# Why is $\mathbb{R}$ not hereditarily countable set?

Edit: I think I should rather ask a different question to dismantle my confusion.

So, why is $\mathbb{R}$ not hereditarily countable set? It seems obvious, but I just can't seem to think of any possible proof... does this just rely on the fact that the set cannot be well-ordered in its original order?

Edit: Right. Sorry for asking a stupid question. Just a slight moment of confusion, forgive me.

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Do you know what the transitive closure of a set is? – Brian M. Scott Sep 18 '12 at 3:46
It's a set which is countable, all its elements are countable, all the elements of its elements are countable, and so on. An example of a countable set which is not hereditarily countable is $\{\mathbb{R}\}$. – Yuval Filmus Sep 18 '12 at 3:49
You didn’t read Yuval’s comment carefully enough: the set in question is $\{\Bbb R\}$, not $\Bbb R$. $\Bbb R$ is obviously not hereditarily countable, since it’s not even countable. $\{\Bbb R\}$ is countable, since it has only one element, the set $\Bbb R$, but since that element is not itself a countable set, $\{\Bbb R\}$ is not hereditarily countable. – Brian M. Scott Sep 18 '12 at 4:04

First comes the question how do you define the real numbers. In many cases the real numbers are thought of as simply $2^\omega$ or $\omega^\omega$, cardinality-wise there is no difference. However you can also take a set of very large sets, and assume that one is the real numbers.
So there comes a question of natural definition. I will assume that we mean binary $\omega$-sequences. Now what is a sequence? It is a function whose domain is $\omega$ and range included in $\{0,1\}$. It is not too hard to see that this is indeed a hereditarily-countable set.
However Cantor's theorem tells us that $2^\omega$ has cardinality strictly larger than $\aleph_0$. Therefore the real numbers fail to satisfy this requirement of being a countable set to begin with and thus it is not a hereditarily countable set.
What does $\omega$ denote here? – Rasmus Sep 18 '12 at 8:17