Unique representation as a linear combination Let $V$ be an infinite dimensional vector space and $B$ be a basis. I would like to prove that every element of $V$ has a unique representation as a linear combination of elements of $B$. 
I can visualize why this is true (heuristically: take the first representation minus the second and get linear dependence), however I seem unable to do the technical work. It is tricky. The finite dimensional case is trivial and it can't be generalized.
Can someone show me how to do this?
 A: Definition: Let V be a vector base, then a set of vectors B is a basis if every $v\in V$ is a finite linear combination of elements in B and every finite subset of B is linearily independent.
Now assume by contradiction there exists two different representations
$$v=\sum_{i\in I_1} \beta_i b_i$$
$$v=\sum_{i\in I_2}^m \alpha_j b_i$$
Where $b_i$ are elements in B and $\alpha_i,\beta_i$ scalars, $I_1,I_2$ are finite sets of indexes.
Now take the first representation minus the second we have 
$$0=\sum_{i\in I_1-I_2} \beta_i b_i - \sum_{i\in I_2-I_1} \alpha_i b_i +\sum_{i\in I_1\cap I_2} (\beta_i-\alpha_i) b_i$$
This is a linear combination of 0 and therefore we have


*

*For all $i\in I_1-I_2$ we have $\beta_i=0$

*For all $i\in I_2-I_1$ we have $\alpha_i=0$

*If $i\in I_1\cap I_2$ we have $\alpha_i=\beta_i$
It follows that the two representations above are the same representation.
A: Let $\beta$ be a basis of a vector space $V$. Suppose
$$
\sum_{v\in \beta}a_v\cdot v=\sum_{v\in\beta}b_v\cdot v\tag{1}
$$
where all but finitely many $a_v$ and $b_v$ are zero. Equation (1) is equivalent to
$$
\sum_{v\in\beta}(a_v-b_v)\cdot v=\vec 0\tag{2}
$$
But $\beta$ is linearly independent since $\beta$ is a basis, so equation (2) implies that $a_v-b_v=0$ for each $v\in V$. Thus $a_v=b_v$ for each $v\in V$. This proves that the linear combinations in (1) are the same.
