5
votes
1answer
66 views

Existence of a real 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 ...
4
votes
2answers
110 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
213 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 ...
2
votes
1answer
67 views

$\aleph(X) < \aleph(\mathcal{P}(\mathcal{P}(\mathcal{P}(X))))$

Prove: $\aleph(X) < \aleph(\mathcal{P}(\mathcal{P}(\mathcal{P}(X))))$ With $W(X)=\{\langle A,R\rangle: A \in \mathcal{P} (X),R \in \mathcal{P}(X \times X)$ and $ R $ wellorders $ A \}$ And ...
6
votes
2answers
127 views

In ZF, how would the structure of the cardinal numbers change by adopting this definition of cardinality?

In ZFC, a good way of ordering sets by cardinality is by leveraging the notion of an injection. We define: $$X \lesssim Y \leftrightarrow \mbox{ there exists an injection } X \rightarrow Y.$$ ...
4
votes
2answers
414 views

The Continuum Hypothesis & The Axiom of Choice

Does anyone here know of a reference to an analysis on a proposed relationship between The Continuum Hypothesis and The Axiom of Choice?
6
votes
0answers
105 views

AC iff $P(\delta)$ can be well-ordered for all $\delta\in{\bf On}$

I'm trying to do the following exercise: Exercise. Assume ZF. Then AC is equivalent to the statement that $P(\delta)$ can be well-ordered for all $\delta\in{\bf On}$. I'm struggling with the ...
5
votes
1answer
157 views

In ZF, does there exist an ordinal of provably uncountable cofinality?

Question is in the title. In ZFC, one can prove that $\aleph_{\alpha+1}$ is regular, so there is a large source of cardinals with uncountable cofinality, but in ZF, it is consistent that ${\rm ...
5
votes
2answers
94 views

Reference request, self study

I'm looking for references (books/lecture notes) for : Cardinality without choice, Scott's trick; Cardinal arithmetic without choice. Any suggestions ? Thanks in advance.
6
votes
2answers
252 views

Proving existence of a surjection $2^{\aleph_0} \to \aleph_1$ without AC

I'm quite sure I'm missing something obvious, but I can't seem to work out the following problem (web search indicates that it has a solution, but I didn't manage to locate one -- hence the ...
15
votes
1answer
279 views

How to formulate continuum hypothesis without the axiom of choice?

Please correct me if I'm wrong but here is what I understand from the theory of cardinal numbers : 1) The definition of $\aleph_1$ makes sense even without choice as $\aleph_1$ is an ordinal number ...
11
votes
2answers
172 views

The relationship of ${\frak m+m=m}$ to AC

Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.) Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$? If not, what is the ...
3
votes
1answer
114 views

(Non) equivalence of regular cardinal definitions

The usual definition of a regular cardinal is "$\kappa$ is regular if $cf(\kappa) = \kappa$", which, assuming the axiom of choice, is equivalent to this definition: "$\kappa$ is regular iff it cannot ...
2
votes
2answers
64 views

Mapping on cardinal without Axiom of Choice

Define $|A|\le|B|$ iff there exists injective mapping $A \to B$. If Axiom of Choice is assumed then this is equivalent as $|A|\le|B|$ iff there exists surjective mapping $B \to A$. But: If Axiom of ...
6
votes
1answer
123 views

How to prove that a regular cardinal cannot be expressed as a union of sets with less cardinality?

My question is about the following: Using the Axiom of Choice show that: If $\kappa\ge\omega$ is a regular cardinal, $\gamma\le\kappa$, and $\langle A_\alpha\mid\alpha\lt\gamma\rangle$ is a ...
1
vote
1answer
93 views

Axiom of choice , Hartogs ordinals, well-ordering principle

I'm trying to prove the following: If it holds that if for any two sets $A$ and $B$, $A$ can be injected into $B$ or $B$ can be injected into $A$, then every set can be well-ordered (axiom of choice ...
4
votes
1answer
76 views

Clarification of a proof in Herrlich

In Herrlich on page 5 he gives a proof of $\textbf{AC} \implies \textbf{WOT}$: He does not give a definition of cardinality $|X|$ before this proof and I searched the index for a definition but ...
2
votes
3answers
88 views

Cardinals of set operations without AC

Given info: $|A|=\mathfrak{c}$ , $|B|=\aleph_0$ in ZF (no axiom of choice). Prove: $|A\cup B|=\mathfrak{c}$ If $B \subset A\implies|A \backslash B|=\mathfrak{c}$? I have found several places ...
10
votes
1answer
462 views

Do you need the Axiom of Choice to accept Cantor's Diagonal Proof?

