A proof about the limit infimum of a bounded sequence I tried to find a proof for this statement: If we have a bounded sequence $x_n$, then the limit infimum is defined as $a=\liminf_n x_n$ such that $a$ is the largest of real numbers which have the property that for all $a' < a$ we have finitely many $x_n < a'$.
Intuitively I can see that this is valid: If we have a plot of a bounded sequence $x_n$ we draw a horizontal line $y=a' < a$ and there must be a finite number of $x_n$ points in the below of the line. But as soon as that line passes $y=a$ then the number $x_n$ points under the line becomes infinite. I have miserably failed to put this thought into a mathematically valid statement, so I turned here for help.
Thanks in advance.
 A: This is one of the definitions of $\lim \inf$. There are several we may choose from, I will use the following definition of $\lim \inf$: Let $x_n$ be a bounded sequence and let $E \subset \mathbb{R}$ be the collection of all subsequential limits of $x_n$. We define $$a=\lim\inf x_n = \inf \{ x \in E \}$$
Now suppose there is an $a' < a$ for which there are infinitely many $n$ such that $x_n < a'$. This means there is a subsequence of $x_n$, call it $x_{n_k}$ for which $x_{n_k} < a'$ for all $k$. This subsequence is bounded, which means there is some subsequential limit. This limit can be at most $a'$, which is strictly less than $a$. This is a contradiction.

For the definition you requested it can be proved in a similar fashion. Suppose that there is an infinite number of $n$ for which $x_n < a' < a$. This means that for each $n$ there is a $k>n$ for which $x_k < a'$. Hence $y_n = \inf_{k>n} x_k < a'$ for all $n$. Now this sequence $y_n$ is upper bounded by $a'$, and $\lim\inf x_n = \lim_{n\to \infty} y_n \le a'$ which is a contradiction.
