Question

Let Xn be a bounded sequence of real numbers. Prove that the minimal subsequential limit of Xn is computed by the formula
supK (inf(n>=K) An)

Hint: Prove first that it is a subsequential limt.

Answer 1

$$A = \lim \inf x_n = \lim_{n \to \infty} \left(\inf_{m \geq n} x_m\right)$$ $$a_n = \inf_{m \geq n} x_m$$

1) $\forall m : a_n \leq x_m$

2) $\forall e > 0 \exists m : x_m < a_n + e$

$$\forall e > 0 \exists N \in \mathbb{N} \forall n > N : |a_n - A| < e \Rightarrow -e < a_n - A < e \Rightarrow A - e < a_n$$ $$\forall n : a_n \leq a_{n+1} \leq \dots \leq A$$ $$A = \sup a_n = \sup_n \left(\inf_{m \geq n} x_m\right) = \liminf x_n$$