Construct a sequence of real numbers whose set of subsequential limits is precisely $[0, 1]$.

Question

Construct a sequence of real numbers whose set of subsequential limits is precisely $[0, 1]$. (From https://ocw.mit.edu/courses/18-100c-real-analysis-fall-2012/142d99f1707e9d98b024b955382d94e7_MIT18_100CF12_finlpractice.pdf )

Partial Solution: The sequence 0, 1, 0, 1/2, 1, 0, 1/4, 1/2, 3/4, 1, 0, 1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8, 1... would satisfy it:

a. Clearly no subsequential limits are outside the interval $[0,1]$.

b. For any $k \in [0,1]$, for any $\epsilon > 0$, there are infininitely many elements within $\epsilon$ of $k$.

Is this correct? How do I prove this more rigorously? To get started: What's a convenient way of writing the sequence I described above (for each power of 2, we need that many elements before going on the next power of 2).

Yes, this sequence is fine. You just show that every real number in $[0, 1]$ is the limit of some subsequence, and you can take the subsequence corresponding to its binary digits. This is already the most convenient way of writing this sequence, trying to explicitly describe the $n^{th}$ term as a function of $n$ is both annoying and unnecessary. — Qiaochu Yuan, Oct 23 '22 at 02:15
@Feng That needs a proof which is more difficult than if one uses a sequence like that of OP. — coffeemath, Oct 23 '22 at 02:45
@coffeemath I agree. Thanks for pointing that out. Sine function is the first thing that came into my mind when I saw this problem. So I left that comment. I didn't claim that my example is easier to prove. Just a suggestion for the OP. Maybe he/she will learn some new things from the link in my earlier comment :) — Feng, Oct 23 '22 at 02:48
@Feng Certainly your sequence has a short definition. But the sine function is a bit involved just in terms of its definition. I agree it is good to go through a proof of density of $|\sin n|.$ — coffeemath, Oct 23 '22 at 03:01
See Given closed $C \subseteq \mathbb{R}$ find a sequence with subsequences convergent to every point in $C$ and nowhere else AND Any closed subset of $\mathbb C$ is the set of limit points of some sequence AND Existence of sequence whose set of subsequential limits is $[0,1].$ — Dave L. Renfro, Oct 23 '22 at 08:01
Alternatively, take a sequence that includes all rationals in [0,1]. — DanielWainfleet, Oct 23 '22 at 17:59

Construct a sequence of real numbers whose set of subsequential limits is precisely $[0, 1]$.

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