1
$\begingroup$

Let $S$ denote an enumerated infinite collection of absolutely normal (i.e. normal in all integer bases greater than or equal to $2$) real numbers (with no repetitions) such that any natural number corresponds to a unique element of $S$.

Let $F(S)$ denote a real number such that $0 < F(S) < 1$ and each $(2^x \times (2y + 1) - 1)$-th bit of the base-$2$ representation of the fractional part of $F(S)$ is equal to an $y$-th bit of the base-$2$ representation of the fractional part of an $x$-th element of $S$.

Question: does there exist an example of $S$ such that the corresponding $F(S)$ is not absolutely normal? If yes (or no), is it possible to prove this?

This question is not about an explicit example of $S$ satisfying the described property, it is only about the existence of such $S$ (and provability of the existence).

Improve this question
$\endgroup$
4
  • $\begingroup$ I expect the answer to the question in the title to be no (and to the one in the body to be yes). Let $s=0.s_1s_2s_3...$ be absolutely normal. Then $s'=0.s_2s_4s_6...$ will again be normal base $2$, and for generic $s$ it should be absolutely normal, but I don't know how to prove it formally. At any rate if $S$ has $s$ as its zeroth element and $s'$ as its first, then $F(S)$ won't be normal, since its $4k+1$st and $4k+2$nd bits will always match. $\endgroup$
    – Wojowu
    Commented Dec 30, 2020 at 11:39
  • $\begingroup$ @Wojowu That's almost right, except that "generic" should be "random" - generic reals are far from normal. $\endgroup$
    – Noah Schweber
    Commented Dec 30, 2020 at 16:57
  • $\begingroup$ @NoahSchweber Apologies, I have never seen "generic" used in the sense of comeager (which I think is what you are implying this term means). $\endgroup$
    – Wojowu
    Commented Dec 30, 2020 at 20:09
  • $\begingroup$ @Wojowu Yes, that's the standard usage in logic at least; I was under the impression that that was standard elsewhere, but I could be wrong. $\endgroup$
    – Noah Schweber
    Commented Dec 30, 2020 at 20:10

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .