Let $\Bbb N$ be equipped with the cofinite topology: call this space $X$.
Definition: a topological space $Y$ is weak-Hausdorff (WH) iff. for every compact Hausdorff space $K$ and continuous $\nu:K\to Y$, $\nu(K)$ is closed in $Y$.
This is quite a mysterious property (in my opinion) which seems hard to verify in general. But we are lucky here: unless I’m crazy, $X$ is in fact a compact space and as such it is compactly generated (CG):
Definition: a space $Y$ is compactly generated iff. a $U\subseteq Y$ is closed iff. for all compact Hausdorff spaces $K$ and continuous $\nu:K\to Y$, $\nu^{-1}(U)$ is closed in $K$ (this property of such $U$ is called “$k$-closed”.)
And it turns out that, given a CG space $Y$, we can canonically produce a (CG)WH space from $Y$ using the weak-Hausdorffification functor:
Let $q:\mathsf{CG}\to\mathsf{CGWH}$ be the weak-Hausdorffification. That is, for a space $Y$, let $R\subseteq Y\times Y$ be the minimal closed equivalence relation. Note: $Y\times Y$ is here given the product topology of the category $\mathsf{CG}$, i.e. we have “$k$-ified” the usual product by adjoining all complements of $k$-closed sets to the topology. $R$ induces a relation $\sim$ on $Y$: define $q(Y)$ to be the quotient space $Y/_{\sim}$. It can be shown that $q(Y)$ is indeed always a WH space on the proviso $Y$ is CG (and it can be shown that $q$ is functorial).
We know $X\times X$ is compact, hence CG, and by the tube lemmas we know both projections $\pi_{1,2}:X\times X\to X$ are closed maps. To compute $q(X)$, I note that $X\times X$ is its own $k$-ification and accordingly we only seek closed equivalences $E$ to be closed in the usual product topology. $\pi_{1,2}(E)$ must also be closed in $X$, which is true iff. $E$’s projections are finite sets, or the whole space. But $E$ must contain $(x,x)$ for $x\in X$ and its projections cannot be finite. Therefore, the only such $E$ is $X\times X$. We conclude that $q(X)$ is the one-point space: in particular, $q(X)\neq X$. Because it can be shown that $q(Y)=Y$ for any CGWH space $Y$, this is proof that $X$ is not a weak-Hausdorff space.
The question. The above proof is quite abstract. Is it possible to exhibit a counterexample, i.e. to find a compact Hausdorff $K$ and a continuous $\nu:K\to X$ with $\nu(K)$ not closed? Such a $K$ would have to be infinite. The only nice infinite compact spaces that come to mind are the friendly and familiar compact Euclidean subspaces, but as the only continuous functions $\nu:[0,1]\to X$ are constant these examples don’t do the job. Indeed, no continuum $K$ has any non-constant map to $X$: any counterexample is necessarily disconnected. In fact, any counterexample must be totally disconnected, I’m fairly certain (I use the Sierpinski continuum theorem). We also can’t fiddle with the Stone-Čech compactification $\beta\Bbb N$ for examples since $X$ is certainly not Hausdorff, so the universal property can’t be exploited.
Moreover, I actually started writing this question under the belief $X$ is WH. I thought I’d found a good example of a not-Hausdorff-yet-WH space; sadly not. Could someone point me to nice examples of the above? Thanks.
