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Let $H$ be a separable Hilbert Space (WLOG, we may assume $H=\ell_2(\mathbb{N})$ is the space of square summable sequences). Can there exist an uncountable chain of closed subspaces? In other words, if we have a family of closed subspaces $\mathcal{F} $ such that for each $V, W \in \mathcal{F}$, either $V \subset W$ or $W \subset V$; then is it necessary that $\mathcal{F}$ is countable?

Attempts: I think the answer is yes, as certain literature seems to suggest so. It is interesting to ask about the following stronger statement: Does there exists a function $\psi: \mathcal{F} \to H$ such that $\psi(V)$ is an orthonormal basis for $V$ and $\psi$ is monotonic (i.e $V \subset W \implies \psi(V) \subset \psi(W)$), this would clearly imply yes to the original question.

Mathemagician
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1 Answers1

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It is known that there is an uncountable chain in $\mathcal P(\mathbb N)$. (In fact, we have a chain of cardinality $\mathfrak c$.) See: Chain of length $2^{\aleph_0}$ in $ (P(\mathbb{N}),\subseteq)$

Now if $\mathcal C$ is such a chain and $e_i$ is an orthonormal basis of $\ell_2$ we can take $$F(C):=\overline{\operatorname{span}\{e_i; i\in C\}}$$ for any $C\in\mathcal C$ and $$\mathcal F=\{F(C); C\in\mathcal C\}.$$

Now if $C_1\subseteq C_2$, then clearly $F(C_1)\subseteq F(C_2)$. Moreover, if $i\in C_2\setminus C_1$, then $$\langle e_i,e_i\rangle=1$$ and $$(\forall x\in F(C_1)) \langle e_i, x \rangle=0.$$ So the continuous map $x\mapsto \langle e_i,x\rangle$ shows that $e_i$ is not in the $F(C_1)$. So we get $$C_1 \subsetneq C_2 \implies F(C_1)\subsetneq F(C_2).$$

This shows that $\mathcal F$ is a chain and it has the same cardinality as $\mathcal C$.

We can take $\psi \colon F(C_i) \mapsto \{e_i; i\in C\}$ to get a map with required properties.

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Martin Sleziak
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  • Thanks, this is a very nice and simple answer to my question. Though I've heard it stated many times in the literature that such a chain does NOT exist; but I think I have now figured out what is really meant: If we define for each countable ordinal, say w, a closed subspace V_w in a monotonic way (i.e. w'<w implies that V_{w'} is a strict subspace of V_{w}), then we run into a contradiction. I think I've just figured out how to prove this using transfinite induction. :) – Mathemagician Aug 18 '14 at 11:00
  • Since you write that you have heard it stated many times in the literature, could you give at least one reference. And I agree that having a chain of cardinality $\mathfrak c$ is not the same thing as having $V_\alpha\subsetneq V_\beta$ for each $\alpha<\beta<\mathfrak c$. – Martin Sleziak Aug 18 '14 at 12:01
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    For instance: (1) Proofs of Furstenberg's multiple recurrence theorem which use transfinite induction. For example see Remark 1 of http://terrytao.wordpress.com/2008/03/05/254a-lecture-15-the-furstenberg-zimmer-structure-theorem-and-the-furstenberg-recurrence-theorem/ (2) Remark 6 of http://terrytao.wordpress.com/2009/02/21/245b-notes-11-the-strong-and-weak-topologies/ (3) Some other papers which were sloppy and didn't mention ordinals explicitly (but can't find them at the moment) – Mathemagician Aug 18 '14 at 12:51
  • The link you've given indeed says: induces an uncountable proper chain of closed subspaces on $V^$, which is not compatible with separability*. To be honest, I am somewhat confused now. – Martin Sleziak Aug 18 '14 at 14:38
  • @Mathemagician One of the comments to the link (2) says: Indeed, if V is not separable, then by induction we can construct a well-ordered strictly increasing sequence of closed subspaces of order type $\omega_1$. So it does not speak about chain, but about increasing transfinite sequence. – Martin Sleziak Aug 18 '14 at 16:27
  • However, I did not find anything in errata. (This remark appears in his book, too.) – Martin Sleziak Aug 18 '14 at 16:28