4
votes
0answers
72 views

$\aleph_\alpha^{\aleph_1} = \aleph_\alpha^{\aleph_0}\cdot 2^{\aleph_1}$ for all $\omega \le \alpha < \omega_1$

I'd like someone to check my proof that $\aleph_\alpha^{\aleph_1} = \aleph_\alpha^{\aleph_0} \cdot 2^{\aleph_1}$ for all $\omega \le \alpha < \omega_1$. First, let's prove that ...
4
votes
1answer
63 views

Question about cardinals in ZF

In Lévy's basic set theory refers a theorem of Bernstein as exercise problem: Theorem. (Bernstein) Let $\mathfrak{a,b}$ be cardinals. If $\mathfrak{a+a=a+b}$, then $\mathfrak{b\le a}$. I try to ...
2
votes
0answers
44 views

A problem with an assumption in a previous lemma for the proof of Silver´s Theorem on SCH in Jech´s “Set Theory”

In the Jech´s textbook proof of Silver´s Theorem, specifically in the Lemma 8.15, there are an assumption in the beginning of the proof. First, the Lemma 8.15 says that, under the assumption that ...
3
votes
1answer
82 views

$\aleph_\omega$ and large ordinals?

Assuming the GCH, each time you take the powerset of an aleph number, the subscript increases by one. So there is $\aleph_0, \aleph_1, \aleph_2\ldots \aleph_\omega\ldots \aleph_{\epsilon_0}\ldots$ and ...
7
votes
1answer
112 views

Constructing a Banach space of cardinality $\beth_{\omega+1}$

This is related to yesterday's question Constructing a vector space of dimension $\beth_\omega$; it's the next exercise (I.13.35 (a)) in Kunen's Set Theory. Let $B_0 = \ell^1$ and let $B_{n+1} = ...
4
votes
2answers
106 views

Constructing a vector space of dimension $\beth_\omega$

I'm trying to solve Exercise I.13.34 of Kunen's Set Theory, which goes as follows (paraphrased): Let $F$ be a field with $|F| < \beth_\omega$, and $W_0$ a vector space over $F$ with $\aleph_0 ...
-1
votes
1answer
40 views

Functions with cardinals as domain

How do we actually define a function with cardinals as domain? For example, take domain of the function as $\aleph_1$. Do we define it for all cardinal numbers strictly less than $\aleph_1$?
0
votes
1answer
33 views

Cardinal Arithmetic, Regular Cardinals, and Exponentiation

I cannot solve a simple cardinal exponentiation/regularity exercise. Let $\kappa$ be a regular cardinal. Why is the cardinal $\kappa^{\lt \kappa}=\sum_{\alpha<\kappa}\kappa^\alpha$ regular as ...
2
votes
1answer
47 views

Cardinal Arithmetic 2: $\beth_{\omega_1}$

This is question is simple to write but (I think) hard to solve. Does the following equality hold? $$\bigcup_{{\beta}<\beth_{\omega_1}}\beth_{\omega+1}(|\beta+\omega|) =\beth_{\omega_1} $$ Where ...
1
vote
1answer
72 views

Partitioning an infinite cardinal number

Can we partition an infinite cardinal greater than aleph null, into countable number of cofinal subsets? Can we have restriction on the cardinality of cofinal subsets? For example, suppose $\alpha$ ...
5
votes
1answer
63 views

Cardinal Arithmetic: $\beth_\omega$

Is the following equality independent of ZFC: $$\bigcup_{\aleph_{\beta}<\beth_{\omega}}2^{\aleph_{\beta}}=\beth_\omega $$ Consistent: Now that the equality is consistent with ZFC since it holds ...
2
votes
1answer
44 views

Is the cardinality of the continuum weakly Mahlo?

Is $2^{\aleph_0}$ a weakly Mahlo cardinal? Can it be? That is, are there conditions (such as the negation of the continuum hypothesis or something) under which it is, and other conditions under which ...
0
votes
0answers
22 views

Collection of sets with a given cardinality $\kappa$ is not set [duplicate]

Show that collection of all sets with cardinality $\kappa\neq0$, is not set. I'll state my approach and I need to see whether this idea is precise/precisable or not : First let $K$ be the set ...
4
votes
1answer
38 views

Extension of ZFC models preserves cardinals

Let $M \subseteq N$ be countable transitive ZFC set models. Assume that this extension preserves cardinals, i.e. if $\alpha$ is an ordinal number (this notion is absolute) such that $(\alpha \text{ is ...
5
votes
1answer
70 views

Partial order on cardinalities without the axiom of choice

Cardinality can still be defined without choice, e.g. as equivalence class of equipotent sets, see Defining cardinality in the absence of choice. Injections define partial order on cardinalities by ...
18
votes
2answers
285 views

Covering $\mathbb R^2$ with function graphs

Suppose we have a countable family of function graphs (each function is $\mathbb R\to\mathbb R$, not necessary continuous). Obviously, they cannot cover the whole plane $\mathbb R^2$, because they ...
2
votes
0answers
92 views

Why are large cardinal axioms actually axioms?

