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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".)

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    $\begingroup$ 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." $\endgroup$ Aug 29, 2023 at 19:55
  • $\begingroup$ 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. $\endgroup$ Aug 30, 2023 at 13:56

2 Answers 2

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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:

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    $\begingroup$ 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. $\endgroup$ Aug 31, 2023 at 0:07
  • $\begingroup$ 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 $\endgroup$
    – FShrike
    Aug 31, 2023 at 11:52
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I'll use this answer to extend things that @PatrickR didn't already demonstrate in his answer.


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:

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

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  • $\begingroup$ 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? $\endgroup$
    – PatrickR
    Aug 31, 2023 at 19:48
  • $\begingroup$ 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 $k_t$? $\endgroup$
    – PatrickR
    Aug 31, 2023 at 19:51
  • $\begingroup$ Took your suggestion to replace $0,1$ for $t,1-t$. Thinking about the first issue you pointed out. $\endgroup$ Aug 31, 2023 at 20:13
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    $\begingroup$ And now I think I've fixed that issue. $\endgroup$ Aug 31, 2023 at 20:21
  • $\begingroup$ It's turning out pretty nicely! $\endgroup$
    – PatrickR
    Aug 31, 2023 at 20:34

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