Let $A_0 \supset A_1 \supset A_2 \supset \cdots$ - sequence of embedded Banach spaces and $B_0 \supset B_1 \supset B_2 \supset \cdots$ - suquence of linear spaces such that $B_i$ dense in $A_i$, it is true that $\bigcap_{k=0}^\infty B_k = 0 \Rightarrow \bigcap_{k=0}^\infty A_k = 0$ ?
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Sorry are you meaning to write that they're equal to 0? Or that they are empty? – Phillip Hamilton Mar 25 '16 at 15:25
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1Is everything here a proper subset? If not, then a simple counterexample is given by $A_i=\mathbb{R}$ for all $i$ and $B_i=\mathbb{Q} \setminus { q_0,\dots,q_i }$ where ${ q_j }$ is an enumeration of $\mathbb{Q}$. – Ian Mar 25 '16 at 15:29
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Phillip, every linear space always contains 0, so, it is impossible that intersection of linear spaces will be $\emptyset$. – Mykola Pochekai Mar 25 '16 at 15:31
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lan, $B_i$ is not linear spaces, in your example – Mykola Pochekai Mar 25 '16 at 15:32
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Oh, you're right, my mistake. – Ian Mar 25 '16 at 15:52
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No. Let $\{q_m\}$ be an enumeration of rationals in $[0,1]$, and define $B_n$ to be the space of continuous functions that vanish at $q_1,\dots,q_n$. Then $B_1\supset B_2\supset \dots$ and $\bigcap_n B_n = \{0\}$.
On the other hand, each $B_n$ is dense in $L^2([0,1])$, so one can take $A_n=L^2([0,1])$ for every $n$.