Statement in an Erdos paper about primitive sets I was reading a paper from Erdos where he proved the following: let $(a_n)$ be a sequence such that if $a_n$ divides $a_m$ then $m=n$, and let $p_n$ be the greatest prime factor of $a_n$ then $\sum_{k=1}^\infty \frac{1}{a_n \log(a_n)}$ converges, and there's something about the proof that I don't get.
The thing I don't understand is that in the proof he shows that $\sum_{k=1}^\infty \frac{1}{a_n}\prod_{p\leq p_k}(1-\frac{1}{p})\leq 1$ (this I understand), but he goes on to say that this implies that $\sum_{k=1}^\infty \frac{1}{a_n \log(a_n)}$ converges because there exists some $c$ such that: $$ \prod_{p\leq p_n} \left(1-\frac{1}{p}\right) > \frac{c}{\log p_n}\geq \frac{c}{\log a_n}$$
Why does this $c$ exist? The second inequality is trivial but I don't see where the first one comes from.
 A: The statement is that
$$\prod_{p\leq x}\left(1-\frac 1p\right)=\Omega\left(\frac1{\log x}\right),$$
where the product runs over primes. This follows from (i.e. is a strictly weaker version of) Mertens's third theorem. There's a sketch of Mertens's original proof of this theorem (actually, of a closely related result) towards the bottom of the linked page.

For completeness, here's a proof, taking a few shortcuts because we only need to prove one direction. (This gives a bit more detail than the Wikipedia sketch, and is mostly here to prevent this answer being a link-only answer.) Since
$$1-\frac1p\geq e^{-\frac1p-\frac1{p^2}}$$
for all $p\geq 2$, we have
$$\prod_{p\leq x}\left(1-\frac 1p\right)\geq \exp\left(-\sum_{p\leq x}\frac1p-O(1)\right),$$
so we need to show that
$$\sum_{p\leq x}\frac 1p\leq \log\log x+O(1).$$
(We have gone from Mertens' third theorem to his second.) Now, for any positive integer $n$, $n!$ has at least $\lfloor n/p\rfloor\geq n/p-1$ factors of $p$ for each prime $p\leq n$. This means
$$n\log n>\log n!>\sum_{p\leq n}\left(\frac np-1\right)\log p=-\sum_{p\leq n}\log p+n\sum_{p\leq n}\frac{\log p}p.$$
For every $m\geq 0$, every prime between $n2^{-m-1}$ and $n2^{-m}$ divides
$$\binom {\lceil n2^{-m}\rceil}{\lceil n2^{-m-1}\rceil}<2^{n2^{-m}+1},$$
so
$$\sum_{p\leq n}\log p=\log\left(\prod_{p\leq n}p\right)\leq \log\left(\prod_{m=0}^{\log_2 n}2^{n2^{-m}+1}\right)\leq \log\left(2^{2n+\log_2 n}\right)=n\log 4+\log n.$$
This means
$$\sum_{p\leq n}\frac{\log p}p<\log(4n)+\frac{\log n}n.$$
Define $f(x)=\sum_{p\leq x}\frac{\log p}p$. We have by partial summation
$$\sum_{p\leq x}\frac1p=\frac{f(x)}{\log x}+\int_2^x\frac{f(t)dt}{t\log^2 t}\leq 1+o(1)+\int_1^x \frac{\log t+\log 4+\frac{\log t}t}{t\log^2 t}dt.$$
The first term in the integrand integrates to $\ln\ln x+O(1)$, and the other two terms are both $O(1)$, giving the desired result.
