Let \(Y \subset X\); let \(X\) and \(Y\) be connected. Show that if \(A\) and \(B\) form a separation of \(X - Y\), then \(Y \cup A\) and \(Y \cup B\) are connected.
Solution
Because \(A\) and \(B\) form a separation of \(X-Y\), they are both open and closed in the subspace \(X-Y\). Hence \[ \overline A^{\,X}\cap B=\varnothing, \qquad \overline B^{\,X}\cap A=\varnothing. \tag{1} \]
We prove \(Y\cup A\) is connected. Suppose instead that \[ Y\cup A=P\cup Q \] is a separation. Since \(Y\) is connected, it lies in one side, say \(Y\subseteq P\). Then \(Q\subseteq A\).
The set \(Q\) is closed in \(Y\cup A\), so \[ \overline Q^{\,X}\cap(Y\cup A)\subseteq Q. \] Also \(Q\subseteq A\), so by (1), \(\overline Q^{\,X}\cap B=\varnothing\). Since \[ X=(Y\cup A)\cup B, \] it follows that \(\overline Q^{\,X}\subseteq Q\); hence \(Q\) is closed in \(X\).
Similarly, \(P\) is closed in \(Y\cup A\), while (1) gives \(\overline B^{\,X}\cap A=\varnothing\). Therefore \[ \overline{P\cup B}^{\,X}\subseteq P\cup B, \] so \(P\cup B\) is closed in \(X\). Thus \(Q=X-(P\cup B)\) is also open in \(X\).
Hence \(Q\) is a nonempty proper clopen subset of connected \(X\), a contradiction. So \(Y\cup A\) is connected. Interchanging \(A\) and \(B\) proves that \(Y\cup B\) is connected as well.