I am having trouble understanding the proof of Theorem 6.1.16 in the lecture notes of Vaughan Jones on von Neumann algebras (which can be found here). The theorem states that $$ tr(\mathcal{P}(M)) = [0,1]$$ where $M$ is a $II_1$ factor, $tr: M \rightarrow \mathbb C$ is its unique faithful normal tracial state and $\mathcal{P}(M)$ is the set of projections in $M$.
The proof goes as follows: For any $r \in [0,1]$ we consider the partially ordered set $S = \{p \in \mathcal{P}(M) \mid tr(p) \leq r\}$. Every chain $(p_\alpha)$ in $S$ has an upper bound (namely $p = \bigvee_\alpha p_\alpha$) which is again in $S$, so by Zorn's lemma, $S$ has a maximal element $q$. Now comes the part which I don't understand: Jones argues that if $tr(q)$ were strictly less than $r$, then by comparability we would have $q \prec p$. (Why is this the case? What is $p$ here, is it the same as above?) Now choose $q' \sim q$ with $q' < p$. Since $M$ is a II$_1$ factor, so is $(p-q')M(p-q')$, thus there is a projection strictly between $q'$ and $p$.
Can someone please clarify the details for me here?
Note: I am familiar with the proof using the halving lemma and dyadic rationals to construct a projection $p$ with $tr(p) = r$ (in particular I am familiar with this post), but I want to understand this alternative proof.
The proof is not very well-written, and I don't immediatelly see a way to save it, since (as you say) it is not clear what $p$ would be. The argument would work if $p$ were a projection with $\operatorname{tr}(p)=r$, but the whole point of the proof is to find such $p$.
So let me show you a way to do the proof with the results at hand in Jones' text. Once you get $q$ maximal with $\operatorname{tr}(q)\leq r$, you could have $\operatorname{tr}(q)=\operatorname{tr}(1)=r$, in which case you are done. Otherwise, suppose that $\operatorname{tr}(q)<r$ and consider the II$_1$-factor $(1-q)M(1-q)$. By 6.1.14 and 6.1.15, there is a projection $q'$ in $(1-q)M(1-q)$ with $\operatorname{tr}(q')<r-\operatorname{tr}(q)$. Then $q+q'$ is a projection with $\operatorname{tr}(q)< \operatorname{tr}(q+q')\leq r$, contradicting the maximality of $q$. Thus $\operatorname{tr}(q)=r$, which is what we wanted to prove.