special extremally disconnected spaces with only finite isolated points

We Know that a cardinal $\kappa$ is measurable if there is a set $X$ with cardinal $\kappa$ and a {0,1}-measure $\mu: P(X) \rightarrow${$0,1$} so that for all $x \in X$, $\mu(x)=0$ and $\mu(X)=1$. also a cardinal which is not measurable, is called non measurable.

Also we Know this is unprovable to find a set with measurable cardinal in "ZFC".

In topology an extremally disconnected space is a topological space in which all open subsets has open closure.

Also we call a topological space to be a P-space if all it's $G_{\delta}$- sets are open.

There is a well-Known theorem that says every extremally disconnected P-space with non measurable cardinal is discrete.

From the aforesaid summaries a question could be posed that:

Question: If we suppose that a measurable cardinal exists, can we construct an extremally disconnected P-space with only a finite number of isolated points.

• What do you mean it is unsettled? We know exactly that from the axioms of ZFC it is unprovable that a measurable exists. In fact from ZFC and many other large cardinal axioms it is unprovable that a measurable exists. However from assumptions like "There is a Woodin cardinal" or "There is a supercompact cardinal" or "AD holds in $L(\mathbb R)$" we can prove that a measurable exists. Nov 22 '12 at 11:42
• (A minor point. I dont see the difference between finding "a cardinal number with measurable cardinal" and finding "a measurable cardinal". According to the definition you give, they're meant to be the same, right? ) Nov 22 '12 at 13:00
• AliReza, the term "unsettled" usually means "an open problem", for example it is unsettled if the nontrivial zeros of the Riemann zeta function lie on the line $Re(z)=\frac12$. Unprovable is a form of settled. It is an answer. We cannot prove the existence of measurable cardinals, nor we can disprove it. Nov 22 '12 at 14:31
• Well, you probably know that you cannot prove the consistency of ZFC from the axioms of ZFC. The theory "ZFC+There exists a measurable cardinal" is a very strong theory compared to ZFC, in particular it proves the consistency of ZFC (if we assume a measurable exists then we can prove that ZFC is consistent). So assuming a measurable is a stronger assumption than assuming that there are no measurable cardinals. There is a lot of delicacy in those arguments, especially if you are unfamiliar with consistency results. But if there is a measurable, then there is a least measurable cardinal. Nov 22 '12 at 17:45
• Your title "On the existence of measurable cardinals" is extremally non-descriptive. I suggest "extremally disconnected spaces without isolated points" instead. Nov 22 '12 at 19:24

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.

• I don't think the sets $O_{s,F}$ really form a neighborhood basis for $s$. It seems like a way to get around this is to only take increasing sequences from $\kappa$ and assume that the ultrafilter is a normal ultrafilter. Or you can just take another basis. Nov 25 '12 at 0:36
• You are right. Thank you. For some stupid reason I thought that there are only countably many finite sequences. Thinking too much about Laver trees. Nov 25 '12 at 1:29
• Whenever $s\in X$, and $\bar A=(A_t:s \le t)$ is a family of sets in $U$, then $\bar A$ naturally defines a clopen neighborhood $O=O_{\bar A}$ of $s$, and these neighborhoods form a base: $s\in O$, all elements of $O$ extend $s$, and for each $t\in O$ we have $(t,i)\in O$ iff $i\in A_t$. Nov 25 '12 at 1:39
• This is a really nice topological space which is very much related to Prikry forcing. Let's take a normal ultrafilter $M$ on a measurable cardinal $\kappa$ and define a topology on $[\kappa]^{<\omega}$ where a subset $U\subseteq[\kappa]^{<\omega}$ is open if and only if for each $(a_{1},...,a_{n})\in U$ there is some $A\in M$ where if $a\in A,a>a_{n}$, then $(a_{1},...,a_{n},a)\in U$ as well. Then $[\kappa]^{<\omega}$ is extremally disconnected, so the Boolean algebra of clopen sets of $[\kappa]^{<\omega}$ is a complete and it is the completion of the Prikry forcing. Nov 26 '15 at 23:03
• Furthermore, the extremal disconnectedness of $[\kappa]^{<\omega}$ is equivalent to the Prikry lemma which states that for every statement $\sigma$ and every stem $s$ (the stems are the elements in the space $[\kappa]^{<\omega}$) there is some $A\in M$ where $(s,A)$ decides $\sigma$. I have not seen any papers or investigations where this relation between Prikry forcing and general topology has been seriously investigated before even though it seems very interesting. Nov 26 '15 at 23:03