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Let $A$ and $B$ be connected subsets of a metric space. Prove that at least one of $A \cup B$ or $A \cap B$ is connected

My two attempts:

  1. Assume neither $A \cup B$ nor $A\cap B$ are connected (hence they are disconnected, then we can partition them $$A\cup B \rightarrow \{U_1,U_2\}\quad A \cap B \rightarrow \{V_1,V_2\} $$ for some open, disjoint sets $U_1,U_2,V_1,V_2$. From here I'm not sure where to continue, and I suppose I need to arrive to a contradiction showing that either $A$ or $B$ is disconnected, which can't be as they are given to be connected but I'm not sure how to reach there.
  2. I tried to go about it by showing that if one of $A \cup B$ or $A \cap B$ is disconnected the other one must be connceted. Let $A \cup B$ be disconnected, then it can be partionted into two open disjoint sets $\{U_1,U_2\}$. Now fix $x\in A \cap B$ and then see that $x \in U_1$ or $x \in U_2$. WLOG choose the former. So we have $x \in A$ and $x\in B$ and $x \in U_1$. But once again I'm not sure how to use this to arrive to the fact to show that $A \cap B$ is connected

Any help would be appreciated.

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  • $\begingroup$ That depends on your definition of connectedness (does it allow the empty set to be connected?). $\endgroup$ – YuiTo Cheng Mar 20 at 10:00
  • $\begingroup$ @YuiToCheng Yes this is taken with the usual notion that $\emptyset$ is connected $\endgroup$ – Hushus46 Mar 20 at 10:02
  • $\begingroup$ related question and another related question. $\endgroup$ – drhab Mar 20 at 10:20
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There are two possibilities:

  1. $A\cap B=\emptyset$: then $A\cap B$ is connected.
  2. $A\cap B\neq\emptyset$: then $A\cup B$ is connected, because if $f\colon A\cup B\longrightarrow\{0,1\}$ (with $\{0,1\}$ endowed with the discrete topology) is continuous then, if $p\in A\cap B$, $f(A)=\bigl\{f(p)\bigr\}=f(B)$, and therefore $f$ is constant.
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    $\begingroup$ I was about to type the very same answer but you are superfast! $\endgroup$ – Kavi Rama Murthy Mar 20 at 10:01
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    $\begingroup$ Same story for me: again you are superfast. $\endgroup$ – drhab Mar 20 at 10:04
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    $\begingroup$ This answer is perfect, just waiting for a couple of minutes before I'm allowed to accept it. Thanks $\endgroup$ – Hushus46 Mar 20 at 10:04
  • $\begingroup$ I know I accepted this question, but now looking back at it im slightly confused. How can you justify/prove that $f(A)$ and $f(B)$ are both equal to the singleton $\{f(p)\}$? Sorry to bother $\endgroup$ – Hushus46 May 2 at 11:49
  • $\begingroup$ Since $A$ is connected and $f$ is continuous, $f(A)$ is connected. But the only non-empty connected subsets of $\{0,1\}$ are $\{0\}$ and $\{1\}$. $\endgroup$ – José Carlos Santos May 2 at 12:03

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