Hello, could you please help me to solve this problem:

let A is a von Neumann algebra, and ($\displaystyle p_i $) - projections on A, $\displaystyle i\in I$. Then inf $\displaystyle p_i$, sup $\displaystyle p_i$ - projections on A.

Thank you!

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- Apr 21st 2011, 02:10 PMkarkushaProjections on Neumann algebra
Hello, could you please help me to solve this problem:

let A is a von Neumann algebra, and ($\displaystyle p_i $) - projections on A, $\displaystyle i\in I$. Then inf $\displaystyle p_i$, sup $\displaystyle p_i$ - projections on A.

Thank you! - Apr 22nd 2011, 11:52 AMOpalg
If $\displaystyle p$, $\displaystyle q$ are two projections in $\displaystyle A$, then $\displaystyle p\vee q$ is the range projection of $\displaystyle p+q$, and is therefore in $\displaystyle A.$ By induction, the sup of any finite family of projections in $\displaystyle A$ is also in $\displaystyle A.$

Given an infinite family, the directed net of sups of finite subfamilies (ordered by inclusion) is increasing, and bounded above by the identity. Therefore it converges strongly to a limit which is a projection and is therefore the sup of the whole family.

The result for infs follows by taking orthogonal complements. - Apr 22nd 2011, 12:29 PMkarkusha
Thank you,

**Opalg**. You explained that sup is in $\displaystyle A$, but why is sup projection on $\displaystyle A$? - Apr 22nd 2011, 12:43 PMOpalg
The sup of an increasing directed net of projections is equal to the sup projection (Kadison and Ringrose, Proposition 2.5.6).