Choosing a net of projections from a given collection

Question

Let $\mathcal M$ be a von Neumann algebra. Let $S$ be a nonempty set. Consider $\Omega$ to be the collection of all finite subsets of $S$. Suppose for all $\omega\in\Omega$ we have $I_\omega$ a collection of projections of $\mathcal M.$ Moreover, we have that for any $\omega_1\subseteq \omega_2$ we have $I_{\omega_2}\subseteq I_{\omega_1}$. Is it possible to choose a net (or subnet) of increasing projections $(e_\omega)_{\omega\in\Omega}$ such that $e_\omega\in I_\omega$ for all $\omega\in \Omega$?

Answer 1

Disclaimer: I don't know anything about von Neumann algebras. But if I've understood the definitions correctly, I think I can show that the answer is no.

For one thing, the answer is trivially no if don't assume that each $I_\omega$ is non-empty. So let's make that assumption.

Now let $(e_n)_{n\in \mathbb{N}}$ be a countable family of minimal projections (corresponding to $1$-dimensional subspaces of $\mathcal{M}$). Let $S = \mathbb{N}$. For each $\omega\in \mathcal{\Omega}$, let $I_\omega = \{e_n\mid n\notin \omega\}$. Then if $\omega_1\subseteq \omega_2$, $I_{\omega_2}\subseteq I_{\omega_1}$, and each $I_{\omega}$ is non-empty.

Since each $e_n$ is minimal, if $e_m \leq e_n$, then $e_n = e_m$. So an increasing net of projections selected from the sets $I_\omega$ is in fact constant (since the poset $\Omega$ is connected). But it can't be constant with value $e_n$ for any $n\in \mathbb{N}$, because $e_n\notin I_{\{n\}}$.

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A very weak positive result: If any set $I_\omega$ is finite, then we can find a net of increasing projections $(e_\omega)_{\omega\in \Omega}$, and in fact a constant one, with all $e_\omega$ equal.

Suppose we have $\omega\in \Omega$ with $I_\omega$ finite. I claim that there is some $e\in I_\omega$ such that $e\in I_{\omega'}$ for all $\omega'\in \Omega$ (so that the constant net $(e)_{\omega'\in \Omega}$ satisfies your conditions).

If not, then writing $I_\omega = \{e_1,\dots,e_n\}$, for each $1\leq i \leq n$, there is some $\omega_i\in \Omega$ such that $e_i\notin I_{\omega_i}$. Let $\omega_* = \omega\cup \bigcup_{1\leq i \leq n}\omega_i$. Then $\omega_*\in \Omega$, and since for all $1\leq i \leq n$, $I_{\omega_*}\subseteq I_{\omega_i}$, we have $e_i\notin I_{\omega_*}$. But also $I_{\omega_*}\subseteq I_{\omega}$, so $I_{\omega_*} = \varnothing$, contradicting our hypothesis that all the sets are non-empty.