Proposition:
Let $(X,d)$ be compact metric space, and $Y$ be Borel subset of $X$.
Suppose $A$ is homeomorphic to $Y$. Then, uniformly continuous function $f:A \to \mathbb{R}$ is bounded function.
I cannot prove this proposition.
I know $X$ is totally bounded. So, if domain of $f$ is $X$, I can prove it is bounded function. But domain of $f$ is only homeomorphic to Borel subset of $X$.
Best Answer
You cannot prove the proposition because it's false.
Take $X=[0,1]$ with the usual metric. Now let $Y=(0,1)$ this is a Borel set. It is well known that $A=\mathbb{R}$ with the usual metric is homeomorphic to $(0,1)$ 1.
But clearly there exists uniformly continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that are not bounded. For instance $f(x)=x$.
1. $\tan(x)$ is an homeomorphism from $(-\frac{\pi}{2},\frac{\pi}{2})$ to $\mathbb{R}$, you can modify this to work with $(0,1)$.