Limsup and Liminf are Limits of Bounds

Theorem
Let $$\left \langle {x_n} \right \rangle$$ be a sequence in $\mathbb{R}$.

Let $$\left \langle {x_n} \right \rangle$$ be bounded.

Let $$\overline l = \limsup_{n \to \infty} x_n$$ be the limit superior and $$\liminf_{n \to \infty} x_n$$ the limit inferior of $$\left \langle {x_n} \right \rangle$$.

Then:
 * $$\overline l = \limsup_{n \to \infty} x_n = \lim_{n \to \infty} \left({\sup_{k \ge n} x_k}\right)$$;
 * $$\underline l = \liminf_{n \to \infty} x_n = \lim_{n \to \infty} \left({\inf_{k \ge n} x_k}\right)$$.

Proof

 * First we show that $$\limsup_{n \to \infty} x_n = \lim_{n \to \infty} \left({\sup_{k \ge n} x_k}\right)$$:

Let $$M_n = \sup_{k \ge n} \left({x_k}\right)$$.

The sequence $$\left \langle {M_n} \right \rangle$$ decreases because we are taking the supremum of a smaller and smaller set for each $$k$$.

Any lower bound for $$\left \langle {x_n} \right \rangle$$ is clearly a lower bound for $$\left \langle {M_n} \right \rangle$$.

So from the Monotone Convergence Theorem it follows that $$\left \langle {M_n} \right \rangle$$ converges.

Suppose $$M_n \to M$$ as $$n \to \infty$$.

From the Bolzano-Weierstrass Theorem there exists a convergent subsequence of $$\left \langle {x_n} \right \rangle$$.

Let $$L$$ be the set of all numbers which are the limit of some subsequence of $$\left \langle {x_n} \right \rangle$$.

Let $$\left \langle {x_{n_r}} \right \rangle$$ be a convergent subsequence of $$\left \langle {x_n} \right \rangle$$ such that $$x_{n_r} \to l$$ as $$r \to \infty$$.

Then $$\forall n_r \ge n: x_{n_r} \le M_n$$.

Hence $$l \le M_n$$ by Lower and Upper Bounds for Sequences and hence (from the same theorem) $$l \le M$$.

This is true for all $$l \in L$$, so $$\overline l = \limsup_{n \to \infty} x_n \le M$$.

Now, from Terms of Bounded Sequence Within Bounds, we have that $$\forall \epsilon > 0: \exists n: \forall k \ge n: x_k < \overline l + \epsilon$$.

Thus $$\overline l + \epsilon$$ is an upper bound for $$\left\{{x_k: k \ge n}\right\}$$.

So $$M \le M_n \le \overline l + \epsilon$$.

Thus from Real Plus Epsilon, $$M \le \overline l$$

Thus we conclude that $$M = \overline l$$ and hence $$\overline l = \limsup_{n \to \infty} x_n = \lim_{n \to \infty} \left({\sup_{k \ge n} x_k}\right)$$.


 * $$\liminf_{n \to \infty} x_n = \lim_{n \to \infty} \left({\inf_{k \ge n} x_k}\right)$$ can be proved using a similar argument.