Let $(X,d)$ be a sequentially compact metric space and let $(Y,d')$ be a metric space. Let $f: X \to Y$ be a continuous function. Then,$f$ is uniformly continuous.

Just to be sure, by sequentially compact I mean: From any sequence in $X$, there is a convergent subsequence.

I feel this is so simple, but still I'm missing something. We let $f$ be indeed continuous, and we assume $f$ is not uniformly continous. By the proof, it just looks like we are showing "continuity implies uniform continuity".

$f$ is not uniformly continuous:

$\exists \epsilon \gt 0$ : $\forall n \gt 0, \exists x_{n}, y_{n} \in X, d(x_{n},y_{n}) \lt \frac{1}{n}, d'(f(x_{n}),f(y_{n})) \gt \epsilon$.

Then, we've built in the mean time two sequences $(x_{n})_{n}$ and $(y_{n})_{n}$ , converging both to say $l$.

Yet, $d'(f(x_{n}),f(y_{n})) \gt \epsilon, \forall n \in \mathbb N $, which contradicts the continuity of $f$.

How is it important to know $(X,d)$ is sequentially compact ? It seems to me we are not using this assumption at all. We just take the "discrete" definition of uniform continuity to get two convergent sequences leading us to the contradiction. So this would show continuity is anytime equivalent to uniform continuity (of course not).

Any help appreciated.

  • $\begingroup$ Something is wrong with your negation of uniformly continuous. $\endgroup$
    – Idonknow
    Oct 12, 2017 at 9:03
  • $\begingroup$ When you wrote $d'(f(x_n,f(y_n))<\epsilon$, my guess is that you meant to write $d'(f(x_n),f(y_n))>\epsilon$. $\endgroup$ Oct 12, 2017 at 9:04
  • $\begingroup$ Yes ! Thank you. $\endgroup$ Oct 12, 2017 at 9:06

3 Answers 3


You made an unjustified jump when you asserted that the sequences $(x_n)_{n\in\mathbb N}$ and $(y_n)_{n\in\mathbb N}$ converge. Why?

Take a subsequence of $(x_n)_{n\in\mathbb N}$ which converges to some $l$. Then the similar subsequence of $yx_n)_{n\in\mathbb N}$ will converge to $l$ too. What I mean by “similar” here is this: if your sebsequence of $(x_n)_{n\in\mathbb N}$ is $(x_{n_k})_{k\in\mathbb N}$, take the sequence $(y_{n_k})_{k\in\mathbb N}$.

The rest is more or less right. You should justify why there is a contradiction with the continuity of $f$. And when you deny that $f$ is uniformly continuous, you should have written $\geqslant\epsilon$ instead of $>\epsilon$.

  • $\begingroup$ That is, we have $(x_{n})$ and $(y_{n})$ two random sequences. Since X is sequentially compact, they both have two subsequences $(x_{ni})$ and $(y_{ni})$ converging both to say $l$. Since we know: $\forall n,\exists \delta_{n}$, st $d(x_{n},y_{n}) < \delta_{n}$, we know that $(x_{n})$ and $(y_{n})$ are Cauchy. Therefore, $(x_{n})$ and $(y_{n})$ converge both to $l$. Is this right ? $\endgroup$ Oct 12, 2017 at 11:13
  • $\begingroup$ It might be useful to add that $(\delta_{n})$ is a bounded decreasing sequence ? $\endgroup$ Oct 12, 2017 at 11:16
  • $\begingroup$ @AudreyI No, it is not right. You have no reason to suppose that $(x_n)_{n\in\mathbb N}$ and $(y_n)_{n\in\mathbb N}$ are Cauchy sequences. $\endgroup$ Oct 12, 2017 at 13:25
  • $\begingroup$ How can I show that $(x_{n})$ and $(y_{n})$ are convergent then ? All I know is that the two sequences are linked by $\delta_{n}$ and they each have a convergent subsequence. Maybe by using the continuity of $f$ ? Do I know if the limit of $f(x_{ni})$ is unique ? $\endgroup$ Oct 12, 2017 at 15:37
  • $\begingroup$ @AudreyI What makes you think that they converge? They don't have to converge. $\endgroup$ Oct 12, 2017 at 15:40

You need the sequential compactness. The fact that $d(x_n, y_n) < \frac1n$ doesn't say anything about covergence of the sequences. It just means they are “equiconvergent” – if one converges to $l$, than the other converges to $l$ as well. You have to use sequential compactness to move to a convergent sumsequences $x_{n_k}$, $y_{n_k}$.

  • $\begingroup$ So it just show that they have the same nature. How can I show then that since X is sequentially compact, these two sequences converge ? This neither tell us if these sequences are bounded or ... ? $\endgroup$ Oct 12, 2017 at 11:03
  • $\begingroup$ As I say, these sequences may not converge, but they have convergent subsequences by the sequential compactness. Sequential compactness means exactly this – every sequence have convergent subsequence – there is no boundedness assumption. $\endgroup$
    – user87690
    Oct 13, 2017 at 10:12

When you build these two sequences $x_n$ and $y_n$ you will need the compactness of $X$. Otherwise they do not have to converge. Think for example of $x_n=n$ and $y_n=n+\frac{1}{n}$


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.