A cardinal is an isomorphism class in ZFC, or a representative of one. I'm not asking what the significant uses of large cardinals are, or why we would want to find them or construct them; I'm ...
2
votes
3answers
210 views

Is there a largest large cardinal?

In ZFC, a cardinal is an isomorphism class of sets. However ZFC doesn't explicitly have classes; NBG, which is a conservative extension of ZFC does. There is no largest cardinal by Cantors Theorem ...
2
votes
2answers
66 views

On those behaviors of continuum function which imply the axiom of choice

It is a folklore fact that within $\text{ZF}$ the generalized continuum hypothesis ($\text{GCH}$) implies the axiom of choice ($\text{AC}$), namely: $$ZF+\forall \kappa\in ...
1
vote
2answers
32 views

The product of cardinals

Let $\gamma_i$ be infinite cardinal numbers for $i=1,2,3$ such that $\gamma_i<\gamma_3$ for $i=1,2$. Is it true that $\gamma_1.\gamma_2<\gamma_3$?
7
votes
2answers
143 views

Is it possible that $2^n=3^n$ for some Dedekind-finite cardinal $n\gt0$?

Is it possible that $2^n=3^n$ for some Dedekind-finite cardinal $n\gt0$? I think the question speaks for itself, but let me try and satisfy the "quality standards" algorithm by padding it. Yes, I ...
4
votes
2answers
129 views

CH imples the existence of a function

I was studying an article and the author stated that CH implies that there exists a function from $\omega_1 \setminus \omega$ onto the set of all countable subsets of $\omega_1$ such that for each ...
1
vote
2answers
42 views

Cardinal of $V_{\omega+\alpha}$

I do not understand why $card(V_{\omega+\alpha})=\beth_\alpha$. The steps in the recursion for $0$ and succesor ordinals are quite easy, but I do not manage to prove it for limit ordinals. I see that ...
10
votes
0answers
114 views

Sets of Cardinals without choice

I have a theory that the axiom of choice is equivalent to the statement that sets of distinct cardinals are well ordered by cardinality. I can prove that the axiom of choice implies this. However I am ...
5
votes
1answer
91 views

Landing between $\beth_\lambda$ and $\beth_{\lambda+1}.$

Main Question. Is it consistent with ZFC that there exists a limit ordinal $\lambda$ and a cardinal number $\nu$ satisfying $$\beth_\lambda < \beth_\lambda^\nu < \beth_{\lambda+1}?$$ I am also ...
3
votes
1answer
45 views

Is it true that for all infinite cardinal numbers $\nu,$ the set $\{\kappa \mid \kappa^\nu > \kappa\}$ is unbounded?

Bind all lowercase greek letters to cardinal numbers. Question. Is it true that for all infinite cardinal numbers $\nu,$ we have that $\{\kappa \mid \kappa^\nu > \kappa\}$ is unbounded? For ...
0
votes
1answer
31 views

A question about Choice Functions.

Assume the axioms of ZFC. Suppose that X is an infinite set of infinite (and pairwise disjoint) sets, none of which has a cardinal number greater than that of X. Is the cardinal number of the set of ...
5
votes
2answers
129 views

What is the value of $\beth_{\omega_1}^{\aleph_0}$?

It is well known that $\beth_\omega^{\aleph_0} = \beth_{\omega+1}$. This follows since for strong limit $\kappa$, we have $\kappa^\kappa = \kappa^{\mathrm{cf}(\kappa)}.$ Question. To the extent that ...
1
vote
1answer
43 views

Is it true that if $\kappa < \kappa^\nu$, then ${\mathrm{cf}(\kappa)} \leq \nu$?

Let $\kappa$ denote an infinite cardinal number. Then we know the following. $$\kappa<\kappa^{\mathrm{cf}(\kappa)}$$ Question. Is it true that if $$\kappa < \kappa^\nu,$$ then ...
3
votes
1answer
67 views

What are the fixed points of cardinal exponentiation?

Whenever $\kappa$ is an infinite cardinal number, write $\mathrm{cl}_\kappa$ for the unique function $\mathrm{Card} \rightarrow \mathrm{Card}$ given by $\mathrm{cl}_\kappa(\nu) = \nu^\kappa.$ It ...
3
votes
1answer
87 views

Proving König's lemma (technical problems)

the aim of my exercise is to give a proof of the König's lemma. So, let $\kappa, \lambda$, be cardinals such that $cf(\kappa)\leq \lambda$. My professor's suggested us to prove that there exists a ...
5
votes
1answer
87 views

Do we need choice to prove that $|\mathbb{N} \times A| = |A|$ for all infinite sets $A$?

I can't think of any way to prove it without choice.
1
vote
1answer
43 views

How can we show that $\omega_1$ is a regular cardinal?

