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June 6

Why does the generalized continuum hypothesis implies the following equality for cardinal exponentiation?

The article about the continuum hypothesis states that the the generalized continuum hypothesis implies the following:

$\aleph _{\alpha }^{\aleph _{\beta }}=\aleph _{\alpha }$ when β+1 < α and $\aleph _{\beta }<\operatorname {cf} (\aleph _{\alpha })$, where cf is the cofinality operation.

How does one prove that?

Thanks!

Dan Gluck (talk) 12:59, 6 June 2020 (UTC)

See Talk:Continuum hypothesis#Proofs of certain consequences of GCH where I justify the claims in the article. Here I extract the relevant portion:

$\aleph _{\alpha }=\aleph _{\alpha }^{1}\leq \aleph _{\alpha }^{\aleph _{\beta }}\,.$

Since we are in the case that $\aleph _{\beta }<\operatorname {cf} (\aleph _{\alpha })$ where cf is the cofinality operation, then any function from $\aleph _{\beta }\,$ to $\aleph _{\alpha }\,$ must be bounded above by some $\gamma <\aleph _{\alpha }\,.$

And γ has a cardinality $\vert \gamma \vert =\aleph _{\delta }\,$ where δ < α. :The cardinality of the set of functions so bounded by γ is

$\aleph _{\delta }^{\aleph _{\beta }}\leq 2^{(\aleph _{\delta }\cdot \aleph _{\beta })}=\aleph _{\max(\delta ,\beta )+1}\leq \aleph _{\alpha }\,.$

Adding these together for the $\aleph _{\alpha }\,$ possible values of γ gives

$\aleph _{\alpha }^{\aleph _{\beta }}\leq \aleph _{\alpha }\cdot \aleph _{\alpha }=\aleph _{\alpha }\,$

which means

$\aleph _{\alpha }^{\aleph _{\beta }}=\aleph _{\alpha }\,.$

OK? JRSpriggs (talk) 04:42, 7 June 2020 (UTC)

June 11

The deduction theorem in provability logic

Does the deduction theorem holds in GL logic? Specifically, when Necessitation is defined as it is here, in that it can only be applied with no assumptions before the $\vdash$ (so from $\{\phi \}\vdash \phi$ one may not deduce $\{\phi \}\vdash \square \phi$). The article indicates the answer might be yes, but I have been told by someone whom I am inclined to believe that the answer is no. For some context, mean that GL logic is in some way the same as talking about provability in PA, which (I believe) does have the deduction theorem. — crh23 (Talk) 08:52, 11 June 2020 (UTC)

(I hope the following is not foot-in-mouth, as I am straying from the areas I'm comfortable with.) For the deduction theorem to be applicable, you need a logic that allows assumptions to the left of the turnstyle. Starting with the assumption $p$, unless we have a perverse logic, it should then be the case that

$p\vdash p.$

From the necessitation rule of GL logic given in the Provability logic article, we obtain

$p\vdash \Box p.$

So, if the deduction theorem holds, we obtain now

$\vdash p\rightarrow \Box p.$

Is this a theorem of GL logic? If not, the deduction theorem does not hold. I see no rule for condition introduction other than the distribution axiom and Löb's axiom, and either one requires a box in the antecedent of the conditional, so if I take the definition of GL logic in the Provability logic article for gospel, this cannot be a theorem.  --Lambiam 11:13, 11 June 2020 (UTC)

I think the normal consensus is that $\vdash p\rightarrow \Box p$ is not a theorem in GL logic. However, the necessitation rule does not allow the deduction you state, since it may only be used when there is no assumption to the left of the $\vdash$: the rule is specifically written

• Necessitation: From $\vdash$ p conclude $\vdash$ □p

rather than

• Necessitation: From p conclude □p

(which is how the rule for modus ponens is denoted) --- see section 2 of .— crh23 (Talk) 13:39, 11 June 2020 (UTC)

If $p\rightarrow \Box p$ means what it looks like, it can't be right, because of incompleteness. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 20:49, 11 June 2020 (UTC)

To confirm the argument here: the (proof of the) second incompleteness theorem states that if $G$ is such that $\vdash G\leftrightarrow \neg \square G$ then for any $X$ we have $\square G\rightarrow \square X$. As such, if $\vdash G\rightarrow \square G$ we get $\vdash \neg G$ so $\vdash \square X$, i.e. everything is provable. Is that the full argument? — crh23 (Talk) 21:38, 11 June 2020 (UTC)

The incompleteness theorems do not apply to all logics. For example, propositional calculus and first-order predicate logic are (semantically) complete. For the proof of the incompleteness results tto go through you need to be able to express natural number arithmetic in the logic, something you can’t do here.  --Lambiam 21:45, 11 June 2020 (UTC)

Indeed, but does arithmetic soundness and completeness (which I used very carelessly) not show that it is applicable here? — crh23 (Talk) 08:14, 12 June 2020 (UTC)

In fact I am pretty sure it is, see this proof (sorry, I couldn't be bothered to turn it into wikitext (I'm much faster at LaTeX)) — crh23 (Talk) 09:42, 12 June 2020 (UTC)