0
votes
1answer
42 views

Dedekind finite set and a special well ordered set

In ZFC, Dedekind finite set and finite set are same things. So I have a set say A(which is equal to N in ZFC) all Dedekind finite set are equivalent to proper subsets of A and A is well ordered set. ...
2
votes
1answer
69 views

Infinite Set has greater or equal cardinality that of N

For any infinite set, we can find a 1-1 function (not necessary onto) from N (set of natural no.) to that set. The proof of this theorem I know using axiom of choice. Can we prove it without using ...
2
votes
2answers
116 views

Is the Axiom of Choice necessary to prove $\mathbb R \approx \mathcal P(\omega)$?

I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876), and I'm getting quite confused about cardinal arithmetic without AC. Here (Which sets are well-orderable ...
5
votes
1answer
78 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 ...
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 ...
8
votes
2answers
153 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 ...
1
vote
1answer
49 views

factorial of infinite Cardinals

Let $S_A$ be set of all bijections over $A$ such that $Card(A)=\kappa$. Define foctorial as $\kappa!:=Card(S_A)$. Show that if $\kappa$ is infinite, then : $\kappa!=2^\kappa$ First, I've ...
2
votes
1answer
36 views

If $|X|<|Y|$ then $|Y|=|Y-X|$ (with $Y$ infinite)

Like the title says, I would like to prove that if $|X|<|Y|$ then $|Y|=|Y-X|$. (with $Y$ infinite) I know I have to use the axiom of choice, but I've no idea about how to proceed. Any help is ...
5
votes
1answer
88 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.
5
votes
1answer
96 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 ...
4
votes
2answers
124 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
233 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
75 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
147 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: $$X \lesssim Y \leftrightarrow \mbox{ there exists an injection } X \rightarrow Y.$$ ...
3
votes
2answers
595 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
111 views

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

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
176 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
109 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
278 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
308 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
187 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
130 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
67 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
140 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
104 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
80 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 ...
11
votes
1answer
523 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
321 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
599 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
123 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
176 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
342 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
86 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
51 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
116 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
117 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
118 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
292 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
132 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
342 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
721 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
131 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
363 views
3
votes
3answers
313 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
176 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
142 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
213 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 ...
4
votes
2answers
548 views

some elementary questions about cardinality

I know little about set theory and while reading some Algebra proof I had difficulty on some details. So my questions are : If $X$ is an infinite set and $Y$ is the set of all finite subsets of $X$, ...
8
votes
1answer
348 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 ...