Generalized Sum Preserves Inequality

== Theorem ==
Let $\family {a_i}_{i \mathop \in I}, \family {b_i}_{i \mathop \in I}$ 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 $\ds \sum \set {b_i: i \in I}$ converges.
Then:
:$\ds \sum \set {a_i: i \in I} \le \sum \set {b_i: i \in I}$
In particular, $\ds \sum \set {a_i: i \in I}$ converges.
== Proof ==
First, it is proven that $\ds \sum \set {a_i: i \in I}$ converges.
Then, the inequality $\ds \sum \set {a_i: i \in I} \le \sum \set {b_i: i \in I}$ 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:
:$\ds \sum_{i \mathop \in G} b_i > \sum \set {b_i: i \in I} - 2^{-n}$ for all finite $G$ with $F_n \subseteq G \subseteq I$
By passing over to $\ds F'_n = \bigcup_{i \mathop = 1}^n F_i$ if necessary, it may be arranged that $F_n \subseteq F_m$ for $n \le m$.
Next, define the sequence $\sequence {a_n}_{n \mathop \in \N}$ by $a_n := \ds \sum_{i \mathop \in F_n} a_i$.
To show $\sequence {a_n}_{n \mathop \in \N}$ 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 $\map d {a_m, a_n} < \epsilon$.
It follows that $\sequence {a_n}_{n \mathop \in \N}$ is a Cauchy sequence.
By Real Number Line is Complete Metric Space, this implies there exists an $a \in \R$ such that $\ds \lim_{n \mathop \to \infty} a_n = a$.
Having identified a candidate $a$ for the sum $\ds \sum \set {a_i: i \in I}$ 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 open balls centered at $a$.
Now let $\epsilon > 0$.
We want to find a finite $F \subseteq I$ such that:
:$\map d {\ds \sum_{i \mathop \in G} a_i, a} < \epsilon$, for all finite $G$ with $F \subseteq G \subseteq I$
Now let $N \in \N$ such that for all $n \ge N$, $\map d {v_n, v} < \dfrac \epsilon 2$ (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 the 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:
:$\ds \map d {\sum_{i \mathop \in G} a_i, a} < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$
By definition of convergence of a net, it follows that:
:$\ds \sum \set {a_i: i \in I} = a$
=== Proof of Inequality ===
 $\ds \sum \set {a_i: i \in I} > \sum \set {b_i: i \in I}$.
Then, as the sums converge, there exists a finite $F \subseteq I$ such that:
:$\ds \sum_{i \mathop \in F} a_i > \sum \set {a_i: i \in I} - \epsilon$
for every $\epsilon > 0$. 
So by picking a suitable $\epsilon$, it may be arranged that:
These inequalities together constitute a contradiction, and therefore:
:$\ds \sum \set {a_i: i \in I} \le \sum \set {b_i: i \in I}$
Category:Generalized Sums