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Is there any convex set $A\subset X$,where $X$ is a normed Banach space, such that:

$$int(A)\neq int(\overline{A})$$

?

There is a theorem which says that if $A$ is a convex series-closed (cs-closed) then:

$$int(A)=int(\overline{A})$$

where $int(A)$ is the usual interior of the set $A$.

It was a hope here, but there is not a concrete example: Are convex sets solid?

Bogdan
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  • Please, give some context and definitions ("$int$" means interior? $X$ is complete/finite dimensional or separable,...?). The intuitive answer would be "no", have you tried to prove that? You might be interested in this question. –  Dec 11 '17 at 09:12
  • I think that there should be an example in an infinite dimensional space – Bogdan Dec 11 '17 at 09:21
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    From your link, it's very easy to derive a concrete example: take $X=C([0,1])$, and $A$ the set of polynomials. Then, $A$ is convex, $\overline{A}=X$ by the Weierstrass approximation theorem, but $A$ has empty interior. –  Dec 11 '17 at 10:47
  • Interior of $C$ is also empty – QED Dec 11 '17 at 11:24
  • Thanks a lot! Now I see! – Bogdan Dec 11 '17 at 13:32

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