There are any closed subspaces for $\ell^{2}$ such that $H_{s} \subset H_{t}$ such that $s < t$ for $0 \leq s, t \leq 1$?

Question

There are exists closed subspaces for $\ell^{2}$ such that $H_{s} \subset H_{t}$ such that $s < t$ for $0 \leq s, t \leq 1$? where $\ell^{2}$ is a Hilbert space with the norm $(\sum_{n} |\alpha_{n}|^{2})^{1/2}$ for a sequence $\{ \alpha_{n} \}_{n=1}^{\infty}$.

Answer 1

Suppose $A\subseteq\mathbb{N}$. Denote $H_A$ the subspace of $\ell^2$ of vectors such that all coordinates with indices outside $A$ are zero, and inside $A$ are arbitrary. For any $A$ such subspace is closed and $A\subset B\iff H_A\subset H_B$. Now it suffices to find such subsets $A_t\subseteq\mathbb{N}, t\in[0,1]$ that $A_s\subset A_t\iff s<t$. This can be done as follows: start with $A_0=\emptyset,A_1=\mathbb{N}$. So $A_{p/2^n}$ is now defined for $n=0$ and all possible integer $p$.

Suppose on $n$th step we already defined $A_{p/2^n}$, and for $p<q$ we have $A_{p/2^n}\subset A_{q/2^n}; A_{q/2^n}\backslash A_{p/2^n}$ infinite. For each possible $p$ then on step $n+1$ we define $A_{p/2^{n+1}}$. It is already done for even $p$, and for odd $p$ we have $B=A_{(p-1)/2^{n+1}}$ and $C=A_{(p+1)/2^{n+1}}$ already defined, $B\subset C$ and $C\backslash B$ countable. Pick a set $D=A_{p/2^{n+1}}$ such that both $D\backslash B$ and $C\backslash B$ are countable. Then all conditions hold after $(n+1)$th step, and we are done for all numbers of the type $p/2^n$. For others we can just take $A_t=\bigcup_{s=p/2^n<t}A_s$.

Answer 2

Theorem: There exists a collection $\{A_s:s\in [0,1]\}$ of nonempty subsets of $\mathbb N$ such that $s<t$ implies $A_s \subsetneq A_t.$

Assuming the theorem, define $H_s$ to the set of sequences in $l^2$ that are supported in $A_s.$ Then $\{H_s:s\in [0,1]\}$ has the desired property.

Proof of the theorem (sketch): Since $\mathbb N$ and $\mathbb Q\cap [0,1]$ have the same cardinality, it suffices to prove the theorem with $\mathbb Q\cap [0,1]$ in place of $\mathbb N.$ The sets $A_s = \{q\in \mathbb Q\cap [0,1]: q\le s\}$ do this.