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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.

Please help. Thanks.

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2 Answers 2

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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.

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  • $\begingroup$ Please explain: "$A \cap B \subset C \implies B \cap D = \phi$". $\endgroup$
    – user226490
    Mar 26, 2015 at 1:21
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    $\begingroup$ could someone please explain where the assumption that $A$ and $B$ are closed is playing a role? $\endgroup$
    – Vidit
    Sep 1, 2020 at 13:45
  • $\begingroup$ $C,D$ 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. $\endgroup$
    – Moya
    Sep 17, 2020 at 12:18
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    $\begingroup$ 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 $B=[1,3)$. $\endgroup$
    – bof
    Jan 1, 2021 at 5:57
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    $\begingroup$ @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 $Y=A\cup B$. $\endgroup$
    – Moya
    Jul 13, 2021 at 19:50
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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.

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