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Let $\kappa$ be a measurable cardinal with $\sigma$-complete ultrafilter $U$. Let $X$ be the set of all finite sequences from $\kappa$. For $s\in X$, $i\in \kappa$, we write $(s,i)$ for the sequence you get by appending $i$ to $s$, similarly $(s,i,j)$, etc. We call a subset $A \subseteq X$ closed if it has the following property:

• Whenever $s$ in $X$, and almost all successors of $s$ are in $A$, then also $s$ is in $A$:

More precisely: If the set $\{i \in \kappa: (s,i)\in A\}$is in $U$, then also $s\in A$.

[EDITED:] In other words: A set $O$ is open if for all $s\in O$ also almost all successors of $s$ are in $O$. (Using the countable completeness of $U$, a neighborhood base of $s$ is given by the sets $O_{s,F}:=\{s\}\cup \{(s,i,j,\ldots, k): i,j,\ldots ,k\in F\}$, for $F\in U$. Using the fact that $U$ is non-principal one can show that these sets are clopen.)

We check that $X$ is extremally disconnected: If $O$ is open, and $A$ is the closure of $O$, we claim that $A$ is open. So let $s\in A$. If $s\in O$, then $s$ has a neighborhood in $A$, and we are done. So assume that $s$ is not in $O$. Then almost all successors of $s$ must also be in $A$, otherwise $A\setminus \{s\}$ is closed. So $A$ is open.

The fact that $X$ is a p-space follows from the countable closure of $U$. It is clear that $X$ has no isolated points.

Let $\kappa$ be a measurable cardinal with $\sigma$-complete ultrafilter $U$. Let $X$ be the set of all finite sequences from $\kappa$. For $s\in X$, $i\in \kappa$, we write $(s,i)$ for the sequence you get by appending $i$ to $s$, similarly $(s,i,j)$, etc. We call a subset $A \subseteq X$ closed if it has the following property:

• Whenever $s$ in $X$, and almost all successors of $s$ are in $A$, then also $s$ is in $A$:

More precisely: If the set $\{i \in \kappa: (s,i)\in A\}$is in $U$, then also $s\in A$.

[EDITED:] In other words: A set $O$ is open if for all $s\in O$ also almost all successors of $s$ are in $O$. (Using the countable completeness of $U$, a neighborhood base of $s$ is given by the sets $O_{s,F}:=\{s\}\cup \{(s,i,j,\ldots, k): i,j,\ldots ,k\in F\}$, for $F\in U$. Using the fact that $U$ is non-principal one can show that these sets are clopen.)(DELETED, see Joseph's comment below.)

We check that $X$ is extremally disconnected: If $O$ is open, and $A$ is the closure of $O$, we claim that $A$ is open. So let $s\in A$. If $s\in O$, then $s$ has a neighborhood in $A$, and we are done. So assume that $s$ is not in $O$. Then almost all successors of $s$ must also be in $A$, otherwise $A\setminus \{s\}$ is closed. So $A$ is open.

The fact that $X$ is a p-space follows from the countable closure of $U$. It is clear that $X$ has no isolated points.

Let $\kappa$ be a measurable cardinal with $\sigma$-complete ultrafilter $U$. Let $X$ be the set of all finite sequences from $\kappa$. For $s\in X$, $i\in \kappa$, we write $(s,i)$ for the sequence you get by appending $i$ to $s$. We call a subset $A \subseteq X$ closed if it has the following property:

• Whenever $s$ in $X$, and almost all successors of $s$ are in $A$, then also $s$ is in $A$:

More precisely: If the set $\{i \in \kappa: (s,i)\in A\}$is in $U$, then also $s\in A$.

In other words: A neighborhood base of $s$ is given by the sets $O_F:=\{s\}\cup \{(s,i): i\in F\}$, for $F\in U$.

We check that $X$ is extremally disconnected: If $O$ is open, and $A$ is the closure of $O$, we claim that $A$ is open. So let $s\in A$. If $s\in O$, then $s$ has a neighborhood in $A$, and we are done. So assume that $s$ is not in $O$. Then almost all successors of $s$ must also be in $A$, otherwise $A\setminus \{s\}$ is closed. So there is a neighborhood of $s$ contained in $A$.

The fact that $X$ is a p-space follows from the countable closure of $U$. It is clear that $X$ has no isolated points.

Let $\kappa$ be a measurable cardinal with $\sigma$-complete ultrafilter $U$. Let $X$ be the set of all finite sequences from $\kappa$. For $s\in X$, $i\in \kappa$, we write $(s,i)$ for the sequence you get by appending $i$ to $s$, similarly $(s,i,j)$, etc. We call a subset $A \subseteq X$ closed if it has the following property:

• Whenever $s$ in $X$, and almost all successors of $s$ are in $A$, then also $s$ is in $A$:

More precisely: If the set $\{i \in \kappa: (s,i)\in A\}$is in $U$, then also $s\in A$.

[EDITED:] In other words: A set $O$ is open if for all $s\in O$ also almost all successors of $s$ are in $O$. (Using the countable completeness of $U$, a neighborhood base of $s$ is given by the sets $O_{s,F}:=\{s\}\cup \{(s,i,j,\ldots, k): i,j,\ldots ,k\in F\}$, for $F\in U$. Using the fact that $U$ is non-principal one can show that these sets are clopen.)

We check that $X$ is extremally disconnected: If $O$ is open, and $A$ is the closure of $O$, we claim that $A$ is open. So let $s\in A$. If $s\in O$, then $s$ has a neighborhood in $A$, and we are done. So assume that $s$ is not in $O$. Then almost all successors of $s$ must also be in $A$, otherwise $A\setminus \{s\}$ is closed. So $A$ is open.

The fact that $X$ is a p-space follows from the countable closure of $U$. It is clear that $X$ has no isolated points.

Let $\kappa$ be a measurable cardinal with ultrafilter $U$. Call sets in $U$ "large"; others, small. Let $X$ be the set of all finite sequences from $\kappa$. We call a subsets $A \subseteq X$ closed if it has the following property:

• Whenever $s$ in $X$, and almost all successors of $s$ are in $A$, then also $s$ is in $A$:

More precisely: If the set $\{i \in \kappa: (s,i)\in A\}$is in $U$, then also $s\in A$.

Let $\kappa$ be a measurable cardinal with $\sigma$-complete ultrafilter $U$. Let $X$ be the set of all finite sequences from $\kappa$. For $s\in X$, $i\in \kappa$, we write $(s,i)$ for the sequence you get by appending $i$ to $s$. We call a subset $A \subseteq X$ closed if it has the following property:

• Whenever $s$ in $X$, and almost all successors of $s$ are in $A$, then also $s$ is in $A$:

More precisely: If the set $\{i \in \kappa: (s,i)\in A\}$is in $U$, then also $s\in A$.

In other words: A neighborhood base of $s$ is given by the sets $O_F:=\{s\}\cup \{(s,i): i\in F\}$, for $F\in U$.

We check that $X$ is extremally disconnected: If $O$ is open, and $A$ is the closure of $O$, we claim that $A$ is open. So let $s\in A$. If $s\in O$, then $s$ has a neighborhood in $A$, and we are done. So assume that $s$ is not in $O$. Then almost all successors of $s$ must also be in $A$, otherwise $A\setminus \{s\}$ is closed. So there is a neighborhood of $s$ contained in $A$.

The fact that $X$ is a p-space follows from the countable closure of $U$. It is clear that $X$ has no isolated points.