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Suppose we have a locally convex space $(V,P)$, where $V$ is a topological vector space and $P$ is a family of seminorms defined on $V$ such that for each nonzero $x\in V$ there is a $p\in P$ such that $p(x)\neq 0$.

My question is about the topology that $P$ induces on $V$. I know the definition: a set $U$ is open in $V$ if for each nonzero $x\in V$ there exists $p_{1}, ... , p_{n} \in P$ and $\epsilon_{1}, ... \epsilon_{n} > 0$ such that

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

My question is this:

Is this equivalent to saying that a net $x_{\alpha}$ converges to $x$ in $V$ if and only if $p(x_{\alpha})\to p(x)$ for every $p\in P$?

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    @KyleSchlitt can you check my question bout locally convex TVS?2015-01-09

1 Answers 1

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No. For example, $\mathbb{R}$ with its usual topology is a locally convex space: take $P$ consisting of the single seminorm $p(x) = |x|$. Now consider the sequence $x_n = (-1)^n$. We don't have $x_n \to 1$, but we do have $p(x_n) \to p(1)$.

(Remember: seminorms are not linear.)

What is true, however, is that $x_\alpha \to x$ in $V$ if and only if $p(x_\alpha - x) \to 0$ for every $p \in P$. This should not be too hard to prove from the definitions.

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    @Nate Eldredge can you check my question http://math.stackexchange.com/questions/1098160/doubt-concearning-the-definition-of-a-locally-convex-space-structure-through-sem?2015-01-09