Math people: It is my understanding that set theorists/logicians compare cardinalities of sets using injections rather than surjections. Wikipedia defines countable sets in terms of injections. ...
14
votes
1answer
298 views

Cardinality of a locally compact Hausdorff space without isolated points

I am interested in the following result: Theorem. A locally compact Hausdorff topological space $X$ without isolated points has at least cardinality $\mathfrak{c}$. To prove it, one can find two ...
15
votes
3answers
561 views

For any two sets $A,B$ , $|A|\leq|B|$ or $|B|\leq|A|$

Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally: $$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$ If this statement is true, ...
1
vote
2answers
112 views

Defining strong limit cardinals in $ZF$

I do not understand the following passage/footnote in the book I am currently reading: An initial ordinal $\lambda$ is called a strong limit cardinal if $2^\kappa < \lambda$ for every $\kappa ...
3
votes
3answers
156 views

Defining cardinals without choice

According to Wikipedia if we assume AC we define a cardinals number to be an ordinal that is not in bijection with any smaller ordinal. Without AC, one takes the cardinality of a set $X$ to be the ...
12
votes
2answers
322 views

What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that…

... $\aleph_1$ is the immediate successor of $\aleph_0$? I was reading the wiki article on $\aleph_1$ where a distinction is made between the two. If there's isn't a cardinal between $\aleph_1$ and ...
2
votes
1answer
81 views

Is $\kappa^\lambda=2^\lambda$($2 \le \kappa<\lambda$,$\lambda$ infinite) valid in set models of ZF?

Let $2 \le \kappa<\lambda$(both cardinal numbers), in which $\lambda$ is infinite. Then these formula as follows hold where in ZFC: $\lambda+\kappa=\lambda$ $\lambda\cdot\kappa=\lambda$ ...
1
vote
1answer
49 views

$A+\alpha\sim A$ when $\omega\le\alpha<h(A)$

I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha<h(A)$, in which $h(A)$ is the Hartogs number of $A$. If there is $\alpha$ s.t. $\omega\le\alpha<h(A)$, we can get $\alpha ...
1
vote
2answers
107 views

Is there any Dedekind-infinite set can be split to two smaller Dedekind-infinite sets?

I have been considering about $A+\alpha\sim A$ when $\omega\le\alpha<h(A)$, in which $h(A)$ is the Hartogs number of $A$. If there is $\alpha$ s.t. $\omega\le\alpha<h(A)$, we can get $\alpha ...
4
votes
1answer
108 views

Is $\aleph_0$ the minimum infinite cardinal number in $ZF$?

$\aleph_0$ is the least infinite cardinal number in ZFC. However, without AC, not every set is well-ordered. So is it consistent that a set is infinite but not $\ge \aleph_0$? In other words, is it ...
4
votes
1answer
115 views

Is this theory $\kappa$-categorical when $\kappa$ is infinite and not $\le \aleph_0$?

Let $\mathcal L$ be a language with only a binary predicate $E$, and let $T$ be a theory of structures in which $E$ is an equivalence relation which partitions the structure into two infinite ...
0
votes
1answer
206 views

Infinite Dedekind Finite sets

I realized that i have used argument below many times before and I'm not sure if it is true. Let $A=\{n\in \omega|\Phi(n)\}$. Then $A\preceq \aleph_0$. (i)Suppose $A$ is dedekind-infinite and find ...
1
vote
4answers
129 views

Which kind product of non-zero number non-zero cardinal numbers yields zero?

Let $I$ be a non-empty set. $\kappa_i$ is non-zero cardinal number for all $i \in I$. If without AC, then $\prod_{i \in I}\kappa_i=0$ seems can be true(despite I still cannot believe it). But what ...
6
votes
1answer
281 views

Can an infinite cardinal number be a sum of two smaller cardinal number?

Let $\kappa$ be an infinite cardinal number. My question is whether there are $\lambda$ and $\mu$ such that both $<\kappa$ but $\lambda+\mu=\kappa$? If AC holds, then the answer is definitely ...
0
votes
2answers
494 views

Definition of denumerable (countable) set

When we say that a set $S$ is denumerable, that is, there is a bijection $S \to \omega$, do we mean that there exists such a bijection or do we mean that we have one and are talking about a pair ...
2
votes
2answers
119 views

Possible inaccuracy in Wikipedia article about initial ordinals

I quote from the Wikipedia article: "So (assuming the axiom of choice) we identify $\omega_\alpha$ with $\aleph_\alpha$, except that the notation $\aleph_\alpha$ is used for writing cardinals, and ...
4
votes
4answers
246 views
3
votes
3answers
281 views

Which set is unwell-orderable?

