When is a P-space the set of P-points of a compact Hausdorff space

general-topology

Let $X$ be a Hausdorff space. A point $x\in X$ is a P-point if every countable intersection of open neighbourhoods of $x$ is a neighbourhood of $x$, and $X$ is a
P-space if every point is P-point, equivalently if every countable intersection of open sets is open.

The set of P-points in a compact Hausdorff space is thus a completely regular P-space. My question is: is every completely regular P-space equal to the set of P-points of some compact Hausdorff space?

If not, is there a characterisation of those completely regular P-spaces which are equal to the set of P-points of some compact Hausdorff space?

Example: a countable discrete set $N$ is a P-space and is equal to the set of P-points of its one-point compactification $N\cup\{\infty\}$.

Sub-question: is every completely regular P-space $X$ equal to the set of P-points of its Stone-Cech compactification $\beta X$?

Best Answer

Here is a complete answer for the sub question:

Let $X$ be completely regular.

  1. If $X$ is realcompact and $x_0$ is a P-point of $\beta X$, then $x_0 \in X$.

Proof: Assume $x_0 \notin X$. Then there is a continuous $h: \beta X \rightarrow [0,1]$ such that $h(x_0) = 0$ and $h(x) > 0$ for any $x \in X$ (Engelking, General Topology, Theorem 3.11.10). Since $x_0$ is P-point, there is an open $U$ such that $x_0 \in U \subseteq \bigcap_{n \in \textbf{N}} h^{-1} ([0, \frac{1}{n}[)$. Hence there is an $x \in U \cap X \Rightarrow h(x) = 0$. Contradiction!

  1. Let $X$ be a P-space. Then: $X$ is realcompact $\Leftrightarrow$ $X = \{x \in \beta X: x$ P-point in $ \beta X\}$

Proof: "$\Rightarrow$": "$\subseteq$": Each P-point of $X$ is P-point in $\beta X$, since $X$ is dense in $\beta X$. "$\supseteq$": 1.
"$\Leftarrow$": By Engelking, General Topology, Theorem 3.11.11 we have to show that an ultrafilter $\mathfrak{F}_0$ in the zero-sets of $X$, which has the countable intersection property, has non-empty intersection. Since in a P-space each zero-set is clopen, hence the complement is zero-set again, and each countable intersection of zero-sets is a zero-set, it is easy to see that in fact each countable intersection of $\mathfrak{F}_0$ is an element of $\mathfrak{F}_0$.
Now consider the standard construction of the Cech-Stone compactification $\beta X = \{\mathfrak{F}: \mathfrak{F}$ zero-set ultrafilter in $X$}. Since complements of zero-sets are zero-sets, $\{ [A] : A$ is zero-set in $X \}$ is a (clopen) base of $\beta X$, where $[A] := \{ \mathfrak{F} \in \beta X: A \in \mathfrak{F} \}$.
$\mathfrak{F}_0$ is P-point in $\beta X$: Let $(U_n)_{n \in \textbf{N}}$ be a sequence of neighbourhoods of $\mathfrak{F}_0$. Then for each $n$ pick a zero-set $A_n$ of $X$ such that $\mathfrak{F}_0 \in [A_n] \subseteq U_n$. As stated above, $A := \bigcap_{n \in \textbf{N}} A_n \in \mathfrak{F}_0$. Hence $\mathfrak{F}_0 \in [A] \subseteq \bigcap_{n \in \textbf{N}} [A_n] \subseteq \bigcap_{n \in \textbf{N}} U_n$ and $[A]$ is open.
By the prerequisite, $\mathfrak{F}_0 \in X$, hence $\bigcap \mathfrak{F}_0 = \{x\}$ for some $x \in X$.

Notes

  1. Of course, this also gives an affirmative answer to the original question in case $X$ is realcompact.

  2. But there are non-realcompact P-spaces, for instance:
    a) any discrete space of measurable cardinality
    b) Example 9L in the book of Gillman, Jerison, which is $\{\alpha < \omega_2:$ cf $(\alpha) \neq \omega\}$. Hence, this provides a ZFC counter-example to the sub question.

By the above, for these spaces $\beta X$ does not provide the required compactification. But for a), the one-point compactification of the discrete space does the job. However, in b) $\omega_2$ is a new P-point in the "obvious" compactification $\omega_2 + 1$. Hence, this might be a candidate for a counter-example also of the original question.

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