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There are some statements, whose consistency (or the consistency of their negation) require the existence of large cardinals (in the sense that if the statement (or its negation) is consistent, then it is consistent that there are some large cardinals). I know many of such examples inside set theory.

Question. What are examples of statements in mathematics (other than set theory), where it is known we need large cardinals to prove their consistency (or the consistency of their negation)?

For each example, giving what kind of large cardinals are sufficient and what kind of large cardinals are necessary is appreciated. Please also give references.

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    $\begingroup$ It might be hard to distinguish "set theory" and "mathematics other than set theory". Also I guess "require the existence of large cardinals" is rather "require the consistency of the existence of large cardinals". $\endgroup$
    – YCor
    Commented Jan 15, 2020 at 6:04
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    $\begingroup$ Accessible categories come to mind. $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 15, 2020 at 7:07
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    $\begingroup$ Near-duplicate: mathoverflow.net/q/129575/30186 (it has narrower scope but its answers also answer this question). The short answer is: Harvey Friedman's work $\endgroup$
    – Wojowu
    Commented Jan 15, 2020 at 10:19
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    $\begingroup$ What about Solovays model where all sets in $\mathbb{R}$ are Lebesgue measurable. It is not only set theory. See: Jech (1978), Set Theory, p. 537 ff. Must also be in the 3rd edition. $\endgroup$
    – Dieter Kadelka
    Commented Jan 15, 2020 at 12:04
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    $\begingroup$ The normal Moore space conjecture in general topology is a classical example of such a statement. Its consistency has been proven from the consistency of a strongly compact cardinal (Nyikos) and implies an inner model with a measurable cardinal (Fleissner). But there are many more examples...maybe this question should be community wiki. $\endgroup$
    – Santi Spadaro
    Commented Jan 15, 2020 at 13:04

3 Answers 3

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The dual of an abelian group $A$ is defined to be the group $\text{Hom}(A,\mathbb Z)$ of homomorphisms to the infinite cyclic group. As usual with such dualities, there's a canonical homomorphism from $A$ to its double dual $$ A\to A^{**}:a\mapsto(h\mapsto h(a)). $$ If this is an isomorphism, $A$ is said to be reflexive.

Question: Are all free abelian groups reflexive?

Answer: Yes if and only if there are no measurable cardinals.

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    $\begingroup$ This is amazing! Where is this proved? $\endgroup$
    – Timothy Chow
    Commented Jan 17, 2020 at 3:53
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    $\begingroup$ @TimothyChow I'm sure it's in the book "Almost Free Modules" by Eklof and Mekler, but I can't check immediately because I'm away from home. The result is, if I remember correctly, due to Los, and it says in more detail that the free abelian group on a set $S$ of generators is reflexive iff there is no measurable cardinal $\leq|S|$. $\endgroup$
    – Andreas Blass
    Commented Jan 17, 2020 at 6:01
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    $\begingroup$ It's the same situation with "All discrete spaces are realcompact", probably for the same reasons! $\endgroup$
    – Todd Eisworth
    Commented Jan 17, 2020 at 23:14
  • $\begingroup$ Another related fact with a similar proof is that a product of $\kappa$ bornological locally convex spaces is bornological iff $\kappa$ is less than the first measurable cardinal. A locally convex space $E$ is bornological iff for all locally convex spaces $F$ and bounded linear maps $f : E \rightarrow F$, $f$ is continuous. The proof goes by reducing to products of $\mathbb{R}$ (see Schaefer's Topological Vector Spaces, Chapter II, Exercise 19). The statement in II.8 only mentions inaccessible cardinals, I think because Hanf's work was very recent when it was published. $\endgroup$
    – Robert Furber
    Commented Jan 18, 2020 at 11:19
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    $\begingroup$ This is the shadow of the following more precise statement: the free abelian group $\mathbf{Z}^{(I)}$ is non-reflexive iff $I$ carries a $\{0,1\}$-valued probability defined an all subsets and vanishing on singletons. (This means $I\ge\kappa_0$ for $\kappa_0$ the first measurable cardinal, and that $\kappa_0$ exists.) So this is a result where the set $I$ is part of the input. $\endgroup$
    – YCor
    Commented Jan 18, 2020 at 16:53
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Recall that the character of a point in a topological space is the smallest cardinality of a local base for that point. (So, for example, "first countable" = "every point has character $\aleph_0$".)

Question: Is there a compact Hausdorff space, containing more than one point, in which any two different points have different character?

Steve Watson found an answer to this question [S. Watson, "Using prediction principles to construct ordered continua," Pacific Journal of Mathematics 186, pp. 251-256 (link)] by showing that

Theorem: There is such a space if and only if there is a cardinal $\kappa$ and a set of cardinals $E \subseteq \kappa$ such that $\diamondsuit_\kappa(E)$ holds.

Watson points out in his paper that this condition is implied by $V=L$ plus the existence of a Mahlo cardinal, and it implies the existence of a weakly inaccessible cardinal. Thus

the consistency strength of a positive answer to the above question lies somewhere between an inaccessible cardinal and a Mahlo cardinal.

I do not know whether better bounds for this statement about $\diamondsuit_\kappa(E)$ have been found in the last 34 years. If you know of any improvements to the bounds Watson gives, please feel free to edit this post.

Finally, Watson shows from ZFC alone that there is an infinite $\sigma$-compact Hausdorff space in which any two different points have different character. Thus removing "compact" from the question makes it much easier, and removes the need for any large cardinals.

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It was a traditional question of descriptive set theory (a question which can be formulated in the language of second order arithmetic) whether all projective sets are Lebesgue measurable. This remained an open problem for many decades, and for a good reason: it turned out that the statement is independent even of the full ZFC set theory (see Solovay 1970). Only by postulating the existence of some extremely large cardinals (so-called Woodin cardinals) can the hypothesis that all projective sets are Lebesgue measurable be proved (this was achieved as a consequence of their work on so-called projective determinacy by Woodin, Martin and Steel; see Woodin 1988; Martin & Steel 1988, 1989).

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    $\begingroup$ I didn't downvote, but I suspect that this may not meet the "outside set theory" criterion. $\endgroup$
    – Noah Schweber
    Commented Jan 16, 2020 at 18:22
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    $\begingroup$ "Descriptive set theory" is a field of mathematics related to topology and was initiated by the French semi-intuitionists (Lebesgue, Baire, Borel). It studies sets which possess relatively simple definitions (in contradistinction to the ideas of arbitrary sets and various higher power-sets, which the semi-intuitionists rejected as meaningless) called projective or analytic sets. It is common to count it, in the literature on the foundations of mathematics, the name notwithstanding, as "ordinary mathematics", in contradistinction to set theory. $\endgroup$
    – Panu Raatikainen
    Commented Jan 18, 2020 at 7:03

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