# How are k-Hausdorff and weakly Hausdorff distinct?

In this pull request to the pi-Base database, we encountered this situation.

A space $$X$$ is said to be weakly Hausdorff provided for every compact Hausdorff space $$K$$ and every continuous $$f:K\to X$$, $$f[K]$$ is closed in $$X$$.

A space is $$X$$ said to be k-Hausdorff provided its diagonal $$\Delta=\{(x,x):x\in x\}\subseteq X\times X$$ is "k-closed".

In Quotients of k-semigroups a set is said to be k-closed if its complement is k-open, that is, its intersection with every compact set is open in the subspace. Call this $$k_1$$-closed, and its corresponding version of k-Hausdorff $$k_1H$$. There it is proven that every $$k_1H$$ space has the property that compact sets are closed, which in turn implies the space is weakly Hausdorff.

On the other hand, in Compactly Generated Spaces, a set is said to be k-closed if given any continuous map of a compact Hausdorff $$K$$ into the space, the set's inverse image is closed in $$K$$. Call this $$k_2$$-closed, and its corresponding version of k-Hausdorff $$k_2H$$. There it's shown that every weakly Hausdorff space is $$k_2H$$.

However, it seems weakly Hausdorff and k-Hausdorff aren't actually equivalent. So what gives?

(Hints: of course we have two different definitions of k-closed here. Additionally, some care should be given to checking when "compact" means "compact and Hausdorff".)

• I don't know exactly what gives but the compactly generated condition is typically assumed in the proof of the equivalence of weakly Hausdorff and kH2 - I like Strickland's notes. For instance, he uses the CG condition on $X$ to prove that "$X$ is kH2 implies $X$ is weakly Hausdorff." Aug 29, 2023 at 19:55
• My hunch would be that the two kHs are equivalent under some form of "compactly generated", which would roll in KC and wH as well. Aug 30, 2023 at 13:56

The bottom line is that the two notions of k-Hausdorff are not equivalent, and as usual with k-spaces/compactly generated spaces, different authors use the same terminology to mean different things, which can be confusing.

In general, the notions of k-closed set, k-open set, k-ification of a topological space, k-space (or compactly generated space, depending on the author), k-Hausdorff property, etc all depend on which family of "test" maps is used to define the corresponding final topology. The wikipedia article on compactly generated space has a good explanation of the general setup. See also Unraveling the various definitions of $k$-space or compactly generated space.

So given a topological space $$(X,\tau)$$, common choices are:

• (Definition 1 = CG1 in Wikipedia, this in pi-base]) Take $$\mathcal F_1$$ to be the family of all inclusions from compact subspaces of $$X$$, or take $$\mathcal F'_1$$ to be the family of all continuous maps from arbitrary compact spaces to $$X$$. The final topology $$\tau_1$$ on $$X$$ with respect to $$\mathcal F_1$$ or with respect to $$\mathcal F'_1$$ is the same. Let's write $$k_1X$$ for that k-ification of $$X$$. This is the setup in [LB]. A k-closed-1 set is then a subset of $$X$$ that is closed in $$k_1X$$, or a $$k_1$$-closed set for short

• (Definition 2 = CG2 in Wikipedia, this in pi-base]) Take $$\mathcal F_2$$ to be the family of all continuous maps from arbitrary compact Hausdorff spaces to $$X$$. Let's write $$k_2X$$ for the corresponding k-ification of $$X$$, namely $$(X,\tau_2)$$ with $$\tau_2$$ the final topology on $$X$$ with respect to $$\mathcal F_2$$. This is the setup in [R] and [S], more commonly used in algebraic topology. A k-closed-2 set is then a subset of $$X$$ that is closed in $$k_2X$$, or a $$k_2$$-closed set for short.

• (Definition 3 = CG3 in Wikipedia, this in pi-base]) Take $$\mathcal F_3$$ to be the family of all inclusions from compact Hausdorff subspaces of $$X$$. Let's write $$k_3X$$ for the corresponding k-ification of $$X$$, namely $$(X,\tau_3)$$ with $$\tau_3$$ the final topology on $$X$$ with respect to $$\mathcal F_3$$.

Because $$\mathcal F_3\subseteq\mathcal F_2\subseteq\mathcal F'_1$$, we have inclusions of topologies $$\tau\subseteq\tau_1\subseteq\tau_2\subseteq\tau_3$$ and continuous identity mappings between the various k-ifications $$k_3X\to k_2X\to k_1X\to X.$$

So closed subsets of $$X$$ are $$k_1$$-closed, $$k_1$$-closed sets are $$k_2$$-closed, etc. And if a space $$X$$ is CG3 (i.e., $$k_3X=X$$), it is CG2 (i.e., $$k_2X=X$$); and so on.

