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The question in its original form deals with the problem of deciding whether the set $T$ of all irrational numbers in the set $[0,1]$ such that they have only digits $0$ and $1$ in their decimal expansion is countable or uncountable?

I have been thinking along this lines and this form of the problem deals with the question in the title, if we look at the set $S$ of all functions $f: \mathbb N \to \{ 0,1 \}$ then this set of functions describe all numbers in $[0,1]$ which have $0$ and $1$ in their decimal expansion, and if $S$ is countable then obviously $T$ is also countable as a subset of $S$ but if $S$ is uncountable then $T$ is uncountable because $R=S\setminus T$ is the set of all rational numbers in the $[0,1]$ that have $0$ and $1$ in their decimal expansion so obviously $R$ is countable as a subset of $\mathbb Q$, which is countable.

So how to decide whether $S$ is countable or uncountable?

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  • $\begingroup$ So far so good. Have you seen Cantor's diagonal proof? It is directly applicable to your $S$. $\endgroup$ – hmakholm left over Monica Apr 6 '13 at 13:21
  • $\begingroup$ @Henning Makholm Just now have finished reading it, and I do not like it, it looks to me that it hardly depends on the additional assumption, that there is no cardinality between that of $\mathbb N$ and that of $\mathbb R$. But that is exactly the Continuum Hypothesis. Oh, now I understand where did that hypothesis come from, thank you! $\endgroup$ – user67878 Apr 6 '13 at 13:41
  • $\begingroup$ Um, no, that must be a mistake. Cantor's proof doesn't depend on the continuum hypothesis. All it shows is that the cardinality if $\mathbb R$ (or $\mathcal P(\mathbb N)$ or $\{0,1\}^\mathbb N$) is strictly larger than the cardinality of $\mathbb N$, that is, it is uncountable. It doesn't tell you which larger cardinality you get. $\endgroup$ – hmakholm left over Monica Apr 6 '13 at 13:49
  • $\begingroup$ (One can show that $\mathbb R$ has the same cardinality of $\mathcal P(\mathbb N)$ without needing any diagonalization or fancy hypotheses.) $\endgroup$ – hmakholm left over Monica Apr 6 '13 at 13:51
  • $\begingroup$ @Henning Makholm Thank you, I feel that I must clarify some things related to set theory. $\endgroup$ – user67878 Apr 6 '13 at 14:05
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The set $S$ is often denoted as $2^\Bbb N$, and it is not hard to show that $2^\Bbb N$ has the same cardinality as $\mathcal P(\Bbb N)$, by mapping a subset to its indicator function.

Use Cantor's theorem to conclude that $S$ is uncountable.

One can actually show further that $|[0,1]|=|\mathcal P(\Bbb N)|$, and that in fact $T$ has the same cardinality as $[0,1]$ as well.

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  • $\begingroup$ Can you edit your answer to include the mapping from $2^\Bbb N$ to $\mathcal P(\Bbb N)$ or show me where can I find the proof using that indicator function? $\endgroup$ – user67878 Apr 6 '13 at 13:46
  • $\begingroup$ @Thus: math.stackexchange.com/a/41007/622 $\endgroup$ – Asaf Karagila Apr 6 '13 at 13:48

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