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In Awfully sophisticated proof for simple facts, we are asked for examples of complex proofs of simple results. To quote from the questioner's post, we are asked for proofs that are akin to "nuking mosquitos." In set theory, a natural "nuke" with respect to a certain result is a large cardinal axiom with unnecessarily high consistency strength (i.e. applying to a much stronger collection of axioms than is required to provide a proof of the possibility of the result in question).

A research focus in set theory is a search for large cardinal axioms with the weakest consistency strength that can be used to prove the possibility of a certain result. My question is of an opposing nature:

Can you think of results that can be proven in a different manner by appealing to a large cardinal axiom with unnecessarily large consistency strength?

There are plenty such examples where the proofs become less technical (e.g., using a $\kappa^{++}$-supercompact cardinal $\kappa$ to show that the GCH can fail at a measurable cardinal is much more than is required), but I'm thinking of examples where the original proof was accomplished without such a strong large cardinal hypothesis or any large cardinal hypothesis at all. For example (from my post to the aforementioned question):

Theorem (ZFC + "There exists a supercompact cardinal."): There is no largest cardinal.

Proof: Let $\kappa$ be a supercompact cardinal, and suppose that there were a largest cardinal $\lambda$. Since $\kappa$ is a cardinal, $\lambda \geq \kappa$. By the $\lambda$-supercompactness of $\kappa$, let $j: V \rightarrow M$ be an elementary embedding into an inner model $M$ with critical point $\kappa$ such that $M^{\lambda} \subseteq M$ and $j(\kappa) > \lambda$. By elementarity, $M$ thinks that $j(\lambda) \geq j(\kappa) > \lambda$ is a cardinal. Then since $\lambda$ is the largest cardinal, $j(\lambda)$ must have size $\lambda$ in $V$. But then since $M$ is closed under $\lambda$ sequences, it also thinks that $j(\lambda)$ has size $\lambda$. This contradicts the fact that $M$ thinks that $j(\lambda)$, which is strictly greater than $\lambda$, is a cardinal.

For the people who are unfamiliar with large cardinal embeddings, let me mention that the critical point of an embedding $j$ is the first ordinal $\kappa$ that is moved (i.e., $j(\alpha) = \alpha$ for all $\alpha$ less than the critical point $\kappa$ and $j(\kappa) > \kappa$.) A cardinal $\kappa$ is $\theta$-supercompact if there exists an elementary embedding $j: V \rightarrow M$ into a transitive (proper class) $M$ with critical point $\kappa$ such that $M^{\theta} \subseteq M$ and $j(\kappa) > \theta$. A cardinal is supercompact if it is $\theta$-supercompact for all $\theta$.

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    $\begingroup$ The use of inaccessible cardinals to prove the existence of derived functor cohomology in SGA surely counts? MacLarty has shown finite-order arithmetic suffices. $\endgroup$
    – David Roberts
    Feb 19, 2018 at 6:16

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This may not be the sort of thing you had in mind, but here goes anyway: The easiest way to prove Borel determinacy (which is a theorem of ZFC) is to assume there's a measurable cardinal and prove analytic determinacy. (Both results are due to Tony Martin. The proof of analytic determinacy from a measurable cardinal came well before the proof of Borel determinacy in ZFC. The exact consistency strength of analytic determinacy is the existence of sharps of all reals.)

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  • $\begingroup$ +1: Regardless of the order, it is always interesting to find out how a large cardinal axiom was used to simplify the proof of an important ZFC theorem. Thank you. $\endgroup$
    – Jason
    Dec 16, 2010 at 5:42
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    $\begingroup$ @Andreas: This is also the best example I could think of. $\endgroup$ Mar 14, 2011 at 23:31
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I think this example given at Richard Borcherds's blog would qualify, no?

