Where does "A family of compact sets with the finite intersection property has nonempty intersection" lie between $T_1$ and $T_2$?

Question

This question is a follow-up of sorts to a recent question by Steven Clontz about the property that (arbitrary) intersections of compact sets are compact (abbreviated "$IKK$" in that question).

I am interested in a stronger property ("$KIP$" defined below), and where $KIP+T_1$ lies between $T_1$ and $T_2$.

I'll give definitions and some results, then conclude with my question and attempts.

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Definition.

Throughout, we say a family of sets $\mathcal F$ has the finite intersection property ($FIP$) if every finite intersection $F_1\cap\dots\cap F_n$, with each $F_k\in \mathcal F$, is nonempty.

Let $X$ be a topological space. We say $X$ has the compact intersection property (abbreviated "$KIP$") if every family of compact subsets of $X$ with the $FIP$ has nonempty intersection.

Below, we shall often abuse grammar and say things like "$X$ is $KIP$" do indicate $X$ has the compact intersection property.

It is immediate that any space in which compact subsets are closed ("$KC$") must be $KIP$. In particular, this holds for all Hausdorff spaces.

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Consequences.

Lemma 1. $KIP\implies IKK$

Proof.

Let $X$ be $KIP$, and suppose $\mathcal K$ is a family of compact sets. We wish to show $\bigcap \mathcal K$ is compact.

To this end, suppose $\mathcal C$ is a family of (relatively) closed subsets of $\bigcap \mathcal K$, having the finite intersection property. Let $\mathcal S$ be the family of all closed subsets $S\subseteq X$ such that $S\cap \bigcap\mathcal K\in \mathcal C$. Then for any finite collections $S_1,\dots,S_n\in \mathcal S$ and $K_1,\dots,K_n\in \mathcal K$, we have $$\bigcap_{i=1}^n(K_i\cap S_i)\supseteq (S_1\cap\dots\cap S_n)\cap \bigcap \mathcal K\neq \emptyset,$$ since the latter intersection is a finite intersection of members of $\mathcal C$. Therefore the family $\mathcal K':=\{K\cap S\mid K\in\mathcal K,S\in\mathcal S\}$ is a family of compact sets with the $FIP$, and thus by $KIP$ has nonempty intersection. But then $$\emptyset\neq \bigcap \mathcal K'=\left(\bigcap \mathcal K\right)\cap \left(\bigcap \mathcal S\right)=\bigcap \mathcal C.$$ Thus every family of closed subsets of $\bigcap \mathcal K$ with the $FIP$ has nonempty intersection, hence $\bigcap \mathcal K$ is compact.

Lemma 2. $KIP+T_1\implies US$ ("Convergent sequences have unique limits.")

Proof.

Suppose $X$ is $T_1$, and not $US$. Let $x_n\in X$ be a sequence with $x_n\to x$ and $x_n\to y$, $x\neq y$. Since $X$ is $T_1$, $x_n$ must contain infinitely many distinct members, and so we may suppose with no loss of generality that the sequence $x_n$ consists of distinct elements not equal to $x$ or $y$. Letting $S_N=\{x_n\mid n\geq N\}\cup \{x\}$ and $T_N=\{x_n\mid n\geq N\}\cup\{y\}$, we see that $\mathcal K:=\{S_N\mid N\in \mathbb N\}\cup\{T_N\mid N\in \mathbb N\}$ is a family of compact sets with the $FIP$, yet $\bigcap \mathcal K=\emptyset$, so $X$ is not $KIP$.

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Equivalent formulation.

A perhaps more tangible formulation of $KIP$ is the following:

Lemma 3. $X$ is $KIP$ if and only if the following two conditions are satisfied:

1. Pairwise intersections of compact subsets are compact ("$IKK_2$").
2. Every family of nonempty compact subsets of $X$, directed under reverse inclusion, has nonempty intersection ("Nonempty nested compact intersections”, or "$NKI$").

Proof.

If $X$ is $KIP$ then the first property follows from Lemma 1, and the second is immediate from the definition, since any family of nonempty subsets directed under reverse inclusion must have the $FIP$.

