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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?

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    What is $A$? And, your definitions of connectedness is with _disjoint_ open sets, I assume.2012-02-25
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    sorry about the ambiguity, edited. my definition of connectedness deals with open sets disjoint in $B_r(x_o)$.2012-02-25

4 Answers 4