Closure of interior and interior of closure in a topological vector space 
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.
 A: 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.
A: 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.
