TITLE: sequences of real numbers
QUESTION [14 upvotes]: Let $\lbrace x_i\rbrace_{i=1}^\infty$ be a sequence of distinct numbers in $(0,1)$. For any $n$ after deleting $x_1,...,x_n$ from $[0,1]$ we get $n+1$ subintervals. Let $a_n$ be the maximum length of these subintervals. Is there any sharp lower bound for $\limsup_n na_n$ ?

REPLY [11 votes]: Here is how to achieve $1/\ln 2$. It is a modification of Harald's construction. We take the sequence $1/2$, $1/4$, $3/4$, etc. and apply the transformation $x \to \ln (1+x)/\ln 2$.
So the sequence is $\ln(3/2)/\ln(2), \ln(5/4)/\ln (2),\ln(7/4)/\ln(2),\dots$
The general terms is $x_n$ for $n=2^a+b$ with $b<2^a$ is
$$\ln \left( \frac{2^{a+1} + 2 b+1}{2^{a+1}}\right)$$
or, equivalently
$$\frac{\ln( 2^{a+1}+2 b + 1)}{\ln(2)} - a-1$$
Then after $n = 2^a + b$ steps for $b<2^a$, we will have just added the cut $(2 b+1 ) /(2^{a+1})$, the largest interval will be
$$\left[\frac{\ln( 2^{a}+2 b + 1)}{\ln(2)} - a, \frac{\ln( 2^{a}+2 b + 2)}{\ln(2)} - a\right] $$
whose length is:
$$\frac{ \ln \left(1 + \frac{1}{2^{a}+b+1}\right)}{\ln 2}$$
If we multiply by $2^a+b+1$, the $\lim\sup$ is $1/\ln 2$.
To show this is optimal, form a binary tree where the vertices are intervals that appear at any point in a sequence. Each interval at some point splits into two intervals, which gives the tree structure. Label each vertex with the step at which it splits. For any $\beta>\lim\sup n a_n$, for all but finitely many vertices, the length of the interval labeled $n$ is at most $\beta/n$. The sum of the lengths of the intervals on each row is $1$, so the sum of the first $k$ rows is $k$.
Thus $k$ is at most the sum over $2^k-1$ distinct numbers $n$ of $\beta/n$ plus a constant coming from the finitely many vertices where the length is not at most $\beta/n$. So $k$ is at most the sum over the first $2^k-1$ numbers of $\beta/n$ plus a constant, which is at most $\beta k \ln 2$ plus a constant. So $\beta\geq 1/\ln 2$.
So $\lim\sup n a_n \geq 1/\ln 2$.