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Fix any number $\delta>0$ and put $A = \{x \in \mathbb{R}: \left|x-3\right|<\delta\}$ and $B = \{x \in \mathbb{R}: \left|x-3\right|>\delta \}$. Prove that $C=A \cup B$ is not a connected set.

Definition of connected: Let $E$ be a subset of a metric space $(S,d)$. The set $E$ is disconnected if there are disjoint open subsets $U_1$ and $U_2$ in $S$ such that $E\subseteq U_1 \cup U_2$, $E\cap U_1\neq\emptyset$ and $E\cap U_2 \neq \emptyset$.

A set $E$ is connected if it is not disconnected.

Intuitively, this obviously makes sense. I just don't get how to prove this is proof form.

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What's your definition of a connected set? – EuYu Nov 8 '12 at 15:56
I added the definition I was taught. – Joakim Nov 8 '12 at 16:05
Can you see why $C$ is, by definition, a disconnected set? – EuYu Nov 8 '12 at 16:15
up vote 3 down vote accepted

A connected set is a set that cannot be partitioned into two nonempty subsets which are open in the relative topology induced on the set. But $A$ and $B$ are both open.

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I'm not sure I understand the definition. Say, $[0,1]\cup[2,3]$ is obviously disconnected. But I couldn't think of any partition $A$ and $B$ such that they are both open. – Leo Jun 21 '13 at 4:59
@mathusiast In $[0,1]\cup[2,3]$ both $[0,1]$ and $[2,3]$ are open. To see this, try to apply the definition of open: pick a point $x$ in $[0,1]$ and try to find an open ball around it that is contained in $[0,1]$. Of course this is possible for all points in $(0,1)$... – Rudy the Reindeer Jun 21 '13 at 7:17

First, note $\overline{A} = \{x\in \mathbb{R}: \left|x-3\right|\leq\delta\}$ and $\overline{B} = \{x\in \mathbb{R}: \left|x-3\right|\geq\delta\}$. Now, if we take $\overline{A} \cap B$ we get the empty set. Similarly, if we take $A \cap \overline{B}$ we get the empty set. Since $\overline{A} \cap B = \emptyset = A \cap \overline{B}$, we know that the set $C$ is not connected by definition.

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