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Let $\tau_X$ denote the collection of open subsets of a topological space $X$ and let $\mathsf{RO}(X)$ be the subset of $\tau_X$ made up of regular open subsets. With this terminology, the inequality $|\mathsf{RO}(X)|\leq |\tau_X|$ is obvious. Furthermore, if $X$ is a semiregular space, it is easy to see that $|\tau_X|\leq 2^{|\mathsf{RO}(X)|}$.

It is well-known (see for example section 10 of Hodel's chapter in the Handbook) that every perfectly normal Hausdorff space satisfies the equality $|\mathsf{RO}(X)|=|\tau_X|$. My question goes in two directions:

  • What are some examples of Tychonoff spaces $X$ such that $|\mathsf{RO}(X)|< |\tau_X|$?

  • Can the "perfect normality" condition be relaxed in such a way that the equality remains true? E.g., is it true that every perfect (closed sets are $G_\delta$) Tychonoff space verifies the equality $|\mathsf{RO}(X)|=|\tau_X|$?

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2 Answers 2

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The compact Hausdorff space $X = \beta\mathbb{N}$ is another example. Every regular open subset is the closure of a subset of $\mathbb{N}$ and there are only $\frak{c}$ such subsets but $X$ has $2^{\frak{c}}$ points, and for each such point $p$, the set $X \setminus \{p\}$ is an open set.

For the second question, a scattered separable space of cardinality $\frak{c}$ and scattered height $2$ (such as a Cantor Tree, or a $\Psi$-space of size $\frak{c}$) gives a space in which every closed subset is a G$_\delta$ but in which the equality fails for the same reason that it fails in the tangent-disk space.

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  • $\begingroup$ How curious, not even one of the natural weakenings of the statement is true. Furthermore, if you weaken "perfect" to "countable pseudocharacter" the statement does not hold either: the square of Alexandroff-Urysohn double arrow, $\mathbb{A}^{2}$, is a compact, $T_4$ and first countable space with $|\mathsf{RO}(\mathbb{A}^{2})|<|\tau_{\mathbb{A}^{2}}|$. Perhaps there are other stronger and different combinations of properties that do imply the desired equality, we just don't know them yet. $\endgroup$
    – Peluso
    Commented Feb 26, 2022 at 21:32
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An answer to the first question: the Niemytzki plane, $N$, is an example. Note first that, in general, if $U$ and $V$ are regular open and $D$ is dense then $U=V$ iff $U\cap D=V\cap D$. From this we find that $|\mathsf{RO}(X)|\le2^{d(X)}$. Since $N$ is separable and has a closed discrete subset of cardinality $\mathfrak{c}$ (the $x$-axis) we find that $|\mathsf{RO}(N)|=\mathfrak{c}$ and $|\tau_N|=2^\mathfrak{c}$.

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  • $\begingroup$ Very nice. For the same reasons, Sorgenfrey's plane is also an answer to the first question. $\endgroup$
    – Peluso
    Commented Feb 25, 2022 at 21:00

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