A question on hereditary C*-subalgebra. From: An introduction to the classification of amenable C*-algebras.
Lemma 3.5.8. Let $A$ be a C*-algebra satisfying the condition that every hereditary C*-subalgebra contains at least two mutually orthogonal nonzero positive elements. Then for any nonzero elements $a,b\in A^+$ there are nonzero positive elements $a_1\in \text{ Her}(a)$ and $b_1\in \text{ Her}(b)$ such that $a_1b_1= 0$.
Proof. We may assume that $x = ab \neq 0$. Suppose that $x = v|x|$ is the polar decomposition of $x$ in $A''$. Fix $0 < \epsilon < \|x\|$. By the assumption,
there are mutually orthogonal nonzero positive elements $d_1,d_2$ such that
$f_\epsilon(|x|)d_i = d_i$, $i = 1,2$. Then $c_i = vd_iv^*\in\text{ Her}(|x^*|)$ and $f_\epsilon(|x^*|)c_i = c_i$,$i = 1,2$ . Since $c_2vd_1d_1v^*C_2 = 0$, $c_2vd_1 = 0$.
Let $g(t) = t^{-l}f_\epsilon(t)$. Then $g \in C_0((0, \|x\|])^+$. Set $z_1 = g(|x|)d_1$. Then
$xz_1 = v|x|g(|x|)d_1 = vf_\epsilon(|x|)d_1 = vd_1$.
Therefore $c_2xz_1= c_2vd_1 = 0$. Now set $b_1 = bz_1b$ and $a_1= ac_2a$. Note that
$b_1\in\text{Her}(b)^+$, $a_1\in \text{ Her}(a)^+$ and both are nonzero. However,
$a_1b_1 = ac_2abz_1b = a(c_2xz_1)b = 0$.

$f_\epsilon(t)$ is the continuous monotonic function such that $f_\epsilon(t)=0$ on $[0,\epsilon/2]$ and $f_\epsilon(t)=1$ on $[\epsilon,\infty)$.
I see the existence of $d_1,d_2$, but I do not see why $b_1=bz_1b$ is positive.
Also, in the next lemma, lemma 3.5.9
Let $A$ be a unital simple C*-algebra with property (SP).
Then for any nonzero positive elements $a,b \in A$ there is $u \in U(A)$ such
that $u\text{Her}(a)u^* \cap \text{Her}(b) \neq {0}$.
It says, It follows from the above lemma that we assume that $ab = 0$. I see why we can assume $ab=0$, but if lemma 3.5.8 is not true, how do I prove this without lemma 3.5.8?
 A: There might be tons of ways to argue in favor of  the positivity of $b_1$ but, since $b_1=b^*z_1b$ (recall that $b$ is self-adjoint),
I suppose  the most natural way is to try  proving that $z_1$ is positive.  However,  if one assumes nothing except   that
$d_1$ is positive,  and that $f_\varepsilon (|x|)d_1=d_1$, it is not possible to  deduce that
$z_1$ is positive.
Here is a counter-example:  working within $M_3(\mathbb {C})$,
let
$x=\text{diag}(1,2,3)$, so that $|x|=x$.
Setting $\varepsilon =2$, we have that  $f_\varepsilon (1)=0$, while $f_\varepsilon (2)=f_\varepsilon (3)=1$,  so
$$
  f_\varepsilon (|x|)=\text{diag}(0, 1, 1).
  $$
On the other hand, with $g(t)=t^{-1}f_\varepsilon (t)$, we have that $g(1)=0$, $g(2)=1/2$,  and $g(3)=1/3$, and therefore
$$
  g(|x|)=\text{diag}(0,1/2,1/3).
  $$
In order for "$f_\varepsilon (|x|)d_1=d_1$" to hold we may thus take  $d_1$ to be any positive matrix of the form
$$
  d_1 = \pmatrix{
    0 & 0 & 0 \cr
    0 & * & * \cr
    0 & * & *}.
  $$
It is therefore  easy to choose one which does not commute with   $g(|x|)$  (any  non-diagonal matrix will do),  in which case
the product
$$
  z_1:=g(|x|)d_1
  $$
will fail to be self-adjoint,  much less positive.
