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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?


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I think you need to appeal to the fact that $\kappa<\kappa^{cf(\kappa)}$. It follows that $cf(2^\omega)>\omega$, and the result is an easy corollary. – GME Jan 19 '14 at 15:57
It also doesn't seem to matter whether the $A_n$'s are disjoint. – GME Jan 19 '14 at 16:06
@user104955 That's what the question say... – Hila Jan 19 '14 at 16:08
Yep, but I'm claiming (tentatively) that the theorem would still be true if we weakened the antecedent by dropping the disjointness requirement. – GME Jan 19 '14 at 16:10
(Your version is true if you assume the sets $A_n$ to be Borel.) – Andrés E. Caicedo Jan 19 '14 at 16:53
up vote 14 down vote accepted

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.

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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.

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Where did I assume that $2^{\aleph_0} = \aleph_1$? – Hila Jan 19 '14 at 15:57
It looks like you assumed that $|A_n|\lt|\mathbb R|$ implies $|A_n|\le\aleph_0$, which is more or less the same thing as assuming $2^{\aleph_0}=\aleph_1$, at least if you're working in ZFC. If that's not what you assumed, how did you get the cardinality of that union to be at most $\aleph_0$? – bof Jan 19 '14 at 16:01
Ok, I understand what you're saying, thanks :) – Hila Jan 19 '14 at 16:02

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