I work in the following question:
Let $E$ be a space Banach, $\{x_n\}_{n=1}^{\infty} \subset E$, $x \in E$ and \begin{equation*} K_{n}= \overline{con(\bigcup_{i=n}^{\infty}\{xi\})}. \end{equation*} prove: If $E$ is finite dimensional and $\bigcap_{n=1}^{\infty}K_{n} = \{ x \}$, then $x_{n} \to x$.
The prove: We may asume that $x = 0$. consider the recesion cone \begin{equation*} C_{n} = \bigcap_{\lambda > 0}\lambda K_{n} \end{equation*} Since $C_{n}\subset K_{n}$ we deduce that $ \cap_{i=1}^{\infty}C_{n} = \{0\} $. Let $S= \{ x \in E: \|x\| = 1 \}$ the sequence $C_{n}\cap S$ is decrecent and $\bigcap_{n=1}^{\infty}(C_{n}\cap S) = \emptyset $. Thus, by compactness $C_{n_{0}}\cap S = \emptyset$ for some $ n_{0} \in \mathbb{N} $. therefore $C_{n_{0}} = \{ 0 \}$ and consequently $K_{n_{0}}$ is bounded. Hence $(x_{n})$ is bounded. (So we are in the conditions previous item of this same exercise and it´s just conclude )
But. i don´t understand why $C_{n_{0}} = \{ 0 \}$? And why $C_{n_{0}} = \{ 0 \}$ implies $K_{n_{0}}$ is bounded?
(There is another way to solve the problem: suppose that $\{x_{n}\}$ is unbounded but i couldn´t do it this way.)
thanks.
If $K_{n_0}$ is unbound, then $\forall n>0,\, \exists \lVert x_n\rVert>n,$ s.t. $$\frac{x_n}{\lVert x_n\rVert}\in(\cap_{\lambda\ge \frac{1}{m}}\lambda K_{n_0})\cap S,\, \forall m\le n.$$ Denote $y_n=\frac{x_n}{\lVert x_n\rVert}$. $\{y_n\}$ is bounded and $S$ is compact, therefore there exists y such that $\lim_{n\to \infty}y_n=y$ . Because $(\cap_{\lambda\ge \frac{1}{n}}\lambda K_n)\cap S$ is compact, $y\in (\cap_{\lambda\ge \frac{1}{n}}\lambda K_{n_0})\cap S,\forall n$. Hence, $y\in (C_{n_0})\cap S = \emptyset$.