Cauchy sequence and uniform continuity 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
 A: 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.
A: 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. $\hspace{2cm}\square$ 
