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$\ a:= \liminf_{n\to \infty}$ $a_n$ = $lim_{n\to \infty}$ $inf_{k\geq n}$ $a_k$

$a$ $\in$ $R$

I am trying to show this by using cases and assuming the equality doesn't hold- to come to a contradiction showing neither side is greater. To imply that they are equal.

In Case 1 I assumed $\liminf_{n\to \infty}$ $a_n$ = L > $lim_{n\to \infty}$ $inf_{k\geq n}$ $a_k$ = l

$=>$ -C < l < L < ... < C where C $\in$ $R$

And used a weak (I think) argument that $a_n$ is bounded:

$=>$ L = $inf${subsequantial limit set}.

$=>$ L $\leq$ $a_n$ for any n.

Then implied that since $a_k$ is a subset of $a_n$ every element of $a_k$ should be greater or equal to L leading to a contradiction..

L $\leq$ l

Is this approach at all valid? Should I use a similar argument for Case 2?

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The part about boundedness is not correct : $L \leq a_n$ for all $n$ is not true either. The rest of the argument is somewhat fuzzy, not clear.


You want to show that $\liminf a_n$ is equal to the infimum of the set of limit points of $a_n$.

Suppose that $a_n$ is bounded, first. Then, the set of limit points $S$ is non-empty by Bolzano Weierstrass. Let $|a_n| \leq M$ for all $n$. It follows that $|\inf S| \leq M$ as well. Let $K = \inf S$. Note that $S$ is closed, so $K \in S$.

Now, we want to show that the sequence $\inf_{k \geq n} a_k$ converges to $K$. For this, first we need to show that the limit exists, but this is clear, since $\inf_{k \geq n} a_k \leq M$ for all $M$, and a bounded increasing sequence converges. Let the limit be $L$.

First, take $s \in S$. Let $a_{n_k}$ be a subsequence of $a$ converging to $s$. Then, note that $\inf_{l \geq n_k} a_{l} \leq a_{n_k}$ for all $k$. Taking the limit as $k \to \infty$ gives $L \leq s$, and since $s$ is arbitrary, $L \leq K$.

For the other direction, suppose that $L < K$ is true. Let $2\epsilon = K-L > 0$. Then, since $L$ is the limit of an increasing sequence, it follows that the sequence $\inf_{k \geq n} a_k < K - \epsilon$ also happens for all $n$. Contradict $K \in S$ with this inequality.


Handling the unbounded case is similar, but you will have to be careful of infinities. I leave it to you.

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