What's "the catch" in this question?

Question

I am solving old exam questions and I came across this question:

Let $\langle A_n \mid n < \omega\rangle$ disjoint sets such that $\bigcup_{n < \omega}A_n = \mathbb{R}$. Prove that there exists $n < \omega$ such that $|A_n| = |\mathbb{R}|$

It seems too easy - if the statement is false, then the cardinality of $\bigcup_{n < \omega}A_n$ is at most $\aleph_0$, which is less than $2^{\aleph_0}$, which is the cardinality of $\mathbb{R}$.

What am I missing?

Thanks!

Answer 1

You can't assume that all the sets are countable. I mean, $\aleph_\omega$ is uncountable and can be written as a countable union of smaller sets.

The point is to use König's theorem, and prove that can't be true. That is, $\operatorname{cf}(2^{\aleph_0})>\aleph_0$.

As Arthur points out, this is false without the axiom of choice, as it is consistent that the real numbers are a countable union of countable sets.

Answer 2

What you are missing is that the cardinal number $|\mathbb R|=2^{\aleph_0}$ is not necessarily equal to $\aleph_1$. How do you know that $2^{\aleph_0}$ is not equal to $\aleph_\omega$? It's not, but that's the point of that old exam question. The assumption that $2^{\aleph_0}=\aleph_1$ is called the continuum hypothesis.