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

, can we construct an extremally disconnected P-space with only aexistsof isolated points.finite number

unprovablethat 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. $\endgroup$isa form of settled. It is an answer. We cannot prove the existence of measurable cardinals, nor we can disprove it. $\endgroup$fromthe 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$7more comments