Let $(x_n)$ be a bounded sequence. Show that $\lim \inf x_n$ is an adherence value of $x_n$.
My proof:
Let's define $a:=\lim \inf x_n$. Let $\epsilon > 0$ and $N \geq 1$. We'll show that $|x_{kn} - a| \leq \epsilon$ for $kn \geq N$.
Let's first define our subsequence $(x_{kn})$ as $x_{kn} = \lim \inf \{x_{kn}, x_{kn + 1} \cdots\}$.
Since $x_{kn} \leq a$ it follows that $|x_{kn} - a| = a - x_{kn}$.
But from our definition of $x_{kn}$ it follows that $(x_{kn})$ is a monotonic non decreasing sequence, therefore $x_{kn} \leq x_{kn + 1} \leq \cdots \leq a \rightarrow a -x_{kn} \geq a -x_{kn + 1} \geq \cdots \geq 0$
Since $(x_n)$ is bounded, $(x_{kn})$ is also bounded and therefore $|x_{kn}| \leq L$ for some $L \in \mathbb{R}$. Hence we can see that: $$ -L \leq x_{kn} \leq L \rightarrow a + L \geq a-x_{kn} \geq a - L $$
Now consider the sequence $s_{kn} = a - x_{kn}$. From what we've previsouly concluded, $s_{kn}$ is monotnic non decreasing, bounded and $s_{kn} \geq 0 $ therefore it'll converge to $0$, and that proves that a is an adherence value of $x_n$.
I've seen a simpler proof of this fact using an equivalent statement of adherence value, without the need to show a subsequence... But that's what I initially did without that equivalent statement. Can someone please check my proof?
Should I be more rigorous to show that $s_{kn} \rightarrow 0$?
Thank you. Any critics and comments about my work are highly appreciated.
Let $a:=\liminf x_{n}$ and $y_{n}:=\inf_{k\geq n}x_{k}$ so that $y_{1}\leq y_{2}\leq\cdots\leq a$ and $a=\lim_{n\to\infty}y_{n}$.
If $a$ is not an adherence point of sequence $\left(x_{n}\right)_{n}$ then some open set $U$ must exist with $a\in U$ and such that $\left\{ n\mid x_{n}\in U\right\} $ is a finite set.
Then some integer $m$ must exist with $x_{n}\notin U$ for every $n\geq m$ and consequently $y_{n}\notin U$ for every $n\geq m$.
This however contradicts that $\lim_{n\to\infty}y_{n}=a$.
So we conclude that $a$ must be an adherence point of sequence $\left(x_{n}\right)_{n}$.