Show that an open ball in $\mathbb{R^n}$ is a connected set.
Attempt at a Proof: Let $r>0$ and $x_o\in\mathbb{R^n}$. Suppose $B_r(x_o)$ is not connected. Then, there exist $U,V$ open in $\mathbb{R^n}$ that disconnect $B_r(x_o)$. Without loss of generality, let $a\in B_r(x_o)$: $a\in U$. Since $U$ is open, for some $r_1>0$, $B_{r_1}(x_o)\subseteq U$. Since $(U\cap B_r(x_o))\cap (V\cap B_r(x_o))=\emptyset$, $a\not\in V$. Thus, $\forall b\in V, d(a,b)>0$. But then for some $b'\in V: b'\in B_r(x_o)$ and some $r>o$, $d(a,b')>r$. Contradiction since both $a$ and $b'$ were in the ball of radius $r$.
Is this the general idea?
Best Answer
An easy way to see this : a convex set is by construction path-connected, since $\forall x, y \in C$ with $C$ convex, $\lambda x + (1-\lambda)y \in C$ by convexity (so that you can choose the line between $x$ and $y$ as a path). Therefore since the unit ball is convex (show it if you wish to), and since path-connected sets are also connected, you're done.
I'm not quite sure your proof is correct though. Why would $d(a,b') > r$? I don't see an explicit reason for this.
Hope that helps,