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Suppose $V$ is a finite dimensional (real) topological vector space. The first lemma in these notes says that

Every vector subspace of a tvs with the induced topology is a topological vector space in its own right.

But no proof is offered. With Michael Greinecker's answer, it makes sense to me why any subspace of a tvs is a tvs.

I want to further ask, if the topology on $V$ is such that every linear functional is continuous, does the induced topology on $S$ also have that all linear functionals on $S$ are continuous? I think the answer is yes. If $f_S$ is a linear functional on $S$, then $f_S$ can be extended to a linear functional $f$ on $V$. Then for $O\cap S$ open in $S$, $$ f_S^{-1}(O\cap S)=f^{-1}(O)\cap S $$ which is open in $S$? Have I made sense or nonsense? If it is nonsense, could someone give a brief proof so I can see how it is done properly?

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A continuous function defined on a topological space is continuous when restricted to a subspace of the domain, so this follows from this general principle.

In a locally convex topological vector space, you can extend every continuous linear functional on a subspace to the whole space, so in that case, the induced family of continuous linear functionals is the same. But the assumption that all linear functionals are continuous is in general extremely restrictive. It never holds for infinite-dimensional Banach spaces for example.

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Oh ok, thanks. And what about the linear functionals? – Tiffany Hwang Dec 16 '12 at 8:32
@TiffanyHwang I add a remarrk about that. – Michael Greinecker Dec 16 '12 at 12:58

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