Conversely, if $X$ satisfies the preceding two properties, and $\mathcal K$ is a family of compact subsets of $X$ with the $FIP$, then the family $\mathcal K'$ of finite intersections of members of $\mathcal K$ is a directed system of nonempty compact sets, and so $\bigcap \mathcal K=\bigcap\mathcal K'\neq \emptyset$.

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In $ZFC$.

So far, everything we have said is valid in $ZF$. Now, even in $ZFC$, a family of nonempty sets directed under reverse inclusion need not have a cofinal chain (consider, as a counter-example, the family of cofinite subsets of any uncountable set). Nonetheless, in $ZFC$ we may replace the directed families in Lemma 3 with well-ordered chains:

Lemma 4. In $ZFC$, $X$ is $KIP$ if and only if the following conditions are satisfied:

1. Pairwise intersections of compact subsets are compact.
2. Every chain of nonempty compact subsets of $X$, well-ordered under reverse inclusion, has nonempty intersection.

Proof.

From Lemma 3, $KIP$ spaces certainly satisfy the preceding properties. Conversely, we suppose the preceding properties are satisfied, and argue by transfinite induction that for every cardinal $\kappa$, and every family $\mathcal K$ of compact subsets of $X$, having the $FIP$, and cardinality at most $\kappa$, $\bigcap \mathcal K$ is nonempty and compact.

To see this, note the claim is true for finite cardinals. If it is true for all cardinals $\kappa'<\kappa$, and $\mathcal K$ has cardinality $\kappa$, then let $\lambda\mapsto K_\lambda$ be a bijection $\kappa\to \mathcal K$, and for each $\lambda<\kappa$ define $J_\lambda:=\bigcap_{\alpha\leq \lambda}K_\alpha$. Then by inductive assumption $\{J_\lambda\}$ is a well-ordered chain of nonempty compact subsets, so by the second property $\bigcap \mathcal K=\bigcap_{\lambda<\kappa} J_\lambda\neq \emptyset$.

Moreover, if $\mathcal C$ is a family of closed subsets of $\bigcap \mathcal K$ with the $FIP$, then as in the proof of Lemma 1, we let $\mathcal S$ be the family of all closed subsets $S\subseteq X$ such that $S\cap \bigcap \mathcal K\in \mathcal C$. Then for each $\lambda<\kappa$, and any finite collection $S_1,\dots,S_n\in\mathcal S$, we have $$S_1\cap\dots\cap S_n\cap J_\lambda\supseteq S_1\cap\dots \cap S_n\cap \bigcap \mathcal K\neq \emptyset,$$ so the family $\mathcal S_\lambda:=\{S\cap J_\lambda\mid S\in \mathcal S\}$ has the finite intersection property. Since $J_\lambda$ is compact, we have $J_\lambda\cap \bigcap \mathcal S=\bigcap \mathcal S_\lambda\neq \emptyset$. Therefore $\{J_\lambda\cap \bigcap \mathcal S\}$ is a well ordered chain of nonempty compact sets under reverse inclusion, and so $$\bigcap \mathcal C=\left(\bigcap \mathcal S\right)\cap \left(\bigcap \mathcal K\right)= \left(\bigcap \mathcal S\right)\cap \left(\bigcap_{\lambda<\kappa} J_\lambda \right)\neq \emptyset.$$ Since this holds for any family $\mathcal C$ of closed subsets of $\bigcap \mathcal K$ satisfying the $FIP$, $\bigcap \mathcal K$ is compact.

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Question.

In general, $KIP$ spaces need not be $T_1$, and in fact it is easy to see that every finite topological space is $KIP$. Moreover, even in a $T_1$ space, the properties $IKK_2$ and $NKI$ from Lemma 3 are insufficient, separately, to guarantee $US$: If only $T_1$ and $IKK_2$ are assumed (or even $IKK$ itself), then the cofinite topology on an infinite set will give an example that is not $US$. If only $T_1$ and $NKI$ are assumed, $[0,\omega]$ with a doubled endpoint gives an example that fails $US$.

