# Prob. 12, Sec. 23 in Munkres' TOPOLOGY, 2nd ed: How are these unions connected?

Let $Y \subset X$; let $X$ and $Y$ be connected. Suppose that $A$ and $B$ form a separation of $X - Y$.

Then how to show that $Y \cup A$ and $Y \cup B$ are connected?

My effort:

Now since $A$ and $B$ form a separation of $X - Y$, we know that $A$ and $B$ are non-empty, disjoint subsets of $X-Y$ such that $A$ and $B$ are both open in $X-Y$ and $$X-Y = A \cup B.$$ Furthermore, $$\overline{A} \cap B = \emptyset = A \cap \overline{B}.$$ Here $\overline{A}$ denotes the closure of $A$ in $X$.

Let's consider $Y \cup A$ first.

Let's suppose that $C$ and $D$ form a separation of $Y \cup A$. That is, let's suppose that $C$ and $D$ are non-empty, disjoint subsets of $Y \cup A$ such that $C$ and $D$ are both open in $Y \cup A$ and $$Y \cup A = C \cup D.$$

Since $C$ and $D$ form a separation of $Y \cup A$, we have $$\overline{C} \cap D = \emptyset = C \cap \overline{D}.$$

Then $$X = B \cup C \cup D.$$

We would arrive at our desired contardiction if we could show that one of the sets $C$ and $D$ is both open and closed in $X$.

Now since $Y$ is connected, $Y$ lies entirely in either $C$ or $D$. Suppose that $Y \subset C$.

Then since $D \subset Y \cup A$, $Y \subset C$, and $D \cap C = \emptyset$, we must have $D \subset A$.

So $\overline{D} \cap B \subset \overline{A} \cap B = \emptyset$ and hence $\overline{D} \cap B = \emptyset$.

Thus we have seen that $$\overline{D} \cap B = \emptyset = \overline{D} \cap C.$$ But since $X = B \cup C \cup D$, we must have $\overline{D} = D$, showing that $D$ is closed in $X$.

Now since $\overline{C} \cap D = \emptyset$, $\overline{B} \cap A = \emptyset$, and since $D \subset A$, therefore we have $\overline{B} \cap D = \emptyset$.

Now since $X = B \cup C \cup D$, we must have $$\overline{B \cup C} = \overline{B} \cup \overline{C} = B \cup C,$$ showing that $B \cup C$ is closed in $X$ and hence $D$ is open in $X$ as well.

Thus we have shown that $D$ is a non-empty subset of $X$ that is both open and closed in $X$. But $X$ is connected. So we have a contradiction. Hence $Y \cup A$ must be connected.

Is this proof correct?

If so, then by interchanging the roles of $A$ and $B$, we can conclude that $Y \cup B$ is also connected.

Is the above reasoning correct? If not, then how to prove our desired assertion?

If this proof is correct, can we give a proof that is more elegant than this one?

This proof is a bit too long-winded, isn't it?

• Hint: You only need to shouw $Y\cup A$ is connected. – Thomas Andrews May 7 '15 at 15:09
• @ThomasAndrews, can you please have a look at my editted post now? – Saaqib Mahmood May 7 '15 at 17:49
• @Brian M. Scott, can you please come to my rescue? – Saaqib Mahmood May 7 '15 at 19:03
• Yes the proof is correct. But it could be written in a shorter way. See this. – user 170039 Nov 9 '16 at 17:59