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If $Y$ is a subset of topological vector space $X$ and is compact and convex show that $\overline{Y^\circ} = \overline{Y}$ and $\overline{Y}^\circ = Y^\circ$.

I tried this way but I am not sure:

$Y$ is compact so $Y = \overline{Y}$. Then it follows that $Y^\circ = (Y)^\circ = (\overline{Y})^\circ$. And for another I stuck so could some help.

One more thing: What happens if $Y$ is not compact and convex the proof is true for this case too.

This is not a homework question.

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Y is compact. This does NOT imply that Y is closed, so your atempt to proof this theorem is not ok from the starting point on. – user127.0.0.1 Nov 30 '12 at 18:20
A topological vector space is, in almost all applications, assumed to be Hausdorff, so in that case $Y$ compact does imply $Y$ closed, if I recall my general topology right. – Christopher A. Wong Nov 30 '12 at 21:07
@motu: Which part of my question is not clear? Do you know what "Hausdorff" means in the context of topological spaces? The reason I ask this is that in a Hausdorff space every compact set is closed, as was already remarked by Christopher. – Nils Matthes Nov 30 '12 at 23:57
$[0,1]\times\{0\}$ is a compact and convex subset of $\mathbb R^2$, but its interior is empty. – Jonas Meyer Dec 1 '12 at 4:25
@JonasMeyer Game over. Or $Y=\{0\}$ in $\mathbb R$. – Did Dec 1 '12 at 10:25

Both parts (under the assumption of nonempty interior, which was pointed out in comments) follow from the more general statement:

Suppose $Y\subseteq X$ is convex with nonempty interior. Then $\operatorname{int}(Y)$ is dense in $Y$, and $\operatorname{int}(Y^c)$ is dense in $Y^c$.

Proof. If $a\in Y$ and $b\in \operatorname{int}(Y)$, then for any $t\in (0,1)$ we have $ta+(1-t)b\in \operatorname{int}(Y)$, considering the convex combinations of $a$ with neighbors of $b$.

Similarly, if $a\notin Y$, then there is an open half-space disjoint from $Y$ on whose boundary $a$ lies. Pick a point $b$ from this half-space and argue that $ta+(1-t)b\in \operatorname{int}(Y^c)$ as above. $\quad\Box$

Thus, if $Y$ is closed then $\overline{\operatorname{int}(Y)}=Y$. And if $a$ is not an interior point of $Y$, then it's not an interior point of $\overline{Y}$ either.

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Of course, as previously mentioned in the above comments, the condition that $Y^\circ$ is non-empty is needed. In this case $Y$ need not be compact. You can find the proof in Holmes's "Geometric functional analysis ..." book on p. 59.

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Do you have a link of this book if please give me link – math Dec 3 '12 at 21:49
Holmes, Richard B. Geometric functional analysis and its applications. Graduate Texts in Mathematics, No. 24. Springer-Verlag, New York-Heidelberg, 1975. x+246 pp. – Neutral Element Dec 3 '12 at 23:29

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