Now for the definitions of k-Hausdorff. A space $$X$$ is k-Hausdorff is the diagonal $$\Delta$$ is k-closed in $$X\times X$$ (usual product space), i.e., if $$\Delta$$ is closed in the k-ification $$k(X\times X)$$. Of course that depends on the specific k-ification we are talking about. So there is the notions of $$k_1$$-Hausdorff (used in [LB]) and $$k_2$$-Hausdorff (used in [R]). We could even have $$k_3$$-Hausdorff. And because of the relationship between closed, $$k_1$$-closed, $$k_2$$-closed sets above, we have these implications between the properties for the space $$X$$:

$$\text{Hausdorff}\implies k_1\text{-Hausdorff}\implies k_2\text{-Hausdorff}.$$

More can be said. [LB] proves (Theorem 2.1) that $$X$$ being $$k_1$$-closed is equivalent to all compact subsets of $$X$$ being Hausdorff, and that implies that $$X$$ is KC (= all compact subsets are closed). On the other hand, [R] proves (Prop. 11.2) that weak Hausdorff implies $$k_2$$-Hausdorff. So we can insert those in the chain of implications already recorded in pi-base (namely $$T_2$$ implies KC implies weak Hausdorff implies US implies $$T_1$$, with US = "unique sequential limits"), and we get:

$$T_2\implies k_1\text{-Haus.}\implies KC\implies\text{weak Haus.}\implies k_2\text{-Haus.}.$$

I haven't given a thought about whether $$k_2$$-Hausdorff implies US, and the position of $$k_3$$-Hausdorff in this.

For an example of a space that is $$k_2$$-Haudorff and not $$k_1$$-Hausdorff, one can choose a space that is weak Hausdorff and not KC. A search in pi-base gives the square of the one-point compactification of the rationals: $$X=\mathbb Q^*\times\mathbb Q^*$$ with this reference.

Another example is the one-point compactification of the rationals itself: $$X=\mathbb Q^*$$. It is KC, hence weak Hausdorff and $$k_2$$-Hausdorff. But it is not $$k_1$$-Hausdorff, because it is compact but not Hausdorff (see [LB] Theorem 2.1). On the other hand, it is Frechet-Urysohn, hence sequential, hence CG2. So it is an example of CGWH space that is not $$k_1$$-Hausdorff.

One can also ask if any of the implication above become equivalences if we make an additional k-space assumption. An important case is the combination CGWH (= CG2 + weak Hausdorff). This is equivalent to CG3 + weak Hausdorff (Lemma 1.4(c) in [S]). Also, CG2 + $$k_2$$-Hausdorff imply weak Hausdorff ([R] Proposition 11.4 or [S] Proposition 2.14) and CG2 + weak Hausdorff imply KC (see here). So under the assumption of CG2, the three properties KC, weak Hausdorff, and $$k_2$$-Hausdorff are equivalent, and they are also equivalent to the combination of CG3 with any of them.

Does CGWH imply $$k_1$$-Hausdorff? No. See the example above of the one-point compactification of the rationals.

References:

• So if $k_2T_2$ is added to pi-Base, T194 can be improved to $CG_2$+$k_2T_2$ implies $KC$. And we'd want an example of $k_2T_2$ not weakly Hausdorff (which cannot be $CG_3$) - I think S165 is the only current candidate. Aug 31, 2023 at 0:07
• hmm, I've only ever seen CG2 as the "correct" notion in use. We get Cartesian closure with this one, not for CG1 and maybe not for CG3, not sure Aug 31, 2023 at 11:52

Theorem: $$k_2H$$ implies $$US$$.

Proof: We use the characterization from Rezk 4.2.4 that a $$k_2H$$ space $$X$$ has the property that for every continuous function $$f:K\to X$$ from a compact Hausdorff $$K$$ and $$k_0,k_1\in K$$ with $$f(k_0)\not= f(k_1)$$, there exist open neighborhoods $$U_0,U_1$$ of $$k_0,k_1$$ (resp.) with $$f[U_0]\cap f[U_1]=\emptyset$$.

Let $$l_0,l_1$$ be limits of $$x_n\in X$$. Let $$K=(\omega+1)\times 2$$, and let $$f:K\to X$$ be defined by $$f(n,i)=x_n$$, $$f(\omega,0)=l_0$$, and $$f(\omega,1)=l_1$$. To show this is continuous, we need only observe that inverse images of open subsets of $$X$$ that contain $$(\omega,i)$$ contain a cofinite subset of $$\omega\times\{i\}$$. This follows as for an inverse open image to contain $$(\omega,i)$$, the open set must contain the limit $$l_i$$ of $$x_n$$, and thus contain a final sequence of $$x_n$$, and thus the inverse open image contains a cofinite subset of $$\omega\times 2$$.

Finally, if $$l_0=f(\omega,0)\not=f(\omega,1)=l_1$$, then there would exist open neighborhoods $$U_0,U_1$$ of $$(\omega,0),(\omega,1)$$ with $$f[U_0]\cap f[U_1]=\emptyset$$. But this is impossible as there exists $$n<\omega$$ with $$(n,0)\in U_0$$ and $$(n,1)\in U_1$$, and $$f(n,0)=x_n=f(n,1)$$. Thus $$l_0=l_1$$, showing limits are unique.

