I'm self studying analysis right now, and I ran into a problem. Most of the proofs I've seen for this problem are done through delta-epsilon. I was wondering if my solution was correct or not.
Problem: Sps. $f$ maps from $(0,1)$ to the complex plane and is uniformly continuous. Prove $f$ is bounded.
What I did Suppose $f$ is unbounded. This means that there exists a sequence $x_n$ converging to $x$ in $(0,1)$ such that $f(x_n)$ diverges to either infinity or negative infinity. However, since $x_n$ is convergent, it is thus Cauchy since the reals are complete. This, as $f$ is uniformly continuous $f(x_n)$ is Cauchy as well. Since Cauchy sequences are always bounded, this means that $f(x_n)$ is bounded, and thus cannot diverge. Thus, $f$ is bounded.
The main idea is fine, although it's unclear what you mean by $x$, and it certainly does not have to be the case that $x\in (0,1)$, but the proof can still work. A slight alteration is:
If $f$ were unbounded, then there exists a sequence $x_n \in (0,1)$ such that $f(x_n) \rightarrow \infty$ or $-\infty.$ Since the sequence $x_n$ is bounded it has a convergent subsequence $x_{n_k}$, and also for this subsequence $f(x_{n_k}) \rightarrow \infty$ or $-\infty$. But since $x_{n_k}$ converges, it is Cauchy, and thus $f(x_{n_k})$ is Cauchy, since the image of a Cauchy sequence under a uniformly continuous function is Cauchy. However Cauchy sequences are bounded, which contradicts the assumption that $f$ was unbounded.