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given that $m = \inf a_n$ and that $\forall n, a_n \not= m $ and $\exists M , |a_n| \leq M $

Prove that $\lim \inf a_n = \inf a_n$ ?

I proved by contradiction that $\inf a_n \not > \lim \inf a_n$

but i don't know how to prove $\inf a_n \not < \lim \inf a_n $

Also if you have a proof not by contradiction, please post it.

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  • $\begingroup$ You seem to say $\forall n \Big( \exists M |a_n|\le M\Big),$ and I wonder if you mean $\forall n \Big( \exists M \big( \text{if } n\ge M\text{ then } |a_n|\le M \big) \Big). \qquad$ $\endgroup$ – Michael Hardy Dec 17 '17 at 0:06
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We have a bounded sequence which never equals its infimum. This implies that it is always greater than its infimum.

We must show that $$\lim \inf a_n = \lim_{k \to \infty} \left(\inf_{n > k} a_n\right) = \inf_{n > 0} a_n$$

It suffices to show that $\inf_{n > k} a_n = \inf_{n > 0} a_n$ for arbitrary $k$. Clearly $$\inf_{n > 0} a_n \leq \inf_{n > k} a_n$$ hence it remains to show $$\inf_{n > k} a_n \leq \inf_{n > 0} a_n$$

To prove this, note that if $\inf_{n > k} a_n > \inf_{n > 0} a_n$ we would have $\min(a_1, a_2 \cdots a_k) < \inf_{n > k} a_n $ which would in turn imply that $\min(a_1, a_2 \cdots a_k) = \inf_{n > 0} a_n$. This contradicts the second sentence of this answer.

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Let's do it directly. There is by definition for every $\varepsilon>0$ an element $a_j$ such that $a_{j}<\inf a_n +\varepsilon$.

We do the same story multiple times, there are $a_{n_1}, a_{n_2},a_{n_3},...$ such that: \begin{align} a_{n_1} &< \inf a_n + 1\\ a_{n_2} &< \inf a_n + \frac{1}{2}(a_{n_1}-\inf a_n)\\ a_{n_3} &< \inf a_n + \frac{1}{3}(a_{n_2}-\inf a_n)\\ \vdots\\ a_{n_k} &<\inf a_n + \frac{1}{k}(a_{n_{k-1}}-\inf a_n) \end{align} Verify yourself that all these numbers $a_{n_k}$ are distinct! Furthermore $\inf a_n < a_{n_k}$ for all $k$ hence $|a_{n_k}-\inf a_n|<\frac{1}{k}(a_{n_{k-1}}-\inf a_n)\leq \frac{1}{k}(M-\inf a_n)$ for all $k\in \mathbb{N}$. However we do not know if $n_1<n_2<n_3<...$. That can be achieved by taking $n_1$ and letting the next one be the one that is larger than $n_1$. There must exist an index larger than $n_1$ because there are $n_1-1$ that are less than it and after $2n_1$ there will be one and repeat this process (infinitely many times). So we get a subsequence $a_{n_{k_j}}$ such that for all $j$ we have: \begin{align} |a_{n_{k_j}}-\inf a_n|<\frac{1}{k_j}(M-\inf a_n) \end{align} So when $j\to\infty$ we have $k_j\to\infty $ and hence $a_{n_{k_j}}\to \inf a_n$ as $j\to \infty$. Now we have found a subsequence converging to the infimum hence: \begin{align} \liminf a_n = \inf a_n \end{align}

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$inf a_n \lt \lim\inf a_n$ is perfectly possible, unless $a_n$ is a monotone decreasing sequence. Simple example $a_1=-1, a_n \gt 0,\ n\gt 1$.

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  • $\begingroup$ we are given that for all $n$ that $a_n \not = \inf a_n$ so we can not take this example as valid given these conditions. $\endgroup$ – Ahmad Dec 16 '17 at 23:21

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