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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
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
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
  • 4,827
2 votes
0 answers
142 views

Does worldly cardinal exist if $\mathsf{ZFC}$ is consistent?

A worldly cardinal is a cardinal $\kappa$ such that $V_\kappa$ is a model of $\mathsf{ZFC}$. Please forgive me if this is very silly, but if $\mathsf{ZFC}$ is consistent (so there exists a model of $\...
Jianing Song's user avatar
  • 1,923
0 votes
0 answers
277 views

The Axioms of 'Fictional Googology'

There have been questions on whether proper classes have cardinality (some say yes). However, I have my own axioms about it. Firstly, we define the 'cardinality' of a proper class as the conglomerate ...
3-1-4-One-Five's user avatar
0 votes
0 answers
187 views

What is the power set of an inaccessible cardinal?

I've been reading about set theory and the difference between small and large cardinals. since taking the power set of small cardinals (alephs) allows us to create larger cardinals/alephs I know that ...
Adithya's user avatar
  • 11
0 votes
0 answers
119 views

"Proof" that "ZFC + there exists an inaccessible cardinal" is consistent

I have a proof that this theory is consistent using this theory itself. I want to know what's wrong with this proof: "ZFC + there exists an inaccessible caridnal" proves that "ZFC is ...
Ryder Rude's user avatar
  • 1,437
0 votes
1 answer
95 views

Are there any proposed operations to actually construct an inaccessible set?

When we postulate the smallest infinite set, we define it using an interative process involving iterations on the empty set. When we postulate the existence of a set of continuum cardinality, we again ...
Ryder Rude's user avatar
  • 1,437
1 vote
1 answer
60 views

elementary embeddings $j$ in set theory with $V$ and $M$

I'm confused by a variety of elementary non-trivial elementary embedings $j$ we might have. There are 9 "syntactical" possiblities;Here $M$ is a transitive model. I'll name them with a wish ...
user122424's user avatar
  • 3,978
1 vote
2 answers
117 views

Quick question about inaccessible cardinals

I am trying to find a source that states, it is consistent with ZFC that weakly inaccessible cardinal does not exist. Can I please get some sources? It was quite hard for me to find such. Indeed, I ...
mathlearner98's user avatar
3 votes
3 answers
90 views

Compact cardinal cannot be successor?

This is a follow-up question to $\kappa$ is compact $\implies$ $\kappa$ is regular. The definition I'm using for "compact" is the same as there. I am trying to show if $\kappa$ is compact, ...
Pascal's Wager's user avatar
2 votes
1 answer
49 views

Axiom schema of Replacement analogue for Von Neumann stage

I am working without Choice for the time being, so "cardinal" means "well-founded cardinal." Let $\kappa$ be a strongly inaccessible cardinal. I want to show that the Axiom Schema ...
Pascal's Wager's user avatar
2 votes
1 answer
184 views

Name of infinite cardinals which has nonprincipal $\sigma$-complete ultrafilters?

The book "General Topology" by Engelking defines non-measurable cardinals as cardinals admitting no nonprincipal $\sigma$-complete ultrafilters. And then it claims that the discrete space of ...
Ris's user avatar
  • 1,292
0 votes
1 answer
71 views

For a measurable cardinal $κ$, show that $cf(γ)≠κ$ implies $j_U(γ)=\sup\{j_U(δ):δ<γ\}$ ($U$ $κ$-complete ultrafilter, $j_U$ associated embedding) [closed]

For : $κ$ a measurable cardinal, $U$ a $κ$-complete ultrafilter over $κ$ $j_U$ the elementary embedding of $V$ into the ultrapower of $V$ to $U$ How to show that : If $\operatorname{cf}(γ)≠κ$ ...
Cantor's turtle's user avatar
0 votes
0 answers
213 views

Books on infinite sets

I'm a senior undergraduate Pure Math student. I'm looking for books/ problems about large/infinite cardinals and infinite sets. Unsolved or challenging problems are very much welcome. It shouldn't be ...
davinci_07's user avatar

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