# Why should I believe the Singular Cardinal Hypothesis?

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?

• Did anyone say you should believe SCH? Or that it "should ... seem reasonable"? Jul 1, 2011 at 20:37
• @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. Jul 1, 2011 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. Jul 1, 2011 at 23:03
• I second Oliver's request, even though I'm a few months late... Oct 3, 2011 at 20:53
• @Andrés: Six and a half years later... :) Jan 17, 2018 at 11:02

• @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. Oct 3, 2011 at 21:11
$SCH$ follows from $MM$ (Foreman-Magidor-Shelah), in fact from $PFA$ (Viale). One reason why one might not want to believe $MM$ is that there is another maximality principle'' which also implies $2^{\aleph_0}=\aleph_2$, namely $(*)$, and the logical relation between $MM$ and $(*)$ is somewhat of a mystery. I proposed an axiom, though, called $MM^{*,++}$, which implies both $MM$ and $(*)$, see my "Woodin's axiom (*), or Martin's Maximum, or both?", in: Foundations of mathematics, essays in honor of W. Hugh Woodin's 60th birthday, Harvard University (Caicedo et al., eds.), pp. 177-204, available at https://ivv5hpp.uni-muenster.de/u/rds/paper_for_woodin_volume_new_version.pdf . It's a really cool axiom to believe in! The mathematical challenge is to prove the consistency of $MM^{*,++}$ relative to anything which is believed to be consistent. In short, you should believe $SCH$ because you should believe $MM^{*,++}$.