Example: S37 of pi-Base, $$\omega_1+1$$ with a duplicate $$\omega_1'$$, is $$US$$ but not $$k_2H$$. PatrickR pointed out that this space was $$US$$ but not weakly Hausdorff in this answer. To see that it fails $$k_2H$$, consider the set $$(\omega_1+1)\times 2$$ and the map $$(\alpha,i)\mapsto \alpha$$ for $$\alpha<\omega_1$$, $$(\omega_1,0)\mapsto\omega_1$$, and $$(\omega_1,1)\mapsto\omega_1'$$. This map is continous, $$f(\omega_1,0)\not=f(\omega_1,1)$$, but there are no open neighborhoods $$U_0,U_1$$ of $$(\omega_1,0),(\omega_1,1)$$ with $$f[U_0]\cap f[U_1]=\emptyset$$.

Theorem: The one-point compactification of a $$KC$$ space is $$k_2H$$.

Proof: Let $$X^+=X\cup\{\infty\}$$ be the one-point compactification of $$X$$. Consider a compact Hausdorff $$K$$ with $$f:K\to X^+$$, and $$k_0,k_1\in K$$ with $$f(k_0)\not= f(k_1)$$. In the case that $$f(k_t)\not=\infty$$ for $$t\in\{0,1\}$$, consider the closed subset $$f^\leftarrow[\{\infty\}]$$ of $$K$$. Since $$K$$ is regular, we may choose open $$U$$ with $$\{k_0,k_1\}\subseteq U\subseteq cl(U)\subseteq K\setminus f^\leftarrow[\{\infty\}]$$. Then $$f\upharpoonright cl(U):cl(U)\to X$$ is a continuous map from compact Hausdroff $$cl(U)$$ to $$KC$$ and therefore $$k_2H$$ $$X$$, so there exist open (in $$cl(U)$$) neighborhoods $$V_0,V_1$$ of $$k_0,k_1$$ with $$f[V_0]\cap f[V_1]=\emptyset$$. It follows that $$U_0=V_0\cap U,U_1=V_1\cap U$$ are open neighborhoods of $$k_0,k_1$$ in $$K$$, and $$f[U_0]\cap f[U_1]\subseteq f[V_0]\cap f[V_1]=\emptyset$$.

Otherwise we have, say, $$k_0$$ with $$f(k_0)=\infty$$. Since $$X^+$$ is $$T_1$$, $$\{\infty\}$$ is closed and $$f^\leftarrow[\{\infty\}]$$ is closed. Thus $$V_{1}=K\setminus f^\leftarrow[\{\infty\}]$$ is an open neighborhood of $$k_{1}$$. Since $$K$$ is regular, choose $$U_{1}$$ open with $$k_{1}\in U_{1}\subseteq cl(U_{1})\subseteq V_{1}= K\setminus f^\leftarrow[\{\infty\}]$$. Then $$cl(U_{1})$$ is compact and thus $$f[cl(U_{1})]$$ is compact and misses $$\infty$$, and thus is a compact and closed subset of $$X$$. It follows that $$W_0=X^+\setminus f[cl(U_{1})]$$ is an open neighborhood of $$\infty=f(k_0)$$, so $$U_0=f^\leftarrow[W_0]$$ is an open neighborhood of $$k_0$$.

So we have open neighborhoods $$U_0,U_{1}$$ for $$k_0,k_{1}$$, and $$f[U_0]\cap f[U_{1}]\subseteq f[cl(U_0)]\cap(X^+\setminus f[cl(U_0)])=\emptyset$$.

Example: S165 of pi-Base, the one-point compactification of the Arens-Fort space, is $$k_2H$$ but not $$wH$$. Denote the Arens-Fort space by $$X=\omega\cup\{\infty'\}$$ and its one-point compactification by $$X^+=\omega\cup\{\infty',\infty\}$$. $$wH$$ fails as $$X^+\setminus\{\infty'\}=\omega\cup\{\infty\}$$ is a copy of $$\omega+1$$ (compact subsets of $$X$$ are finite and closed so neighborhoods of $$\infty$$ are cofinite) which is not closed in the space. But S165 is $$k_2 H$$ by the above theorem: $$X$$ is Hausdorff.

In summary, we have the following implications, none of which reverse:

$$T_2\Rightarrow k_1H\Rightarrow KC\Rightarrow wH\Rightarrow k_2H\Rightarrow US\Rightarrow T_1$$

• In the first paragraph of the second theorem: it may not be the case that $f$ maps $K$ into $X$, so probably some type of restriction to avoid the inverse image of $\infty$ is necessary? Aug 31, 2023 at 19:48
• and for the second paragraph, would be easier to read if just take, say $f(k_0)=\infty$ and not have to work with $t$ and $1-t$. I imagine $y_t$ is $k_t$? Aug 31, 2023 at 19:51
• Took your suggestion to replace $0,1$ for $t,1-t$. Thinking about the first issue you pointed out. Aug 31, 2023 at 20:13
• And now I think I've fixed that issue. Aug 31, 2023 at 20:21
• It's turning out pretty nicely! Aug 31, 2023 at 20:34