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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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$\mathbf{0}$ is a member of $V$ such that for all members $p$ of $P$, $\;\; p(\mathbf{0}) \: = \: p(0\cdot \mathbf{0}) \: = \: |0|\cdot p(\mathbf{0}) \: = \: 0\cdot p(\mathbf{0}) \: = \: 0 \;\;$. $\;\;\;$ –  Ricky Demer Jan 28 '12 at 19:07
    
Thanks. I fixed the question to reflect this. :) –  Kyle Schlitt Jan 28 '12 at 19:21

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up vote 6 down vote accepted

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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I was finally able to verify this. Thanks for your push in the correct direction (namely making sure I was trying to prove the right thing!). –  Kyle Schlitt Jan 28 '12 at 21:28

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