TITLE: Why is this sum well-defined?
QUESTION [5 upvotes]: Let $$S=\mathbb{N}[1/2]$$ be the set of rational numbers greater than $0$ which has a power of $2$ as its denominator. Let $R$ be any commutative ring. Let us consider $R^S,$ the infinite direct product. It is clear that we can write the elements as $\sum_{s\in S } a_s x_s$ with $a_s \in R.$ Consider the $R$-submodule $M$ which isformed by the elements $\sum a_s x_s$ which satisfy the following property:  

(*) For every real number $r >0,$ there is an $\epsilon >0 $ such that $a_s = 0$ if $r-\epsilon \leq s <r.$

Let now $\alpha = \sum a_s x_s$ and $\beta = \sum b_s x_s$ be two elements of $M.$ I want to show that (*) implies that for any $s \in S$ the products $$a_{s'}b_{s''}$$ with $s'+s''=s$ are almost all zero.  
Does anyone have a short proof of this fact?

REPLY [2 votes]: Assume there are infinitely many $s',s''$ with $s'+s''=s$ and $a_{s'}\ne 0$ and $b_{s''}\ne 0$.
We can pick a monotonic sequence from the set of such $s'$. If this sequence is strictly increasing, we obtain a contradiction to (*) if we let $r$ be the limit/supremum of this sequence. If on the other hand the sequence is strictly decreasing, we obtain a corresponding increasing sequence of $s''$ and again a contradiction to (*).