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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?

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

up vote 5 down vote accepted

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]

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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
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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.

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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. –  Feanor 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
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