6
$\begingroup$

I'm thinking about results of the form: Under assumption $A$, if $X$ is a compact Hausdorff space and $C$ is a cover of $X$ by closed $G_\delta$ sets, then there is a subcover of cardinality $\leq\kappa$.

Obviously in general we can take $\kappa\leq|X|$ and in some cases this is the best we can do such as in Cantor space in which every point is a closed $G_\delta$ set.

The strongest result I've found so far is this: If $X$ is a scattered compact Hausdorff space and $C$ is a cover of $X$ by closed $G_\delta$ sets, then there is a countable subcover.

Proof: Since $X$ is scattered it is totally disconnected, and we can represent each $F\in C$ as $F=\bigcap_{n<\omega}Q_{F,n}$, where $Q_{F,n}$ is a nested sequence of clopen sets. Partition $X$ into a finite collection $X_0, \dots ,X_{n-1}$ of clopen sets such that each $X_i$ has a unique point $x_i$ such that $CB(x_i) = CB(X)$ (where $CB$ is the Cantor-Bendixson rank). In each $X_i$ choose some $F_i\in C$ such that $x_i \in F$ and then consider $X_i \setminus F_i$, which can be written as a countable union of clopen sets. Iterate this procedure in each such clopen set. This forms a well-founded forest which is countably branching and thus countable. The subcover is the $F_i$'s chosen at each node of the forest.

This is certainly optimal for the general scattered case, as you can construct a cover that saturates the bound in any infinite scattered compact Hausdorff space.

A weak generalization is this: If $X$ is a compact Hausdorff space with perfect core $P$, and $C$ is a cover of $X$ by closed $G_\delta$ sets, then for any $C_0 \subseteq C$ covering $P$, $C$ has a subcover of cardinality $\leq \omega^{|C_0| + 1}$.

Proof: If $P$ is empty we can use the previous result, so assume that $P$ is non-empty. Let $C_0$ be a cover of $P$, so in particular $P \subseteq \bigcup_{F\in C_0}\bigcap_{n<\omega}U_{F,n} = \bigcap_{g:C_0\rightarrow\omega}\bigcup_{F\in C_0}U_{F,g(F)}$. By compactness, for each $g:C_0 \rightarrow \omega$, there is a finite set $C_{0,g}\subseteq C_0$ such that $P\subseteq \bigcup_{F\in C_{0,g}}U_{F,g(F)}$. So let $R_g=X\setminus \bigcup_{F\in C_{0,g}}U_{F,g(F)}$. Each $R_g$ is closed and scattered as a subspace of $X$, so we can run the previous argument to get a countable $C_g \subseteq C$ that covers $R_g$. This means that in order to cover the rest of $X$ we need at most $\omega^{|C_0|}\times \omega = \omega^{|C_0|+1}$ more sets, so since $\omega^{|C_0|+1} \geq |C_0|$ we need at most $\omega^{|C_0|+1}$ all together.

This is only clearly non-trivial in cases where $P$ is much smaller than $X$. The question is how much better can you do? In particular is there a universal bound for all compact Hausdorff spaces? Can you get a better bound in terms of $|C_0|$? Can you get bounds in terms of other cardinal invariants of $P$ or $X$, such as the weight or density character? Can you say anything non-trivial about perfect spaces? What is $\kappa$ for $\beta \omega \setminus \omega$ or $2^\lambda$ for various cardinals $\lambda$?

EDIT: In the negative direction I suspect you can show that if $X$ is a compact Hausdorff space with non-empty perfect core, then $X$ has a cover by closed $G_\delta$ sets such that no subcover has cardinality $< 2^\omega$.

EDIT2: Suppose that $X$ is a compact Hausdorff space with non-empty perfect core $P$. Construct a tree of closed sets $\{F_{\sigma}\}_{\sigma \in 2^{<\omega}}$ with the following properties: Every $F^\circ_{\sigma}$ has non-empty intersection with $P$. For every $\sigma \in 2^{<\omega}$, $F_{\sigma\frown 0},F_{\sigma\frown 1} \subset F_\sigma^\circ$ and $F_{\sigma\frown 0}\cap F_{\sigma \frown 1} = \varnothing$. (We can do this since $P$ is perfect.) By these conditions, for any $\alpha \in 2^\omega$, the set $G_\alpha = \bigcap_{n<\omega} F_{\alpha \upharpoonright n}$ is a closed $G_\delta$ set. If $\alpha,\beta \in 2^\omega$ with $\alpha \neq \beta$, then $G_\alpha \cap G_\beta = \varnothing$, and $H = \bigcup_{\alpha \in 2^\omega} G_\alpha$ is a closed $G_\delta$ set. Since $H$ is a closed $G_\delta$ set, we can find a sequence of closed $G_\delta$ sets $\{D_n\}_{n<\omega}$ such that $\bigcup_{n<\omega} D_n = X \setminus H$. So then the cover $\{D_n\}_{n<\omega} \cup \{ G_\alpha \}_{\alpha \in 2^\omega}$ is a cover by closed $G_\delta$ sets with no subcover of cardinality $< 2^\omega$.

