Linked Questions

94
votes
1answer
6k views

How do we know an $ \aleph_1 $ exists at all?

I have two questions, actually. The first is as the title says: how do we know there exists an infinite cardinal such that there exists no other cardinals between it and $ \aleph_0 $? (We would have ...
24
votes
3answers
2k 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 ...
12
votes
1answer
894 views

There's non-Aleph transfinite cardinals without the axiom of choice?

I can't find anything on this anywhere. The book I'm largely using at the moment is based around ZFC, so it makes no mention of anything other than the Aleph numbers, but according to Wikipedia on the ...
5
votes
4answers
626 views

Does every ordinal have cardinality no greater than $\aleph_\mathbb{0}$?

My notes say that the ordinals $\omega + 1, \omega + 2, ... , 2 \omega, ... , 3 \omega, ... \omega^2, ... $ are all countable, and hence have cardinality equal to $\omega = \aleph_\mathbb{0}$. So I ...
6
votes
1answer
496 views

“Real” cardinality, say, $\aleph_\pi$?

Is there any meaningful definition to afford for $\aleph_r$ (as in cardinality) where $r\in\mathbb{R}^+$? $r\in\mathbb{C}$? What about $\aleph_{\aleph_0}$? Can we iterate this? $\aleph_{\aleph_{\...
3
votes
2answers
303 views

Well-orderings of $\mathbb R$ without Choice

The question is about well-ordering $\mathbb R$ in ZF. Without the Axiom of Choice (AC) there exists a set that is not well-ordered. This could occur two ways: a) there are models of ZF in which $\...
1
vote
1answer
370 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 ...
1
vote
2answers
189 views

Injections from all ordinals into a set $X$

We are working in $\mathsf{ZF}$. Let $X$ be a set. Let $A$ be the class of all injections $f: \alpha \to X$ for arbitrary ordinals $\alpha$. I am quite sure that, in fact, $A$ is a set, since if not,...
0
votes
2answers
68 views

$|A|:=\{B|B \sim A\}$ is a set?

let: http://en.wikipedia.org/wiki/Equinumerosity let $A$ a set, I define the cardinality of $|A|:=\{B|B \sim A\}$, but $|A|$ is a set? Thanks in advance!!
0
votes
1answer
67 views

If AC is false, does that mean there exist a set $A$ which has different cardinality from any ordinals?

If a set $A$ has the same cardinality as an ordinal $\alpha$, then there exists a bijection $f:\alpha\to A$, so $A$ is indexed by $\alpha$ and hence well-ordered. Therefore a choice function $g:\...