Which of the following facts on cardinals is provable in ZF

Like what the title said, I'm interested to know which of the following is provable in ZF. As much as possible I will give my thoughts and reasoning for each choice.

1.) If $$\kappa$$ is an infinite cardinality then $$\aleph_0\leq \kappa$$

2.) If $$\kappa$$ is an infinite cardinal number then $$\aleph_0\leq \kappa$$

3.) If $$\kappa$$ is an infinite cardinality, then $$\kappa \cdot\kappa = \kappa$$

For 3.) I just followed my gut feeling and say it is not provable without choice since it involves some form of cardinal arithmetics, though I do wish for a more formal reason.

My struggle is between 1. and 2.

For 1.) If $$\kappa$$ is an infinite cardinality, then it means there is an injection from $$\omega = \aleph_0$$ into $$\kappa$$, hence the inequality. This sounds right.

For 2.) While it is true we need Choice to assign cardinal number to a infinite set $$A$$, but if we already know $$\kappa$$ is some $$\aleph$$'s, then it also makes sense for the inequality to hold.

So I really wish to know where in my reasoning is wrong, and an explanation for the right options.Of course further insights on similar question is more than welcome.

Cheers

1 Answer

The question is what do you mean by "cardinal number". One interpretation is that cardinal numbers are all ordinals. Another is that every set has an associated cardinal number, which may or may not be an ordinal number.

It is generally false that every infinite set has a countably infinite subset. But it is true that every infinite ordinal has a countably infinite subset, quite literally since $$\omega$$ is itself a subset of every infinite ordinal.

It is generally false that $$A\times A$$ has a bijection with $$A$$ for every infinite set, in fact it is equivalent to the Axiom of Choice, but it is still true for $$A$$ which is well-ordered (in fact, much more, but at least that much is easy to formulate).

So if your "cardinal numbers" are all ordinals, choice is sort of irrelevant. If they are arbitrary, then this is more difficult.

• I’m guessing based on how 1 and 2 are phrased that “cardinal number” means aleph and “infinite cardinality” is not necessarily well-orderable. – spaceisdarkgreen Jan 3 at 21:41
• I also guess that. I am just very adamant against that sort of use of terminology... – Asaf Karagila Jan 3 at 21:42
• Hmm, let me summarize my understanding of your answer. Because it is generally false that every infinite set has a countably infinite subset, thus if we view $\kappa$ as some dumb-set, 1. is no longer true. By your third comment; we need choice for 3. to be true. Finally, if we view cardinal numbers to be ordinals (albeit special kind), then this would make 2. true – user16319 Jan 3 at 21:42
• @user16319 Yes, that is correct. – Asaf Karagila Jan 3 at 21:43
• @user16319: Just as a note that in order to prove that every infinite set has a countably infinite subset you can use Dependent Choice for a straightforward proof; Countable Choice for a slightly more complicated proof; and in fact the principle itself is even weaker than Countable Choice itself. – Asaf Karagila Jan 3 at 22:35