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I almost have a homework problem solved but I've used a claim that might be dubious.

The setting is this: Let $(V,Q)$ be a locally convex space ($Q$ is the family of seminorms inducing the topology on $V$). And let $q\in Q$.

Claim:

For any neighborhood $U$ of $0$ in $V$. There exists a sufficiently small $\epsilon > 0$ such that $q^{-1}([0,\epsilon))\subset U$.

The reason this claim helps me is that I need to prove something about all neighborhoods of $0$ in $V$, and it greatly simplifies things if I can simplify my situation to sets of the form $q^{-1}([0,\epsilon))$.

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I think the answer is false. If there is an $x\notin U$ such that $q(x) = 0$, then I cannot do this since $x\in q^{-1}([0,\epsilon))$ for any $\epsilon >0$. –  Kyle Schlitt Jan 29 '12 at 19:25
    
In fact it can be false even if $U$ contains $\{q=0\}$ as the answer below shows. –  Davide Giraudo Jan 29 '12 at 20:45

1 Answer 1

I think there is a more simple example. Take $V=\mathbb{R}^2$, $Q=\{p_1,p_2\}$ where for all $x\in V$ we have $p_1(x)=|x_1|$ and $p_2(x)=|x_2|$. Then $(V,Q)$ is a locally convex space. Take $U=p_2^{-1}([0,1))=\{x\in V: |x_2|<1\}$, then for all $\varepsilon>0$ the neighborhood of zero $U$ is not contained in $p_1([0,\varepsilon))$

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Indeed, it's really more simple! –  Davide Giraudo Jan 29 '12 at 21:05

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