# Prove that the liminf of a sequence is equal to its smallest subsequential limit

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

## $$a$$$$\in$$$$R$$

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?

## 1 Answer

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.