Show that $\langle \{e_i \mid i \in \Bbb N\}\rangle \ne \Bbb Z^\Bbb N$ that is the set of all such sequences does not generate $\Bbb Z^\Bbb N$. 
Let $\Bbb Z^\Bbb N$ be the set of all infinite sequences $(a_0, a_1, \dots)$ where $a_i \in \Bbb Z$ for $i \in \Bbb N$. This set is a group when we consider addition component-wise. If $e_i=(0, \dots,1,0,\dots)$ is a sequence with $1$ on the $i$'th index show that $$\langle \{e_i \mid i \in \Bbb N\}\rangle \ne \Bbb Z^\Bbb N$$ that is the set of all such sequences does not generate $\Bbb Z^\Bbb N$.

The inclusion $\langle \{e_i \mid i \in \Bbb N\}\rangle \subset \Bbb Z^\Bbb N$ will probably hold as adding component-wise these $e_i$'s will give me elements of $\Bbb Z^\Bbb N$ however I fail to see how can $\Bbb Z^\Bbb N$ contain an element which cannot be made from summing $e_i$'s? Is it $(0,0, \dots)$ that cannot be made from any $e_i$'s?
 A: If $e_i = (0,\ldots,0,1,0,\ldots)$ with a one in the $i$-th entry, and we add finitely many such elements, we will only ever have finitely many nonzero entries in the result. In other words, it is impossible to write
$(1,1,\ldots,)$, the element with all ones, as a finite sum of the $e_i$'s.
A group is generated by a set of elements if anything in the group can be written as a finite sum of the given elements.
A: What is the definition of $\langle \{e_i \mid i \in \Bbb N\}\rangle$ ?
Quoting from the first line of wikipedia:

In abstract algebra, a generating set of a group is a subset of the
group set such that every element of the group can be expressed as a
combination (under the group operation) of finitely many elements
of the subset and their inverses.

In other words, every element of $\langle \{e_i \mid i \in \Bbb N\}\rangle$ is a sequence with finitely many non-zero entries.
For example, $(1,1,1,...)$ cannot be expressed as a addition of finitely many $e_i,$ and therefore $(1,1,1,...) \not\in\langle \{e_i \mid i \in \Bbb N\}\rangle$
