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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

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    $\begingroup$ No reason for $\epsilon$ stuff in the first part. It isn't wrong, but easier to prove: If there are infinitely many $x_i<a'$ then for any $n$ there is a $k>n$ such that $x_k<a'$. So $y_k=\inf_{k\geq n} x_k<a'$, and hence $\lim y_k \leq a'$. $\endgroup$ – Thomas Andrews Jul 24 '14 at 12:52
  • $\begingroup$ Thanks for the comment. Is the rest of the proof true? $\endgroup$ – Ufuk Can Bicici Jul 24 '14 at 13:01
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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.

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  • $\begingroup$ I said that "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 $y_n < a -\epsilon$ for $n < N$.". I thought saying that the sequence is nondecreasing and converging is enough to imply that the only $y_n$ with $y_n < a - \epsilon$ are the ones with $n <N$ Isn't this valid? $\endgroup$ – Ufuk Can Bicici Jul 24 '14 at 13:19
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    $\begingroup$ You didn't say it, and you could have, so it was confusing. You used $N-\epsilon$ in an odd way, did not highlight it, and did not put that condition on $N$. The letters $N-\epsilon$ don't imply the other fact - you still have to state it. $\endgroup$ – Thomas Andrews Jul 24 '14 at 13:21
  • $\begingroup$ Ah I saw your point. I edited it and added that only $y_n$ with $y_n < a - \epsilon$ are the ones with $n < N$ under the given conditions of nondecreasing - bounded sequence. $\endgroup$ – Ufuk Can Bicici Jul 24 '14 at 13:26

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