# $\mathbb{R}$ represented using an infinite union of finite sets containing reals?

If $S_1$, $S_2$, $\dots$ are sets of real numbers and if $\bigcup_{j=1}^{\infty}{S_j} = \mathbb{R}$ then one of the sets $S_j$ must have infinitely many elements.

I believe at least one of the $S_j$ must be an infinite set, but I can't work out a proof. What's the trick I'm missing?

-
If all $S_j$ were finite, then their sum would be countable. – enzotib Sep 16 '12 at 9:10

Suppose that all sets were finite, the union of countably many finite sets is countable, but the real numbers are not.

[This argument uses the axiom of choice, however it is true without the axiom of choice that a countable union of finite sets of real numbers is countable]

-
Ahha, I incorrectly believed an infinite union of sets is UNcountable. I read up on countable sets and this makes sense. Thank you – ljdelight Sep 16 '12 at 9:35

In fact something stronger is true: at least one of $S_i$ has to be uncountable.

It is a standard fact that a countable union of countable sets is countable. You could for instance enumerate elements of $S_i$ so that: $$S_i = \{ S_i^{j} \}_{j \in \mathbb{N} }$$ and then arrange $\{ S_i^{j}: i,j \in \mathbb{N} \} = \bigcup_{i} S_i$ ordering first according to $i+j$ and then $i$. Since $\mathbb{R}$ is not countable, it can't be that all $S_i$ are countable, and you are done.

-
Of course this is not something that matters for most people, but it is consistent without the axiom of choice that the real numbers are a countable union of countable sets. – Asaf Karagila Sep 16 '12 at 9:23
@Asaf Karagila - thanks for this rather surprising insight. Mathematics without AC is apparently even stranger than I thought. – Jakub Konieczny Sep 16 '12 at 11:37
@Feanor: Yeah, sometimes it is even stranger than mathematics with AC :) – Hagen von Eitzen Sep 16 '12 at 12:51