# non-P-points a Baire space

Let $X$ be a compact Hausdorff space. A P-point in $X$ is a point which does not lie in the boundary of the cozero set of a continuous real-valued function on $X$.

Question. Suppose that $X$ has no isolated points. Is the set of P-points a meagre subset of $X$? Failing this, is the set of non-P-points at least a Baire space?

-

The set of non-P-points is a Baire space:

I´ll work with the usual definition of P-point: $p \in X$ is a P-point if the intersection of countably many neighborhoods of $p$ is again a neighborhood of $p$. This is equivalent to the definition in the question for compact $X$.

Fix a compact $X$, let $P \subseteq X$ be the set of all P-points and $N=X \setminus P$. The first thing to notice is that if $A$ is a $G_\delta$ subset of $X$ and $A \subseteq P$ then every $p \in A$ is isolated. This is because (using that $p$ is a P-point and regularity of $X$) there would be a closed (hence compact) neighborhood of $p$ consisting only of P-points. Since compact P-spaces are finite, we get that $p$ is isolated. It follows that if $X$ has no isolated points then $P$ contains no $G_\delta$ and in particular $N$ is dense in $X$.

Now if $\left< U_n : n\in \omega \right>$ is a sequence of open dense subsets of $N$, then each $U_n=V_n \cap N$ for some $V_n$ open in $X$. Since $N$ is dense in $X$ we have that $V_n$ must also be dense in $X$. Then $\bigcap_{n \in \omega}V_n$ is dense in $X$ and since $P$ contains no $G_\delta$, we get that $\bigcap_{n \in \omega}U_n=N \cap \bigcap_{n \in \omega}V_n$ is dense in $N$.

-
Great! Many thanks. – Douglas Somerset Apr 26 '12 at 17:10

The answer to the first half of your question is no, though, at least consistently.

Let $\mathbb{N}^*=\beta \mathbb{N} \setminus \mathbb{N}$ be the remainder of the Stone-Cech compactification of the integers. Under the Continuum Hypothesis, the set of all $P$-points is a dense subset of $\mathbb{N}^*$. But $\mathbb{N}^*$ has no dense meager subset. Indeed, let $\{N_n: n \in \mathbb{N} \}$ be a sequence of nowhere dense sets. Construct a decreasing sequence of open sets $\{U_n: n \in \mathbb{N} \}$ such that $U_n \cap D_n =\emptyset$ and $\overline{U_{n+1}} \subset U_n$ for every $n \in \mathbb{N}$. The set $\bigcap_{n \in \mathbb{N}} \overline{U_n}=\bigcap_{n \in \mathbb{N}} U_n$ is non-empty because of compactness. Moreover, every countable intersection of open subsets of $\mathbb{N}^*$ has non-empty interior (see, for example, Jan van Mill's chapter on $\beta \mathbb{N}$ in the Handbook of Set-theoretic Topology). Thus $Int(\bigcap_{n \in \mathbb{N}} U_n)$ is a non-empty open set missing every $N_n$ and hence $\bigcup_{n\in\mathbb{N}} N_n$ is not dense.

-