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Can this happen?

We know the theorem that a union of countably infinite sets is countable, but can we, for example, take a union of countably infinite sets (not necessarily mutually exclusive) and get $\mathbb{R}$, which is uncountably infinite?

I can surmise no way, because I know that $\mathbb{Q} \cup \mathbb{I}=\mathbb{R}$. Any suggestions?

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  • $\begingroup$ The irrationals aren't countable so that is not a union of countable sets. $\endgroup$ – fleablood Nov 28 '16 at 4:13
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It's very easy to write $\mathbb{R}$ as a union of countable sets:

$$ \mathbb{R} = \bigcup_{x \in \mathbb{R}} \{ x \} $$

If you really want to insist on each set being infinite, you can do something like

$$ \mathbb{R} = \bigcup_{x \in \mathbb{R}} \left( \{ x \} \cup \mathbb{Z} \right)$$

You can even arrange for all of the individual sets to be disjoint! e.g.

$$ \mathbb{R} = \bigcup_{x \in [0,1)} \{ x + n \mid n \in \mathbb{Z} \} $$


You're misremembering the theorem: it's only a countable union of countable sets that's guaranteed to be countable. Each of the examples above is an uncountable union of countable sets.

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  • $\begingroup$ Beat me, +1!${}{}{}$ $\endgroup$ – Noah Schweber Nov 28 '16 at 4:02
  • $\begingroup$ Awesome, exactly what I was looking for. Thanks! $\endgroup$ – Xpert Nov 28 '16 at 6:43
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Your question is a bit confused. It is not true that a union of countable sets need be countable: only that a countable union of countable sets is countable. For example, the reals can be written as a union of countable (indeed, finite) sets: $$\mathbb{R}=\bigcup_{x\in\mathbb{R}}\{x\};$$ the point is that we need uncountably many sets to do so.


Interestingly, we need the axiom of choice to show that a countable union of countable sets is countable; but that's another story.

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A countable union of countable sets is countable, which can be proven by a diagonalization argument. An arbitrary union of countable sets may not be countable though.

Consider $\cup \{x\}, \forall x \in \mathbb{R}$

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    $\begingroup$ "which can be proven by a diagonalization argument" Huh? It's not really diagonalization in any sense that I'm aware of . . . $\endgroup$ – Noah Schweber Nov 28 '16 at 4:02
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    $\begingroup$ @NoahSchweber The enumeration of pairs $(a,b)$ of natural numbers in order of increasing $a+b$ and, for equal $a+b$, in order of increasing $a$ is often called "Cantor's first diagonal argument", because it proceeds along (short) diagonals in the array $\mathbb N\times\mathbb N$. In this connection, what we usually call a diagonal argument is called "Cantor's second diagonal argument". $\endgroup$ – Andreas Blass Nov 28 '16 at 5:05
  • $\begingroup$ @AndreasBlass Thanks - I did not know that! I retract my previous comment (but I'll leave it up so that your response, which I think is useful and interesting, makes sense most easily). $\endgroup$ – Noah Schweber Nov 28 '16 at 5:26
  • $\begingroup$ @AndreasBlass But don't the words "diagonalize" and "diagonalization" refer only to Cantor's second diagonal argument? $\endgroup$ – bof Nov 28 '16 at 5:54
  • $\begingroup$ @bof I don't think "diagonalize"is so precisely defined. I use it only in the sense of Cantor's second diagonal argument, but I don't complain if other people use it in a broader sense. $\endgroup$ – Andreas Blass Nov 28 '16 at 16:22

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