I'm stuck in a proof: given that $E\subset \mathbb{R}^n$ is convex, prove that it is connected. I know that this can be proved easier using the concept of path-connectedness, but that's not I'm "allowed" to do. I need to use another definition: $E$ is connected if and only if it cannot be separatedby a pair of two relatively open sets.
My attempt:
Pick any $x, y\in E$. Since $E$ is convex, $tx+(1-t)y \subset E$ for $t\in [0,1]$. Suppose that $E$ is not connected. Then there exist non-empty sets $U$ and $V$, such that $U\cap V =\emptyset$, $U\cup V=E$, and $U$ and $V$ are relatively open in $E$, which implies that there exist open sets $A$ and $B$ such that $U=A\cap E$ and $V=B\cap E$.
Let $x \in U$ and $y \in V$.
(i) $U$ and $V$ are clearly not empty.
(ii) $U\cup V$…
Intuitively I understand that what needs to be shown is that $U \cup V \ne E$, but I'm stuck in trying to show it. Would appreciate some help.
Best Answer
Try to use the contrapositive of the statement.