# closure of nets in von-Neumann algebras

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.

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 $$1$$.