I would like to show that if $f : (a, b) \rightarrow \mathbb{R}$ is uniformly continuous, then $f$ is bounded. I have the following proof, which is probably relevant:
But, I cannot invoke the sequential compactness theorem, as that theorem states for two numbers $a$ and $b$, if $a < b$, then the interval $[a, b]$ is sequentially compact. I have been struggling with this problem for a while, and I would really appreciate any help.
There is a better way than using that theorem. Suppose $f(x_n) \to \infty$ for some sequence $\{x_n\}$. Since $\{x_n\}$ is a bounded sequence of real numbers it has a convergent subsequence, say $\{x_{n_k}\}$. Then $\{x_{n_k}\}$ is Cauchy sequence in $(a,b)$ , so $|f(x_{n_k})-f(x_{n_j})|\to 0$ as $j,k \to \infty$ (because of uniform continuity). This is a contradiction because any Cauchy sequence of real numbers is (convergent) hence bounded.