EDIT3: There's a much better bound in the second case that I missed: If $X$ is a compact Hausdorff space with perfect core $P$, and $C$ is a cover of $X$ by closed $G_\delta$ sets, then for any $C_0 \subseteq C$ covering $P$, $C$ has a subcover of cardinality $\leq |C_0|+\omega$.

Proof: Same proof as before, just notice that while there are $\omega^{|C_0|}$ many functions $g:C_0 \rightarrow \omega$, there are only $(|C_0|\times \omega)^{<\omega} = |C_0|+\omega$ many finite subsets of $\{U_{F,n}\}_{F\in C,n<\omega}$, so there are only at most $|C_0|+\omega$ many $R_g$'s to consider, and we get that $C$ has a subcover of size at most $|C_0|+(|C_0|+\omega)\times\omega = |C_0|+\omega$.

By the same argument as in the scattered case, this is basically optimal relative to the size of $|C_0|$, so now the questions really is about perfect spaces.

EDIT4: Combining the content of Taras and Anonymous's comments, you get a more refined bound in terms of generalized Cantor-Bendixson ranks (these probably have a name somewhere), specifically for any cardinal $\kappa$ we can can define

$$X^\prime_{\kappa}=X \setminus \bigcup \{U\subseteq X : U\text{ open, has density character }\leq\kappa\}$$

$$X^{(\alpha)}_{\kappa} = \bigcap_{\beta < \alpha} (X^{(\beta)}_{\kappa})^\prime _{\kappa}$$

where $\bigcap\varnothing=X$, for any $\alpha \in \text{Ord}\cup\{\infty\}$. And $CB_{\kappa}(X)$ is the smallest ordinal $\alpha$ for which $X^{(\alpha + 1)}_{\kappa}=\varnothing$, if it exists, and $\infty$ otherwise. Then if we let $S(X)$ be the smallest $\kappa$ such that $CB_{\kappa}(X)<\infty$, we should get a better bound in terms of $S(X)$.

If the density character of $X$ is $\kappa$, then every closed $G_\delta$ cover has a subcover of cardinality $\leq 2^\kappa$, since there are only at most that many closed $G_\delta$ sets in $X$.

Assume that for some cardinal $\kappa$ and some ordinal $\alpha$, we've shown that if $CB_{\kappa}(X)<\alpha$, then every closed $G_\delta$ cover has a subcover of cardinality $\leq 2^\lambda$ for some $\lambda < \kappa$. Let $X$ be a compact Hausdorff space such that $CB_{\kappa}(X) = \alpha$ and let $C$ be a cover of $X$ by closed $G_\delta$ sets. This implies that the density character of $P = X^{(CB_{\kappa}(X))}_{\kappa}$ is $\leq\kappa$, so there is some set $C_0 \subseteq C$ of cardinality $\leq 2^\kappa$ such that $C_0$ covers $P$. By the same argument as before, $X\setminus \bigcup C_0$ can be written as the union of at most $2^\kappa$ many closed subspaces, each of which has strictly smaller $CB_{\kappa}$. So by the induction hypothesis each of these has a subset of $C$ of cardinality $\leq 2^\kappa$ that covers it, so overall $X$ is covered by a subset of $C$ of cardinality $\leq 2^\kappa$.

So by induction, if $CB_{\kappa}(X)<\infty$, then every closed $G_\delta$ cover of $X$ has a subcover of cardinality $\leq 2^\kappa$. So in general any closed $G_\delta$ cover of a compact Hausdorff space has a subcover of cardinality $\leq 2^{S(X)}$.

You can probably get slightly tighter bounds at singular cardinals using a CB derivative for density characters $<\kappa$ rather than just $\leq \kappa$.

$\endgroup$
  • 1
    $\begingroup$ Your question is about the Lindelof number of the $G_\delta$-modification of a topological space. You can consider scatteredness with respect to the property $P$ you are interested in: a topological space is P-scattered if each closed non-empty subasace contains a non-empty relatively open subspace with property $P$. In your case the property P is the existence of small subcover of any cover by $G_\delta$-sets. $\endgroup$ – Taras Banakh Feb 4 at 6:55
  • 2
    $\begingroup$ For $\beta \omega \setminus \omega$ the cardinal $\kappa$ is $2^\omega$. The reason is that for compact Hausdorff spaces, closed $G_\delta$ sets are the same as zero sets and a compact subset of a separable space has only $2^\omega$ continuous functions and, therefore, only $2^\omega$ zero sets. $\endgroup$ – Anonymous Feb 4 at 14:48
3
$\begingroup$

Your question is related to a pair of old questions of A.V.Arhangel'skii. First of all note that in a regular space, for every point $x$ and every $G_\delta$ set G containing $x$ there is a closed $G_\delta$ $H$ contained in $G$ such that $x \in H$. So the topology generated by the closed $G_\delta$ sets of a compact Hausdorff space $X$ coincides with the "$G_\delta$-topology" $X_\delta$ (that is, the topology generated by the $G_\delta$ subsets of $X$).

The Lindelof degree of $X$ ($L(X)$) is defined as the minimum cardinal $\kappa$ such that every open cover of $X$ has a $\leq \kappa$-sized subcover and the weak Lindelof degree of $X$ ($wL(X)$) is defined as the minimum cardinal $\kappa$ such that every open cover of $X$ has a $\leq \kappa$ subcollection whose union is dense in $X$.

Question (A.V. Arhangel'skii, 1970): Let $X$ be a compact Hausdorff space.

  1. Is it true that $L(X_\delta) \leq 2^{\aleph_0}$?
  2. Is it true that $wL(X_\delta) \leq 2^{\aleph_0}$?

Various partial positive answers to these questions can be found in the literature. For example (see the references below):

  • In 1972 Juhász proved that $wL(X_\delta) \leq 2^{\aleph_0}$ for every compact ccc space $X$.
  • In 1974 Fleischmann and Williams proved that $L(X_\delta) \leq 2^{\aleph_0}$ for every compact linearly ordered space $X$.
  • In 1985 Pytkeev proved that $L(X_\delta) \leq 2^{\aleph_0}$ for every compact space of countable tightness $X$.
  • In 2016 I proved that $wL(X_\delta) \leq 2^{\aleph_0}$ for every compact space $X$ where player II has a winning strategy in the "weak Lindelof game of length $\omega_1$" (which among other things implies Juhász's 1972 result).

However both questions have a negative answer. A construction of a compact space $X$ such that $wL(X_\delta) > 2^{\aleph_0}$ can be found in my paper with Szeptycki (item 5 from the reference list). To find a compact space $X$ such that $L(X_\delta) > 2^{\aleph_0}$ it's enough to take the Cantor cube $X=2^{\mathfrak{c}^+}$ (just note that $X$ is homeomorphic to $(2^\omega)^{\mathfrak{c}^+}$, so $X_\delta$ is homeomorphic to the $G_\delta$ topology on $D^{\mathfrak{c}^+}$, where $D$ is a discrete space of size continuum and a $G_\delta$ cover without subcovers of size continuum for the latter space is given by $\{[\sigma]: \sigma \in Fn(\mathfrak{c}^+, D, \omega_1) \wedge \sigma$ is not one-to one on its domain $\}$ where $[\sigma]=\{f \in D^{\mathfrak{c}^+}: f \supset \sigma \}$).

Finally, to address your question regarding the existence of a bound, Toshimichi Usuba very recently proved that an $\omega_1$-strongly compact cardinal is a precise upper bound on both the Lindelof degree and the weak Lindelof degree of the $G_\delta$-topology on a compact space. So it's consistent that there is no bound to $wL(X_\delta)$ for $X$ compact!

References:

  1. Juhász, István, On two problems of A.V.Arkhangel’skij, General Topology Appl. 2, 151-156 (1972). ZBL0237.54002.

  2. Williams, Scott; Fleischman, William, The $G_\delta$-topology on compact spaces, Fundam. Math. 83 (1974), pp. 143-149. ZBL0278.54021.

  3. Pytkeev, E. G., About the $G_\lambda$-topology and the power of some families of subsets on compacta, Topology theory and applications, 5th Colloq., Eger/Hung. 1983, Colloq. Math. Soc. János Bolyai 41, 517-522 (1985). ZBL0604.54007.
  4. Spadaro, Santi, Infinite games and chain conditions, Fundam. Math. 234 (2016), pp. 229-239. ZBL1360.54010.
  5. Spadaro, S.; Szeptycki, P., $G_{\delta}$ covers of compact spaces, Acta Math. Hung., 154 (2018) pp. 252-263. ZBL06850215.
  6. T. Usuba, "$G_\delta$-topology and compact cardinals", preprint, arXiv:1709.07991.

P.S.: A related question is: "how big can a partition of a compact space by closed $G_\delta$s be?". In this case the continuum is a bound (see my answer to the following Mathoverflow question: A generalization of the Arhangelskii Theorem).

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.