# Real Analysis

Connected Sets

# Connected Sets

Two sets \(A\) and \(B\) in a space \(X\) are *separated* if both \(A \cap \overline{B}\) and \(\overline{A} \cap B\) are empty.

A set \(E\) is *connected* if \(E\) is not the union of two separated sets. Or, equivalently, a set \(E\) is *connected* if \(E\) is not the union of two non-trivial relatively open sets in \(E\). Or, equivalently, a set \(E\) is *connected* if \(E\) is not the union of two non-trivial relatively closed sets in \(E\).

Two separated sets are called a *separation*.

For example,

- In \(\mathbb{R}^2\), \(E = \{(x, y) : x, y \in \mathbb{Q}\}\). One separation for \(E\) is \(A = \{(x, y) : x, y \in \mathbb{Q}, x < \pi\}\) and \(B = \{(x, y) : x, y \in \mathbb{Q}, x \geq \pi\}\).

**Theorem.** Any closed interval \([a, b]\) is connected.

*Proof (by contradiction).* If not, then there exists a separation \(A\) and \(B\) with \(a \in A\). Let \(s = \sup A\). Then, \(s \in \overline{A}\). Hence, \(s \notin B\). So, \(s \in A\). Thus, \(s \notin \overline{B}\). Then, there exists \((s - \epsilon, s + \epsilon)\) containing no point of \(B\). Hence, \((s - \epsilon, s + \epsilon) \subset A\), which contradicts the fact that \(s\) is the supremum of \(A\). QED.