Cfrgtkky

$ a:= liminf_{nto infty}$ $a_n$ = $lim_{nto infty}$ $inf_{kgeq n}$ $a_k$

$a$ $in$ $R$

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 $liminf_{nto infty}$ $a_n$ = L > $lim_{nto infty}$ $inf_{kgeq n}$ $a_k$ = 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?

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 $2epsilon = K-L > 0$. 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.