Let $V$ be a separable infinite dimensional Banach space over $\mathbb{C}$


Let $B \subset V$ be a subset of $V$ such that:


1) $B$ is linearly independent and closed


2) $\overline{\operatorname{span} B} = V$


3) For all $X$ and $Y$ closed subsets of $B$: $X \cap Y = \emptyset \Longrightarrow \overline{\operatorname{span} X} \cap \overline{\operatorname{span} Y} = \{0\}$


I would like to know if is it true that:


$X \cap Y = Z \Longrightarrow \overline{\operatorname{span} X} \cap \overline{\operatorname{span} Y} = \overline{\operatorname{span} Z}$



In Banach spaces is $X \cap Y = Z \Rightarrow \overline{{span} X} \cap \overline{{span} Y} = \overline{{span} Z}$
I'm sorry for the last answer where I misunderstood the problem. Now I get it:


So I still think this isn't the case and suggest:$\require{begingroup}\begingroup\newcommand{\span}{\operatorname{span}}$


$V = l\_{2}$.


$B = \{ \sum\_{i=0}^{N} \left((\frac{1}{i} e\_{2i + 1}) + e\_{2i} \right) : i \in \mathbb{N} \} \cup \{ e\_{2i} : i \in \mathbb{N}\}$. For simplicity lets define $f\_N = \sum\_{i=0}^{N} \left((\frac{1}{i} e\_{2i + 1}) + e\_{2i} \right)$ and $Z = \{ e\_{2i} : i \in \mathbb{N}\}$.


$X = \{f\_i : i \equiv 0 \space \text{mod} \space 2 \} \cup Z $. and $Y = \{f\_i: i \space \equiv 1 \space mod \space 2\} \cup Z$


first lets check that **1)**$\overline{B} = B$ and that $B$ is linearly independent.


i) $\overline{B} = B$. To see this, first notice that $\langle e\_{2n}, f\_k\rangle =e\_{2n}^{\*}(b) \geq 1$ for each $k \in \mathbb{N}$ and each $n \leq \frac{k}{2}$. Assume $(b\_l)\_{l \in \mathbb{N}} \subseteq B$ is a non-constant sequence that converges to $v \in V$. Either $(b\_l)\_{l \in \mathbb{N}} $ has an infinite amount of $e\_{2i}$ which contradicts that it converges, or it has an infinite amount of $f\_k$ which would force $\langle e\_{2n}, v\rangle \geq 1 $ for each $n \in \mathbb{N}$ which is also a contradiction.


ii) $B$ is linearly independent. Take 


$\sum\_{i = 0}^N \lambda\_i f\_i + \sum\_{j=0}^M \mu\_je\_{2j} = 0$.


By using that the only $\lambda$ (or $\mu$) that is affecting $e\_{2N + 1}$ is $\lambda\_N$, we get 


$\frac{\lambda\_N}{N} = 0$. Then we have that $\lambda\_N = 0$ so that gives us


$\sum\_{i = 0}^{N - 1} \lambda\_i f\_i + \sum\_{j=0}^M \mu\_je\_{2j} = 0$. Where we repeat the argument until we get that all $\lambda\_i = 0$ for each $ i \leq N$, which leaves us with


$ \sum\_{j=0}^M \mu\_je\_{2j} = 0$ which implies $\mu\_j = 0$ for each $j \leq M$. So linear independence is established.


Now **2)** $\overline{\span(B)} = V$,


To see this just notice:


$Z \subset \overline{\span(B)}$ and $\sum\_{i=0}^{N} \left((\frac{1}{i} e\_{2i + 1}) + e\_{2i} \right) \in \overline{\span(B)}$ for each $N \in \mathbb{N}$ implies


$\sum\_{i=0}^{N} \frac{1}{i} e\_{2i + 1} \in \overline{\span(B)}$ for each $N \in \mathbb{N}$. which gives us


$\frac{1}{2N +1} e\_{2N + 1} = \sum\_{i=0}^{N} \frac{1}{i} e\_{2i + 1} - \sum\_{i=0}^{N-1} \frac{1}{i} e\_{2i + 1} \in \overline{\span(B)}$ for each $N \in \mathbb{N}$. So that takes care of 2).


To see **3)** we are gonna have to check some properties of $\overline{\span(B\_1)}$for all $B\_1 \subseteq B$. Basically, we are gonna assign every such $B\_1$ a partition of the natural numbers in the following way:


Take $\{2n + 1: f\_{n} \in B\_1\}$, so the subset of all the "last odd numbers that each $f\_n \in B\_1$ knows, meaning the last odd number such that $e\_i^{\*}(f\_n) \neq 0$", lets denote them in the ascending order as a posibly finite sequence (or even empty sequence)


$(n\_1, n\_2, \cdots, n\_k , \dots)$


and make the partition, starting with $n\_0 = 0$, $P\_0 = [0, n\_{1}]$ if $n\_{1}$ exists and $P\_0 = [0, \infty)$ if it doesn't exist, and then recursively $P\_{k+1} = [n\_k + 1, n\_{k+1}]$ if $n\_{k+1}$ exists and $P\_{k+1} = [n\_{k} +1 , \infty]$ if $n\_{k+1}$ doesn't exists.


