Trying to understand the definition of a free Abelian group. 
Let $B$ be a subset of an (additive) abelian group $F$.  Then $F$ is free abelian with basis $B$ if the cyclic subgroup $\langle b \rangle$ is infinite cyclic for each $b \in B$ and $F=\sum_{b \in B} \langle b \rangle $ (direct sum).  A free abelian group is thus a direct sum of copies of $\Bbb Z$.  A typical element $x \in F$ has a unique expression $$x = \sum m_b b$$ where $m_b \in \Bbb Z$ and almost all $m_b$ (all but a finite number) are zero.

I understand that by $F = \sum_{b \in B} \langle b \rangle$ (direct sum), they mean an external direct product of the infinite cyclic subgroups $\langle b \rangle$.
But what does $x=\sum m_bb$ mean?  
Since $F = \sum_{b \in B} \langle b \rangle = \{ ...,-b_1, 0, b_1, ...\} \oplus \dots \oplus \{\dots, -b_n, 0, b_n, \dots\}$ and $x \in F$, then $x=(m_1b_1, \dots, m_nb_n)$.  But each $b$ is an element of $F$ and therefore by the same logic shouldn't each $b$ be of the form $(j_1b_1, \dots, j_nb_n)$?  And therefore $x=(m_1(j_1b_1, \dots, j_nb_n), etc \dots)$?
What exactly am I misunderstanding here?  This seems like a circular definition.
 A: 
I understand that by $F = \sum_{b \in B} \langle b \rangle$ (direct sum), they mean an external direct product of the infinite cyclic subgroups $\langle b \rangle$.

No. They mean an internal direct sum of these cyclic subgroups.

But what does $x=\sum m_bb$ mean?

It means that every $x$ may be written as an integral linear combination of the elements of $B$. Moreover, because the sum is direct, the coefficients are uniquely determined by $x$.
This is just the generalisation of the concept of a vector space with a fixed basis if we allow the coefficients (scalars) to lie in the ring $\mathbb{Z}$ instead of in a field. The price to pay is Abelian groups are generally not free, in opposition with vector spaces (which are always free).

Example. Consider $F = \mathbb{Z}^2$, and $B = \{b_0,b_1 \} = \{ (1,0),(-1,1)\}$. Then for example $$x = (2,-1) = b_0 - b_1 = 1b_0 + (-1)b_1,$$ so $m_{b_0} = 1$, $m_{b_1} = -1$ in this case. Note that the addition operation $+$ is the one in $F$ (hence the internal direct sum).
A: There is problema already when you write that $F = \sum_{b \in B} \langle b \rangle = \{\ldots,-b_1, 0, b_1,\ldots\} \oplus \dots \oplus \{\dots, -b_n, 0, b_n, \dots\}$, because you assume that $B$ is countable. But the greatest problem comes later. What follows from this is that each element of $F$ is a sum of an element of $\{\ldots,-b_1, 0, b_1,\ldots\}$ with an element of $\{\ldots,-b_2, 0, b_2,\ldots\}$ with an element of $\{\ldots,-b_3, 0, b_3,\ldots\}$ and so on, where all but finitely many of the summands is $0$.
A: The coordinate expression of $b$,
$$ b = \sum_{c \in B} m_c c $$
is the one where
$$ m_c = \begin{cases} 1 & c = b \\ 0 & c \neq b \end{cases} $$
The formula for $F$ is actually an internal direct sum.
However, in a related construction of the external direct sum, (when it doesn't lead to confusion) one usually uses the same letter $b$ both for the element of the given set and its image in the group (i.e. is the tuple with a $1$ in the $b$-th coordinate and $0$ elsewhere).
