User:Lord Farin/Sandbox

This page exists for me to be able to test out features I am developing. Also, incomplete proofs may appear here.

Feel free to comment.


 * I've been in two minds about posting up multiple proofs. If two proofs have a radically different approach, then I feel both ought to be posted. However, if two proofs have basically the same approach but one is more elegant and / or precise than another, then I would rather there were just the one proof. --prime mover 18:53, 27 January 2012 (EST)

= Alternative (simpler) proof for Absolutely Convergent Generalized Sum Converges =

That $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ converges absolutely means that $\displaystyle \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$ converges.

Now, for all $n \in \N$, let $F_n \subseteq I$ be finite such that:


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

It may be arranged that $n \ge m \implies F_m \subseteq F_n$ by passing over to $F'_n = \displaystyle \bigcup_{m=1}^n F_m$ if necessary.

Define $v_n = \displaystyle \sum_{i \in F_n} v_i$.

Next, it is to be shown that the sequence $\left({v_n}\right)_{n \in \N}$ is Cauchy.

So let $\epsilon > 0$, and let $N \in \N$ be such that $2^{-N} < \epsilon$.

Then for $m \ge n \ge N$, have:

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 $d \left({v_m, v_n}\right) < \epsilon$.

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

As $V$ is a Banach space, this implies there exists a $v \in V$ such that $\displaystyle \lim_{n \to \infty} v_n = v$.

Having identified a candidate $v$ for the sum $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ 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 a Topology ensures that we can limit the choice of opens $U$ containing $v$ to neighborhoods of $v$.

Now let $\epsilon > 0$.

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


 * $d \left({\displaystyle \sum_{i \in G} v_i, v}\right) < \epsilon$, 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:


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

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


 * $\displaystyle \sum \left\{{v_i: i \in I}\right\} = v$