$A \cup B$ and $A \cap B$ connected $\implies A$ and $B$ are connected

Let $X$ be a topological space. Let $A$ and $B$ be two subsets of $X$ such that: $A$ and $B$ are closed, $A \cup B$ is connected, $A \cap B$ is connected

Prove that $A$ and $B$ are connected.

Thoughts:

Suppose that $A$ is disconnected. Since $A$ is closed, there are two non-empty, disjoint, closed sets $E$ and $F$ in $X$ such that $A = E \cup F$. Then I attempted many things but all were not really useful.

Let $X=A\cup B$. Suppose $A$ is disconnected. In particular, $A=C\cup D$ a separation, where $C,D\subset A$ are closed. Thus, they are closed in $X$. Now, consider $A\cap B\subset A$. By a common theorem taught in topology courses, we know $A\cap B\subset C$ or $A\cap B\subset D$, but not both. If $A\cap B\subset C$, then $B\cap D=\emptyset$, so $(B\cup C)\cup D=X$ is a separation. If $A\cap B\subset D$, then $B\cap C=\emptyset$ and $(B\cup D)\cup C=X$ is a separation. Contradiction. So $A$ must be connected.

Hope that helps.

Edit: Suppose $A\cap B\subset C$. Let $x\in B\cap D\subset B\cap A$ since $D\subset A$. But $x\in B\cap A\subset C$ by supposition. But $x\in B\cap D\subset D$. So $x\in C\cap D=\emptyset$ since this is a separation. Contradiction. So $B\cap D=\emptyset$. The other case is equivalent.

• Please explain: "$A \cap B \subset C \implies B \cap D = \phi$".
– user226490
Mar 26, 2015 at 1:21
• could someone please explain where the assumption that $A$ and $B$ are closed is playing a role? Sep 1, 2020 at 13:45
• $C,D$ are closed in the subset topology on $A$ by definition of a separation (essentially, see here). Since $A$ is closed in $X$, it follows that $C,D$ are closed in $X$. Moreover, without this, the prop isn't true: $\mathbb{R} = \mathbb{R}\setminus\{0\}\cup\{0\}$ satisfies the criteria but $\mathbb{R}\setminus \{0\}$ isn't connected.
– Moya
Sep 17, 2020 at 12:18
• But $(\mathbb R\setminus\{0\})\cap\{0\}=\emptyset$ which isn't connected, or at best is an edge case of connectedness. Maybe a better example would be something like $A=[0,2]\cup[3,4]$ and $B=[1,3)$.
– bof
Jan 1, 2021 at 5:57
• @BhargavKale yes, poor choice of notation on my part. $X$ is the full space in this question. I was just saying $X=A\cup B$ for simplification. Actually should have said $Y=A\cup B$.
– Moya
Jul 13, 2021 at 19:50

We may use prove by contrapositive. If $$A$$ or $$B$$ is disconnected, with loss of generality, assume $$A$$ is disconnected. Then by definition, $$\exists E, F\not=\emptyset$$ such that $$A= E\cup F, E\cap F= \emptyset$$. Then, $$A\cap B= (E\cup F)\cap B=(E\cap B)\cup (F\cap B)$$. Notice that $$(E\cap B)\cap (F\cap B)=\emptyset$$ If $$E\cap B\not=\emptyset$$, and $$F\cap B\not=\emptyset$$, then $$A\cap B$$ is disconnected. (If $$E\cap B=\emptyset$$ or $$F\cap B\not=\emptyset$$, then we just need to continue the process above), then we are done.