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A space is called Frechet if it is complete metrizable locally convex space. Suppose $Y$ is a Frechet space, and $K$ is a compact subset of $Y$. We let $V$ denote the closed convex envelope of $K$. Is it true that $V$ could be separated by a countable family of functionals in $Y^*$? If yes, how to prove the proposition?

Any hint will be greatly appreciated!

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  • $\begingroup$ Can you recall a definition what is the envelope of $K$ and what means "a set $V$ is separated by a countable family of functionals"? $\endgroup$ – Alex Ravsky Dec 2 '17 at 9:12
  • $\begingroup$ @AlexRavsky In the book I am using the terminology is "envelope". I think that closed convex envelope is the same as closed convex hull. On the other hand, my question is that if there exists a countable family $\{l_n\}$ of continuous linear functionals, such that for any $x \neq y$, there exists $l_n$ so that $l_n(x) \neq l_n(y)$. $\endgroup$ – Dormire Dec 2 '17 at 14:01
  • $\begingroup$ For any $x\ne y$ belonging to $V$? $\endgroup$ – Alex Ravsky Dec 2 '17 at 14:50
  • $\begingroup$ @AlexRavsky Yes exactly. $\endgroup$ – Dormire Dec 2 '17 at 14:55
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For the sake of simplicity I’ll assume that $Y$ is a real locally convex space. Let $X$ be a smallest closed linear subspace of $Y$ containing $K$. Then $X\supset V$. Since $K$ is a metrizable compact, it is separable. Let $D$ be a countable dense subset of $K$. Then a set $$D'=\{q_1d_1+\dots+q_nd_n: q_i\in\Bbb Q, d_i\in D\}$$ is a countable dense subset of the space $X$. Let $\mathcal B$ be a countable base at the $0$ consisting of convex open subsets of $Y$. Let $$\mathcal S=\{(d+U, d’+U’): d,d’\in D', U,U’\in V, (d+U)\cap(d’+U’)=\varnothing\}.$$ By Hahn-Banach separation Theorem for each pair $S=(d+U, d’+U’)\in\mathcal S$ there exists a continuous linear functional $l_S$ on $Y$ separating sets $ d+U $ and $d’+U’$. It is easy to check that a countable family $\{l_S:S\in\mathcal S\}$ separates the points of $X$, and hence the points of $V$.

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