Suppose \( X \) is a locally connected space . Then every component of \( X \) is open.
Proof
Given a component \( A \subset X \). For \( p \in A \),
there is a connected neighborhood \( U \), and it must lie entirely in \( A \).
Otherwise, \( A \) is not maximal.
Thus every point in \( A \) has a neighborhood in \( A \), so \( A \) is open.