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I believe the following little theorem is valid, but the premises are ugly, and I'd like some advice on improving them:

Theorem: Let $G$ be a totally ordered, Archimedean, and abelian group with identity $0_G$, which is first-countable when considered under the order topology.

Let $\sum \{ x_i \mid i \in I \}$ be a convergent generalized sum of non-zero elements of $G$.

Then $I$ is countable.

Proof: By the Archimedean property, for any $g>0_G$, there can only be finitely many $i \in I$ such that $|x_i| \ge g$. Since $G$ is first-countable, there is a sequence, $\{g_i\}$, of positive elements that approaches $0$. Taking the countable union of the finite sets of $x_j$ with $|x_j|>g_i$ shows that there are only countably many elements. So, $I$ is countable.
$\blacksquare$

I just don't care for "Archimedean + first-countable" as a premise. Is there a prettier way to make things work, or a name for structures where things do work for one reason or other?

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Every totally ordered Archimedean group is isomorphic to a subgroup of $\mathbb R$, so you could omit the "abelian" and "first countable" hypotheses! The proof of this claim isn't too hard, but here's a reference: http://books.google.com/books?id=XQXiSHXkQDcC&pg=PA77

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  • $\begingroup$ Well then, I will prove that now. $\endgroup$ – dfeuer Aug 3 '13 at 18:42

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