# Prob. 12, Sec. 23, in Munkres' TOPOLOGY, 2nd ed: If $Y\subset X$, $X, Y$ are connected, and $A, B$ form a separation of $X-Y$,

Here is Prob. 12, Sec. 23, in the book Topology by James R. Munkres, 2nd edition:

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.

My Attempt:

As sets $A, B$ form a separation of $X-Y$, so, by Lemma 23.1 in Munkres, $A$ and $B$ are non-empty subsets of $X-Y$ such that $X-Y = A \cup B$, and also $\overline{A} \cap B = \emptyset = A \cap \overline{B}$.

We show that $Y \cup A$ is connected. Let us suppose that $Y \cup A$ is not connected. Then, again by Lemma 23.1 in Munkres, there exist non-empty subsets $C$ and $D$ of $Y \cup A$ such that $Y \cup A = C \cup D$, and $\overline{C} \cap D = \emptyset = C \cap \overline{D}$.

As $X-Y = A \cup B$, $Y \cup A = C \cup D$, and $X = Y \cup (X-Y)$, so $$C \cup D \cup B = Y \cup A \cup B = Y \cup (X-Y) = X,$$ which we rewrite as $$X = ( B \cup C) \cup D. \tag{1}$$

Now as $C, D$ form a separation of $Y \cup A$ and as $Y$ is connected, so, by Lemma 23.2 in Munkres, $Y$ must be contained either in $C$ or in $D$, but not both. Let's assume without any loss of generality that $Y \subset C$. Then $$\emptyset \subset Y \cap D \subset C \cap D \subset \overline{C} \cap D = \emptyset,$$ and so $Y \cap D = \emptyset$.

If $d \in D$, then $d \in C \cup D = Y \cup A$, but $d \not\in C$ and as $Y \subset C$, so $d \not\in Y$ either, which implies that $d \in A$. Therefore $D \subset A$.

Now as $D \subset A$, so $$\emptyset \subset D \cap \overline{B} \subset A \cap \overline{B} = \emptyset,$$ which implies that $D \cap \overline{B} = \emptyset$.

But by our choice of $C$ and $D$ in the second paragraph of this proof, we also have $D \cap \overline{C} = \emptyset$.

Now we can show that $$\overline{B \cup C} = \overline{B} \cup \overline{C}.$$ So $$D \cap \overline{B \cup C} = D \cap \left( \overline{B} \cup \overline{C} \right) = \left( D \cap \overline{B} \right) \cup \left( D \cap \overline{C} \right) = \emptyset.$$ This last equality together with (1) implies that $\overline{B \cup C} = B \cup C$, which shows that $B \cup C$ is closed and $D$ is open.

On the other hand, since $D \subset A$, therefore $\overline{D} \subset \overline{A}$. Then, as $\overline{A} \cap B = \emptyset = \overline{D} \cap C$, so we can conclude that $$\emptyset \subset \overline{D} \cap (B \cup C) = \left( \overline{D} \cap B \right) \cup \left( \overline{D} \cap C \right) \subset \left( \overline{A} \cap B \right) \cup \emptyset = \emptyset,$$ which implies that $$\overline{D} \cap (B \cup C) = \emptyset,$$ and this together with (1) implies that $\overline{D} = D$, and so $D$ is also closed in $X$. Thus $B \cup C$ is open in $X$.

Then the sets $B \cup C$ and $D$ are non-empty disjoint open subsets of $X$ such that $X = (B \cup C) \cup D$, which contradicts our hypothesis that $X$ is connected. Hence $Y \cup A$ is connected.

Similarly (by interchanging $A$ and $B$ in the above proof), we can show that $Y \cup B$ is also connected.

Is each and every step in this proof correct and clear enough? If not, then where is it lacking?

• relevant – Andres Mejia Dec 16 '17 at 22:51
• Suggestion to shorten the proof: Let $A^*, B^*$ be open in $X$ with $A=A^*$ \ $Y$ and $B=B^*$ \ $Y.$ Suppose $Y\cup A$ is disconnected. Let $C^*, D^*$ be open in $X$ with $C=C^*\cap (Y\cup A)$ and $D=D^*\cap (Y\cup A),$ where $\{C,D\}$ is a separation of $Y\cup A.$ ... Since $Y$ is connected, one of $C^*,D^*$ covers $Y$ and one of $C^*, D^*$ is disjoint from $Y.$ so WLOG let $C^*\subset Y$ and $D^*\cap Y=\phi.$ ... Now show that $\{C^*\cup B^*, D^*\cap A^*\}$ is a separation of $X.$ – DanielWainfleet Apr 4 '18 at 18:30