1
$\begingroup$

Let $r_n$ be an enumeration of the rationals in $[0,1]$, does the sequence $\{B_{\frac{1}{n}}(r_n)\}_{n=k}^m$ cover $[0,1]$ for $m-k$ finite?

This came up while trying to solve a different problem, and it strikes me as a deep question, one which I think is true but haven't been able to answer in the affirmative.

First I thought well what about the sequence of rationals $\frac{1}{3n}$, but then I realized, this is not an enumeration of the rationals! For $r_n$ to be an enumeration of the rationals, we need the number of steps in the sequence between any two rationals in $[0,1]$, to be finite, and thus since the rationals are dense in $[0,1]$, we can get arbitrarily close to any point in $[0,1]$ in a finite number of steps. Furthermore, $\sum\frac{1}{n}$ diverges, and thus no matter how far down I choose $k$, I can find an $m > k$ such that $\sum_{n=k}^m\frac{1}{n}$ is as large as I like. And so no matter how far down $k$ is, we never lose at least the possibility of covering an interval of any size in a finite number of steps.

So the question seems to be, given that an acceptable enumeration of the rationals is actually more restrictive than we might at first expect, is the placement of these intervals of length $\frac{1}{n}$ 'diffuse' or 'random' enough to cover every point in $[0,1]$?

One of my reservations, however, is let $x\in [0,1]$. Then given $\varepsilon > 0$, I know I can find some $m>k$ such that $|x - r_m|<\varepsilon$. But what if for this particular $x$, every time I find such an $m$, we have that $\frac{1}{m} < |x - r_m|$? My hunch is that somehow the fact that the harmonic series diverges saves us from this, and if we were to consider intervals of length $\frac{1}{n^2}$, this would no longer be true!

Anyways, can anyone make sense of all this? Thanks.

$\endgroup$

1 Answer 1

3
$\begingroup$

It depends very much on the enumeration $(r_n)$.

If $(r_n)$ is an enumeration obtained by choosing $r_{n+1}$ in the interval $(r_n+1/(2n),r_n+1/n)$ if $r_n\lt1-1/(2n)$, and in the interval $(0,1/n)$ otherwise, then the answer is yes.

If $(r_{6^n})$ enumerates the rational numbers in $[0,\frac12]$ and $(r_n)$, $n$ not a power of $6$, enumerates the rational numbers in $(\frac12,1]$, then $[0,\frac12]$ is not entirely covered hence the answer is no.

$\endgroup$
3
  • $\begingroup$ why wouldn't $[0,\frac{1}{2}]$ be covered? $\endgroup$
    – Set
    Commented Oct 29, 2012 at 18:03
  • 1
    $\begingroup$ Because the sum of the lengthes of the intervals based on rational numbers $r\lt\frac12$ is twice $\sum\limits_{n\geqslant1}6^{-n}$, which is $\frac25\lt\frac12$. $\endgroup$
    – Did
    Commented Oct 29, 2012 at 18:06
  • $\begingroup$ ohhh, I see, clever, ok I'll have to think this one through some more, thanks. $\endgroup$
    – Set
    Commented Oct 29, 2012 at 18:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .