# How to prove this countable intersection is empty?

Let $$(l_p,\|\cdot\|_p)$$ be a normed space, for some fixed $$p \in [1,\infty)$$.

Let $$\{e_i\}_{i=1}^{\infty}$$ be standard basis and $$E_n=\{e_i\}_{i=n}^{\infty},\ n=1,2,3,...\ .$$

Then how can I prove $$\bigcap \limits_{n=1}^{\infty} E_n=\emptyset$$?

• Absolutely nothing to do with FA or $l^{p}$ spaces. it is just basic set theory. – Kavi Rama Murthy Jun 24 at 5:32
• I think you mean $E_n = \mathop{\rm span}(\{e_i\colon i\ge n\})$. – Carsten S Jun 24 at 14:52

This has nothing to do with $$l^{p}$$ space and bases. For any sequence of distinct points $$a_1,a_2,...$$ (in any set $$S$$) the set $$\cap_n \{a_n,a_{n+1},a_{n+2},...\} =\emptyset$$ . This is because LHS is contained in $$\{a_1,a_2,a_{n+2},...\}$$ , so if there is some pint in LHS it must be one of the points $$a_k$$. But $$a_k$$ does not belong to $$\{a_{k+1},a_{k+1},a_{n+2},...\}$$ so it cannot belong to the intersection.
The set $$\bigcap \limits_{n=1}^{\infty} E_n$$ is the limit superior of the sequence of pairwise disjoint sets $$\left(\{e_k\}\right)_{k\geqslant 1}$$. A limit superior of pairwise disjoint sets is empty because a element of the limsup belons to infinitely many sets in the sequence.
Notice that $$\forall n \in \mathbb{N}$$, $$e_n \in E_k$$ for $$k \leq n$$ and $$e_n \notin E_k$$ for $$k > n$$. In particular, $$e_n \notin E_{n + 1}$$. Therefore, if $$\exists x \in \bigcap\limits_{n = 1}^{\infty} E_n$$, then in particular $$x \in E_1 = \left\lbrace e_i \right\rbrace_{i = 1}^{\infty}$$ so that the element is one of the basis element. Let it be $$e_{i_0}$$ for some $$i_0 \in \mathbb{N}$$. Since it is in the intersection, $$\forall n \in \mathbb{N}$$, $$e_{i_0} \in E_n$$ which is a contradiction to the fact that $$e_{i_0} \notin E_{i_0 + 1}$$.