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    $\begingroup$ @Jason: Note that Laver's result was first proved using large cardinals. At that time, no ZFC proof was known. But once you gain some confidence that something is true, it usually becomes easier to actually prove it. I am not sure that Laver actually tried to proof his result and then had the idea to use large cardinals. I would guess it was the other way around, namely that he observed some structure that you get from certain large cardinals, and that this structure actually solves a problem about left-distributive algebras. $\endgroup$ Dec 10, 2010 at 7:19
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    $\begingroup$ @Stefan: I knew that Laver's result preceded the theorem of Dehornoy. My point was that our confidence that a statement is probably relatively consistent with ZFC because we can prove it from one of the strongest known large cardinal notions (not known to be inconsistent with ZFC) might be running a little too high. My concern is what becomes of our research assuming the existence of strong large cardinal notions if it all comes crashing down one day. $\endgroup$
    – Jason
    Dec 16, 2010 at 5:25
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    $\begingroup$ @Jason:I completely disagree with this. I think this is a case where the large cardinals led us to discover a proof (and an algebraic structure) that may not have been discovered or appreciated otherwise. We have about 50 years of experience telling us that a rank to rank is a safe hypothesis. Maybe double that for ZFC. And probably the last 50 years is when we really started to deepen our understanding of CON(ZFC), largely as a consequence of studying large cardinals. So why do we view them with such skepticism? To me, Laver's result is one of the most remarkable of modern mathematics. $\endgroup$ Mar 14, 2011 at 23:30
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    $\begingroup$ The author has deleted the blog post, so it no longer exists. $\endgroup$ Aug 28, 2013 at 0:58
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    $\begingroup$ I am supposing that the blog post was referring to the result of Dehornoy exposed here: ams.org/journals/tran/1994-345-01/S0002-9947-1994-1214782-4/… $\endgroup$
    – Todd Trimble
    Dec 1, 2016 at 20:25
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There is a fantastic (and not too well-known) result of Shelah stating that $L({\mathcal P}(\lambda))$ is a model of choice whenever $\lambda$ is a singular strong limit of uncountable cofinality.

This is a consequence of a more general theorem that can be found in 4.6/6.7 of "Set Theory without choice: not everything on cofinality is possible", Archive for Math Logic 36 (1997) 81-125.

(Understanding this argument is in my "immediate" to-do list. Alas, the list is longer each day.)

Woodin has a nice, short argument when the cofinality of $\lambda$ is a Woodin cardinal, using stationary tower techniques. (I do not think Woodin's argument is published anywhere, though.) It certainly gives you an idea that the result is plausible, and that an analysis of ideals seems to be relevant.

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  • $\begingroup$ +1: This is more of the type of result I had in mind: a proof that proceeded the original one using stronger assumptions. I think this pattern is typical: Shelah comes up with an amazing general result, and his work is then clarified by considering a weaker one. $\endgroup$
    – Jason
    Dec 10, 2010 at 1:36
  • $\begingroup$ I read this before as Shelah's result coming first, but I now realize that you didn't write this. Out of curiosity, do you know which one came first? $\endgroup$
    – Jason
    Dec 10, 2010 at 2:18
  • $\begingroup$ @Jason: Shelah's result appeared first (the paper is dated 1997). I am not sure how Woodin heard of it, but he mentions it in his recent manuscript on "suitable extender models". I believe his argument is fairly recent (he showed it to me in October). $\endgroup$ Dec 10, 2010 at 2:21
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    $\begingroup$ @Andres: OK, thanks. I think this may be another trend: Woodin has so many nice results that he doesn't get a chance to publish them all. $\endgroup$
    – Jason
    Dec 10, 2010 at 2:36
  • $\begingroup$ This result is important because it implies a form of weak covering: $∃S⊂Ord \, (|S|=λ) \, (λ^+)^{HOD[S]} = λ^+$ (and even $∃S⊂λ \, P(λ)∈HOD[S]$). For all we know, such weak covering might fail for all other infinite $λ$. Also, the axiom of choice fails for $L(\mathcal{P}(λ))$ in $\mathrm{Coll}(λ, <\mathrm{inaccessible})$ (hence the condition that $λ$ is singular), or given an $\mathrm{I}_0$ embedding $j$, for $λ=j^ω(\mathrm{crit}(j))$ (hence the condition $\mathrm{cf}(λ)>ω$; much weaker hypotheses suffice for consistency of a counterexample for $\mathrm{cf}(λ)=ω$). $\endgroup$ Dec 30, 2018 at 3:26
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This may not be quite the sort of thing you were looking for, but some of Harvey Friedman's examples of $\Pi^0_1$ statements unprovable in ZFC but provable using large cardinals can be used to produce $\Pi_0^0$ statements (i.e., statements whose truth can be verified with a finite computation) that have large-cardinal proofs that require at most (say) a million symbols to write down, but which have no ZFC-proofs less than 101000 symbols long. See this post from the Foundations of Mathematics mailing list for example.

