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Questions tagged [large-cardinals]

Large cardinals are such cardinals whose existence cannot be proved within ZFC, and requires stronger axioms to be added to ZFC, they are often used to measure the consistency strength of a certain statement in the language of set theory.

5 votes
1 answer
113 views

Absoluteness of inaccessible cardinals

I'm studying large cardinals and I'm hoping to fully understand the proof that says ZFC is not able to prove the existence of inaccessibles (given ZFC is consistent, of course). I've already fully ...
Darsen's user avatar
  • 3,630
0 votes
0 answers
36 views

Aleph sequence fixed points and the least inaccessible [duplicate]

Using the properties of the least inaccessible cardinal (being a regular fixed point of the aleph function) I was able to prove that the least inaccessible $\kappa$ is the $\kappa$-th fixed point of ...
Arianit Niti Gashi's user avatar
7 votes
1 answer
167 views

Use of (weak forms of) AC for elementary embeddings proof

I encounter this issue when going through equivalent characterizations of measurable cardinals. For completeness, let me reproduce the statement: For ordinal $\kappa$, the following are equivalent. ...
Raczel Chowinski's user avatar
1 vote
0 answers
46 views

What is $j_\mathcal{U}(\kappa)$ if $2^{\kappa}<j_\mathcal{U}(\kappa)<(2^{\kappa})^{+}$?

Let $\mathcal{U}$ a non-principal $\kappa$-complete ultrafilter on $\kappa$ and let $j_\mathcal{U}\colon V\to M$ its associated elementary embedding of the universe $V$ in $M=Ult_\mathcal{U}(V)$. In ...
Alberto Caccavale's user avatar
0 votes
0 answers
56 views

On the Singular Cardinal Hypothesis

I'm trying to find the proof of this result. If for each $\lambda\geq2^\omega$, $\lambda^\omega\le\lambda^+$, then the SCH holds. I'm not sure where to look. So if you have any info about this, please ...
Selena's user avatar
  • 103
7 votes
1 answer
114 views

If $\mathsf{GCH}$ fails on a measure one set must it fail at $\kappa$?

Suppose $U$ is a normal measure on $\kappa$. It is well known that if $\{\alpha<\kappa:2^\alpha=\alpha^+\}\in U$ then $2^\kappa=\kappa^+$. Question: does $\{\alpha<\kappa:2^\alpha=\alpha^{++}\}\...
Lxm's user avatar
  • 998
2 votes
0 answers
48 views

Every weakly compact cardinal $\kappa$ is the $\kappa$th inaccessible cardinal

I'm currently reading Jech's Set Theory book, and in particular, while reading chapter 9, he mentions right after Lemma 9.9 that "We shall prove in Chapter 17 that every weakly compact cardinal $\...
Num2's user avatar
  • 329
6 votes
1 answer
130 views

A metrizable space is realcompact iff it has non-measurable cardinality?

A space is realcompact if its a closed subspace of an arbitrary product of real lines, with product topology. A cardinal $\kappa$ is called measurable if there exists a (countably additive) $\{0, 1\}$-...
Jakobian's user avatar
  • 10.5k
2 votes
1 answer
59 views

How to show that the application of two elementary embeddings is an elementary embedding?

Let $\mathcal L$ be a language of first-order logic. Given two structures $\mathfrak M$, $\mathfrak N$ for $\mathcal L$ with domains $M$, $N$ respectively, an elementary embedding from $\mathfrak M$ ...
C7X's user avatar
  • 1,311
0 votes
0 answers
74 views

Is this property of Mahlo cardinals correct?

Quoting from a paper discussing large cardinals: for any ρ that is a Mahlo cardinal, there must also exist ρ smaller cardinals (call them κ, with all κ < ρ) that are k-inaccessible, hyper k-...
Theorem's user avatar
  • 21
0 votes
1 answer
107 views

What is the relation between the first and second inaccessible cardinals? [closed]

I know how the first inaccessible transcends the other cardinals by being regular and strong limit but what does a second inaccessible mean.I had two possibilities in mind: Either: It is inaccessible ...
Arianit Niti Gashi's user avatar
0 votes
1 answer
78 views

How to prove consistency with choice for large cardinal extensions?

How can we know if an extension of $\sf ZF$ by some large cardinal property that results in a consistency strength beyond $0^{\#}$ is compatible with choice or not? I mean the easiest way to know if ...
Zuhair's user avatar
  • 4,631
1 vote
1 answer
75 views

Why is strongness of a cardinal a different notion than Reinhardtness?

A cardinal $\kappa$ is Reinhardt if there is a nontrivial elementary embedding $j:V\to V$ such that $\kappa$ is the critical point of $j$. Kunen proved (using choice) that this large cardinal axiom is ...
C7X's user avatar
  • 1,311
7 votes
1 answer
272 views

If there is a large cardinal, can GCH also hold?

Let $P$ be a statement saying there is large cardinal of some kind. For example, $P$ can be one of There is a weakly inaccessible cardinal. There is a Mahlo cardinal. There is a weakly compact ...
fantasie's user avatar
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2 votes
1 answer
82 views

If $\kappa$ is weakly inaccessible, then it is the $\kappa$th element of $\{\alpha: \alpha =\aleph_\alpha\}$

This is an exercise from Kunen: Exercise I.13.17 If $\kappa$ is weakly inaccessible, then it is the $\kappa$th element of $\{\alpha: \alpha =\aleph_\alpha\}$. If $\kappa$ is strongly inaccessible, ...
Alphie's user avatar
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