# Uniformly continuous imply bounded

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 $$X$$.

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 $$(0,1)$$.