Thread 16709267 - /sci/ [Archived: 670 hours ago]

Anonymous
6/28/2025, 8:38:55 AM No.16709267
Inaccessible
Inaccessible
md5: 60e02f88262ec181faa9b60278df2698🔍
explain to me inaccessible cardinals. are we just saying they exist and working from there? like an axiom declaring "suppose this cardinal exists"? like with infinity?
Replies: >>16709295 >>16709302 >>16709310 >>16709425 >>16709843
Anonymous
6/28/2025, 9:44:28 AM No.16709295
>>16709267 (OP)
that's a logical problem
Anonymous
6/28/2025, 10:02:50 AM No.16709302
>>16709267 (OP)
https://en.wikipedia.org/wiki/Grothendieck_universe
Anonymous
6/28/2025, 10:14:31 AM No.16709308
the transitive set does or does not influence recursion
Replies: >>16709309
Anonymous
6/28/2025, 10:16:02 AM No.16709309
>>16709308
it's 50/50
Replies: >>16709321
Anonymous
6/28/2025, 10:21:51 AM No.16709310
>>16709267 (OP)
the answer is consciousness
Anonymous
6/28/2025, 10:54:04 AM No.16709321
>>16709309
the other half is a strictly larger number or something else?
Replies: >>16709324
Anonymous
6/28/2025, 11:00:11 AM No.16709324
>>16709321
yeah
Anonymous
6/28/2025, 2:12:10 PM No.16709425
IMG_3389
IMG_3389
md5: 18368184304541e1eeede3e88c361223🔍
>>16709267 (OP)
>like with infinity?
Yes that’s pretty much it. An inaccessible cardinal is an ordinal that you can’t reach by applying basic operations to smaller ordinals, just like you can’t reach [math]\omega[/math] from by applying basic operations to ordinals [math]\alpha_1,\ldots,\alpha_n<\omega[/math] aka natural numbers.
> we just saying they exist and working from there? like an axiom declaring "suppose this cardinal exists"?
Yeah exactly. If ZFC is consistent, then it has models in which there is no inaccessible cardinal, and other models in which there are inaccessible cardinals. So when we want one we just add an extra axiom (or really, an axiom scheme).

That’s how it works pure-mathematically when you don’t worry about philosophical hangups like whether these ordinals “do” exist or “should” exist. Personally I think infinity is fake and gay, but it definitely is a useful abstraction for talking about really large and small numbers. I hope something like that happens with these crazy big cardinals too, but who knows, I’m not sitting through some crazy Shelahshit or Woodinshit for 1000pp just to find out the answer is “wait, you wanted this to be even a little bit useful? why??”
Replies: >>16709708 >>16709776
Anonymous
6/28/2025, 2:28:47 PM No.16709434
>these countable numbers exist
>we can never construct them
>but we know they exist
yet another problem with infinity. starting to thing wildburger is correct.
Replies: >>16709435 >>16709667
Anonymous
6/28/2025, 2:34:44 PM No.16709435
>>16709434
inaccessibles are supposed to be uncountable, but yes you will also get ones that are countable not “constructible” too
Anonymous
6/28/2025, 9:06:59 PM No.16709667
>>16709434
>starting to thing
you sure do, you sure do...
Anonymous
6/28/2025, 10:14:07 PM No.16709708
>>16709425
Just wait, Hamkins will finish Ultimate-L someday.
Anonymous
6/28/2025, 10:29:34 PM No.16709715
The correct way to think about it might be the brainlet take:
If you can't encounter it, what good is it?

There is a actual answer to that, but good luck having a competent enough math teacher and a well geared enough curriculum to avoid some wishy washy hand waving to try to magic away the problem.
Replies: >>16709854
Anonymous
6/29/2025, 1:49:18 AM No.16709776
>>16709425
>If ZFC is consistent, then it has models in which there is no inaccessible cardinal, and other models in which there are inaccessible cardinals.

Are you sure about the second part of this? I'm pretty sure it isn't possible because ZFC + inaccessible proves Con(ZFC) which according to you would then prove Con(ZFC + inaccessible).
Replies: >>16709841
Anonymous
6/29/2025, 4:04:54 AM No.16709841
>>16709776
ZFC + inaccessible implies Con(ZFC) because ZFC + inaccessible implies the existence of a model of ZFC V_k, in which k is an inaccessible, and ZFC (and any theory in general) is consistent if and only if it has a model. I don't think it implies Con(ZFC + inaccessible), however. In order for a proposition to prove, say, Con(ZFC + inaccessible), a new (and stronger) Large Cardinal axiom is needed.
Replies: >>16709964
Anonymous
6/29/2025, 4:11:44 AM No.16709843
>>16709267 (OP)
Basically, with the added assumption that your axioms are consistent with ZFC. Typically they are given via means of critical points of elementary embeddings.
Replies: >>16709985
Anonymous
6/29/2025, 4:43:18 AM No.16709854
>>16709715
who says you can’t encounter it? just go look inside the Grothendieck universe, you’re bound to come across one.
Replies: >>16709940
Anonymous
6/29/2025, 7:23:56 AM No.16709940
>>16709854
>it? just go look inside the Grothendieck universe
Do I need string theory and wormholes to get there
Anonymous
6/29/2025, 8:31:10 AM No.16709964
>>16709841
Yes, that was my point. If the earlier post's claim were true then ZFC + inaccessible would prove its own consistency which is not possible if it is consistent.
Anonymous
6/29/2025, 8:57:48 AM No.16709985
>>16709843
>assumption that your axioms are consistent with ZFC.
is there no way to prove additional axioms are consistent with ZFC?
Replies: >>16710051
Anonymous
6/29/2025, 11:38:01 AM No.16710051
>>16709985
There is the method of forcing which essentially constructs extra models of ZFC in which certain axioms hold, but forcing is used more often to show the independence of an axiom.