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I have discovered a "proof" of the fact, that $\omega_1$ can be countable, contradicting its definition. It mustn't assume axiom of choice, so I guess there are some parts of this construction which can't be made without it. Anyway, here is my argument:

It's relatively consistent with ZF that $\omega_1$ is countable union of countable ordinals (here you can find this claim). Let's say that $\omega_1=\bigcup\alpha_n$. For every countable ordinal $\alpha$ we can find subset $S_\alpha$ of real numbers with order type $\alpha$. Let's construct sets $S_{\alpha_n}$, and without loss of generality assume that $S_{\alpha_n}\subset(n,n+1)$. Now, if we take union of all sets $S_{\alpha_n}$ we get a subset of real numbers with order type $\omega_1$ (from our initial assumption on $\alpha_n$). But we can show that every well ordered subset of reals is countable: let $r_\alpha$ be $\alpha$-th number in this set. Rationals are dense in real numbers, so, as $r_\alpha<r_{\alpha+1}$, we can find rational $q_\alpha$ with $r_\alpha<q_\alpha<r_{\alpha+1}$. From this it follows that we can inject $\omega_1$ on $\mathbb{Q}$ by using map $\alpha\rightarrow q_\alpha$, proving $\omega_1$ subset of countable set, thus countable.

Where is the flaw in that argument?

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Wojowu
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2 Answers2

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Note that you are using the axiom of choice twice here. The first is to choose $S_{\alpha_n}$. There is no canonical choice for such subset. If you just want it to be a subset of $(n,n+1)$ it might be a subset of $(n,n+\frac12)$ or it might be a subset of $(n+\frac12,n+1)$.

You could argue, however, that you fix one enumeration of the rational numbers and then embed it into $(n,n+1)\cap\Bbb Q$ using that enumeration. However for this to be possible you need to fix an enumeration of $\alpha$ as well (or once again you lose the ability to make canonical choices).

Finally, the countable union of countable sets of real numbers need not be countable without the axiom of choice. In the classical example for "$\omega_1$ is singular" we actually have more, we have that $\Bbb R$ is a countable union of countable sets. Does that mean that $\Bbb R$ is countable in that model? No. It just means that countable union of countable sets are not necessarily countable.

The flaw is that. And you just pushed it around a little bit. If the countable union of countable sets is countable, then you don't need to go through the real numbers, just argue directly that the countable union of countable ordinals is countable, and therefore not $\omega_1$. But as said, you need to choose an enumeration for each ordinal in order to have their union countable, and in this sort of situation you just can't.

I should perhaps point out something of interest here.

The argument may fail on two accounts, either we want that the countable union of countable ordinals is countable, or we want that the countable union of countable sets of real numbers is countable. In both cases the argument fails because [a fragment of] countable choice is needed to choose enumerations of the sets in order to prove the wanted result.

However it is interesting to note that the two are completely independent. It is consistent that the countable union of countable sets of real numbers are countable again; but $\omega_1$ is the countable union of countable ordinals. And it is also consistent that the countable union of countable ordinals is countable, but there is a countable family of countable sets of reals whose union is uncountable!

So while both of these failures consist of the same "backbone argument" (choosing enumerations), they are completely independent of one another.

Asaf Karagila
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The proof of "Union of countable set is countable" requires the axiom of countable choice, but "$\omega_1$ is countable union of countable sets" is relatively consistent with only ZF, not ZF + countable choice.

Note that $\omega_1$ is not embeddable even if you does not assume the choice. The set $\bigcup_n S_{\alpha_n}$ is not order-isomorphic to $\omega_1$. Especially, there is no reason that the cardinality of $\bigcup_n S_{\alpha_n}$ is equal to $\aleph_1$.

Hanul Jeon
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