Proof
(related to Lemma: Sets of Integers CoPrime to a given Integer are DivisorClosed)
 Assume, \(d\ge 0\) is a fixed integer.
 Case \(d=0\):
 Let \(n\in \mathbb Z_0\).
 By definition of $Z_0$ we have \(\gcd(n,0)=1\) only for \(n=1\), it follows that \(\mathbb Z_0=\{1\}\).
 Trivially, \(\mathbb Z_0\) is divisorclosed.
 Case \(d = 1\):
 Let \(n\in \mathbb Z_1\).
 Since we have \(\gcd(n,1)=1\) for all \(n\in \mathbb Z\), it follows \(\mathbb Z_1=\mathbb Z\).
 Trivially, \(\mathbb Z\) is divisorclosed.
 Case \(d > 1\):
 There is a unique factorization \(d=p_1^{e_1}\cdot\ldots\cdot p_k^{e_k}\) with \(e_i\ge 0\) for \(i=1,\ldots,k\) by the fundamental theorem of arithmetic.
 Let \(n\in \mathbb Z_d\) and let \(t\mid n\). We have to show that \(t\in\mathbb Z_d\).
 If \(t=1\), then \(\gcd(t,d)=1\). Thus, \(t\in \mathbb Z_d\).
 If \(t > 1\), then also \(n > 1\) and we have two unique factorizations:
* \(n=q_1^{f_1}\cdot\ldots\cdot q_m^{f_m}\) be with \(0\le f_j\) for \(j=1,\ldots,m\) and
* \(t=q_1^{g_1}\cdot\ldots\cdot q_m^{g_m}\) with \(0\le g_j\le f_j\) for \(j=1,\ldots,m\), since \(t\mid n\).
 Because \(n\in \mathbb Z_d\), we have \(\gcd(n,d)=1\).
 Thus, there is no such index \(l\), \(l=1,\ldots,\min(m,k)\), for which both exponents \(e_l > 0\) and \(g_l > 0\).
 Otherwise, \(\gcd(q_l^{e_l},q_l^{g_l})=q_l^{\min(e_l,g_l)} > 1 \), because \(\min(e_l,g_l)>1\), and we would have found a proper common divisor of \(n\) and \(d\), contradicting \(n\in\mathbb Z_d\).
 For the same reason, $t$ and\(d\) are coprime.
 Thus, \(t\in\mathbb Z_d\).
∎
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References
Bibliography
 Piotrowski, Andreas: Own Research, 2014