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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 $\bigcap \limits_{n=1}^{\infty} E_n=\emptyset$?

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

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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.

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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 $e_{i_0} \notin E_{i_0 + 1}$.

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