In his seminal 1937 paper, Jones [1] proved the following result about Moore spaces:

Theorem. (Jones) If $2^{\aleph_0}<2^{\aleph_1}$ then all separable normal Moore spaces are metrizable.

Then he came up with the idea that maybe the separability condition in the above theorem could be removed. This led him to formulate the following famous conjecture:

Normal Moore Space Conjecture (NMSC). All normal Moore spaces are metrizable.

Long story short, NMSC turned out to be of high large cardinal strength and so far beyond the power of $ZFC$ to decide.

In 1980, Nyikos [2] proved the consistency of NMSC using Product Measure Extension Axiom (PMEA) which is known to be consistent assuming the consistency of strongly compact cardinals. Later a more direct proof of the consistency of NMSC has been presented by Dow et al [3] using the stronger assumption of the consistency of supercompact cardinals.

Interestingly, Fleissner [4] used a core model argument to prove the fact that the consistency of NMSC implies the consistency of measurable cardinals as well. Thus NMSC lies between strongly compacts and measurables in the consistency strength order of the large cardinals hierarchy.

I wonder whether the direct implication power of NMSC is as strong as its large cardinal strength or not.

Question 1. Does NMSC directly imply the existence of any large cardinals, say those of some topological nature such as weakly compacts? To be more precise, let me add that I look for the theorems of the form NMSC $\Rightarrow \exists \kappa$ Large (i.e. at least strongly inaccessible) if there is any.

Update. According to Will's answer, it turned out that NMSC doesn't contain any direct large cardinal strength.

In other direction, one may strengthen NMSC as follows:

Complete Normal Moore Space Conjecture (CNMSC). All normal Moore spaces are completely metrizable.

Here is my second question:

Question 2. Is CNMSC consistent? What are upper and lower bounds for its large cardinal strength? Strongly compacts and measurables or more?


  1. Jones, F. Burton, Concerning normal and completely normal spaces, Bull. Am. Math. Soc. 43, 671-677 (1937). ZBL0017.42902.
  2. Nyikos, Peter J., A provisional solution to the normal Moore space problem, Proc. Am. Math. Soc. 78, 429-435 (1980). ZBL0446.54030.
  3. Dow, Alan; Tall, Franklin D.; Weiss, William A. R., New proofs of the consistency of the normal Moore space conjecture. I & II, Topology Appl. 37, No. 1, 33-51 (1990). ZBL0719.54038.
  4. Fleissner, William G., If all normal Moore spaces are metrizable, then there is an inner model with a measurable cardinal, Trans. Am. Math. Soc. 273, 365-373 (1982). ZBL0498.54025.
  • $\begingroup$ Your notion of direct implication does not seem to make sense. You should probably define it precisely. $\endgroup$ Commented Jul 20, 2018 at 10:42
  • $\begingroup$ @AndrésE.Caicedo Could you please explain the issue? I mean, if there is any theorem of the form $NMSC\Rightarrow \exists \kappa$ large. Is it ok now? $\endgroup$ Commented Jul 20, 2018 at 10:45
  • $\begingroup$ What you wrote about Woodin cardinals is nonsense. $\endgroup$ Commented Jul 20, 2018 at 10:46
  • $\begingroup$ @AndrésE.Caicedo Hmmm... So there is no problem concerning the meaning of the question. Right? Let me add that by the weakness of Woodin cardinals in direct implication order I am pointing out to the facts such as "the first Woodin cardinal is not even weakly compact." Shall I change my phrasing to mention this more explicitly? I thought it is pretty obvious. $\endgroup$ Commented Jul 20, 2018 at 10:51
  • $\begingroup$ No, it is not pretty obvious, because your notion of "direct implication" makes no sense. Sure, change the phrasing to something meaningful. $\endgroup$ Commented Jul 20, 2018 at 10:54

1 Answer 1


If $\text{NMSC}$ is consistent, then so is $\text{NMSC}+\text{"there are no strongly inaccessible cardinals"}$.

This is because if $V \models \text{NMSC}$, then $V_\kappa \models \text{NMSC}$ for any inaccessible cardinal $\kappa$. (A proof sketch is given below.) In particular, if $\kappa$ is the first strongly inaccessible cardinal, then $V_\kappa \models \text{NMSC}+\text{"there are no strongly inaccessible cardinals"}$.

[If you think through the definition of a Moore space, it's not too hard to see that if $(X,\tau)$ is some topological space in $V_\alpha$, then the statement "$(X,\tau)$ is a normal Moore space" can be decided by looking at $V_{\alpha+\omega}$. Likewise, the fact that $(X,\tau)$ is metrizable can be decided by looking at $V_{\alpha+\omega}$. Thus the truth/falsity of the NMSC for $(X,\tau)$ is the same in $V$ and in $V_\kappa$. Because this is so for every topological space, the truth/falsity of the NMSC is the same in $V$ and in $V_\kappa$.]

  • $\begingroup$ (+1) Interesting observation! Thank you, Will! Any thoughts on the second question? $\endgroup$ Commented Jul 20, 2018 at 15:23
  • $\begingroup$ The answer to the second question is negative -- all metrizable spaces are normal Moore spaces, but not all metrizable spaces are completely metrizable. (The Wikipedia article I linked to in my answer mentions that all metrizable spaces are normal Moore spaces, and explains why.) $\endgroup$
    – Will Brian
    Commented Jul 20, 2018 at 15:51
  • $\begingroup$ Oh I see! I glimpsed the Wiki article but somehow missed this very point! Thanks for reminding, anyway! $\endgroup$ Commented Jul 20, 2018 at 15:58

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