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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.

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    $\begingroup$ 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
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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 $x$.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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