Are Projections $p$ and $q$ in a C$^{*}$-algebra $A$ equal if $p\leq q\leq tp$ for some $t\in[0,\infty)$?

Question

I am working on Exercise 3.11 (ii) in Rordam's book. This is where the motivation for my question (which is in the title) comes from. It is the last step I need to finish off the problem. If it is false in general, is there some condition on $t$ or on the projections $p$ and $q$ that make the result true?

Thank you.

Answer 1

Well if $t\in[0,1)$ then $p=0$ (since $p\leq tp$), so $0\leq q\leq 0$ and thus $q=0$. Now if $t\in [1,\infty)$, then by taking successive roots we have $$p\leq q\leq t^\frac{1}{2^n}p$$ for all $n\in \mathbb N$. Hence $p\leq q \leq p$, and thus $p=q$.

Answer 2

If $t=0$, $p=0=q$.

If $t\ne0$, from $q\leq tp$ we have $$ (1-p)q(1-p)\leq t(1-p)p(1-p)=0. $$ So $$ 0=(1-p)q(1-p)=[q(1-p)]^*(1-p)q, $$ implying that $(1-p)q=0$. That is $q=pq$, from there $q\leq p$.