Therefore it seems $KIP+T_1$ spaces may be reasonable objects of study. Where does this property fall among other properties between $T_1$ and $T_2$? Particularly, those discussed in this answer?

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Attempts.

We know that $KC\implies KIP+T_1\implies US$. Moreover, the second implication is strict, as can be seen by the example of $[0,\omega_1]$ with a doubled endpoint, which is $US$, but fails $IKK_2$, hence fails $KI$.

I suspect the first implication is strict as well (it in fact is, see update), with a possible example in $X=\mathbb Q^*\times\mathbb Q^*$, the square of the one-point compactification of the rationals. $X$ is not $KC$ (the diagonal being compact and not closed), but $X$ is $NKI$ (this follows from the fact that $\mathbb Q^*$ is $KC$, hence $KIP$, hence $NKI$, and it is relatively straightforward to show that a product of two $T_1$ $NKI$ spaces is again $NKI$).

However, I ran into trouble trying to determine whether $X$ is $IKK_2$. I expect there should be either a straightforward proof of this, or a straightforward counter-example, but both have so far eluded me.

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Update.

I neglected to consider the one point compactification of the Arens-Fort space, discussed here. Subsets there are compact if and only if they are either finite, or contain the point at infinity. From this, $KIP$ is immediate, yet this space is not $KC$ or even weak Hausdorff, though it is $k_2$-Hausdorff.

Given that the implication $KC\implies KIP+T_1$ is indeed strict, it remains to show how $KIP+T_1$ relates to the aforementioned properties (i.e., does weak Hausdorff imply $KIP$, and is there an implication in either direction for $k_2$-Hausdorff?) or any other weak separation axioms.

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Addendum.

Let me add one additional set of observations that I didn't mention originally. They weren't particularly fruitful for me, but might be of some use that I didn't see:

If $X=(X,\tau)$ is a topological space, let $\tau_{cc}\subseteq \mathcal P(X)$ consist of the co-compact subsets of $X$, together with the empty set. Then

1. $\tau_{cc}$ is a topology if and only if $X$ is $IKK$. In this case $\tau_{cc}$ is always $T_1$.
2. $\tau_{cc}$ is a compact topology if and only if $X$ is $KIP$.
3. $\tau_{cc}\subseteq \tau$ if and only if $X$ is $KC$.
4. $\tau_{cc}\supseteq \tau$ if and only if $X$ is compact.

Answer 1

This is only a partial answer, but it may be of some use in determining separation properties on $\pi$-base so I've decided to post it.

We actually have

Proposition 1. $KIP+T_1\implies k_2\text{-Hausdorff}$.

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Proof.

The proof is similar in spirit to the proof for $KIP+T_1\implies US$, and could be thought of as a generalization, though we will argue more directly this time. Suppose $X=X$ is $T_1$ and $KIP$. Let $f,g\colon Z\to X$ be continuous maps from a compact Hausdorff space $Z=(Z,\tau)$ into $X$, and let $A=\{a\in Z\mid f(a)=g(a)\}$. We argue that $A$ is closed.

To see this, suppose $z\in \overline{A}$. Let $\mathcal K_f=\{f(\overline{U})\mid z\in U\in \tau\}$, define $\mathcal K_g$ similarly, and let $\mathcal K=\mathcal K_f\cup \mathcal K_g$. Then for any sets $f(\overline{U_1}), \dots, f(\overline{U_m}), g(\overline{V_1}), \dots, g(\overline{V_n}) \in \mathcal K$, we have some $a\in A\cap U_1\cap\dots\cap U_m\cap V_1\cap\dots\cap V_n$, (since $U_1\cap\dots\cap U_m\cap V_1\cap\dots\cap V_n$ is a neighborhood of $z\in\overline{A}$). Therefore $f(a)=g(a)\in f(\overline{U_1})\cap\dots\cap f(\overline{U_m})\cap g(\overline{V_1})\cap \dots\cap g(\overline{V_n})$, so $\mathcal K$ has the finite intersection property.

