Existence of sequence whose set of subsequential limits is $[0,1].$ I need help with this problem: 
Does exist a sequence $ \{x_n\} $ such that  $ \lim \inf \{x_n\} = 0 $ and $ \lim \sup \{x_n\} = 1 $ , and for every real $\alpha \in [0,1] $  exists a subsequence of $ \{x_n\} $ that converges to $\alpha$ ? 
 I saw a similar question already posted but it didn't include the lim inf and lim sup restriction.
Thanks in advance
 A: Sure this is possible! There are a number of ways to do this, one of them being to use the Farey sequence. We can recursively build our sequence by starting with $x_0=0$ and $x_1=1$ and then go to the next new rational number in the order that they are in the Farey sequence, breaking ties ($1/4$ and $3/4$ get introduced at the same step) by putting the smaller one first. This sequence slowly "refines" rational estimates, and so since $\mathbb{Q}\cap[0,1]$ is dense in $\mathbb{R}\cap[0,1]$, has a subsequence that converges to every real number. This particular example is very important in Continued Fraction theory. 
Analogously, you can do something similar refining by decimal digit. The first few elements of that sequence are $$0,1,0.1,0.2,0.3,\ldots,0.9,0.01,0.02,\ldots, 0.11,0.12,\ldots0.98,0.99,0.001,\ldots$$
It turns out to be the case that any mapping of $\mathbb{N}$ to $\mathbb{Q}\cap[0,1]$ will do as your sequence. Can you figure out why?
A: I would recommend the sequence $$\frac{1}{2}(\sin(n) + 1)$$ because $\{\sin(n) : n\in \mathbb{N}\}$ is dense in $[-1,1]$.
Edit. Here is a nice an constructive proof of the density property of the sine function. 
A: Take any irrational $\beta$ and let $x_n = n \beta \mod 1$.
A: Density of $\sin n$ is not easy to prove, thus an overkill. Instead I'd suggest you consider the following lemma:

Let $(u_n)$ be a bounded sequence such that $u_{n+1}-u_n\to 0$. Then the set of subsequential limits of $(u_n)$ is a closed interval $[\alpha, \beta]$.

Consider $u_n=\frac{1}{2}(\sin(\ln n) + 1)$
The mean value theorem says $\displaystyle \frac{\sin(\ln (n+1))-\sin(\ln n)}{n+1 -n} = \frac{\cos(\ln(\xi_n))}{\xi_n}\to 0$
The set of subsequential limits of $(u_n)$ is therefore an interval and surely a subset of $[0,1]$. It suffices to prove that $0$ and $1$ are indeed subsequential limits of $(u_n)$. 
To this effect, consider the subsequences given by these indices $\alpha_n=\lfloor \exp(\frac \pi 2+2n\pi) \rfloor$ and $\beta_n=\lfloor \exp(-\frac \pi 2+2n\pi) \rfloor$
