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This is with regard to this question: Topology induced by seminorms and initial topology

I saw somwehere that topology $\mathcal{S}$ is the smallest topology with respect to which all the seminorms are continuous and $V$ is a topological vector space. But I don't have any proof of it.

Is the arbitrary intersection of vector topologies a vector topology. How does one show the "smallest" assertion?

mathable
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  • Doesn't the cited answer say that in general they don't coincide? – Berci Feb 12 '21 at 16:45
  • The cited answer says it is not the weak topology for the seminorms . It doesn't say anything about the topology being the weakest topology such that $V$ is a t.v.s and the seminorms are continuous. T.v.s structure would impose more restrictions – mathable Feb 12 '21 at 16:48

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If $\tau$ is a vector topology such that the seminorm $p$ is continuous, then $$x\mapsto p(x-x_0)$$ is also continuous since it is the composition of $p$ with the translation by $x_0$ (the latter being continuous thanks to the fact that $\tau$ is a vector topology).

Therefore the "ball" $$ B_{x_0, p, r} =\{x \in V : p(x - x_0) < r \} $$ is open relative to $\tau$.

If $\mathscr P$ is a given family of seminorms, and if $\tau$ is a vector topology relative to which all seminorms in $\mathscr P$ are cotinuous, we therefore conclude that the topology induced by $\mathscr P$ is smaller than $\tau$.

Ruy
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