First we show that abundancy index is multiplicative. Let $n \in \N$ and let $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ be its prime factorization. Then the abundancy index of $n$ is: {{begin-eqn}} {{eqn | l = \frac {\map \sigma n} n       | r = \frac {\map \sigma {p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} } } {p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} } }} {{eqn | r = \frac {\map \sigma {p_1^{a_1} } \map \sigma {p_2^{a_2} } \cdots \map \sigma {p_k^{a_k} } } {p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} }       | c = Sigma Function is Multiplicative }} {{eqn | r = \frac {\map \sigma {p_1^{a_1} } } {p_1^{a_1} } \cdot \frac {\map \sigma {p_2^{a_2} } } {p_2^{a_2} } \cdot \cdots \cdot \frac {\map \sigma {p_k^{a_k} } } {p_k^{a_k} } }} {{end-eqn}} so abundancy index is multiplicative. {{qed|lemma}} Next we show that $M$ exists. Note that Sum of Reciprocals of Primes is Divergent. By Divergent Sequence with Finite Number of Terms Deleted is Divergent: :the sum of reciprocals of all elements of $\mathbb P^-$ is also divergent. Observe that: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \frac {\ln \frac {p_n} {p_n - 1} } {\frac 1 {p_n - 1} }       | r = \lim_{\frac 1 {p_n - 1} \mathop \to 0} \frac {\map \ln {1 + \frac 1 {p_n - 1} } } {\frac 1 {p_n - 1} } }} {{eqn | r = 1       | c = Derivative of Logarithm at One }} {{end-eqn}} By Limit Comparison Test: :$\displaystyle \sum_{n \mathop = 1}^\infty \ln \frac {p_n} {p_n - 1}$ is divergent as well. By Logarithm of Divergent Product of Real Numbers: :$\displaystyle \prod_{n \mathop = 1}^\infty \frac {p_n} {p_n - 1}$ diverges to infinity. Hence: :$\exists N \in \N: \forall n \ge N: \displaystyle \prod_{n \mathop = 1}^\infty \frac {p_n} {p_n - 1} > r$ Therefore $\displaystyle \set {m \in \N: \displaystyle \prod_{i \mathop = 1}^m \frac {p_i} {p_i - 1} > r} \ne \O$. By the Well-Ordering Principle, $M$ exists. {{qed|lemma}} Finally, we prove our claims $(1)$ and $(2)$. Let $n$ be a number formed with fewer than $M$ distinct prime factors in $\mathbb P^-$. Let $n = q_1^{a_1} q_2^{a_2} \cdots q_k^{a_k}$ be its prime factorization, where $q_i \in \mathbb P^-$ and $k < M$. Then: {{begin-eqn}} {{eqn | l = \frac {\map \sigma n} n       | r = \prod_{i \mathop = 1}^k \frac {\map \sigma {q_i^{a_i} } } {q_i^{a_i} } }} {{eqn | r = \prod_{i \mathop = 1}^k \frac {q_i^{a_i + 1} - 1} {q_i^{a_i} \paren {q_i - 1} }       | c = Sigma Function of Power of Prime }} {{eqn | r = \prod_{i \mathop = 1}^k \frac {q_i - q_i^{-a_i} } {q_i - 1} }} {{eqn | o = <       | r = \prod_{i \mathop = 1}^k \frac {q_i} {q_i - 1}       | c = as $q_i^{-a_i} > 0$ }} {{eqn | o = \le       | r = \prod_{i \mathop = 1}^{M - 1} \frac {p_i} {p_i - 1} }} {{eqn | o = \le       | r = r       | c = by minimality condition on $M$ }} {{end-eqn}} This proves $(1)$. {{qed|lemma}} Now define $M \bar \sharp = \displaystyle \prod_{i \mathop = 1}^M p_i$ (an analog of the primorial for $\mathbb P^-$). Consider the sequence of abundancy indices of $\paren {M \bar \sharp}^n$, where $n$ is a strictly positive integer. We have: {{begin-eqn}} {{eqn | l = \frac {\map \sigma {\paren {M \bar \sharp}^n} } {\paren {M \bar \sharp}^n}       | r = \prod_{i \mathop = 1}^M \frac {\map \sigma {p_i^n } } {p_i^n} }} {{eqn | r = \prod_{i \mathop = 1}^M \frac {p_i - p_i^{-n} } {p_i - 1}       | c = similar to above }} {{end-eqn}} This product is strictly increasing and tends to $\displaystyle \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}$ as $n \to \infty$, which is strictly greater than $r$. From the definition of convergence to a limit: :$\displaystyle \exists N \in \N: \forall n \ge N: \size {\frac {\map \sigma {\paren {M \bar \sharp}^n} } {\paren {M \bar \sharp}^n} - \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}} < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1} - r$ Since $\displaystyle \frac {\map \sigma {\paren {M \bar \sharp}^n} } {\paren {M \bar \sharp}^n} < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}$ for all $n$: :$\displaystyle r < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1} - \size {\frac {\map \sigma {\paren {M \bar \sharp}^N} } {\paren {M \bar \sharp}^N} - \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}} = \frac {\map \sigma {\paren {M \bar \sharp}^N} } {\paren {M \bar \sharp}^N}$ Therefore $\paren {M \bar \sharp}^N$ is a number formed with $M$ distinct prime factors in $\mathbb P^-$ with abundancy index at least $r$. This proves $(2)$. {{qed}}