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Can continuum be a countable union of sets with cardinality, less than continuum? I can prove it for a finite union by mathematical induction from this:

$\mathbb R = A_1 + A_2 => |\mathbb R| = |A_1 + A_2| = max (|A_1| + |A_2|) => |A_i| = |\mathbb R|$

But how to prove it for a countable union?

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3 Answers 3

Assuming the Axiom of Choice, the real line cannot be a union of countably many sets each of size less than continuum. To prove this, one needs to know that the continuum has cofinality strictly greater than $\omega$. (The cofinality of an limit ordinal $\delta$ is the least cardinality of a set $X$ of ordinals strictly less than $\delta$ such that $\delta = \sup (X)$.)

This resolves you question because if $\{ A_n : n \in \omega \}$ is a family of subsets of $\mathbb{R}$ each with size strictly less than $2^{\omega}$, then setting $\kappa = \sup \{ |A_n| : n \in \omega \}$ we have that $\kappa < 2^\omega$. It then follows that $$| \textstyle{\bigcup_{n < \omega}} A_n | \leq \sum_{n < \omega} | A_n | \leq \sum_{n < \omega} \kappa = \omega \cdot \kappa = \max \{ \omega , \kappa \} < 2^{\omega},$$ and therefore $\bigcup_{n < \omega} A_n$ cannot equal all of $\mathbb{R}$.

Without the Axiom of Choice, it is possible that the continuum is a countable union of countable sets. (Note that this does not contradict the well-known fact (provable without Choice) that $\mathbb{R}$ is uncountable because in such models countable unions of countable sets need not be countable.)

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This is the right approach, to be sure - I had been assuming a union of sets of the same cardinality, but this handles cases like looking at $\displaystyle\cup_{n\in\omega}\aleph_n$ when $\displaystyle\mathfrak{c} = \aleph_{\omega+1}$, which my argument doesn't. –  Steven Stadnicki Sep 19 '12 at 18:50
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To add to what Arthur said, one way to show (assuming AC) that $2^{\aleph_0}$ has uncountable cofinality is to use König's theorem

( http://en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_%28set_theory%29, )

which is a generalization of Cantor's theorem. An instance of König's theorem says that given a countable sequence of sets $(A_i : i<\omega)$ such that $|A_i|<|\mathbb{R}|$ for all $i$, we have

$\sum_{i<\omega} |A_i| < \prod_{i<\omega} |\mathbb{R}|$.

The right hand side is equal to $|\mathbb{R}|$ because countable sequences of reals can be coded by reals.

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Hi Trevor, I understand that you're in the Fields semester now. I was wondering whether or not you attended Shelah's talk two days ago about the axiom of choice. The title seems interesting but there are no abstracts anywhere. –  Asaf Karagila Sep 30 '12 at 15:28
    
@AsafKaragila Unfortunately I was out of town last week and could not attend Shelah's talk. (I also cannot find an abstract anywhere.) –  Trevor Wilson Oct 3 '12 at 16:33
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As remarked by others, assuming the axiom of choice this is false.

However there are several interesting cases when the axiom of choice fails:

  1. The real numbers may be a countable unions of countable sets.
  2. The real numbers may be the union of two sets both of cardinality strictly less than the continuum, furthermore at least one of these sets can be split into two smaller sets, so we can even split the continuum into three sets of smaller cardinality; and so on.
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