# Any Real uniformly continuous function on bounded set is bounded

What I think is as follows: $E$ is bounded set therefore by Bolzano Weierstrass property it has convergent sub sequence which must be Cauchy. The function $f$ is uniformly continuous function, so it preserve Cauchy sequence under it.

Let $x_{n_k}$ be a Cauchy sub sequence then $f(x_{n_k})$ is also Cauchy sequence. On contrary assume that given function is unbounded Then can I write directly $f(x_{n_k})$ must not be Cauchy then. Because Unbounded sequence have no Cauchy sub sequence.

But I was not convincing with this case as there is sequence {1,1,1,2,1,3,...} which is unbounded still had Cauchy sub sequence.

Any help will be appreciated.

Thanks A lot

• I means that sequence has Cauchy sub sequence – MathLover Apr 30 '18 at 5:37

You gave an example of an unbounded sequence with a convergent, hence Cauchy, subsequence. This is irrelevant. No such sequence is in the bounded set $E$.
Furthermore, the proof simply relies on the fact that if $f$ were unbounded there must be an unbounded sequence $(f(x_{n}))$ such that any subsequence $(f(x_{n_k}))$ is also unbounded. In particular, there exists a sequence $(x_n)$ in $E$ such that $|f(x_n)| > n$ for all $n$, and for any subsequence $(x_{n_k})$ we have $|f(x_n)| > n_k \geqslant k$ for all $k$.
However, the bounded sequence $(x_n)$ has a Cauchy subsequence $(x_{n_k})$ such that $(f(x_{n_k}))$ is also Cauchy, a contradiction. Again, your counterexample does not enter the picture.