show there is a subsequence that converges to $\lim_{x_k \rightarrow \infty} \inf x_k $

Question

For a sequence $(x_k) \in \mathbb{R}$ ,

$\lim_{k \rightarrow \infty} \sup x_k = \lim_{k \rightarrow \infty}$ $(\sup{x_l| l ≥ k})$

$\lim_{k \rightarrow \infty} \inf x_k$ = $\lim_{k \rightarrow \infty}(\inf{x_l| l ≥ k})$.

Show there is a subsequence of $(x_k)$ that converges to $\lim_{x_k \rightarrow \infty} \sup x_k $, and a subsequence that converges to $\lim_{x_k \rightarrow \infty} \inf x_k $

Im trying to show both of these but I am having trouble explaining why the subsequence would be convergent since it doesn't say the sequence $(x_k)$ is bounded.

Answer 1

If the sequence is unbounded above, then lim sup is infinite.

If the sequence is bounded, let

$$x^* = \limsup x_k$$

and

$$\alpha_k = \sup_{l\geq k}x_l$$

Since $x_k$ is bounded, $\alpha_k$ is bounded below and non-increasing.

Hence,

$$\lim_{k \rightarrow \infty}\alpha_k=\lim_{k \rightarrow \infty}\sup_{l\geq k}x_l= \inf_{k}\sup_{l\geq k}x_l=x^*$$

For any $\epsilon >0$, there exists $N \in \mathbf{N}$ such that for $k \geq N$

$$x^* - \epsilon < \alpha_k < x^* + \epsilon.$$

In particular, given $m \geq N$, since $x^*-\epsilon < \sup_{k\geq m}x_k=\alpha_m$, there exists $k_m \geq m$ such that

$$x^* - \epsilon < x_{k_m} \leq\alpha_m< x^* + \epsilon.$$

Hence, $|x_{k_m} - x^*| < \epsilon$ when $m \geq N$ and the subsequence $(x_{k_m}$) converges to $x^*$.

You can make a similar argument for $\liminf x_k$.