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A subset of a space is $k_1$-closed if its intersection with every compact subset is closed in the subset.

A subset of a space is $k_2$-closed if its intersection with every continuous image of a compact Hausdroff space is closed in the image.

A subset of a space is $k_3$-closed if its intersection with every compact Hausdorff subset is closed in the subset.

Say a space is $k_i$-Hausdorff provided its diagonal is $k_i$-closed in the product.

Here we distinguished the difference between $k_1$-Hausdorff and $k_2$-Hausdorff, namely $k_1H\Rightarrow KC\Rightarrow wH\Rightarrow k_2H\Rightarrow US$ with no arrows reversing.

It's immediate that $k_2H\Rightarrow k_3H$. Does this arrow reverse? Does $k_3H\Rightarrow US$? If so, does that arrow reverse?

Steven Clontz
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  • It looks like $k_2$-closed definition is off, should be, its preimage is closed under any continuous map from a compact Hausdorff space. This question went into the difference between that and what you've got here: https://math.stackexchange.com/q/4846810/1210477 – M W Jan 23 '24 at 04:25

2 Answers2

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Not every $k_3$-Hausdorff space is $T_0$.

Let $2=\{0,1\}$ have the non-$T_0$ indiscrete topology. The compact Hausdorff subspaces of $2^2$ are singletons, so all sets are $k_3$-closed.

Not every $T_0+k_3$-Hausdorff space is $T_1$.

Let $2=\{0,1\}$ have the $T_0$-not-$T_1$ Sierpinski topology. The compact Hausdorff subspaces of $2^2$ are singletons and $\{(0,1),(1,0)\}$. It follows that the diagonal's intersection with each compact Hausdorff subspace is empty or the whole subspace, so it is $k_3$-closed.

In particular, $k_3H$ certainly does not imply $US$, but perhaps $k_3H+T_1$ implies $US$?

Steven Clontz
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[fixed earlier incorrect argument]

Just to hammer the final nail in this:

Not every $T_1+k_3H$ space is $US$.

Let $X$ be an infinite set with the cofinite topology. Then $X$, and $X\times X$, are $T_1$.

Suppose $K\subseteq X\times X$ is compact Hausdorff. If $K$ has more than one point, then choose distinct $(x_1,y_1),(x_2,y_2)\in K$, and we have open rectangles $U_i\times V_i\ni(x_i,y_i)$ for $i=1,2$, so that $(U_1\times V_1)\cap (U_2\times V_2)\cap K=\emptyset$. Since each $U_i$ and $V_i$ are cofinite, $U=U_1\cap U_2$ and $V=V_1\cap V_2$ are cofinite, and we have $(U\times V)\cap K=\emptyset$. Then if $X\backslash U=\{x_1,\dots,x_m\}$ and $X\backslash V=\{y_1,\dots,y_n\}$, we have

$$K\subseteq \left(\bigcup_{i=1}^m \{x_i\}\times X\right)\cup \left(\bigcup_{j=1}^n X\times \{y_j\}\right).$$

Since each of the subspaces $\{x_i\}\times X$ and $X\times \{y_j\}$ have the cofinite topology, $K$ must intersect each of these sets at most finitely in order to be Hausdorff. Therefore $K$ is finite.

Since compact Hausdorff subsets of $X\times X$ are finite, they intersect the diagonal in finite sets, which by the $T_1$ property of $X\times X$, must be closed.

We conclude the diagonal is $k_3$-closed, hence $X$ is $k_3H$.

On the other hand, every sequence in $X$ with infinite range that doesn't frequently return to any point must converge to every point in the space, so $X$ is certainly not $US$.

Remark.

We also do not have a reverse implication, since the space $X=[0,\omega_1]\cup\{\omega_1'\}$ ( "$\omega_1+1$ with a doubled endpoint" ) is $US$, but not $k_3H$, since $K:=\{(x,x)\mid x\in [0,\omega_1]\}$ is a compact Hausdorff subset of $X\times X$ whose intersection with the diagonal (which is still $K$) is not closed.

M W
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