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So I asked this question yesterday, Existence of Non-Trivial, Convex, Open Set in $C_{\mathbb{C}}[0,1]$ Under $L^{0}$ Metric, and it made my start wondering the following...

Suppose the following:

  • $Y$ is a topological vector space
  • X is a linear subspace of Y endowed with the subspace topology
  • X is dense in Y
  • U is open and convex in X.
  • V is an open set in Y s.t. $U=V\cap X$
  • $Conv(V)$ is the convex hull of V

Question: Is it necessary that $U=Conv(V)\cap X=V\cap X$?

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SingularDegenerate
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  • I've been playing with this problem. I am able to construct counterexamples where V need NOT be convex if X is not dense. – SingularDegenerate Apr 16 '14 at 17:25
  • Do you mean $U = V \cap X$ in your last bullet? Note that $V$ is not uniquely determined. In case $U = X$, you can take $V = Y \setminus {y_0}$, where $y_0 \not\in X$. – gerw Apr 16 '14 at 17:48
  • @gerw, ugh, yeah, I noticed that just before I saw your comment. Thanks for the input! – SingularDegenerate Apr 16 '14 at 19:10

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