Let $X$ be a compact Hausdorff space without any isolated point. Show that $X$ is uncountable.
As $X$ is compact Hausdorff, it is normal. then for any two distinct points $x$ and $y$, we have a continuous map $f$ from $X$ to $[0,1]$ such that $f(x)=0$ and $f(y)=1.$
As $X$ is compact, $f(X)$ is closed and bounded in $[0,1]$. Now I want to show that $f(X)$ contains an interval by using the fact that $X$ has no isolated point. Some one Help me.
Best Answer
One way to do this is via a version of the Baire Category Theorem, which states that every locally compact Hausdorff space is a Baire space, i.e. it is not the union of a countable set of nowhere dense subsets. Recall that compact implies locally compact, so $X$ is not the union of a countable set of nowhere dense subsets. But $$X=\bigcup_{x\in X}\{x\}$$ and each $\{x\}$ is nowhere dense since $x$ is not an isolated point, thus $X$ must be uncountable.