# Subprojection of a finite projection

The question is entirely explained here, in that I wonder why every source seems to regard as obvious the claim that subprojections of finite projections are finite.

To me, playing around with the tools and relations at hand doesn't work, and that's what would have to work before I'd make something a remark without further proof. So, I mean, I let $f$ be a subprojection of $e$ when $e$ is finite, and then if $g$ is a strict subprojection of $f$ then I need to show $g$ is not equivalent to $f$, so I notice $g$ is also a strict subprojection of $e$, hence is not equivalent to it, and then everything comes to a grinding halt. I must be missing something really easy.

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Let $M$ be a von Neumann algebra. Suppose that $f\in M$ is a finite projection, i.e., if $e \in M$ is a projection with $e \leq f$, $e \sim f$, then $e=f$.
Let $g \in M$ be a projection such that $g \leq f$. We will show that $g$ is finite. So we consider a projection $h \in M$, such that $h\leq g$ with $h\sim g$, and we seek to show that $h=g$. The basic idea is to move from a statement about $g$ to a statement about $f$ and so we are led to consider the difference $f-g$.
As $g \leq f$, we have that $(f-g)$ is a projection. Note that $(f-g)+h \leq f$. Then we have, as $(f-g)\perp h$ and $(f-g)\perp g$ (*), that $$(f-g) + h \sim (f-g)+g=f.$$ By finiteness of $f$, we have that $(f-g)+h=f$. Thus $g=h$ and $g$ is finite.
Further details on (*): If $vv^*=e \sim e_1=v^*v$, and $uu^*=f \sim f_1=u^*u$, with $e \perp f$, $e_1 \perp f_1$, then $$(u+v)^*(u+v)=e+f \sim e_1+f_1 =(u+v)(u+v)^*.$$ Just keep in mind what the supports of the partial isometries $u$ and $v$ are.