The property that we are looking for is the next one:


Let $x \in \overline{\span(B\_1)}$ for each $k \in \mathbb{N}$ there exist a single $\lambda\_k$ such that 


for each $l \in P\_k$ 


* $e\_l^\*(x) = \frac{\lambda\_k}{l}$ if $l \in 2\mathbb{N} + 1$, that is, if $l$ is odd.
* $e\_l^\*(x) = \lambda\_k$ if $l \in 2 \mathbb{N}$ such that, $e\_{l} \notin B\_1$, that is, if $e\_l \in (Z \setminus B\_1 \cap Z )$


The proof goes by proving this property in $\span(B)$ and then using continuity of all the relevant $e\_l^\* = \langle e\_l, \text{\_\_} \rangle$ to establish it in the closed span. To see that it holds in $\span(B)$ it's enough to notice that every $B\_1 \subseteq B$ has the property, and that everytime you take a linear combination 


$\gamma x + \mu f\_n + \nu e\_{2m}$ with $x$ having the property and $f\_n, e\_{2m} \in B\_1$. Denote by $\{\lambda\_{x,r}: r \in \mathbb{N}\}$, the lambdas for that specific $x$, and for their respective $\{P\_r : r \in \mathbb{N}\}$. Then, $\nu$ doesn't change anything about the property because we purposely took those $2m$ away, and since $f\_n = \sum\_{i=0}^{N} \left((\frac{1}{i} e\_{2i + 1}) + e\_{2i} \right)$ such that $2N +1 = n\_{k\_0}$for some $k\_0 \in \mathbb{N}$. Then, for every $P\_r$ either assign $\lambda\_r = \gamma \lambda\_{x,r} + \mu$ if $P\_r \subseteq [0, n\_{k\_0}] $or $ \lambda\_r = \gamma \lambda\_{x,r}$ if $P\_r \subseteq [n\_{k\_0}+1, \infty]$. Notice that it cannot intersect both because of the definition of the $P\_r$.


Therefore, $\overline{\span(B\_1)}$ has such property.


So finally, take $B\_1, B\_2 \subseteq B$ such that $B\_1 \cap B\_2 = \emptyset$. We are gonna prove that $\overline{\span(B\_1)} \cap \overline{\span(B\_2)} = \{ 0\}$ using the property. Take $x \in \overline{\span(B\_1)} \cap \overline{\span(B\_2)} $, therefore $x$ has the property both for the partition given by $B\_1$ and the partition given by $B\_2$, lets call them $\{P\_{B\_1 , r} : r \in \mathbb{N}\}$ and $\{P\_{B\_2 , r} : r \in \mathbb{N}\}$, notice that the fact that $B\_1 \cap B\_2 = \emptyset$, gives us that 


**a)**$\max P\_{B\_1, r\_1} \neq \max P\_{B\_2, r\_2}$ for any $r\_1, r\_2 \in \mathbb{N}$. 


We are gonna prove that there exists a single $\lambda$ such that


* $e\_l^\*(x) = \frac{\lambda}{l}$ if $l \in 2\mathbb{N} + 1$, that is, if $l$ is odd.
* $e\_l^\*(x) = \lambda$ if $l \in 2 \mathbb{N}$ such that, $e\_{l} \notin B\_1$ or $e\_{l} \notin B\_2$.


and this is done by first checking the bigger of $P\_{B\_1, 0}$ and $P\_{B\_2, 0}$, Suppose this is $P\_{B\_1, 0}$ and take that $\lambda\_{B\_1, 0}$, then whatever $P\_{B\_2, c\_1}$ such that $\max P\_{B\_1, 0} \in P\_{B\_2, c\_1}$, notice that since $x$ has this property with both subsets and $\max P\_{B\_1, 0} \in P\_{B\_2, c\_1} \cap P\_{B\_1, 0}$ is odd, this means $\lambda\_{B\_1, 0} = \lambda\_{B\_2, c\_1}$. Because of $a)$ we have that $\max P\_{B\_2, c\_1} > \max P\_{B\_1, 0} $ so now choose $\max P\_{B\_2, c\_1} \in P\_{B\_1, d\_1}$ and then $P\_{B\_2, c\_2}$ and recursively until you have 


$\mathbb{N} = \bigcup\_{i=0}^{\infty} P\_{B\_1, c\_i} \cup P\_{B\_2, d\_i} $, such that $\lambda\_{B\_1, c\_i} = \lambda\_{B\_2, d\_j}$ for each $i,j \in \mathbb{N}$. Notice that this actually gives us that all $\lambda\_{B\_1, r} = \lambda\_{B\_2, r'}$ for $r, r' \in \mathbb{N}$ (simply check with any of the ones we know to be equal that it intersects).


Since, because they are disjoint, $(B \setminus B\_1) \cup (B \setminus B\_2) \supset B \supset Z $. We actually get that $e^{\*}\_n(x) = \lambda$ for each $n \in 2 \mathbb{N}$, each even $n$. That means that $\lambda = 0$, so $x = 0 $.


And now we just have to check that $X \cap Y = Z$ (that is clear) and


$ \overline{\span(X)} \cap \overline{\span(Y)} \neq \overline{\span{Z}}$, and that is easy to do because


$\sum\_{i=0}^{\infty} (\frac{1}{i} e\_{2i + 1}) \in \overline{\span(X)} \cap \overline{\span(Y)} $ and its clearly not in $\overline{\span(Z)}\endgroup$


