Lower Semicontinuous functions and Countable Compactness

Question

Is it true that, if for any lower semicontinuous function $f : X \rightarrow (0,\infty)$, there exist $\varepsilon > 0$ such that $\varepsilon \leq f(x)$ for all $x \in X$, then $X$ is countably compact?

Answer 1

Yes. If $X$ is not countably compact, let $D=\{x_n:n\in\Bbb N\}$ be a countably infinite closed, discrete subset of $X$. Define $f:X\to(0,\to)$ by $f(x)=1$ if $x\in X\setminus D$, and $f(x)=2^{-n}$ if $x=x_n$; then $f$ is lower semicontinuous, and $\inf f[X]=0$.