Let \(X\) be a connected space with connected subset \(Y\), and suppose that \(X\setminus Y\) is not connected, with \(X\setminus Y=A\cup B\) a separation of \(X\setminus Y\). Show that if \(B\) is open in \(X\), then \(A\cup Y\) is a connected subset of \(X\).
Solution
Put \(Z=A\cup Y\); since \(X\setminus Y=A\sqcup B\), \(Z=X\setminus B\), which is closed in \(X\) because \(B\) is open. Suppose \(Z=C\sqcup D\) with \(C,D\) disjoint and open in \(Z\). Since \(Y\) is connected, it lies in \(C\) or in \(D\); interchange them so that \(Y\subseteq C\). It suffices to show that \(C\) or \(D\) is empty.
1. \(D\) is closed in \(X\), and \(D\subseteq A\).
\(D\) is closed in \(Z\), and \(Z\) is closed in \(X\). \(D\cap Y\subseteq D\cap C=\varnothing\) and \(D\subseteq A\cup Y\), so \(D\subseteq A\).
2. \(D\) is open in \(X\).
Choose open \(U,V\subseteq X\) with \(D=U\cap Z\) and \(A=V\cap(X\setminus Y)\); the latter exists because \(A\) is open in \(X\setminus Y\). Let \(d\in D\); by step 1, \(d\in A\), so \(d\in U\cap V\). Let \(q\in U\cap V\). If \(q\in Y\), then \(q\in U\cap Z=D\), contradicting \(D\cap Y=\varnothing\); so \(q\notin Y\), hence \(q\in V\cap(X\setminus Y)=A\subseteq Z\) and \(q\in U\cap Z=D\). Thus \(U\cap V\) is an open neighborhood of \(d\) contained in \(D\).
3. Q.E.D.
By steps 1 and 2, \(D\) is clopen in the connected space \(X\), so \(D=\varnothing\) or \(D=X\). If \(D=X\), then \(C\subseteq Z\setminus D=\varnothing\). So \(C\) or \(D\) is empty, and \(A\cup Y\) is connected.