I found the following assertion in my class notes, but i did not understood why doe's it holds:
"Suppose that in an infinite-dimensional von Neumann algebra there is an increasing net $\{p_j\}$ of projections with $p_j\nearrow 1$ and $p_j\ne 1$ for all $j$. Then the subspace $P=\overline{\operatorname{span}}\{p_j:\ j\}$ does not contain $1$."
It is a basic fact of von Neumann algebras that any bounded, totally ordered net of self-adjoints has a supremum, and said supremum is the sot (wot, etc) limit of the net. but I can't see why doesn't 1 belongs in P.
Thanks in advance!
Suppose $p_1,..., p_n$ are projections $≤p$ for some projection $p<1$. Then for any scalars $a_1,...,a_n$ you have:
$$\|1-a_1\,p_1+...a_n\,p_n\|≥1.$$
The simplest way to see this is to consider a representation on some Hilbert space. Then since $p<1$ you have some $x\in\mathrm{im}(p)^\perp$, which is necessarily also in $\mathrm{im}(p_i)^\perp$ for all $i$ since $p_i≤p$. You then get:
$$(1-a_1p_1+...+a_np_n)(x)=x,$$ which implies the inequality above. Now in your situation if you have some linear combination of projections from the net there is a common supremum in the net of these projections. This supremum must $<1$ because $1$ is not in the net. Then the above inequality implies that the linear combination has distance $≥1$ from the operator $1$.