infinite product of $Z$ hasn't a basis

Question

I was reading this brief article (I don't know if I am allowed to post links, if don't I apologize) concerning the fact that $\prod_{i=1}^{\infty} \mathbb{Z}$ has no basis and is not a free group. The problem is just before the end of the article. When he says that the last equation has solutions for $n=n_1,n_2, \dots$, shouldn't he stop at $n_i$? Then how can he conclude using the final lemma, if we have only a finite number of integers that allow a solution? (that is exacly what the lemma is saying, and so I shouldn't get an absurd). Probably I am getting something wrong, but I can't figure out where my mistake is. Thanks for the help.

Answer 1

Let $y=(n_1,n_2,\ldots)$ with $\frac{n_{i+1}}{n_i}\in\mathbb{Z}\setminus\{\pm 1\}$. Let $i\geq 1$. To get a solution for $\overline{y} = n_i\cdot x \in G/H$, we use the equation $y\equiv (\underbrace{0,\ldots,0}_{i\text{ entries}},n_{i+1},n_{i+2},\ldots)\mod H$. (Note that $\bigoplus_{i=1}^\infty\mathbb{Z}\subset H$) We set $$x' =\Big(\underbrace{0,\ldots,0}_{i\text{ entries}},\frac{n_{i+1}}{n_i},\frac{n_{i+2}}{n_i},\ldots\Big)$$ By definition of $y$, the fractions in $x'$ are all integers, i.e. $x'\in G$. So by setting $x = \overline{x'}$, the equation $\overline{y}=n_i\cdot x\in G/H$ is satisfied. This construction can be done for all $i$.

Answer 2

The proof needs infinitely many integers, and we have them: namely all $n_i$. To get a solution for $ y =n_i \cdot x$ we use the $y$ as constructed there. We can do this for all $i$, so $ y = nx $ has a solution for all $n = n_i$ and this gives a contradiction from the final lemma.