This is Example 1 in Munkres' section 67. Consider the cartesian product $\mathbb{R}^{\omega}$ being an abelian group under the operation of coordinate-wise addition.
He constructed subgroups of $\mathbb{R}^{\omega}$ whose direct sum is $\mathbb{R}^{\infty}$, in the following way:
Consider the set $G_{n}$ consisting of those tuples $(x_{i})$ such that $x_{i}=0$ for all $i\neq n$. It is really quick to see that $G_{n}$ is a group, and it is isomorphic to $\mathbb{R}$, and thus subgroups of $\mathbb{R}^{\omega}$. Then $G_{n}$ generate the subgroup $\mathbb{R}^{\infty}$ of $\mathbb{R}^{\omega}$; indeed, $\mathbb{R}^{\infty}$ is their direct sum.
I understand $G_{n}\cong\mathbb{R}$ and thus subgroups of $\mathbb{R}^{\omega}$.
However, how could $G_{n}$ generate $\mathbb{R}^{\infty}$? Note that for each $x\in\mathbb{R}^{\infty}$, it is an $\infty$ tuples $$x=(x_{1}, x_{2}, x_{3},\cdots),$$ which is the sum of $$(x_{1},0,0,\cdots)+(0,x_{2},0,\cdots)+\cdots$$ and this sum is the sum from the element of each $G_{n}$. By then definition given by Munkres, groups $G_{\alpha}$ generates $G$ if every element $x\in G$ can be written as a finite sum of elements of the groups $G_{\alpha}$, but the sum above is not even finite sum.
Is there other way to express $x\in\mathbb{R}^{\omega}$ as a finite sum from the element of $G_{n}$? Why does the alternative way give a unique summation?
Thank you!
$\mathbb{R}^{\infty}$ is defined in Exercise 7 in Section 19, as the set of sequences with only finitely many nonzero terms. So if $$ x = (x_1, x_2, x_3, \dots) $$ is in $\mathbb{R}^{\infty}$, then it equals $$ x = (x_1, x_2, x_3, \dots, x_n, 0, 0, 0, \dots) $$ for some $n$, and so is a finite sum of elements from the subgroups $G_i$.