I was studying the notion of series, it states that a series $ \sum \limits_{s \in S} x_s $, in $\mathbb{R}$ over a set $S$, is convergent if there exists a $F\in \mathbb{R}$ such that for every $\epsilon > 0$ there exists a finite set $T_{\epsilon} \subset S$ such that every finite set $T \subset S$ with $T_{\epsilon} \subset T$ satisfies,

$$ |F -\sum \limits_{s \in T} x_s| < \epsilon.$$

Due to the (absolute) convergence will any permutations of the term be convergent and give the same sum. Also every subseries of a (absolute) convergent series will converge. If we rewrite $S = U \cup T$ where $U \cap T = \emptyset$, then we know that subseries will converge (absolutely). My question is the following, is the following equality true,

$$ \sum \limits_{s \in S} x_s = \sum \limits_{t \in T} x_t + \sum \limits_{u \in U} x_u ?$$

The reason why I ask this is because I have seen a proof where they make use of the fact that they can separate positive and negative terms of a sum in a similar way as above. Then my second question is, if the equality is true, how can one prove that the equality holds?


1 Answer 1


Yes. It is true. To prove it, for a given $\epsilon>0$, choose a finite $T_{\frac{\epsilon}{2}}$ and $U_{\frac{\epsilon}{2}}$ as you described; e.g. for finite $Y$, $T\supset Y\supset T_{\frac{\epsilon}{2}}\implies\left|\sum_{t\in Y}x_t-\sum_{t\in T}x_t\right|<\frac{\epsilon}{2}$. Define $S_{\epsilon}=T_{\frac{\epsilon}{2}}\cup U_{\frac{\epsilon}{2}}$. You're done.


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