# $A$ and $B$ connected implies atleast one of $A \cup B$ or $A \cap B$ is connected

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.

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

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.
• 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 – Hushus46 May 2 at 11:49
• 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 $\{1\}$. – José Carlos Santos May 2 at 12:03