If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$.
Lemma 23.1 statement is vague, IMO. Munkres Lemma 23.1. Rephrasing it
$Y$ is connected(usual definition) $\iff$ $\nexists$ separation of $Y$. Property of separation is given in lemma 23.1.
Munkres used two different notion of separation. We can rephrase lemma 23.1 by showing two notion of separation of $Y$ are equivalent.
$A,B\in \mathcal{T}_{Y}$, $A,B\neq \emptyset$, $A\cap B=\emptyset$ and $A\cup B=Y$ $\iff$ $A,B\neq \emptyset$, $A\cap B=\emptyset$, $A\cup B=Y$, $\overline{A}\cap B=\emptyset$ (or equivalently $A’\cap B=\emptyset$) and $A\cap \overline{B}=\emptyset$ (or equivalently $A\cap B’=\emptyset$). In other word, Assume $A,B\neq \emptyset$, $A\cap B=\emptyset$ and $A\cup B=Y$. $A,B\in \mathcal{T}_Y$ $\iff$ $\overline{A} \cap B=\emptyset$ and $A\cap \overline{B}=\emptyset$.
Is my interpretation correct?
Best Answer
Use that $Y=A \cup B$ separated, then $A \cap \overline{B}=\emptyset$ and so $A \subseteq Y\setminus \overline{B}$. It follows that $A$ is open in $Y$ (as the complement of $\overline{B}$). Symmetrically $B$ is open in $Y$ as well (as the complement of $\overline{A}$). So a separation via "separated sets" is the same as a separation by disjoint open sets and vice versa.
So it doesn't matter which separation we use in the definition of (dis)connectedness.