A sequence for which the set of limits points is the interval $[0,1]$.

Question

My professor challenged me to find a sequence with limit points the whole interval $[0,1]$.

Answer 1

Try $$ a_n=\sin^2n. $$ Also, you can enumerate $\mathbb Q\cap [0,1]$: $$ 0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{1}{6}\cdots. $$

Answer 2

The rationals are dense in the interval [0,1]. So the sequence mapping N into $Q\cap [0,1]$ will give a sequence where if L is the set of limit points, min(L) = 0 and max(L) = 1. Andre's explicit constructions above are nice examples, although we can come up with others. Can you?

Answer 3

If you can find a sequence such that:

• $\limsup a_n=1$ and $\liminf a_n=0$
• $\lim (a_{n+1}-a_n) = 0$

The for this sequence the set of limit points will be the interval $[0,1]$. (See: Set of cluster points of a bounded sequence and If a sequence satisfies $\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0$ then the set of its limit points is connected. But if you come up with an example of such sequence, it will probably be easier to verify what the set of limit points of that particular sequence looks like than to check the proof of the general result valid for all sequences with these properties.)

Answer 4

One sequence is to use rational fractions with increasing powers of 2 in the denominators, and where the numerators are the odd numbers from $1$ to $2^n-1$:

$$\frac12,~~ \frac14, \frac34,~~ \frac18, \frac38, \frac58, \frac78,~~ \frac1{16},\frac3{16},\dots$$ $\def\mod{\text{ mod }}$

The fractions $\dfrac{a_n}{b_n}$ can easily be made explicit:

$$\begin{align} a_n &= 1 + 2\cdot (n\mod 2^{d_n}) \\ b_n &= 2^{1+d_n} \\ \end{align}$$

Where $d_n = \lfloor\log_2n\rfloor$, and $u\mod v$ is the remainder of $u/v$ with $0\leqslant (u\mod v) < v$.