we all know that if $X$ is a compact topological space then $F(X)$ is compact for all $F\colon X\to \mathbb{R}$ continuous. I was wondering whether the converse is true? For metric spaces I have found a proof, one can argue by contradiction and use the metric to construct an unbounded continuous function. But the the topological case, I have no clue? Does it still hold? Do we need some strong stuff like Tychonoff?
Thanks a lot
The first uncountable ordinal $\omega_1$, with the order topology is not compact. Actually, an ordinal $\alpha$ is compact in the order topology iff it is of the form $\beta+1$, or equivalently, it has a maximum.
Now, $\omega_1$ has the following property: Every continuous function $f:\omega_1\to\mathbb{R}$ is eventually constant. Hence there exists some $\beta\in\omega_1$ such that for all $\xi\geq\beta$, $f(\xi)=f(\beta)$; in particular, $f[\omega_1]=f[\{0,\dots,\beta\}]=f[\beta+1]$. Since the latter subspace is compact, the range of $f$ is compact in $\mathbb{R}$.
The Wikipedia article about pseudocompactness has some info concerning additional hypothesis.
