[proofplan]
We use the defining implementation of Murray-von Neumann equivalence by partial isometries: $p\sim q$ means that some $v\in M$ satisfies $v^*v=p$ and $vv^*=q$. Reflexivity, symmetry, and transitivity of $\sim$ are proved by using $p$, taking adjoints, and multiplying compatible partial isometries. For subequivalence, we use the definition $p\precsim q$ iff $p$ is equivalent to a subprojection of $q$, and the only point needing care is transitivity: the partial isometry carrying $q$ into $r$ must be restricted to the subprojection of $q$ that is equivalent to $p$.
[/proofplan]
[step:Prove reflexivity of Murray-von Neumann equivalence using the projection itself]
Let $p\in\mathcal{P}(M)$. Since $p=p^*=p^2$, the operator $p\in M$ is a partial isometry and satisfies
\begin{align*}
p^*p=p
\end{align*}
and
\begin{align*}
pp^*=p.
\end{align*}
Thus $p\sim p$. Since $p\in\mathcal{P}(M)$ was arbitrary, $\sim$ is reflexive on $\mathcal{P}(M)$.
[/step]
[step:Prove symmetry of Murray-von Neumann equivalence by taking adjoints]
Let $p,q\in\mathcal{P}(M)$ and suppose $p\sim q$. By definition, there exists a partial isometry $v\in M$ such that
\begin{align*}
v^*v=p
\end{align*}
and
\begin{align*}
vv^*=q.
\end{align*}
Because $M$ is self-adjoint, $v^*\in M$. The adjoint $v^*$ is a partial isometry, and its initial and final projections are
\begin{align*}
(v^*)^*v^*=vv^*=q
\end{align*}
and
\begin{align*}
v^*(v^*)^*=v^*v=p.
\end{align*}
Therefore $q\sim p$. Hence $\sim$ is symmetric.
[/step]
[step:Prove transitivity of Murray-von Neumann equivalence by composing compatible partial isometries]
Let $p,q,r\in\mathcal{P}(M)$, and suppose $p\sim q$ and $q\sim r$. Choose partial isometries $v,w\in M$ such that
\begin{align*}
v^*v=p,\qquad vv^*=q
\end{align*}
and
\begin{align*}
w^*w=q,\qquad ww^*=r.
\end{align*}
Define $a\in M$ by $a=wv$. Since $M$ is a subalgebra, $a\in M$. We compute its initial projection:
\begin{align*}
a^*a=(wv)^*(wv)=v^*w^*wv=v^*qv.
\end{align*}
Since $q=vv^*$, we have
\begin{align*}
v^*qv=v^*vv^*v=p^2=p.
\end{align*}
Similarly, its final projection is
\begin{align*}
aa^*=wvv^*w^*=wqw^*.
\end{align*}
Since $q=w^*w$, we get
\begin{align*}
wqw^*=ww^*ww^*=r^2=r.
\end{align*}
Thus $a^*a=p$ and $aa^*=r$, so $a$ is a partial isometry implementing $p\sim r$. Therefore $\sim$ is transitive.
[guided]
The orientation of the implementing partial isometries is the main point. The relation $p\sim q$ is implemented by a partial isometry whose initial projection is $p$ and whose final projection is $q$. Thus we choose $v\in M$ with
\begin{align*}
v^*v=p
\end{align*}
and
\begin{align*}
vv^*=q.
\end{align*}
Likewise, $q\sim r$ is implemented by a partial isometry $w\in M$ with
\begin{align*}
w^*w=q
\end{align*}
and
\begin{align*}
ww^*=r.
\end{align*}
The correct composition is $wv$, because $v$ moves from the initial projection $p$ to the intermediate projection $q$, and $w$ moves from $q$ to $r$. Define $a\in M$ by $a=wv$. Since $M$ is closed under multiplication, $a\in M$.
We now verify directly that $a$ implements $p\sim r$. First,
\begin{align*}
a^*a=(wv)^*(wv)=v^*w^*wv=v^*qv.
\end{align*}
The equality $q=vv^*$ gives
\begin{align*}
v^*qv=v^*vv^*v=(v^*v)(v^*v)=p^2=p,
\end{align*}
because $p$ is a projection. Hence the initial projection of $a$ is $p$.
For the final projection,
\begin{align*}
aa^*=wvv^*w^*=wqw^*.
\end{align*}
Using $q=w^*w$, we obtain
\begin{align*}
wqw^*=ww^*ww^*=(ww^*)^2=r^2=r,
\end{align*}
because $r$ is a projection. Therefore $a^*a=p$ and $aa^*=r$. This proves that $a$ is a partial isometry implementing $p\sim r$, so Murray-von Neumann equivalence is transitive.
[/guided]
[/step]
[step:Prove reflexivity of Murray-von Neumann subequivalence by choosing the same projection]
Let $p\in\mathcal{P}(M)$. Since $p\le p$ and $p\sim p$ by reflexivity of $\sim$, the definition of Murray-von Neumann subequivalence gives $p\precsim p$. Hence $\precsim$ is reflexive.
[/step]
[step:Prove transitivity of Murray-von Neumann subequivalence by restricting the second partial isometry]
Let $p,q,r\in\mathcal{P}(M)$, and suppose $p\precsim q$ and $q\precsim r$. By definition of $\precsim$, there are projections $q_0,r_0\in\mathcal{P}(M)$ such that
\begin{align*}
q_0\le q,\qquad r_0\le r,
\end{align*}
with $p\sim q_0$ and $q\sim r_0$.
Choose partial isometries $v,u\in M$ such that
\begin{align*}
v^*v=p,\qquad vv^*=q_0
\end{align*}
and
\begin{align*}
u^*u=q,\qquad uu^*=r_0.
\end{align*}
Define $b\in M$ by $b=uq_0$. Since $u,q_0\in M$, we have $b\in M$. Its initial projection is
\begin{align*}
b^*b=q_0u^*uq_0=q_0qq_0=q_0.
\end{align*}
Its final projection is
\begin{align*}
bb^*=uq_0u^*.
\end{align*}
The operator $uq_0u^*$ is a projection, since
\begin{align*}
(uq_0u^*)^2=uq_0u^*uq_0u^*=uq_0qq_0u^*=uq_0u^*
\end{align*}
and
\begin{align*}
(uq_0u^*)^*=uq_0u^*.
\end{align*}
Also,
\begin{align*}
uq_0u^*\le uqu^*=uu^*=r_0\le r.
\end{align*}
Set $r_1=uq_0u^*$. Then $r_1\in\mathcal{P}(M)$, $r_1\le r$, and $b$ implements $q_0\sim r_1$.
Since $p\sim q_0$ and $q_0\sim r_1$, transitivity of $\sim$ gives $p\sim r_1$. Because $r_1\le r$, the definition of subequivalence gives $p\precsim r$. Thus $\precsim$ is transitive.
[/step]
[step:Deduce subequivalence in both directions from equivalence]
Let $p,q\in\mathcal{P}(M)$ and suppose $p\sim q$. Since $q\le q$, the definition of subequivalence gives $p\precsim q$. By symmetry of $\sim$, we also have $q\sim p$, and since $p\le p$, the definition gives $q\precsim p$. This proves the final assertion and completes the proof.
[/step]
