Problem with finding an ordered subsequence of a bounded sequence that converges to lim inf or lim sup I've been reading Mathematical Analysis I by Zorich and I'm stuck with the construction of a subsequence of a bounded sequence that converges to lim inf (limit inferior) or analogously lim sup (limit superior).
I've searched Google and Math Stackexchange and the proof on Subsequence that converges to $\lim \text{inf}$ is essentially what everyone offers.
Let $(x_{k})$ be the sequence and $i_{n}= \inf_{k \geq n} x_{k}$.
We wish to choose $k_{n}\in \mathbb{N}, k_{n}\geq n$ such that $i_{n}\leq x_{k_{n}}<i_{n}+\frac{1}{n}$ so that $(x_{k_{n}})$ forms a subsequence that converges to $\lim_{n \rightarrow \infty}i_{n}$.
However, I can't seem to justify the existence of $x_{k_{n}}, k_{n}>k_{n-1}$ because contrary to what is said in the link, there aren't infinitely many $x_{m}, m \geq n$ such that $i_{n}\leq x_{m}<i_{n}+\frac{1}{n}$. The definition of the infinum only guarantees one such element, which might be the same element used as a previous $x_{k_{n-1}}$.
A simple example is the sequence 5, 5, 1, 5, 5, 5 ...
$i_{1} = i_{2} = i_{3} = 1$ and $x_{k_{1}} = x_{k_{2}} = x_{k_{3}} = 1$ is the only possibility, which does not form a legitimate subsequence. Clearly, the subsequence 5, 5, 5, 5, 5, ... converges to lim inf = 5 but my point is that the construction in the proof has a loophole.
Would appreciate it if someone could find a way to solve this problem.
Thanks!
Edit: thanks RRL! However, isn't the subsequence generated by your method only for a particular ε and if ε is reduced, the subsequence might not fulfil the requirement for convergence to $x^{*}$: being within the ε-neighborhood of $x^{*}$ from a certain N onwards. 
 A: What you may be missing is that, by definition, the limit inferior itself is a limit (or supremum) of the bounded and non-decreasing sequence $(i_n)$.  There are infinitely many terms of the sequence $(i_n)$ within $\epsilon$ of $\liminf x_k$. 
Let
$$x^* = \liminf x_k$$
and 
$$i_n = \inf_{k\geq n}x_k.$$
Since $(x_k)$ is bounded, $i_n$ is bounded above. Furthermore, the sequence $(i_n)$ is non-decreasing, since for $m > n$ we have $\{x_k:\, k \geqslant m \} \subset \{x_k:\, k \geqslant n \}.$
Hence,
$$\lim_{n \rightarrow \infty}i_n=\lim_{n \rightarrow \infty}\inf_{k\geq n}x_k= \sup_{n}\inf_{k\geq n}x_k=x^*$$
Assume now that $i_n < x^*$ for all $n$.  If not,  then the special case where $i_m =x^*$ for some $m$ can be handled in a straightforward way. 
Since $i_1 < x^*$, there exists $k_1 \geqslant 1$ such that $i_1 < x_{k_1} < x^*$. As $i_n \to x^*$, there exists $n_2 > k_1$ such that $x_{k_1} < i_{n_2} < x^*$.  Again by the properties of an infinum, there exists $k_2 \geqslant n_2 > k_1$ such that $i_{n_2} < x_{k_2} < x^*$.
Proceeding inductively, we generate subsequences with the property
$$i_{n_m} < x_{k_m} < x^*.$$
By the squeeze theorem,  it follows that as $ m \to \infty$, the subsequence $(x_{k_m})$ converges to the limit inferior:
$$x_{k_m} \to x^*.$$
