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Results for believe in "V=L"
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110 votes
10 answers
15k views

Analogues of P vs. NP in the history of mathematics

The question is this: When, in the history of mathematics, have problems "like P vs. NP" arisen and then been solved? In those cases, what were the resolutions? … Until Cohen, there were many proven statements about transfinite sets, and then a whole class of other statements---V=L, GCH, CH, AC, Zorn's Lemma, well-orderability... …
54 votes

Arguments against large cardinals

[Reinhardt suggested an ultimate axiom of the form "there is a non-trivial elementary embedding $j:V\to V$". … For some of them, set theory is about all possible (natural) extensions of $\mathsf{ZFC}$, and there are certainly many interesting such extensions (such as $V=L$) that rule out large cardinals. …
Andrés E. Caicedo's user avatar
49 votes

Why not adopt the constructibility axiom $V=L$?

V=L isn't alone in this ...). … V=L, by contrast, is genuinely complicated. …
44 votes

Jean Bourgain's relatively lesser known significant contributions

The special case when d=1 was done in 1870 by Georg Cantor... V. L. … operator is bounded on $L^2$. …
40 votes

Arguments against large cardinals

In the second place, even a Platonist might believe the consistency of some axioms for reasons other than a belief in their truth. … On the other hand, I've never been able to persuade myself of the actual existence of even subtle cardinals (which are so small that they're consistent with $V=L$). …
Andreas Blass's user avatar
28 votes
Accepted

Are proper classes objects?

(For example, $V=L$ abbreviates "all sets are constructible.") … In particular, when I talk about statements being true in $V$, I mean simply that the quantified variables are to be interpreted as ranging over arbitrary sets. …
Andreas Blass's user avatar
28 votes
0 answers
2k views

How did Gauss find the units of the cubic field $\mathbb Q[n^{1/3}]$?

=v, t''= \frac{t'v'}{l'}, t''' = \frac{t''v''}{l''}, etc.$$ He also uses the notation $Div$, which I believe is an abbreviation for "Divisor", since the arithmetical procedure he describes involves the … Secondly, the power of $z$ written there is actually 4; I believe this is just a typo of Gauss. Thirdly, this cubic form has $v=|2^{\frac{1}{3}}|$, and I believe $\epsilon$ is a third root of unity. …
user2554's user avatar
  • 2,099
28 votes
3 answers
3k views

What sorts of extra axioms might we add to ZFC to compute higher Busy Beaver numbers?

A person who believes only in the constructible universe might reject the same ordinal collapsing functions if cardinals incompatible with $V=L$ are used. … So in a way, this "real world" value changes depending on our philosophical stance. …
exfret's user avatar
  • 509
27 votes
Accepted

Are there any good nonconstructive "existential metatheorems"?

In particular, to prove that a particular $\Sigma^1_2$ statement is true in ZFC, it suffices to prove it under the assumption also that V=L, where one also has all kinds of additional structure available … To prove that there is a proof, is a proof, so I believe ultimately there will be no way to avoid the quibbling over whether it is easy or hard to translate the high-level proof into a low level proof, …
27 votes
Accepted

On statements independent of ZFC + V=L

retains its large cardinal property in L, so we get consistency with V=L. … These are all independent of ZFC+V=L, since they are independent of ZFC, and their truth is the same in V as in L. …
Joel David Hamkins's user avatar
26 votes
7 answers
5k views

What "forces" us to accept large cardinal axioms?

$\vdots$ These are a part of arguments which could be used against large cardinal axioms, but many set theorists not only believe in the existence of large cardinals, but also refute every statement like … $V=L$ which is contradictory to their existence. …
user avatar
25 votes

Arguments against large cardinals

The doubts, however, are usually of the form, "I don't feel that there's enough evidence to make me believe in them," rather than any concrete arguments against their existence. … Once in a while you may run into someone who believes $V=L$ strongly enough to use that as a reason to disbelieve in measurable cardinals, but that's pretty rare. …
Timothy Chow's user avatar
  • 82.7k
24 votes
Accepted

Classifying set theories whose standard models sharing the same ordinals are equal

It is known that if the theory $\mathrm{ZF + V = L(0 ^{\#}})$ has transitive model of height $\alpha < \omega_1$, then $\alpha \in U$; coupled with Friedman's above theorem, this implies that $\mathrm{ … ZF + V = L(0 ^{\#}})$ has more than one transitive model of height $\alpha$. …
Ali Enayat's user avatar
  • 17.7k
23 votes
Accepted

Why believe in the existence of large cardinals rather than just their consistency?

Without that existence, we wouldn't have any reason to believe in even the much weaker consistency assertions. … According to Steel, the large cardinal set-theorist can still consider the nature of V=L set theory, without the stronger large cardinals, simply by relativizing set theory to L. …
Joel David Hamkins's user avatar
23 votes
3 answers
3k views

Why believe in the existence of large cardinals rather than just their consistency?

What reasons are there to believe in their truth, rather than merely in their consistency, or at most their truth in some transitive model? … One might be $V = L$, or even the axiom of restriction, but most set theorists dismiss them, as they rule out large large cardinals and inaccessible cardinals, respectively, and instead seek core models …
Jesse Elliott's user avatar

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