By $KIP$, there is some $x\in \bigcap \mathcal K$. We claim we must have $f(z)=x$. Otherwise, as $X$ is $T_1$, we could take a neighborhood $W$ of $f(z)$ avoiding $x$, and by regularity of $Z$ choose a neighborhood $U$ of $z$ with $\overline{U}\subseteq f^{-1}(W)$, obtaining a contradiction with $$x\notin f(\overline{U})\supseteq \bigcap \mathcal K\ni x.$$

Similarly, we have $g(z)=x$, so that $f(z)=g(z)$, whereby $z\in A$. Thus $A$ is closed, and since this holds for arbitrary pairs of maps into $X$, the diagonal is $k_2$-closed. Thus $X$ is $k_2$-Hausdorff.

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Remarks.

We now have $$KC\implies KIP+T_1\implies k_2\text{-Hausdorff}.\tag{1}$$ From this question we also know $$KC\implies \text{weak Hausdorff} \implies k_2\text{-Hausdorff},\tag{2}$$ and from the example of the one point compactification of the Arens-Fort space in the question above, we know that a space can be $KIP$ without being weak Hausdorff. Therefore three possibilities remain:

1. There is a weak Hausdorff space (which a fortiori is also $k_2\text{-Hausdorff}$) that fails to be $KIP+T_1$, so that the implications in (1) and (2) are strict, and neither weak-Hausdorff nor $KIP+T_1$ imply the other.
2. We have $\text{Weak Hausdorff} \implies KIP+T_1\implies k_2\text{-Hausdorff}$, with all implications strict.
3. We have $KIP+T_1\iff k_2\text{-Hausdorff}$

My suspicion is that 1 is most likely, but either 2 or 3 would be quite interesting.

In any case, if $X$ is assumed to be compactly generated ($CG_2$), we know from the answers to the linked question that the three properties in (1) are equivalent, hence from the proposition we see that they are all equivalent to $KIP+T_1$. Thus we have another characterization of $CGWH$ spaces as those $CG_2$ spaces which are $T_1$ and satisfy $KIP$. Or, if a stronger version of compactly generated is used ($CG_3$), we may drop the $T_1$ assumption as it is built into $CG_3$.

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Application.

We also have the following useful application of the above proposition:

Proposition 2. Let $X$ be countable and $US$. Then $X$ is $KIP$, hence $k_2\text{-Hausdorff}$.

Proof. Let $X$ be countable and $US$. From this answer we know that in countable spaces, compactness and sequential compactness coincide. If $K\subseteq X$ is compact, and $x_i\to x\in X$ with each $x_i\in K$, then by sequential compactness, there is a subsequence of $(x_i)$ converging to some $k\in K$, and by $US$, we must have $x=k$. We conclude that compact subsets are sequentially closed. On the other hand, a sequentially closed subset of a (sequentially) compact space is certainly again sequentially compact, so the intersection of any family of compact sets must be compact, i.e., $X$ satisfies condition $IKK$ (and a fortiori $IKK_2$) from the above question.

On the other hand, if $(K_\alpha)$ is a (strictly) decreasing transfinite sequence of non-empty compact subsets of $X$, then by countability of $X$, $(K_\alpha)$ is countable, so after passing to a cofinal chain if necessary, we may re-index the chain by naturals $(K_n)$. Then choosing $x_n\in K_n$, we have some convergent subsequence $y_m\to y$. By sequential closure of compact sets, we must have $y\in \bigcap_n K_n$, so $\bigcap_n K_n\neq \emptyset$. Thus $X$ satisfies the nonempty nested intersection property $NKI$, alongside $IKK_2$, which is an equivalent formulation to $KIP$ as proved in the question above.

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It is worth noting that in the question above, it was not clear whether $\mathbb Q^*\times \mathbb Q^*$ was $KIP$. As a corollary to Proposition 2, we see that it is, though in this example, since $\mathbb Q^*\times\mathbb Q^*$ was already known to be weak Hausdorff, concluding it is $k_2\text{-Hausdorff}$ from Proposition 1 yields nothing new.

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I will leave this question open for now, in the hopes that someone can determine what the precise status of $KIP+T_1$ is, in terms of the three possibilities mentioned above.