for all $n$, then $\{f(x_n)\}$ is the sequence of real numbers with $\{f(x_n)\} = \{1/n^2 + 1/n\}$.

So keeping in mind of this fact, the reason we want to show $\{f(x_n)\}$ is Cauchy is because of the fact that a sequence of real numbers convergences if and only if it is Cauchy. So in order to show that $\{f(x_n)\}$ converges to a limit, it suffices to show it is Cauchy.

Note that showing a sequence is Cauchy does not necessarily give you what the sequence converges to! However, in this case, if you show $\{f(x_n)\}$ is Cauchy, then it must converge to $f(x_0)$ since you know that the subsequence $f(x_{n_k}) \to f(x_0)$ by assumption.

You can apply the lemma mentioned by @Lost1 to show it is in fact Cauchy.