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 $\{ \alpha_{n} \}_{n=1}^{\infty}$.
There are any closed subspaces for $\ell^{2}$ such that $H_{s} \subset H_{t}$ such that $s < t$ for $0 \leq s, t \leq 1$?
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functional-analysis
hilbert-spaces
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0Do you want a strict inclusion? Then the answer should be no from separability of $\ell^2(\Bbb N)$. – 2017-01-28
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0Do you mean $H_s$ for all $s\in[0,1]$? – 2017-01-28
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0@s.harp Yes, a strict inclusion, since $H_{s}$ for all $s \in [0, 1]$ – 2017-01-28
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0@Wolfram Yes!, I do – 2017-01-28
2 Answers
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Theorem: There exists a collection $\{A_s:s\in [0,1]\}$ of nonempty subsets of $\mathbb N$ such that $s 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.
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0Interesting, no wonder I couldn't show that it was false :) – 2017-01-28
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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 Suppose on $n$th step we already defined $A_{p/2^n}$, and for $p