Condition for Connected

Let $E \subseteq R$ . $E$ is connected iff $\forall a < b \in E$ and $\forall c \in R$ if $a < c < b$ then $c \in E$ .

Really this is saying that $E$ is an 'interval', in the more traditional sense.

Proof

#todo/school : Read this proof up 📅 2024-11-05 ✅ 2024-11-06

( $\Rightarrow$ ): Suppose $E$ is connected, and let $a, b \in E$ and $a < c < b$ . Set:

A = (- \infty, c) \cap E; B = (c, \infty) \cap E

Because $a \in A$ and $b \in B$ , then neither set is empty and neither set contains a limit point of the other (see Separated, Disconnected, Connected Sets#Examples to see how that works, namely example 2). If $E = A \cup B$ , then we would have that $E$ is disconnected, which it is not (by our given). It must then be that $A \cup B$ is missing some element of $E$ and $c$ is the only possibility. Thus, $c \in E$ .

( $\Leftarrow$ ): Suppose $\forall a < b \in E$ and $\forall c \in R$ that $(a < c < b \to c \in E)$ . To show that $E$ is connected, assume $E = A \cup B$ for contradiction, where $A, B$ are nonempty and disjoint (being separated allows us to get disjoint for free). We need to show that one of these sets contains a limit point of the other. Pick $a_{0} \in A$ and $b_{0} \in B$ and, suppose (for the sake of argument) that $a_{0} < b_{0}$ . Because $E$ itself is an interval, then interval $I_{0} = [a_{0}, b_{0}] \subseteq E$ . Now, bisect $I_{0}$ into two equal halves. The midpoint of $I_{0}$ must either be in $A$ or $B$ , so choose $I_{1} = [a_{1}, b_{1}] \subseteq E$ to be the half that allows us to have $a_{1} \in A$ and $b_{1} \in B$ . Continue this process, where each $I_{n} = [a_{n}, b_{n}] \subseteq E$ where $a_{n} \in A$ and $b_{n} \in B$ and the length $(b_{n} - a_{n}) \to 0$ . By the Nested Interval Property then $\exists x$ where:

x \in ⋂_{n = 0}^{\infty} I_{n}

It is straightforward to see that $a_{n} \to x$ and $b_{n} \to x$ . But now $x \in E$ must belong to either $A$ or $B$ , thus making it a limit point of the other, contradicting $A, B$ being separated. As such, then $E \neq A \cup B$ so then $E$ must be connected.

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