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Given $E$, a locally convex space (l.c.s.) and $E\subseteq F$ where $F$ is another subspace of a larger vector space. The inclusion is strict since I know there exists a $y\in F\backslash E$.

I have two questions

Q1. Is it a known result that F also has a base of neighborhoods of 0 that are convex (in F) thereby making F also an l.c.s.?

My initial thoughts are that since every open set in $E$ (via the subspace topology) is $V \cap E$, where V is open in F, $V$ itself need not be convex. Yet, is it possible to "extend" the open sets in $E$ and "complete" them in some sense so that they are convex thereby forming a local convex base in F? Is this a known result? If it is, I would appreciate any reference (I have access to Munkres, Rudin's Functional Analysis) where it is proven.

Q2. Is $\overline E = E$? I know that there is a result that states that in a topological vector space, a hyperplane is either closed or dense. But the issue I am having here is that $F$ itself may be a proper subspace of a larger vector space. If $\overline E = E$, then I know for sure that $y\notin \overline E$ and hence I can separate $y$ and E by a closed hyperplane if $F$ is also a l.c.s.

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$\mathbb R$ is a locally convex space. If your statement were true, every real topological vector space would be locally convex.

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  • $\begingroup$ thanks. So, we cannot say anything about the locally convex properties of F if E, its subspace, is locally convex. A question about the converse - just to confirm my understanding, via the subspace topology, if F is locally convex and E is a subspace of F, then E is locally convex. right? $\endgroup$ – F1E2D3 Apr 11 '16 at 16:13
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    $\begingroup$ Yes, a linear subspace of a locally convex space is locally convex, because the intersection of convex sets is convex. $\endgroup$ – Robert Israel Apr 11 '16 at 19:43

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