Since $\Pi_0^0$ statements are finitely checkable, they are in principle provable using ridiculously weak axioms (assuming that they are true). In other words, strong axioms are certainly nonessential. However, large cardinal axioms are required if you want a proof that can actually be written down in practice.

These examples are a little peculiar because if you don't believe large cardinal axioms then you may not believe these $\Pi_0^0$ statements—and yet they could in principle be directly checked if you just had enough computational power…

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  • $\begingroup$ Just came across this, but it is not at all convincing. Let Q be a sentence over ZFC such that ZFC ⊢ ( Q ⇔ ( no proof of "Q" over ZFC has length less than 2^(2^1000) ) ), which exists by the fixed-point lemma. Then ZFC+Con(ZFC) proves Q with proof length less than 1 million, but ZFC (if consistent) cannot prove Q with proof length less than 2^(2^1000). It has nothing to do with large cardinals. This example also show that there is no reason to believe truth of Σ0-sentences derived from large cardinal axioms. What if I call ¬Q a large cardinal axiom? $\endgroup$
    – user21820
    Dec 17, 2021 at 7:11
  • $\begingroup$ To be clear, what I meant to say by "no reason to believe" is "no reason to believe just because it is in principle directly checkable", because it is simply false to claim that it can be directly checkable. For instance, you simply cannot directly check and verify Q, nor can you check and refute ¬Q. $\endgroup$
    – user21820
    Dec 17, 2021 at 8:16
  • $\begingroup$ @user21820 In your example, the blowup is artificially baked in to your statement. The blowup in the large cardinal case isn't artificially baked in; it's a "combinatorial" consequence of the large cardinal axiom. Also, I'm not claiming that the finite checkability is a reason to believe the statements. Believing them is a matter of whether the large cardinal axioms (or at least their 1-consistency) themselves seem believable. $\endgroup$ Dec 17, 2021 at 18:10
  • $\begingroup$ Okay I get what you're saying about "artificially baked", and given your clarification about believing the statements I am now convinced that your example here is a good one for those who believe Con(ZFC+MAH). Thanks! $\endgroup$
    – user21820
    Dec 18, 2021 at 4:14
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I find it hard to believe that "Every set of reals has the Baire Property" was not mentioned yet.

Solovay proved that if we collapse an inaccessible cardinals to be $\omega_1$, then in $V(\Bbb R)$ every set of reals is Lebesgue measurable and has the Baire Property and $\sf DC$ holds.

Shelah later proved that the inaccessible is necessary for the Lebesgue measurability, but it is not necessary for the Baire Property. This proof is different and much more technical than that of Solovay (which is arguably not very difficult once you have a few theorems about forcing under your belt).

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    $\begingroup$ +1. Clarifying for the OP: the model Asaf describes might be more clearly expressed as "$V(\mathbb{R}^{V[G]})$." That is, we let $G$ be appropriately generic and look at the model generated by all the old sets and the new reals (but not, say, the new sets of reals etc.). $\endgroup$ Feb 18, 2018 at 21:38
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When dealing with the singular cardinals hypothesis ($SCH$), one may face with many such examples, let me say a few:

$\star_1:$ The consistency of the failure of $SCH$ was proved first by silver using supercompact cardinals. Later, Woodin reduced it to large cardinals up to strong cardinals, and finally Gitik showed that a measurable cardinal with $o(\kappa)=\kappa^{++}$ is suffices (which is also necessary).

$\star_2:$ Magidor first proved the consistency of $GCH$ below $\aleph_\omega$ with $2^{\aleph_\omega}=\aleph_{\omega+2}$ from a supercompact cardinal and a huge cardinal above it. Later Woodin reduced it to the level of strong cardinals, and finally it turned out that a measurable cardinal with $o(\kappa)=\kappa^{++}$ is suffices.

$\star_3:$ Foreman and Woodin proved the consistency of the total failure of $GCH$ from a supercompact cardinal and infinitely many inaccessibles above it. Later, it turned out that a $(\kappa+3)$-strong cardinal is suffices (and even less is needed).

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