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And in general, what else can we say about the set of normal numbers? We know that a normal number must be irrational, and hence it is possible that the set of normal numbers is countable or uncountable, since the irrational numbers are uncountable. I have been unable to find much literature or research about this topic.

ImJackJ
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    Almost every real number (in the sense of Lebesgue measure) is normal. A set of positive Lebesgue measure has cardinality $c$. – Robert Israel Oct 23 '23 at 02:46
  • I suppose that by "normal" you mean absolutely normal. If you just mean normal in base ten, it is trivial that there are continuum many normal numbers if there is one, since you can alter a "thin" set of its decimal digits at will without affecting normality. – user14111 Oct 23 '23 at 02:51
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    https://en.wikipedia.org/wiki/Normal_number – user14111 Oct 23 '23 at 02:52
  • Verónica Becher authors a fair number of papers about normal and absolutely normal numbers. Google should get you PDFs of many of them. – Brian Tung Oct 23 '23 at 03:29

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Suppose the set of normal numbers is $N$. Then, according to Borel's Normal number theorem $N^c$ is negligible, that is, $N^c$ can be covered by a countable union of intervals $I_k$ : $$ N^c \subset \bigcup_k I_k , $$ such that the sum total of their lengths can be made arbitrarily small: $$\sum_k \text{length}(I_k) < \varepsilon$$ The members of $I_k$ are such real numbers, a finite number of whose first digits aren't normally distributed. This means that there can still be uncountably infinite number of irrational members of $N^c$, even though $\Bbb Q \subset N^c$. This means, even though the Lebesgue measure of $N^c$ is 0, it's cardinality is $c$ as suggested in the comments by @Robert Israel. Thus, we can see that $N^c$ is dense in $\Bbb R$.

Awe Kumar Jha
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    Much stronger results than continuum many non-normal numbers in each nonempty open interval are known. In each nonempty open interval we have: (A) the set of non-normal numbers is maximally large in the sense of Hausdorff dimension -- has Hausdorff dimension $1$ (see here for a stronger result); (B) in the sense of Baire category the set of non-normal numbers is maximally large -- it's the complement of a first category set. For MUCH stronger results still, see my comments to this MSE question. – Dave L. Renfro Jan 12 '25 at 16:32