10 questions linked to/from Non-aleph infinite cardinals
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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,...
### $|A|:=\{B|B \sim A\}$ 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!!
### 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:\...