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The definitions I am using are as follows:

A vector space $V$ equipped with a family $P$ of semi-norms such that $\cap_{p\in P}\{x\in V : p(x) = 0\} = \{0\}$ is called a locally convex vector space.

I am concerned with the topology given by the collections of subsets $U$ of $V$ such that for all $x\in U$ there exists $n\geq 1$ and $p_{1}, ... , p_{n}$ in $P$ and $\epsilon_{1}, ..., \epsilon_{n} > 0$ such that

$x\in \{y\in V : p_{i}(x - y) < \epsilon_{i}$ for $i = 1,... n\}\subset U$.

Showing this topology is closed under finite intersection is very straight-forward. But arbitrary unions is eluding me because if I take a family of such sets, say $\{U_{\alpha}\}$, then I don't get a finite subset of seminorms and epsilons to use to show that the resulting union is still open.

How can I finish this?

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But... You're describing a base for the topology given by the sets of the form $U_{p_1,\ldots,p_n} (x) = \{y \in V : p_i (x-y) \lt \varepsilon_i \text{ for } i = 1,\ldots,n\}$! A set is open if and only if it is a union of such sets. – t.b. Jan 12 '12 at 5:11
Yes I am describing the topology in terms of the basis. I tried to verify the properties of a basis but that turned into an absolute mess (in particular with the 2nd basis axiom) so I went back to following the way my professor did it in the notes, which was just show that the resulting open sets U form a topology. However he left the union axiom as an exercise. – roo Jan 12 '12 at 5:24
up vote 2 down vote accepted

Each $x$ in the union must be in one of the particular sets $U$ appearing in the union, and you get your $n$, $p_i$s and $\varepsilon_i$s from this $U$.

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Thank you. This is definitely a facepalmer. Next time I am going to have to try doing some jumping-jacks to clear my head before I waste 20 minutes typing up a question to which the answer is this trivial. I was thinking about the problem totally backwards. – roo Jan 12 '12 at 5:26

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