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Can there exist an uncountable sequence $R_1 \subset R_2 ...$ of von Neumann algebras all acting on the same separable Hilbert space $H$, with a "limit" algebra $R$ such that $R_\alpha \subset R$ for all countable ordinals $\alpha$; and $R$ and the $R_\alpha$'s are all either type II or III? (Here $\subset$ means proper inclusion.)

I've tried to find a contradiction between the uncountability of the $(R_\alpha)$ sequence and the countable basis for $H$, but haven't been successful and now wonder whether there is any contradiction. In the other direction I've tried to construct a concrete example of a $(R_\alpha)$ sequence using a group-von-Neumann-algebra for $R$, but I have little experience manipulating those things and didn't get anywhere. Any help greatly appreciated!

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No. The von Neumann algebra $B(H)$ is separable (in the sot topology, say); it follows that any von Neumann algebra $R\subset B(H)$ is separable. This is not totally straightforward, but it is simple:

  • A countable dense subset of $B(H)$ is given by the complex-rational linear combinations of matrix units (because its commutant is trivial);
  • Separability can be assessed at the level of the unit ball (because then your algebra is a countable union of balls);
  • The sot is metrizable on the unit ball;
  • In a metric space, a subset of a separable set is separable. .
  • So $R$ is separable.

The claim that the chain $R_1\subset R_2\subset\cdots$ is uncountable implies that all the inclusions are proper (or at least for a subnet, and then we can renumber). We can then construct a net $\{r_\alpha\}$ such that $r_{\alpha}\not\in R_\beta$ for $\beta<\alpha$. Because each inclusion $R_\beta\subset R_\alpha$ is proper, for each $\alpha$ there exists a sot neighbourhood $V_\alpha\subset R_\alpha$ and $r_\beta\not\in V_\alpha$ for $\beta<\alpha$. We then have an uncountable family of disjoint open neighbourhoods, contradicting separability.

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  • $\begingroup$ Thanks! I'd been thinking along those lines but there was one implicit step I couldn't justify. At each stage $\alpha$ we choose $r_\alpha \in R_\alpha$, and a SOT neighborhood $V_\alpha \subseteq R$ centered on $r_\alpha$, such that $V_\alpha$ is disjoint from all $R_\beta$ with $\beta < \alpha$. But can we find this $V_\alpha$ disjoint from each previous $V_\beta$? Each $V_\beta$ is a subset of $R$, not of its respective $R_\beta$, so by some countable step, maybe the $V_\beta$'s will have covered $R$, leaving no "room" to put a new $V_\alpha$. Is there some way to address this? $\endgroup$ – Doug McLellan Jun 29 '16 at 10:41
  • $\begingroup$ (I think the $V_\alpha$'s need to be subsets of $R$ in order to get the desired contradiction.) $\endgroup$ – Doug McLellan Jun 29 '16 at 10:44
  • $\begingroup$ They way I see it, let $R_0=\overline{\bigcup \{R_\beta:\ \beta<\alpha\}}$. Then $R_0\ne R_\alpha$ (if they were equal, you would only have countably different subalgebras), so we can choose $V_\alpha\subset R_\alpha\setminus R_0$. Since both $R_\alpha$ and $R_0$ are closed, $R_\alpha\setminus R_0$ is closed. I agree that then one has to deal with the relative topologies and some issue might arise that way. $\endgroup$ – Martin Argerami Jun 29 '16 at 18:47
  • $\begingroup$ I think that is basically the right idea ... I tried to work it out more formally in this comment but not allowed enough characters, so posted it as an answer below. $\endgroup$ – Doug McLellan Jul 1 '16 at 19:22
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I think we were on the right track. Here is a slightly different approach that I believe will work to get a formal contradiction.

Start with a countable basis $(V_n)$ for the SOT of $B(H)$ (the set of all bounded operators on the separable space $H$). [NOTE it isn't actually legitimate to assume this exists; see comment below.] Then define, for all $\alpha$,

$A_\alpha = \{ n : V_n \cap \mathcal{R}_\alpha \neq \emptyset \}.$

Claim: if $\alpha > \beta$, then $A_\alpha$ is a strictly larger set of natural numbers than $A_\beta$. Proof: $\mathcal{R}_\beta \subset \mathcal{R}_\alpha$ so $A_\beta \subseteq A_\alpha$. To show the converse doesn't hold, take any $r_\alpha \in \mathcal{R}_\alpha \setminus \mathcal{R}_\beta$; there must be a neighborhood $O$ of $r_\alpha$ in the SOT of $B(H)$ that is disjoint from $\mathcal{R}_\beta$, otherwise $r_\alpha$ is a SOT limit point of $\mathcal{R}_\beta$, and $\mathcal{R}_\beta$ is SOT closed (by definition of vN algebra). Now $O$ is a union of some $V_n$'s (since they are a base) so at least one $V_n$ satisfies $r_\alpha \in V_n$ and $V_n \subseteq O$, so $V_n \cap \mathcal{R}_\beta = \emptyset$.

... Then assuming the proof of that claim is right, the uncountable sequence of $\mathcal{R}_n$'s would give you an uncountable increasing sequence of sets of natural numbers, impossible.

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  • $\begingroup$ this answer is wrong because it assumes that separability implies the existence of a countable base. In fact I don't think the SOT even has a local countable base for neighborhoods of a given operator. I'm not sure whether this answer can be patched up by making $(V_n)$ a countable dense set rather than a countable basis (I'm thinking about it). $\endgroup$ – Doug McLellan Aug 1 '16 at 16:31

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