3
$\begingroup$

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 $X$.

$\endgroup$
2
$\begingroup$

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 $(0,1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.