Linked Questions

11
votes
4answers
3k views

Is the class of cardinals totally ordered?

In a Wikipedia article http://en.wikipedia.org/wiki/Aleph_number#Aleph-one I encountered the following sentence: "If the axiom of choice (AC) is used, it can be proved that the class of cardinal ...
4
votes
2answers
1k views

A question about cardinal arithmetics without the Axiom of Choice

Is multiplication of infinite cardinals defined in ZF without Choice?
4
votes
2answers
2k 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 = $({{\...
5
votes
4answers
628 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 ...
3
votes
2answers
463 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 ...
1
vote
2answers
844 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 ...
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
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:\...
-2
votes
1answer
53 views

Relation between limit ordinals and alephs. [duplicate]

I was wondering what the relation is between a limit ordinal and the alephs. Are all limit ordinals alephs and if so can it be proven.