I am trying to prove this statement. I am new to topology and I think the question has something to do with Baire's category theorem, but I can't figure it out.
I tried to come up with a contradiction. I started with the assumption that (X,d) is complete and there exists an open and continuous function. I was planning to prove that X can be written as a union of countable nowhere dense subsets but couldn't succeed. Would appreciate any help.
I am new to topology and the only thing I know about the Baire Category theorem is that every complete metric space is second countable. An alternative proof is provided here but I couldn't grasp it.
Thanks…
Best Answer
$X=\bigcup_{q \in \mathbb Q} f^{-1}(\{q\})$ and $f^{-1}(\{q\})$ is closed for each $q$. By Baire Category Theorem there is a non-empty open set $U$ contained in $f^{-1}(\{q\})$ for some $q$. But then $f(U)$ is a non-empty open set in $\mathbb Q$ which is contained in $\{q\}$. There is no such open set, so $f$ cannot exist as stated.