[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*}[/step]

[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: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]