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.

  • $\begingroup$ lim inf and lim sup don't necessarily converge- but if they don't, neither does the function. You can show that there is a subsequence that diverges in the same way, though (e.g. if $x_n = (-1)^n n$, then $\limsup x_n = \lim x_{2n}$, and $\liminf x_n = \lim x_{2n+1}$) $\endgroup$ – Clinton Bradford Oct 10 '14 at 2:32

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

If the sequence is bounded, let

$$x^* = \limsup x_k$$


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

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


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

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 $\liminf x_k$.

  • $\begingroup$ how can $x_{k_m}$ converge to $\lim \sup x_n$ ? the variables in here are confusing me a little bit. isn't it supposed to be a subsequence of $x_k$ and not $x_n$ where did the $x_n$ come from? $\endgroup$ – Alex Chavez Oct 10 '14 at 7:21
  • $\begingroup$ @AlexChavez: $\limsup_n x_n = \limsup_k x_k$ You can change the index notation as long as you are consistent. Nevertheless you are correct that my notation was confusing. I changed it so that the basic sequence is $(x_k)$ throughout. $\endgroup$ – RRL Oct 10 '14 at 14:56

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