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.

  • 1
    $\begingroup$ So the sequence $(f(x_n))$ admits a subsequence converging to $f(x)$. This does not show yet that $(f(x_n))$ converges to $f(x)$. One sufficient (and obviously necessary) extra condition for that is that $(f(x_n))$ be Cauchy. $\endgroup$ – Julien Apr 10 '13 at 15:58
  • $\begingroup$ @julien why does this mean $f(x_n)$ has a sequence converges to $f(x)$, I agree that it shows it has a subsequence which converges, but why to $f(x)$? $\endgroup$ – Lost1 Apr 10 '13 at 16:03
  • $\begingroup$ @Lost1 I was wondering the same thing $\endgroup$ – user66807 Apr 10 '13 at 16:04
  • $\begingroup$ For your first question, this comes from the fact, every bounded sequence has a monotone subsequence. Since $x_n$ comes from an interval, it is bounded sequence and we have a monotone subsequence $x_{n_k}$ we now wish to evaluate $f(x_{n_k})$ for this monotone sequence, this is a sequence, for example if your function is $x^2$ and your sequence is $1/n$, then $f(x_n)=1/n^2$ $\endgroup$ – Lost1 Apr 10 '13 at 16:05
  • $\begingroup$ Because by assumption, as $x_{n_k}\longrightarrow x$ in a monotone way, $f(x_{n_k}))\longrightarrow f(x)$. $\endgroup$ – Julien Apr 10 '13 at 16:05

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 $x$.

  • $\begingroup$ That's an important lemma, I'll edit to highlight it. Nicely done, +1. Maybe you could add a proof? I coudn't find one on the web and I don't want to edit your answer to that extent... $\endgroup$ – Julien Apr 10 '13 at 16:51
  • $\begingroup$ @julien i have seen the proof this lemma on this site, at least twice. I need to go out shopping, maybe i will link it when i get back. $\endgroup$ – Lost1 Apr 10 '13 at 16:53
  • $\begingroup$ @Lost1 I realized that my book gives a proof by contradiction that seems to use the same lemma: we know that $f$ continuous at $x_0$ and $x_n$ monotone and converges to $x_0$, we have that $\lim f(x_n)=x_0$ by def. Assume that if $x_n$ monotonic and converges to $x_0$ then $\lim f(x_n)=x_0$ but $f$ discontinuous at $x_0$. Then $\exists x_n: x_n \rightarrow x_0$ but $f(x_n)$ doesn't converge to $x_0$. If so then $\exists \epsilon>0: |f(x_n)-f(x_0)| \ge \epsilon$. With subseq $x_{n_k}$ we have $|f(x_{n_k})-f(x_0)| \ge \epsilon$ for all k. $\endgroup$ – user66807 Apr 10 '13 at 17:33
  • $\begingroup$ $x_{n_{k_j}}$ is a monotone subseq, and by assumption $f(x_{n_{k_j}})$ converges to $f(x_0)$ but we also have $|f(x_{n_{k_j}})-f(x_0)| \geq \epsilon$ for all j, which is a contradiction and our assumption of f discontinuous is false. $\endgroup$ – user66807 Apr 10 '13 at 17:34
  • $\begingroup$ @user66807 yes, it looks like they work on the same line. $\endgroup$ – Lost1 Apr 10 '13 at 17:40

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.

  • $\begingroup$ I think he might be wanting a direct proof though, i wonder whether there is a direct proof. If there is, I would quite like to see it. $\endgroup$ – Lost1 Apr 10 '13 at 16:46
  • $\begingroup$ So using what @Lost1 mentioned proves that the sequence $f(x_n)$ is Cauchy, because every sequence that converges is Cauchy. Is there any way to prove it using this way: $|f(x_m)-f(x_n)|<\epsilon \quad for \quad m,n>N_{\epsilon}$ though? $\endgroup$ – user66807 Apr 10 '13 at 16:48

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.