# Weak Hausdorff space not KC

I am stuck with a problem in general topology. First of all, recall that a space $X$ is KC if every compact subset of $X$ is closed, and is weak Hausdorff if for all $u:K\rightarrow X$ continuous (where $K$ is compact Hausdorff) $u(K)$ is closed in $X$.

It's clear that KC implies weak Hausdorff. I tried to find a space that is weak Hausdorff and not KC but it looks tough (I don't even know if that's possible), does anyone have an idea ?

• There was a "few" user accounts that asked a lot of questions about KC spaces like this one some months ago. – Asaf Karagila Jan 9 '14 at 19:07
• I didn't know about it ; actually I am trying to find a counterexample for a semester project. – Amathstudent Jan 10 '14 at 0:01
• @StefanHamcke this could be helpful – Norbert Jan 26 '14 at 23:06
• @Norbert: You could take the idea from that post and turn it into an answer. – Stefan Hamcke Jan 26 '14 at 23:18
• Ok, I'll do it tomorrow. Its too late now. – Norbert Jan 26 '14 at 23:33

I need a couple of preliminary results. First note that every weak Hausdorff space $X$ is $T_1$: the map from the compact singleton space to any point of $X$ has closed range. (Thanks to a comment by johndoe for this observation.)

Lemma $\mathbf 1$: Suppose that $X$ is weak Hausdorff. Let $K$ be a compact Hausdorff spaces and $f:K\to X$ continuous; then $f[K]$ is not just compact, but also Hausdorff.

Proof: Let $x$ and $y$ be distinct points of $f[K]$, and let $H_x=f^{-1}[\{x\}]$ and $H_y=f^{-1}[\{y\}]$; $H_x$ and $H_y$ are disjoint closed sets in the normal space $K$, so there are disjoint open sets $U_x$ and $U_y$ in $K$ such that $H_x\subseteq U_x$ and $H_y\subseteq U_y$. Now let $V_x=f[K]\setminus f[K\setminus U_x]$ and $V_y=f[K]\setminus f[K\setminus U_y]$. Each of the sets $K\setminus U_x$ and $K\setminus U_y$ is compact and Hausdorff, so their images under $f$ are closed, and $V_x$ and $V_y$ are open in $f[K]$. Clearly $x\in V_x$ and $y\in V_y$. Finally, if $z\in V_x\cap V_y$, fix $p\in K$ such that $f(p)=z$; then $p\in U_x\cap U_y$, which is impossible, so $V_x$ and $V_y$ are disjoint open nbhds of $x$ and $y$, respectively, in $f[K]$. $\dashv$

Using this it’s not hard to show that weak Hausdorffness is productive.

Lemma $\mathbf 2$: Let $\{X_i:i\in I\}$ be a family of weak Hausdorff $T_1$ spaces, and let $X=\prod_{i\in I}X_i$; then $X$ is weak Hausdorff.

Proof: Suppose that $K$ is compact Hausdorff, and $f:K\to X$ is continuous. For $i\in I$ let $\pi_i:X\to X_i$ be the canonical projection map, and let $H_i=(\pi_i\circ f)[K]$; each $H_i$ is a closed compact Hausdorff subspace of $X_i$, so $\prod_{i\in I}H_i$ is a closed compact Hausdorff subset of $X$. Finally, $f[K]$ is a compact subset of $\prod_{i\in I}H_i$, so $f[K]$ is closed in $\prod_{i\in I}H_i$ and hence in $X$. $\dashv$

Now let $\Bbb Q^*$ be the Alexandroff extension (i.e., the non-Hausdorff one-point compactification) of the rationals; it’s well-known that $\Bbb Q^*$ is $KC$ and hence weak Hausdorff. Let $X=\Bbb Q^*\times\Bbb Q^*$; in this answer I showed that $X$ is not $KC$: its diagonal is compact, being homeomorphic to $\Bbb Q^*$, but not closed in $X$. By Lemma $2$, however, $X$ is weak Hausdorff.

• Weakly Hausdorff spaces are always $T_1$ (any point is closed being the image of some constant function $u\colon K\to X$). – johndoe Dec 17 '14 at 17:15
• @johndoe: Thanks; I don’t know how I missed that. I’ve incorporated it into the answer. – Brian M. Scott Dec 17 '14 at 20:18