A cardinal $\kappa$ is regular if and only if there is no $\lambda<\kappa$ for which there is a function $f:\lambda\rightarrow\kappa$ with range cofinal in $\kappa$. How can we see in ZFC that ...
2
votes
1answer
41 views

Preserve Cardinals and Adding No Bounded Subsets

In Chapter 15 at the bottom of page 228 of $\textit{Set Theory}$ by Jech, he writes that if $\kappa$ is a cardinal in $V$ and if $\kappa$ has no new bounded subsets in $V[G]$, then $\kappa$ remains a ...
1
vote
1answer
61 views

How many ways can we totally order a set $X$ (but not necessarily the whole set $X$, perhaps just a subset) up to isomorphism?

Here's Attempt #2 at asking this question. Suppose $X$ is a set (not necessarily finite), and let $T$ denote the collection of all totally ordered sets $(A,\leq)$ such that $A \subseteq X.$ Now let ...
5
votes
1answer
95 views

Existence of a regular uncountable $\aleph_{\alpha}$ without $\mathsf{AC}$

Set theory (Jech) $\text{p.}\;27:$ It is an open problem whether one can prove without the axiom of choice that there exists a regular uncountable $\aleph_{\alpha}\;($the informed guess is that ...
2
votes
1answer
44 views

Can we define ordinals such that the following sentences are independent of ZFC?

Can we explicitly define two ordinals $\alpha$ and $\beta$ in the language of $\{\in\}$ such that the following hold? ZFC proves that $\alpha$ and $\beta$ exist. ZFC proves that $\beth_\beta \neq ...
2
votes
2answers
117 views

A property of strong limit cardinal

Suppose $\lambda$ is a strong limit cardinal, i.e. $\forall \alpha<\lambda \ 2^\alpha<\lambda$, and the cofinality of $\lambda$: $cf(\lambda)=\omega$. How do we show that $2^\lambda \leq ...
4
votes
2answers
126 views

about the smallest $k$ that $V_k$ is a model of ZFC

Let $k$ to be the smallest ordinal that $V_k$ is a model of ZFC. I know that $k$ need not to be inaccessible cardinal,and $k$ has confinality $\omega$. Then how big is $k$? How to write down $k$ in ...
5
votes
4answers
147 views

Cardinality of Irrational Numbers

I know and I have proved more than once that the set of irrational numbers ($\mathbb{I}$) is uncountable, but now I'm given to solve this problem: Show that $|\mathbb{I}|=|\mathbb{R}|$, How can I ...
4
votes
2answers
123 views

The regularity of successor cardinal

I was looking at two different proofs of the fact that successor cardinals are regular. It struck me as odd that both proofs used AC. Looking at the concepts involved in defining cofinality I feel as ...
5
votes
3answers
228 views

Uncountable Cardinals without AC

I am doing an exercise, proving that without AC or Replacement that there are uncountable cardinals. As a point of reference I looked at the proof in Kunen's "The Foundations of Mathematics" that ...
1
vote
1answer
70 views

About alephs and beths

If $2^{\aleph_{0}} \ge \aleph_{\omega_1}$, show that $\beth_{\aleph_{\omega}} = 2^{\aleph_{0}}$ , and that $\beth_{\aleph_{\omega_1}} = 2^{\aleph_{1}}$ I don´t know how to start, can you give me a ...
3
votes
1answer
313 views

A question about splitting sets

I've been looking into combinatorics and small cardinals, in particular, the splitting number $\mathfrak{s}$. By definition, a set $X \subseteq \omega$ splits an infinite set $Y \subseteq \omega$ if ...
2
votes
1answer
47 views

Are there ordinals other than the set of natural numbers which satisfy this property?

Let $\alpha$ be an ordinal. We say that $\alpha$ is good iff for every $\beta\in \alpha$, there exists $\gamma\in \alpha$ such that $|\scr{P}(\beta)|\leq |\gamma|$. Question: Is the set of natural ...
2
votes
1answer
46 views

Cofinality assuming GCH

There is this statement that GCH holds iff any pair of regular cardinals $\kappa,\lambda$ such that $\kappa<\lambda$ satisfy that $\lambda^\kappa = \lambda$. Assume we do have two such cardinals. ...
2
votes
1answer
29 views

Identity on singular strong limit cardinals

Let $\lambda$ be a singular strong limit cardinal. Prove that $2^\lambda = \lambda^{\mbox{cf}\lambda}$. It has been a while since I had to prove anything relating to cardinals, and I am not sure ...
1
vote
1answer
98 views

Continuum Hypothesis $\iff ?$?

I have read that CH cannot be proved nor disproved within ZFC, and I was wondering: Which (If any) branches/fields of Mathematics are built upon CH being true? Are there any subjects built upon ...
2
votes
1answer
96 views

$\aleph$ function fixed points below a weakly inaccessible cardinal are a club set

I am throwing yet another one of my solutions out here for the internets to debug and for future set-theory students. Let $\aleph_\delta$ a weakly inaccessible cardinal. Prove that $A =\{\alpha ...
10
votes
2answers
2k views

What's “the catch” in this question?

I am solving old exam questions and I came across this question: Let $\langle A_n \mid n < \omega\rangle$ disjoint sets such that $\bigcup_{n < \omega}A_n = \mathbb{R}$. Prove that there ...