Yesterday I asked in the question at A proof about the limit infimum of a bounded sequence about the proof of the following statement:
Let $x_n$ be a bounded sequence of real numbers. Then the limit infimum $a=\liminf_n x_n$ is the largest real number with the property such that for all real numbers $a'< a$ there are finitely many $x_n < a'$.
In my question a beatiful proof was given as answer, using an alternative definition of $\liminf$. Still, I tried to find another proof of that and I think I have found something:
I am using the $\lim_{n \to \infty} \inf_{k \geq n} x_k = \sup_n \inf_{k \geq n} x_k$ definition.
I first try to prove that for every $a' < a=\liminf_n x_n$ there are finitely many $x_n < a'$. The sequence $y_n = \inf_{k \geq n}x_k$ is non decreasing and converging and since $\lim_{n \to \infty} y_n = a$ there is a $N \in \mathbb{N}$ for each $\epsilon > 0$ such that only $y_n$ with $y_n < a -\epsilon$ are the ones with $n < N$. For each element of the set $y_i \in (y_1,y_2,...,y_{N-1})$ and the corresponding $x_i \in (x_1,x_2,...,x_{N-1})$ it is $x_i \geq y_i$ since $y_i = \inf_{k \geq i} x_k$. This means there are at most $N-1$ $x_n$ such that $x_n < a - \epsilon$. This is the first part of the proof.
In the second part of the proof I need to show that for every $a' > a = \liminf_n x_n$ there are infinitely many $x_n$ such that $x_n < a'$. First I assume that for such an $a'$ there are finitely many $x_n$. Since there are finitely many $x_n$ there must exist a $K$ such that for $k > K$ it is $x_k > a'$. This in turn means $\inf_{k > K} x_k > a$. This is a contradiction since $a$ is defined as $a=\sup_n \inf_{k \geq n} x_k$.
Is this proof correct?
Thanks in advance
The second part of the proof is correct.
The part of the proof using $N$ and $\epsilon$ is wrong. You need to state that the only $y_k<a-\epsilon$ are the $k<N$. You didn't say that if $k\geq N$ then $y_k\geq a-\epsilon$, so you haven't proven that there are finitely many $y_k$. It's not hard to fix.
Lemma: If $y_n$ is a non-decreasing sequence of real numbers with limit $y$. Then for any $\epsilon>0$ there is an $N$ such that $y_n<y'$ if and only if $n<N$.
It is often best to break out a "trick" like this into a lemma to spotlight it, rather than springing it in the middle of a proof, where it confuses things.