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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:

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

But how to prove it for a countable union?

Martin Sleziak
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Zhenya
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3 Answers3

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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.)

user642796
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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.

Trevor Wilson
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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 03 '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.
Asaf Karagila
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  • Is it possible in some "weird" set-theoretic universe with compromised AC to have a version of the Continuum's cardinality which can be split into a set of all and only well-ordered subsets of ω and a separate set of all and only nonwell-ordered subsets of ω? So something that might be represented like $\aleph_{\alpha} + C$, where $C$ is a tag for the choice-free component? Or am I just totally off about this idea??? –  Mar 03 '25 at 03:39
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    @KristianBerry: Surely every subset of $\omega$ is well-orderable, being a subset of $\omega$. So you probably mean something different. – Asaf Karagila Mar 03 '25 at 08:19
  • Hmm... I read through Jech's book about the axiom of choice recentlyish to try to get a better handle on the topic, before that my main study of it was via the SEP article about it. But I'm not good with most of this material, and I don't know why $\mathbb{R}$ would not obviously admit of a well-ordering unless choice is given? Am I failing to understand the difference between $\mathbb{R}$'s elements and its subsets in this case? –  Mar 03 '25 at 13:44
  • Generally, I would want to ask, "In contexts with relativized/compromised choice, can we represent cardinalities that are compounds of well-ordered and nonwell-ordered components, like $|\omega_{\alpha} \cup A|$, for e.g. amorphous or other choice-trespassers A?" So that we might have an alternative to the Continuum Hypothesis like $2^{\aleph_0} = \aleph_{\alpha} + \mathfrak{X}$ for some non-aleph $\mathfrak{X}$? I'm minded to update my Overflow question about the "justifiable" universe accordingly, but I'm not confident in my reasons for an update yet... –  Mar 03 '25 at 13:53
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    @KristianBerry: Well, sometimes, yes, since if $X$ is not well-orderable and $\lambda\geq\aleph(X)$, then $X\cup\lambda$ is such example. But in the case of the continuum, since $|\Bbb{|R\times R|=|R|}$, if $\Bbb R=X\cup\lambda$, then $\lambda\leq|X|$ due to Tarski's lemma. – Asaf Karagila Mar 03 '25 at 14:36
  • Thank you so much! I saw here that you write: "Without the axiom of choice it might not even be an ℵ number." Is there some way to represent $\mathfrak{c}$ as something that might "oscillate" between an alephic and a non-alephic state, like switching its essence back and forth between two set-theoretic worlds, in some "dynamical" multiversal set theory? (So rather than saying it has a composite value "simultaneously.") (I guess this might be more like a quasi-physics question, although I'm even less confident in my physics sense right now...) –  Mar 03 '25 at 15:57
  • As an aside, are Fraktur letters the most commonly used for possibly non-aleph cardinals, does $\beth$ get repurposed sometimes for such effect, or is a different Hebrew letter reserved for this kind of use? I've long wanted to assign the lamedh glyph to some special class of transfinite numbers, maybe not cardinals but something... (Aesthetic argument: the word "lamedh" figures in the title of the legendary Thirty-Six Justifiers of Jewish lore, to which I'm rather partial after reading Schwarz-Bart's tragic novel about that lore.) –  Mar 03 '25 at 18:14
  • Here's one way I've thought of trying to write the idea down now: $\mathfrak{X} \leftrightharpoons \wp(\aleph_0) \leftrightharpoons \aleph_{\alpha}$. I thought I'd seen $\leftrightharpoons$ in a depiction of the function for the Mandelbrot set, somewhere, once upon a time, but now I'm seeing it used just in chemistry for some dynamical-equilibrium relation. So I guess I'd have to define a "new" equivalence relation, at the least, for this symbolism to be even remotely acceptable? (If the idea of a transworld "oscillating" Continuum cardinality is at all acceptable...) Sorry if this all... –  Mar 07 '25 at 21:07
  • ... seems like gibberish. –  Mar 08 '25 at 03:36