How do i write down "unique representation" formally? Let me start my question with an example.
Suppose $\beta$ is a basis for a vector space $V$. Then, for every $v\in V$, if $v=\sum_{i\in I} a_i u_i$ and $v=\sum_{i\in I} b_i u_i$, then for all $i\in I, a_i=b_i$.($I$ is an indexing set for $\beta$) Hence, we can conclude that every vector in $V$ has a unique representation with respect to the basis $\beta$.
This argument seems very shaky to me. How do we write "unique representation" precisely in first-order logic? This seems quite intuitive to me, not abstract. Is there anyway to make "$\sum_{i\in I}$" the actual sum?
For instance, one can actually give a topology on  $R[[X]]$, hence make the formal sum the actual sum.
Just like formal sum, is there anyway to give a topology on a vector space $V$? If it is not worth doing this, then what would be the way to say "unique representation" in first-order logic?
Thank you in advance.
 A: This is an attempt to summarize the discussion in the comments in order to get this off the Unanswered list.
Let $\beta=\{u_i:i\in I\}$. By definition we consider only sums $\sum_{i\in I}a_iu_i$ in which $\{i\in I:a_i\ne 0\}$ is finite. This ensures that $\sum_{i\in I}a_iu_i$ is well-defined: we define
$$\sum_{i\in I}a_iu_i=\begin{cases}
\sum\{a_iu_i:i\in I\text{ and }a_i\ne 0\},&\text{if }\{i\in I:a_i\ne 0\}\ne\varnothing\\\\
0,&\text{otherwise}\;,
\end{cases}$$ 
so it’s always either $0$ or a finite sum. Thus, it really is meaningful to say that

$\qquad\qquad\qquad$if $v=\sum_{i\in I}a_iu_i$ and $v=\sum_{i\in I}b_iu_i$, then $a_i=b_i$ for all $i\in I$.

The actual proof involves a bit of bookkeeping. Let $I_a=\{i\in I:a_i\ne 0\}$ and $I_b=\{i\in I:b_i\ne 0\}$; $I_a\cup I_b$ is finite, and $a_i=b_i=0$ for $i\in I\setminus(I_a\cup I_b)$. For $i\in I_a\cup I_b$ let $c_i=a_i-b_i$, and let $c_i=0$ for $i\in I\setminus(I_a\cup I_b)$; then $\sum_{i\in I}c_iu_i=0$, so $c_i=0$ for all $i\in I$, and in particular $a_i=b_i$ for all $i\in I_a\cup I_b$ and hence for all $i\in I$.
