1

Suppose that I label every point in the real number line with either the label 1 or 0.

My first question is, can this be done so that the following property holds:

Every sequence of the form rn + s for n a natural number, r a fixed arbitrary nonzero real number and s a fixed arbitrary real number, gives the binary digits of a normal number.

My second question is: can one show that the cardinality of the set of numbers labeled 1 is the same as that of the reals. (I think it is straightforward to show the cardinality is not the same as that of the integers, and I am not assuming the continuum hypothesis.)

Bill
  • 265
  • Can you please clarify gives the binary digits of a normal number? – Jonathan Y. Jan 30 '14 at 18:40
  • I mean that if you write out the labels for the numbers in your sequence {rn + s} for n in the natural numbers, starting from n = 1, you get a number, e.g. for r = pi and s = 2 perhaps the labels give me 0.110100100010111100... This number should be normal, in the sense defined in the Wikipedia article on normal numbers, for every choice of r and s with r not zero. – Bill Jan 30 '14 at 18:44
  • Is there any formal content to your assumption that each point has a particular label with such-and-such probability? It looks like all you're looking for is whether there exists some labeling (random or not) such that every arithmetic sequence maps to a normal sequence of bits? – hmakholm left over Monica Jan 30 '14 at 18:47
  • this is probably not what you have in mind: run a two state markov process with stationary initial distribution $= (\frac 12, \frac 12)$. Flip a coin to decide what stat you start in, run the process forward & backward, and label the real according to the state of the process. I'm guessing you want something more like i.i.d. labelling. – mike Jan 30 '14 at 18:51
  • I don't think there is any content. In essence I'm asking, "if I define my labeling to have been done in such a way that the normal property I give holds... and by the way, does that make sense... then what is the cardinality of the set of numbers labeled one. – Bill Jan 30 '14 at 18:53
  • 1
    I'd remove the "probability 1/2" part from the question then. It sounds like you want an additional condition along the lines of "for any interval $I$, the set of points in $I$ labeled $1$ in has half the measure of $I$. – MartianInvader Jan 30 '14 at 19:04
  • mike - I'm unfamiliar with Markov processes. Can a two-state Markov process label all the reals or simply the numbers in one of my sequences? I simply couldn't tell by reading the Wikipedia article. Also, I don't know what i.i.d numbering is. Is there an easy reference? – Bill Jan 30 '14 at 19:05
  • @MartianInvader I think I have changed it as you suggest. I guess now I'm curious whether that automatically implies your measure property. I think possibly the measure property was what I originally had in mind when the question about the cardinality occurred to me, but for now I like the normal property better. – Bill Jan 30 '14 at 19:13
  • As for @MartianInvader's initial reading, note that it is not possible to label the reals such that the numbers labeled 1 in every interval are half of that interval by measure. – hmakholm left over Monica Jan 30 '14 at 19:47
  • @HenningMakholm Yes, it looks like it falls down because of the countable additivity property of a measure. Somewhat counterintuitive to me, but the proof looks solid enough. – Bill Jan 30 '14 at 20:06

1 Answers1

1

Yes, if you have a labeling $f$ with the property you give, then there are exactly continuum many numbers labeled $1$.

For each $x\in(0,1)$, the sequence $f(x), f(x1+1), f(x+2), \ldots, f(x+n), \ldots$ is normal by assumption, and therefore contains at least (!) one $1$. Let $g(x)$ be the least $x+n$ such that $f(g(x))=1$.

All of these sequences are disjoint, so $g$ is an injection from $(0,1)$ into the set of numbers with label $1$. By Bernstein's theorem, then, $f^{-1}\{1\}$ has the cardinality of the continuum.

It is not entirely clear to me that such a labeling exists, though. I think I can see an outline of an argument for it assuming the axiom of choice and the continuum hypothesis, but the details are slightly murky.

hmakholm left over Monica
  • 286,031