4
$\begingroup$

Definition. A topological space $X$ is a $Q$-space if every subset of $X$ is of type $G_\delta$.

It is clear that every $Q$-space has countable pseudocharacter (= all singletons are $G_\delta$) and is perfect (= all closed subsets are $G_\delta$).

It is well-known that Martin's Axiom (MA) implies that every second-countable space of cardinality $<\mathfrak c$ is a $Q$-space.

A standard proof of this fact actually shows more, namely, that every topological space $X$ with countable separating weight and $|X|<\mathfrak c$ is a $Q$-space.

Let us recall that a topological space $X$ has countable separating weight if it admits a countable open cover $\mathcal U$ such that for any distinct points $x,y\in X$ there exists a set $U\in\mathcal U$ that contains $x$ but not $y$.

It is easy to see that every topological space with countable separating weight has countable pseudocharacter. The space $\omega_1$ of ordinals with the order topology is a first-countable but not a $Q$-space.

Question. Let $X$ be a first-countable Lindelof Hausdorff space of cardinality $|X|<\mathfrak c$. Is $X$ a $Q$-space under MA? Is $X$ perfect under MA?

Remark. The "Lindelof" requirement can not be removed from this question as the ordinal $\omega_1=[0,\omega_1)$ with the standard order topology is first-countable but not perfect (because the closed set of countable limit ordinals is not of type $G_\delta$ in $\omega_1$ by Fodor's Lemma) and hence $\omega_1$ is not a $Q$-space.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

I think that the answer to both of your questions is negative.

Let $X$ be a subspace of the real line with its usual topology of size $\omega_1$. Then, clearly, $X$ is first countable, Lindelöf and Hausdorff. Thus, its Alexandroff duplicate, $A(X)$, is also first countable, Lindelöf and Hausdorff (see here The Alexandroff Duplicate and its subspaces ). Now, since every point in $X\times \{1\}$ is isolated, it is clear that $X\times \{1\}$ is a uncountable and discrete subspace of $A(X)$; in particular, $X\times \{1\}$ is not Lindelöf. Since every Lindelöf perfect space is hereditarily Lindelöf, it follows that $A(X)$ is not perfect.

Hence, $A(X)$ is a first countable, Lindelöf and Hausdorff space of cardinality $\omega_1$ which is not perfect (thus, not a $Q$-space). Finally, since $\mathsf{MA}+\neg \mathsf{CH}$ is consistent, this procedure would in turn give an example of a first countable, Lindelöf and Hausdorff space of cardinality $<\mathfrak{c}$ which is not perfect (thus, not a $Q$-space).

Improve this answer
$\endgroup$
5
  • $\begingroup$ Great answer! Thank you. By the way what is the separating weight of the Alexandroff duplicate? Is it uncountable? $\endgroup$
    – Taras Banakh
    Commented May 30, 2022 at 19:42
  • $\begingroup$ I thought this would be a consequence of some simple inequalities with respect to $X$, but I only have managed to show (I think) that $sw(X)\leq sw\left(A(X)\right) \leq sw(X)\cdot |X|$, which do not imply that the separating weight of $A(X)$ is uncountable in this case. I'll have to think about it a bit more. $\endgroup$
    – Peluso
    Commented May 30, 2022 at 22:21
  • $\begingroup$ The separating weight is indeed uncountable because countably many open sets is not sufficient for separating points of $X$ from their duplicates (becuase for any open set $U\subseteq A(X)$ the set $U\cap pr(X_1\setminus U)$ is discrete in $X$, so countably many open sets can separate only countably many points of $X_1=X\times \{1\}$ form points of $X\times\{0\}$). $\endgroup$
    – Taras Banakh
    Commented May 30, 2022 at 22:33
  • $\begingroup$ Did you mean $pr(U)\cap pr(X_1\setminus U)$? $\endgroup$
    – Peluso
    Commented May 30, 2022 at 23:19
  • $\begingroup$ Yes. Some how I first identified $X$ with the subset of $AD(X)$, $\endgroup$
    – Taras Banakh
    Commented May 31, 2022 at 3:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .