Cauchy sequence and uniform continuity

Question

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

Answer 1

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.

Answer 2

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. $\hspace{2cm}\square$