MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does an existence of large cardinals have implications in more down-to-earth fields like number theory, finite combinatorics, graph theory, Ramsey theory or computability theory? Are there any interesting theorems in these areas that can be proved based on assumptions of existence of certain large cardinals? Or they are too far to change anything here?

I know that existence of certain cardinals implies consistency of some axiomatic systems of the set theory, which in turn can be expressed as a statement about certain huge Diophantine equations having no solutions, but it looks unlikely that we could use these Diophantine equations for anything else.

share|cite|improve this question
up vote 20 down vote accepted

Harvey Friedman has recently produced some results in this area. See for example Friedman, Invariant Maximal Cliques and Incompleteness, 2011. There is also a draft of a text book titled Boolean Relation Theory and Incompleteness at also by Harvey Friedman which is apparently also in this area.

In the paper it seems that Friedman has produced a graph theoretic theorem which is somewhat natural and requires a certain large cardinal axiom to prove.

Unfortunately I'm not particularly familiar with either of these works, so I'm not able to give a better explanation (although maybe another poster will be able to put a good explanation in their answer).

share|cite|improve this answer
My favorite example of Friedman's is one that I mentioned in my answer to a related MO question… . It appears at the end of Martin Davis's Notices article: – Timothy Chow May 4 '13 at 0:46
Regarding an "explanation," see Friedman's own succinct summary of the general methodology for coming up with these statements, which I reproduced here:… – Timothy Chow May 4 '13 at 0:52

There's an extremely elementary theorem whose only known proof relies on the existence of a rank-into-rank cardinal (basically the strongest large cardinal axiom not known to contradict ZFC).

Let $R_n = \mathbb{Z}/2^n\mathbb{Z}$.The $n$th Laver table is the unique binary operator $\star : R_n \times R_n \rightarrow R_n$ determined by the following conditions:

  • $p \star 1 \equiv p + 1$
  • $p \star (q \star r) \equiv (p \star q) \star (p \star r)$

Then the function $f_n(q) = 1 \star q$ is obviously periodic with some period $P(n)$ dividing $2^n$.

The existence of a rank-into-rank cardinal implies that $P(n)$ grows without bound.

Reference: Laver, Richard (1995), "On the algebra of elementary embeddings of a rank into itself"

share|cite|improve this answer
A recent short article: – Todd Trimble Mar 27 '15 at 15:09

Applications of Friedman's Jump-free theorem (the proof of which requires large cardinals) to distance functions in lattice graphs is on his website:

It is item 11: Applications of Large Cardinals to Graph Theory, October 23, 1997.

This remarkable paper by Friedman arose when a homework problem presented in an upper division undergraduate course at UCSD-CSE (solvable by a version of Ramsey's theorem) was extended slightly and no one could solve it (including the instructor and other faculty members). In January 1997, Friedman gave a lecture at UCSD where he presented his jump-free theorem. It was just what we needed to solve the homework problem!

Another extension of this homework problem has been shown by Friedman (in the above paper) to itself require large cardinals and is equivalent to the jump-free theorem.

These problems can be viewed as a "toy versions" of the string theory landscape or multiverse concept.

A fanciful presentation for high school level students showing how these simply stated but hard problems arise is at

share|cite|improve this answer
These problems can be viewed as a "toy versions" of the string theory landscape or multiverse concept <- what is this I don't even – David Roberts Jun 2 '15 at 23:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.