I know that any uniformly continuous is fine with it with the particular proof of this theorem I have found.
Suppose that f is uniformly continuous on a bounded set S. Prove that f is bounded on S.
The proof goes like this. (I can’t format it properly, so I decided to use the picture instead.) It was all about Cauchy’s Sequence being bounded and contradiction.
My question is :
Why can’t I just replace the word “Cauchy’s sequence” with “Convergent sequence” so that I don’t need the function to be uniformly continuous but only regular continuous?
The problem is the sequence may not converge to a point of $S$. If $(x_n) \subset S$ and $x _n \to x$ then we can say $x_n-x_m \to 0$ and use uniform continuity to conclude that $f(x_n)-f(x_m) \to 0 $ but continuity can be used only if the limit $x$ is in side $S$.
For example consider the function $\frac 1 x$ on $(0,1)$. The sequence you get from Bolzano-Weistrass Theorem may be $(\frac 1 n)$. Since $f$ is not even defined at $0$ (which is the limit of $\frac 1 n$) we cannot use continuity of $f$.