[proofplan]
The proof is the elementary upper-bound half of Brun's pure sieve. For each $a \in A$, we compare the indicator of the event $\gcd(a,P(z))=1$ with the even truncation of the inclusion-exclusion sum over prime divisors of $\gcd(a,P(z))$. Summing this pointwise inequality over $A$ gives an upper bound in terms of the quantities $|A_d|$, and the assumed asymptotic formula for $|A_d|$ supplies the main term. The remaining contribution is bounded by counting squarefree divisors of $P(z)$ with at most $2k$ prime factors.
[/proofplan]
[step:Dominate the coprimality indicator by an even inclusion-exclusion truncation]
Let $Q_z$ denote the finite set of primes $p<z$, so that $P(z)=\prod_{p \in Q_z}p$. For each $a \in A$, define
\begin{align*}
G(a) := \{p \in Q_z : p \mid a\}.
\end{align*}
Let $m(a):=\#G(a)$. Since $P(z)$ is squarefree, the squarefree divisors $d$ of $P(z)$ that divide $a$ are exactly the products of subsets of $G(a)$. Therefore
\begin{align*}
\sum_{\substack{d \mid P(z), d \mid a, \nu(d) \leq 2k}} \mu(d)
=
\sum_{j=0}^{\min(2k,m(a))}(-1)^j\binom{m(a)}{j}.
\end{align*}
If $m(a)=0$, both sides equal $1$. If $1 \leq m(a)\leq 2k$, the full binomial sum gives
\begin{align*}
\sum_{j=0}^{m(a)}(-1)^j\binom{m(a)}{j}=(1-1)^{m(a)}=0.
\end{align*}
If $m(a)>2k$, the standard alternating binomial identity gives
\begin{align*}
\sum_{j=0}^{2k}(-1)^j\binom{m(a)}{j}=\binom{m(a)-1}{2k}\geq 0.
\end{align*}
Thus, in all cases,
\begin{align*}
\mathbb{1}_{\{\gcd(a,P(z))=1\}} \leq \sum_{\substack{d \mid P(z), d \mid a, \nu(d) \leq 2k}} \mu(d).
\end{align*}
[guided]
The goal is to replace the difficult condition $\gcd(a,P(z))=1$ by a finite divisor sum. Let $Q_z$ be the set of primes below $z$, and for a fixed element $a \in A$ define
\begin{align*}
G(a) := \{p \in Q_z : p \mid a\}.
\end{align*}
The number $m(a):=\#G(a)$ counts how many of the sieving primes divide $a$. Since $P(z)$ is squarefree, every divisor $d$ of $P(z)$ is the product of a subset of $Q_z$, and the extra condition $d \mid a$ says precisely that this subset lies inside $G(a)$.
For a squarefree integer $d$ with $\nu(d)=j$, the Mobius function satisfies $\mu(d)=(-1)^j$. Hence the truncated divisor sum becomes
\begin{align*}
\sum_{\substack{d \mid P(z), d \mid a, \nu(d) \leq 2k}} \mu(d)
=
\sum_{j=0}^{\min(2k,m(a))}(-1)^j\binom{m(a)}{j}.
\end{align*}
We now compare this expression with the indicator of coprimality.
If $m(a)=0$, no prime below $z$ divides $a$, so $\gcd(a,P(z))=1$. The sum contains only the divisor $d=1$, so it equals $\mu(1)=1$, matching the indicator.
If $1\leq m(a)\leq 2k$, the truncation includes every subset of $G(a)$. Therefore the [binomial theorem](/theorems/750) gives
\begin{align*}
\sum_{j=0}^{m(a)}(-1)^j\binom{m(a)}{j}=(1-1)^{m(a)}=0.
\end{align*}
In this case $\gcd(a,P(z))>1$, so the coprimality indicator is also $0$.
Finally suppose $m(a)>2k$. The truncation stops at the even index $2k$. The alternating binomial identity
\begin{align*}
\sum_{j=0}^{r}(-1)^j\binom{m}{j}=(-1)^r\binom{m-1}{r}
\end{align*}
valid for $0\leq r<m$ gives, with $r=2k$ and $m=m(a)$,
\begin{align*}
\sum_{j=0}^{2k}(-1)^j\binom{m(a)}{j}=\binom{m(a)-1}{2k}\geq 0.
