Why is $\lim \sup x_n$ a limit point of $x_n$?

Question

I found an answer trying to construct a subsequence of $x_n$ converging to lim sup $x_n$. But unfortunately, I don't get it. Can someone explain that or give a better/simpler example?

I define lim sup as either of the following:

$\lim \sup x_n = \lim_{n \to \infty} \sup_{k \geq n} x_k$

or

$\lim \sup x_n = \inf_m \sup_{k \geq m} x_k$?

Answer 1

Let $(x_n)$ be a sequence and also let $s=\limsup x_n$. That is:

$$s=\lim_n\sup_{k\geq n}x_k.$$

For $\epsilon_1=1>0$ there exists some $n_1$ such that:

$$s\leq\sup_{k≥n_1}x_k<s+1.$$

Hence, we can find some $k_1≥n_1$ such that $\sup\limits_{k≥n_1}x_k≥x_{k_1}>\sup\limits_{k≥n_1}x_k-1$.

Similarly, we can find a $k_2≥ k_1$ such that:

$$s+\frac{1}{2}>\sup\limits_{k≥n_2}x_k≥x_{k_2}>\sup\limits_{k≥n_2}x_k-\frac{1}{2},$$

and, inductively, find a subsequence $(x_{k_m})$ of $(x_k)$ such that:

$$s+\frac{1}{m}>\sup\limits_{k≥n_m}x_k≥x_{k_m}>\sup\limits_{k≥n_m}x_k-\frac{1}{m}$$

Letting $m\to\infty$, we take:

$$\lim_m x_{k_m}=s.$$

Answer 2

Set $g_n=\sup_{k \geq n} x_k$. Then $(g_n)$ decreases to $s=\lim_{n \to \infty} \sup_{k \geq n} x_k$. Thus means that there is an integer $n_1$ such that $g_{n_1}-s<1/2$ and an $x_{n_1}\in (x_n)$ such that $|x_{n_1}-g_{n_1}|<1/2.$ Combining these inequalities gives $|x_{n_1}-s|<1.$

Now suppose we have chosen integers $n_1<\cdots <n_l$ such that $|x_{n_m}-s|<1/m;\ 1\le m\le l.$ Since $g_n\to s$, there is an integer $N>n_l$ such that $n \ge N\Rightarrow g_n-s<\frac{1}{2(l+1)}$. And since $g_N=\sup_{k \geq N} x_k$, there is an integer $n_{l+1}>N$ such that $|g_N-x_{n_{l+1}}|<\frac{1}{2(l+1)}.$ Then, $n_{l+1}>n_l$ and $|x_{n_{l+1}}-s|<\frac{1}{l+1}$, which completes the induction. It follows that $x_{n_l}\to s.$

Answer 3

This answer just rephrases the accepted answer above.

Let $\lim \sup x_n = K$ and let $s_n = \sup_{k \geq n} x_k$ so that $\lim s_n = K$ from above. Then, for $\epsilon = 1$, there exists an $N_1 \in \mathbb{N}$ such that for all $n \geq N_1$, we have $|s_n - K| < 1$ $\implies $ $K - 1 < s_n < K+1$ $\implies $

$K - 1 < \sup_{k \geq N_1} x_k < K + 1$ $\implies $ there exists $k_1 \geq N_1$ such that $K - 1 \leq x_{k_1} < K+1$. The upper bound is obvious. The lower bound follows from the fact that $\lim s_n = K$. If for all $k_1 \geq N_1$, we have $x_{k_1} < K - 1$, then that would imply $s_n \leq K - 1$ for all $n \geq N_1$ which is a contradiction since $s_n$ is a non increasing sequence with $\lim s_n = K$.

Let $\epsilon = 1/2$, there exists an $N_2 \in \mathbb{N}$ such that for all $n \geq N_2$, we have $|s_n - K| < 1/2$ $\implies $

$K - 1/2 < \sup_{k \geq N_2} x_k < K+ 1/2$. Again, by the same logic as before, there exists $k_2 \geq N_2$ such that $K - 1/2 \leq x_{k_2} < K + 1/2$.

Proceeding similarly, we have $k_{m}$ such that $K - (1/2)^{m-1} \leq x_{k_m} < K + (1/2)^{m - 1}$. Clearly, this gives us a subsequence $(x_{k_m})$ such that $\lim_m x_{k_m} = K$.