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I read somewhere that because uniform continuous function maps Cauchy sequence to Cauchy sequence and Cauchy sequence is bounded, so the function must be bounded. I am not sure if it is correct. My concern is that what happens if the value that I am plugging into the function is not part of the Cauchy sequence, how can I be sure that it is bounded. Can someone verify whether this idea is correct? That uniform continuity imply boundedness using the fact that it maps Cauchy sequence to Cauchy sequence. Thanks

Martin Sleziak
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user10024395
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  • The sequence is bounded, not the function. – pmal Apr 24 '14 at 08:50
  • I've noticed that you have already asked more than 30 questions in April. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, StackExchange software will not allow you to do so.) For more details see meta. – Martin Sleziak Apr 25 '14 at 10:53
  • ok thanks my exam is coming next week, after that i will probably ask less questions. – user10024395 Apr 25 '14 at 10:55

2 Answers2

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Consider $f:\mathbb{R} \rightarrow \mathbb{R}$ given as $f(x)=x$ on $\mathbb{R}$.
It is clear that $f$ is uniform continuous but not bounded.

user134927
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  • then is this wrong? http://math.stackexchange.com/questions/88257/uniform-continuity-and-boundedness – user10024395 Apr 24 '14 at 09:43
  • Your statement "uniform continuous function is bounded" is wrong and $f(x)=x$ is a counter example – user134927 Apr 24 '14 at 14:17
  • have u read the link? what is it proving then? – user10024395 Apr 24 '14 at 14:26
  • @user136266 In the link you provided is the result saying: Every uniformly continuous function $f\colon I\to\mathbb R$ is bounded. Here $I$ denotes a bounded interval. In this case the above function is not a counterexample. (And, as far as I know, this result is correct.) – Martin Sleziak Apr 25 '14 at 08:08
  • @Martin Sleziak what happens if i plug in a value that is not part of the cauchy sequence into the function, how can i be sure that the output from the function is within the bound of the cauchy sequence, the one that the function maps to? – user10024395 Apr 25 '14 at 09:24
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As you have already seen in another answer, a uniformly continuous function $f\colon\mathbb R\to\mathbb R$ is not necessarily bounded.

But this is true:

Claim. If $I\subseteq\mathbb R$ is a bounded interval and $f\colon I\to\mathbb R$ is a uniformly continuous, then $f$ is bounded.

Bounded intervals have several possible forms: $I=(a,b)$, $I=(a,b]$, $I=[a,b)$, $I=[a,b]$. But in any of these cases the closure $\overline I=[a,b]$ is a compact interval.

Let me try to give a proof of the above claim using the fact that $\overline I$ is a compact subset of $\mathbb R$ and your idea to use the fact that $f$ preserves Cauchy sequences.

Proof. Suppose that $f$ is unbounded. This means that there exists a sequence $(x_n)$ such that $x_n\in I$ and $|f(x_n)|\to\infty$.

Since $\overline I$ is compact, the sequence $(x_n)$ has a convergent subsequence $(x_{n_k})$. However, we only know that the limit of this subsequence belongs to $\overline I$. We do not know whether the limit belongs to $I$.

But even the fact that the subsequence $x_{n_k}$ is convergent in some larger space is sufficient to imply that this subsequence is Cauchy.

So we have a Cauchy sequence $(x_{n_k})$ such that the sequence $(f(x_{n_k}))$ is unbounded. Therefore $(f(x_{n_k}))$ is not Cauchy, a contradiction. $\hspace{2cm}\square$

Martin Sleziak
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