\end{align*}
Since $m(a)>0$, the indicator $\mathbb{1}_{\{\gcd(a,P(z))=1\}}$ is $0$, so this non-negative truncated sum still dominates it.
Combining the three cases yields the pointwise inequality
\begin{align*}
\mathbb{1}_{\{\gcd(a,P(z))=1\}} \leq \sum_{\substack{d \mid P(z), d \mid a, \nu(d) \leq 2k}} \mu(d).
\end{align*}
for every $a \in A$.
[/guided]
[/step]
[step:Sum the pointwise bound over the set $A$]
Summing the pointwise inequality over all $a \in A$ gives
\begin{align*}
S(A,P,z) = \sum_{a \in A}\mathbb{1}_{\{\gcd(a,P(z))=1\}} \leq \sum_{a \in A}\sum_{\substack{d \mid P(z), d \mid a, \nu(d) \leq 2k}}\mu(d).
\end{align*}
Since the sums are finite, we interchange their order:
\begin{align*}
\sum_{a \in A}\sum_{\substack{d \mid P(z), d \mid a, \nu(d) \leq 2k}}\mu(d) = \sum_{\substack{d \mid P(z), \nu(d) \leq 2k}}\mu(d)\#\{a \in A : d \mid a\}.
\end{align*}
By the definition of $A_d$, this becomes
\begin{align*}
S(A,P,z) \leq \sum_{\substack{d \mid P(z), \nu(d) \leq 2k}}\mu(d)|A_d|.
\end{align*}
[/step]
[step:Insert the assumed local density formula]
For every squarefree divisor $d$ of $P(z)$, the hypothesis gives
\begin{align*}
|A_d|=\frac{\omega(d)}{d}X+r_d.
\end{align*}
Substituting this into the preceding inequality yields
\begin{align*}
S(A,P,z) \leq X\sum_{\substack{d \mid P(z), \nu(d) \leq 2k}}\frac{\mu(d)\omega(d)}{d} + \sum_{\substack{d \mid P(z), \nu(d) \leq 2k}}\mu(d)r_d.
\end{align*}
The first term is the claimed main term. It remains only to bound the remainder sum.
[/step]
[step:Count the truncated divisor set to bound the remainder]
Let $N:=\pi(z)=\#Q_z$. Since the formalized statement now assumes $z>2$, the prime $2$ lies in $Q_z$, so $N\geq 1$. Each squarefree divisor $d$ of $P(z)$ with $\nu(d)=j$ is obtained by choosing $j$ primes from $Q_z$, so the number of such divisors with $\nu(d)\leq 2k$ is
\begin{align*}
\sum_{j=0}^{2k}\binom{N}{j}.
\end{align*}
Using $N\geq 1$ and keeping $k$ fixed, define
\begin{align*}
C_k := \sum_{j=0}^{2k}\frac{1}{j!}.
\end{align*}
Then
\begin{align*}
\sum_{j=0}^{2k}\binom{N}{j}\leq C_k N^{2k}.
\end{align*}
Because $|\mu(d)|\leq 1$ and $|r_d|\leq 1$ for every squarefree divisor $d$ of $P(z)$, the triangle inequality gives
\begin{align*}
\left|\sum_{\substack{d \mid P(z), \nu(d) \leq 2k}}\mu(d)r_d\right| \leq \sum_{\substack{d \mid P(z), \nu(d) \leq 2k}}1 \leq C_k(\pi(z))^{2k}.
\end{align*}
Therefore
\begin{align*}
\sum_{\substack{d \mid P(z), \nu(d) \leq 2k}}\mu(d)r_d = O_k((\pi(z))^{2k}).
\end{align*}
[/step]
[step:Combine the main term and the error term]
Combining the upper bound after substitution with the divisor-counting estimate gives
\begin{align*}
S(A,P,z) \leq X\sum_{\substack{d \mid P(z), \nu(d) \leq 2k}}\frac{\mu(d)\omega(d)}{d} + O_k((\pi(z))^{2k}).
\end{align*}
This is the claimed crude Brun upper bound.
[/step]
