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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.

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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 22, 2012 at 11:42
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    $\begingroup$ (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? ) $\endgroup$
    – Qfwfq
    Commented Nov 22, 2012 at 13:00
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 22, 2012 at 14:31
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 22, 2012 at 17:45
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    $\begingroup$ Your title "On the existence of measurable cardinals" is extremally non-descriptive. I suggest "extremally disconnected spaces without isolated points" instead. $\endgroup$
    – Goldstern
    Commented Nov 22, 2012 at 19:24

1 Answer 1

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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.)(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.

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    $\begingroup$ 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. $\endgroup$
    – Joseph Van Name
    Commented Nov 25, 2012 at 0:36
  • $\begingroup$ 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. $\endgroup$
    – Goldstern
    Commented Nov 25, 2012 at 1:29
  • $\begingroup$ 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 $i\in A_t$. $\endgroup$
    – Goldstern
    Commented Nov 25, 2012 at 1:39
  • $\begingroup$ 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. $\endgroup$
    – Joseph Van Name
    Commented Nov 26, 2015 at 23:03
  • $\begingroup$ 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. $\endgroup$
    – Joseph Van Name
    Commented Nov 26, 2015 at 23:03

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