basic question of topology involving compactness and convexity

Question

Consider in $R^n$ a compact and convex set $A$ with $int(A) $ nonempty. then $\overline{int(A)} = A$ ?.

i have no idea to prove this. In this direction i only know the following (and hard to prove) fact : If $B$ is a convex set of $R^n$ then $int (\overline{B}) = int(B)$. Someone can give me a help to prove (or give a counter example) for my affirmation ?

thanks in advance

Answer 1

No, in general this is not true. Take $A=\{0\}\subset R^n$ which is convex and compact (and let $n>0$). Then $int(A)=\emptyset$ and so is $\overline{int(A)}$, hence $\overline{int(A)}\neq A$.

Answer 2

Without loss of generality, we can assume that $0 \in \operatorname{int}(A)$. Now, $B = \overline{\operatorname{int}(A)}$ is a compact convex subset of $A$. Suppose that $B \neq A$. Then we have $a \in A\setminus B$, and since $B$ is closed, there is an $\varepsilon > 0$ with $B_\varepsilon(a) \cap B = \varnothing$. Further, since $0 \in \operatorname{int}(A)$, there is an $\eta > 0$ with $B_\eta(0) \subset \operatorname{int}(A)$, and there is a $t\in (0,1)$ such that $t\cdot a \in \partial B$ (since $t\cdot a \in \operatorname{int}(A)$ for small enough $t$, and $t\cdot a \notin B$ for $t$ close to $1$).

By the convexity of $A$, we have

$$A \supset t\cdot a + (1-t)\cdot B_\eta(0),$$

and $t\cdot a + (1-t)\cdot B_\eta(0)$ is an open neighbourhood of $t\cdot a$, so $t\cdot a \in \operatorname{int}(A)$, contradicting $t\cdot a \in \partial B$, which followed from the assumption $a\in A\setminus B$. Hence that assumption must be false, and $B = A$.