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Irrational numbers Cardinality.

The real numbers, $\mathbb{R}$, are uncountable and the rational numbers, $\mathbb{Q}$, are countable. We can write $\mathbb{R} = \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})$. Since $\mathbb{Q}$ ...
Mathstudent123's user avatar
0 votes
1 answer
47 views

For $k$-algebras $B_1, \dots, B_n$, $\# \operatorname{Hom}_k( \prod_{i=1}^nB_i, \Omega) = \Sigma_{i=1}^n \# \operatorname{Hom}_k(B_i, \Omega)$?

Let $k$ be a field with $ k \subseteq \Omega$ a algebraically closed field. Let $B_1 , \dots, B_n$ be ( possibly finite local ) $k$-algebras. Then next equality of cardinals holds $$ \# \operatorname{...
Plantation's user avatar
  • 2,656
0 votes
1 answer
50 views

Proving the Equality of Infinite Cardinal Products and Powers

Theorem: Let $\Xi$ be an infinite set, $\{\kappa_i\}_{i \in \Xi}$ be a family of cardinal numbers, and $\lambda$ be a cardinal number. Then: $\prod_{i \in \Xi} \kappa_i^{\lambda} = \left(\prod_{i \in \...
Chau Long's user avatar
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
2 votes
0 answers
43 views

Lemma 0 in Hajnal's Paper "Embedding Finite Graphs into Graphs Colored with Infinitely Many Colors"

I am looking for a proof of the following lemma. Let $E_0$ be the set of edges of an undirected graph with no loops with vertex set a cardinal $\kappa$. Let $E_1$ be the family of two-element subsets ...
Tri's user avatar
  • 417
2 votes
1 answer
70 views

Assuming GCH holds, calculate $\aleph_{\omega_1}^{\aleph_0}$

I'm working through the book Discovering Modern Set Theory by Just and Weese, and this question comes right after this theorem: Here's what I've worked out so far: I believe the cofinality of $\...
violeta's user avatar
  • 849
4 votes
2 answers
167 views

Does the Cardinal Supremum Commute with the Cardinal Power?

Let $\kappa,\tau$ be two cardinals and $\{\varkappa_\alpha\}_{\alpha<\kappa}$ an indexed set of cardinals. Is it true that $$\sup_{\alpha<\kappa}(\varkappa_\alpha^\tau)=\left(\sup_{\alpha<\...
tripaloski's user avatar
0 votes
1 answer
84 views

For infinite cardinals $\kappa$, we have $\kappa \otimes \kappa = \kappa$.

I am aware that other questions are quite similar to this; however, it seems like the other questions regarding the same statement are looking at proofs that seem somewhat different from the one I am ...
Ben123's user avatar
  • 1,296
1 vote
1 answer
83 views

Can cardinality $\kappa$ exist where $\forall n\in\mathbb{N} \beth_n<\kappa$,$\kappa<|\bigcup_{n\in\mathbb{N}}\mathbb{S}_n|$,$|\mathbb{S}_n|=\beth_n$

The Wikipedia article on Beth numbers defines $\beth_\alpha$ such that $\beth_{\alpha} =\begin{cases} |\mathbb{N}| & \text{if } \alpha=0 \\ 2^{\beth_{\alpha-1}} & \text{if } \alpha \text{ is a ...
SarcasticSully'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
0 votes
0 answers
34 views

For all cardinals $\kappa, \lambda$ with $\lambda \geq cf(\kappa)$ the inequality $\kappa^{\lambda} > \kappa$ holds [duplicate]

I genuinely have no idea why the proposition in the title holds or how to show it. I am kind of new to cardinals and ordinals and very confused. If someone could explain, I would really appreciate ...
metamathics's user avatar
1 vote
0 answers
92 views

Cardinal power towers

I am not an expert on large cardinals. I could not find any reference (and terminology) for the following question: We start with $$\lambda:=\aleph_0 \text{ [tet] } \omega = \aleph_0 ^ {\aleph_0 ^ {\...
Cem Aksak's user avatar
0 votes
0 answers
25 views

Would well-founded Scott cardinals work in ZCA + Ranks?

Does original Zermelo's set theory + Regularity + Ranks, prove that every set is of equal size to some element of a Scott cardinal? The original Zermelo does include an axiom of Choice, and it admits ...
Zuhair's user avatar
  • 4,631
1 vote
1 answer
119 views

The cardinality of specific set $A\subset \mathbb{N}^{\mathbb{N}}$

Let $A$ be a set of total functions from the naturals to the naturals  such that for every $f\in A$ there is a finite set $B_f\subset \mathbb{N}$ , such that for every $x\notin B_f$ , $f(x+1)=f(x)+1$. ...
Daniel's user avatar
  • 83
4 votes
1 answer
82 views

Induction does not preserve ordering between cardinality of sets?

Consider building a binary tree and consider it as a collection of points and edges. Here is one with five levels, numbered level $1$ at the top with $1$ node to level $5$ at the bottom with $16$ ...
jdods's user avatar
  • 6,360

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