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The Singular Cardinal Hypothesis (SCH) is the statement that $\kappa^{cf(\kappa)} = \kappa^+ \cdot 2^{cf(\kappa)}$ for every singular cardinal $\kappa$ (or various equivalent statements).

It is obviously implied by the Generalized Continuum Hypothesis. It is also implied by the Proper Forcing Axiom (under which $2^{\aleph_0} = \aleph_2$). Nonetheless it doesn't seem terribly compelling to me. But I am trying to learn to appreciate it!

Why should I believe SCH? (Now when I say "believe," it isn't clear exactly what I mean by that. There are obviously plenty of models of set theory in which SCH holds, and they are certainly worth studying, but somehow they aren't the models that feel most "realistic" in my little head.) What I want is an answer in the style of Maddy's Believing the axioms, explaining why I should "like" (or not) this hypothesis.

Some thoughts on why SCH isn't so unreasonable:

  • Somehow I find very compelling the result that SCH holds above the first strongly compact cardinal. This is what makes it seem most reasonable to me that SCH should hold everywhere.

  • SCH is implied by various contradictory axioms. Its negation is equiconsistent with the existence of a fairly large cardinal.

  • It is obvious that $\kappa^{cf(\kappa)} \geq \kappa^+ \cdot 2^{cf(\kappa)}$, so SCH is just saying that the cardinality on the left shouldn't be any larger than strictly necessary.

  • It makes cardinal arithmetic much easier! (But maybe this is an argument against SCH too...)

Why else should SCH seem reasonable?

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Did anyone say you should believe SCH? Or that it "should ... seem reasonable"? – Andreas Blass Jul 1 '11 at 20:37
@Andreas Not explicitly, but I always got the impression that other people liked SCH and I never understood why. Maybe I was mistaken. Do you think it "seems reasonable"? Do you know people who do? – Oliver Jul 1 '11 at 22:14
@Oliver: I don't see any particular reason for believing SCH. Of course, if Hugh Woodin eventually convinces us all to believe GCH, then that takes care of SCH. But I don't recall anyone specifically arguing for plausibility of SCH separately from GCH. Let's see what other comments or answers you get. – Andreas Blass Jul 1 '11 at 22:23
@Oliver, @Andreas: I think that SCH is a reasonable statement (as well as the combinatorics that establish it from reflection principles), actually, and part of an argument for (consequences of) forcing axioms could use this as a case study. – Andrés E. Caicedo Jul 1 '11 at 23:03
(As usual I am a bit short of time, but I'll see if I can write something down as an answer.) – Andrés E. Caicedo Jul 1 '11 at 23:06

Solovay proved in ZFC that all instances of the SCH hold above any strongly compact cardinal. Thus, you should believe the eventual SCH to at least the extent that you believe in the very large large cardinals, which I believe are discussed by Maddy in the context of believing in the axioms.

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As I said in the question, I agree that this is a very compelling piece of evidence in favor of SCH. Is it known if "strongly compact" is best possible here? – Oliver Jul 2 '11 at 4:55
One way to interpret your "best possible" question is to ask merely whether it is consistent with the various smaller large cardinals that the eventual SCH does not hold. For this, it seems that with sufficiently many supercompact cardinals, one could force unboundedly many failures of the SCH, while still retaining a partially supercompact cardinal to any desired degree of supercompactness (by forcing the failures only above that degree). And one can similarly violate the SCH above a strong cardinal, given suitable large cardinals above. – Joel David Hamkins Jul 2 '11 at 10:01
@Oliver: Strong compactness can be weakened a little bit: If the singular cardinals hypothesis fails at a singular $\mu$ of countable cofinality, then there can be no uniform countable complete ultrafilter on $\mu^+$. This isn't possible above a strongly compact cardinal $\kappa$, as the compactness of $\kappa$ would let you extend the co-bounded filter on $\mu^+$ to a $\kappa$-complete ultrafilter. This is overkill, as all we need is a countably complete uniform ultrafilter on $\mu^+$ for the contradiction. – Todd Eisworth Oct 3 '11 at 21:11

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