Arhangeleskii's Theorem states the following

For any Hausdorff topological space $X$, $$ |X|\leq2^{\chi(X)L(X)} $$ where $\chi(X)$ is the character of $X$ and $L(X)$ is the Lindelöf degree of $X$.

In the article Paracompactness of Box Products of Compact Spaces, Kenneth Kunen presents a generalization of the countable case of the previous Theorem, also by Arhangelskii. It says

If $Y$ is a compact Hausdorff space and $\mathfrak{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathfrak{F}$ satisfies $$ \forall H\in \mathfrak F (|\{K\in\mathfrak F: H\cap K\neq\emptyset\}|\leq \mathfrak c), $$ then $|\mathfrak F|\leq \mathfrak c$

I would like to prove this result. Kunen says only that the proof to this Theorem is an easy modification of a proof by R. Pol of the original (on the article Short proofs of two theorems on cardinality of topological spaces). However, the steps of the original proof don't translate well to this generalization. When I try to emulate Pol's proof, I can't take the closure of each step, because that why I couldn't ensure that I take less than $\mathfrak c$ elements of $\mathfrak F$. When I tried to use something other than the closure in that step, I can't guarantee that after the induction I have closed sets. I wanted those set to be closed in order to use the compactness of the space. Maybe there is some other way to ensure the compactness, but I don't have any clue of what that could be.

Does the theorem lack some extra hypothesis? There is a proof supposing that $X$ has countable cellularity, but this assumption does not help me.

There is a similar result in the article Box Products, by Scott Williams, in the Handbook of Set Theoretic Topology (Lemma 4.1 on page 184).

No compact Hausdorff space can be partitioned into more than $\mathfrak c$ closed $G_\delta$ sets.

This result would be enough for me, but the proof in the article has a flaw. During the induction step, a function is defined in order to guarantee that each step does not exceed $\mathfrak c$ closed $G_\delta$ sets. However, that function is not always injective. I think that this can be related to the troubles I have with the other version.

  • $\begingroup$ I am not sure if this is the best place for this question, or if it is best suited for mathoverflow. $\endgroup$
    – Ottenbreit
    Jul 5, 2014 at 4:30
  • $\begingroup$ I agree it would be better on mathoverflow. $\endgroup$ Jul 6, 2014 at 15:34

1 Answer 1


I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models.

If $(Y, \tau)$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then $|\mathcal{F}| \leq 2^\omega$.

Let $\theta$ be a large enough regular cardinal and $M \prec H(\theta)$ be a countably closed elementary submodel of $H(\theta)$ of cardinality $\mathfrak{c}$ such that $\mathfrak{c} \subset M$ and $\{Y, \tau, \mathcal{F}, \mathfrak{c}\} \subset M$.

We claim that $\mathcal{F} \subset M$. Suppose this is not the case and let $K \in \mathcal{F} \setminus M$.

Using compactness and regularity of $Y$, we can write every closed $G_\delta$ set $F \subset Y$ in the form $F=\bigcap \{\overline{V_n(F)}: n < \omega \}$, for a suitable decreasing family of open neighbourhoods $\{V_n(F): n < \omega \}$ of $F$.

If $F \in \mathcal{F} \cap M$, then the family $\{V_n(F): n < \omega \}$ may be fixed in $M$, but then we actually have $\{V_n(F): n < \omega \} \subset M$.

Note now that for every $x \in \overline{Y \cap M}$, we can find $F \in \mathcal{F} \cap M$ such that $x \in F$. Indeed, let $x \in \overline{Y \cap M} \setminus Y \cap M$. Then there is $F \in \mathcal{F}$ such that $x \in F$. Let $x_n$ be any point of $U_n(F) \cap Y \cap M$. Using compactness of $F$ and the fact that $F=\bigcap \{\overline{U_n(F)}: n <\omega \}$ we can easily see that the sequence $\{x_n: n < \omega \}$ converges to $F$, in the sense that, every open neighbourhood of $F$ contains a final segment of $\{x_n: n < \omega \}$. Now by $\omega$-closedness of $M$, we have that $\{x_i: i \geq n \} \in M$, for every $n<\omega$ and hence the set:

$$\mathcal{S}=\{K \in \mathcal{F}: \{x_n: n <\omega \} \rightarrow K \}$$

is an element of $M$ (the formula defining that set has all free variables in $M$).

Note that since every compact Hausdorff space is normal, $F$ can be separated from a closed set disjoint from it by a pair of disjoint open sets. Therefore, we have that if $G \in \mathcal{S}$ then $G \cap F \neq \emptyset$. Hence $|\mathcal{S}| \leq 2^{\aleph_0}$ by assumption. But then we actually have that $\mathcal{S} \subset M$ and hence $F \in M$, as we wanted.

For every $F \in \mathcal{F} \cap M$ we have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \in M$, and, since $|\{H \in \mathcal{F} : F \cap H \neq \emptyset \}| \leq \mathfrak{c}$, we also have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \subset M$. So, for every $F \in \mathcal{F} \cap M$, we have $F \cap K=\emptyset$.

Now, by compactness of $K$, we can find for every $F \in \mathcal{F} \cap M$, an integer $n <\omega$ such that $\overline{U_n(F)} \cap K=\emptyset$.

Thus for every $x \in \overline{Y \cap M}$, we can find an open neighbourhood $U_x \in M$ of $x$ such that $U_x \cap K=\emptyset$.

Now $\mathcal{U}=\{U_x: x \in \overline{Y \cap M} \}$ is an open cover of the compact space $\overline{Y \cap M}$ such that $\mathcal{U} \subset M$. So there is a finite set $\mathcal{G} \subset \mathcal{U}$ such that $\overline{Y \cap M} \subset \bigcup \mathcal{G}$. This implies that $M \models Y \subset \bigcup \mathcal{G}$ and hence $H(\theta) \models Y \subset \bigcup \mathcal{G}$. But this contradicts $K \cap \bigcup \mathcal{G}=\emptyset$.

So $\mathcal{F} \subset M$ and hence $|\mathcal{F}| \leq |M| \leq \mathfrak{c}$.

  • 2
    $\begingroup$ Nice argument Santi. Question: Is there a way to topologize $\mathcal{F}$ so that Kunen´s theorem becomes a corollary of one known cardinal invariant inequality? $\endgroup$ Nov 29, 2017 at 18:31

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