Generalized Sum Preserves Inequality

Theorem
Let $\left({a_i}\right)_{i \in I}, \left({b_i}\right)_{i \in I}$ be $I$-indexed families of positive real numbers.

That is, let $a_i, b_i \in \R_{\ge 0}$ for all $i \in I$.

Suppose that for all $i \in I$, $a_i \le b_i$.

Furthermore, suppose that $\displaystyle \sum \left\{{ b_i: i \in I }\right\}$ converges.

Then $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$.

In particular, $\displaystyle \sum \left\{{ a_i: i \in I }\right\}$ converges.

Proof
First, it is proven that $\displaystyle \sum \left\{{ a_i: i \in I }\right\}$ converges.

Then, the inequality $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$ is well-defined, and hence can be proven.

Proof of Convergence
For every $n \in \N$, let $F_n \subseteq I$ be finite such that:


 * $\displaystyle \sum_{i \mathop \in G} b_i > \sum \left\{{ b_i: i \in I }\right\} - 2^{-n}$ for all finite $G$ with $F_n \subseteq G \subseteq I$

By passing over to $\displaystyle F'_n = \bigcup_{i \mathop = 1}^n F_i$ if necessary, it may be arranged that $F_n \subseteq F_m$ for $n \le m$.

Next, define the sequence $\left({a_n}\right)_{n \in \N}$ by $a_n := \displaystyle \sum_{i \mathop \in F_n} a_i$.

To show $\left({a_n}\right)_{n \in \N}$ is a Cauchy sequence, let $\epsilon > 0$.

Subsequently let $N \in \N$ be such that $2^{-N} < \epsilon$, and let $m \ge n \ge N$. Then:

Now to estimate this last quantity, observe:

Finally, as $n \ge N, 2^{-n} < 2^{-N} < \epsilon$ (by defining property of $N$).

Combining all of these estimates leads to the conclusion that $d \left({a_m, a_n}\right) < \epsilon$.

It follows that $\left({a_n}\right)_{n \in \N}$ is a Cauchy sequence.

By Real Number Line is Complete Metric Space, this implies there exists an $a \in \R$ such that $\displaystyle \lim_{n \to \infty} a_n = a$.

Having identified a candidate $a$ for the sum $\displaystyle \sum \left\{{a_i: i \in I}\right\}$ to converge to, it remains to verify that this is indeed the case.

According to the definition of considered sum, the convergence is convergence of a net.

Next, Metric Induces Topology ensures that we can limit the choice of opens $U$ containing $a$ to neighborhoods of $a$.

Now let $\epsilon > 0$.

We want to find a finite $F \subseteq I$ such that:


 * $d \left({\displaystyle \sum_{i \mathop \in G} a_i, a}\right) < \epsilon$, for all finite $G$ with $F \subseteq G \subseteq I$

Now let $N \in \N$ such that for all $n \ge N$, $d \left({v_n, v}\right) < \dfrac \epsilon 2$ (with the $v_n$ as above).

By taking a larger $N$ if necessary, ensure that $2^{-N} < \dfrac \epsilon 2$ holds as well.

Let us verify that the set $F_N$ defined above has sought properties.

So let $G$ be finite with $F_N \subseteq G \subseteq I$. Then:

For the first of these terms, observe:

Using that $2^{-N} < \dfrac \epsilon 2$, combine these inequalities to obtain:


 * $\displaystyle d \left({\sum_{i \mathop \in G} a_i, a}\right) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$

By definition of convergence of a net, it follows that:


 * $\displaystyle \sum \left\{{a_i: i \in I}\right\} = a$

Proof of Inequality
Suppose that $\displaystyle \sum \left\{{ a_i: i \in I }\right\} > \sum \left\{{ b_i: i \in I }\right\}$.

Then, as the sums converge, there exists a finite $F \subseteq I$ such that:


 * $\displaystyle \sum_{i \mathop \in F} a_i > \sum \left\{{ a_i: i \in I }\right\} - \epsilon$

for every $\epsilon > 0$.

So by picking a suitable $\epsilon$, it may be arranged that:

These inequalities together constitute a contradiction, and therefore:


 * $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$