In textbook it says that every well-orderable set is equipotent to an initial ordinal number. However, of course unwell-orderable cannot equipotent to any ordinal number, but is there any set is ...
0
votes
1answer
166 views

If $\kappa$ is a cardinal, $\aleph$ is any $\aleph$-number, and if $\kappa\leq\aleph$ then $\kappa$ can be well ordered as well.

I'm having trouble understanding the statement: If $\kappa$ is a cardinal, $\aleph$ is any $\aleph$-number, and if $\kappa\leq\aleph$ then $\kappa$ can be well ordered as well. I understand the ...
6
votes
2answers
141 views

Sum of cardinals without AC

Let $A$ and $B$ be infinite sets. To show $|A\cup B|=\max\{|A|,|B|\}$ we need AC. Now let us assume $|A|<|B|$. Can we show $|A\cup B|=|B|$ without AC?
0
votes
1answer
189 views

Is cardinality a total order? Is AC necessary? [duplicate]

Possible Duplicate: Is the class of cardinals totally ordered? Intuitively, it seems like for any sets $A,B$ either $\lvert A\rvert\leq \lvert B\rvert$ or $\lvert B \rvert \leq \lvert ...
8
votes
1answer
327 views

Is the axiom of choice needed to show that $a^2=a$?

A comment on this answer states that choice is needed for the statement that $a^2=a$ for all infinite cardinals $a$. In Thomas Jech's Set Theory (3rd edition), his theorem 3.5 proves this statement ...
1
vote
2answers
281 views

Non-aleph infinite cardinals

I'm now confused with a concept of $\aleph$. 1.$\aleph$ is a cardinal number that is well-ordered in ZF.(Defined as an initial ordinal that is equipotent with). Does that mean $\aleph_x$ in ZF may ...
2
votes
2answers
315 views

Cardinality of the complex numbers in ZF

As you all know, cardinality of $\mathbb{R} = 2^{\aleph_0}$ can be proved in ZF, since cardinality of $\mathbb{N} \times \mathbb{N} = \aleph_0$ can be proved in ZF. I know that the statement 'For any ...
6
votes
1answer
200 views

Number of well-ordering relations on a well-orderable infinite set $A$?

Given a well-orderable infinite set $A$, can we always say that the set $$\left\{R\in A\times A:\langle A,R\rangle\, \text{is a well-ordering}\right\}$$ has cardinality $2^{|A|}$? How much Choice is ...
2
votes
1answer
149 views

Equivalences to “D-finite = finite”

By a D-finite set, we mean a set admitting no injection from the natural numbers (or equivalently, a set not in bijection with any proper subset). I have encountered a proof that the following are ...
2
votes
2answers
455 views

A question about cardinal arithmetics without the Axiom of Choice

Is multiplication of infinite cardinals defined in ZF without Choice?
7
votes
2answers
408 views

axiom of choice: cardinality of general disjoint union

I have this exercise involving the axiom of choice, but I don't understand where it's needed: Let $(X_i)_{i \in I}$ and $(Y_i)_{i \in I}$ be pairwise disjoint sets with $|X_i| = |Y_i|$. Prove, using ...
2
votes
1answer
570 views

Cardinality of an infinite union of finite sets

This follows Arturo's answer of this question. Let $I$ an infinite set and $\{E_i\}_{i\in I}$ a family of finite sets. It is said that there exists an injection $$\bigcup_{i\in I} \ ...
3
votes
2answers
729 views

Cardinality of union of ${{\aleph }_{0}}$ disjoint sets of cardinality $\mathfrak{c}$

I have a home work question which is: " what is the cardinality of the union of ${{\aleph }_{0}}$ disjoint sets of cardinality $\mathfrak{c}$?" I believe somehow we can get to: cardinality = ...
17
votes
2answers
957 views

For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice

how to prove the following conclusion: [for any infinite set $S$,there exists a bijection $f:S\to S \times S$] implies the Axiom of choice. Can you give a proof without the theory of ordinal ...
15
votes
2answers
756 views

Defining cardinality in the absence of choice

Under ZFC we can define cardinality $|A|$ for any set $A$ as $$ |A|=\min\{\alpha\in \operatorname{Ord}: \exists\text{ bijection } A \to \alpha\}. $$ This is because the axiom of choice allows any ...