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I'm having some trouble solving the following exercise:

Let $A$ and $B$ be connected subspaces of a topological space $(X,\tau)$. If $A\cap B \neq \emptyset$, prove that the subspace $A\cup B$ is connected.

I was trying to do a proof by contradiction. I assumed that : $\exists D,F \in \tau_{A\cup B}: D\cap F = \emptyset \wedge D\cup F = A \cup B$.

Because $D,F \in \tau_{A\cup B }$, then $\exists D',F'\in \tau: D = D' \cap (A\cup B) \wedge F=F' \cap (A\cup B)$, But I don't know how to proceed from now on. How can I proceed with my proof?

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1 Answers1

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You should only resort to proofs by contradiction if all simpler approaches fail, like writing down the definitions and trying to prove that the conditions of the definitions are fulfilled. Your aim is to prove that if $A\cup B=F\dot\cup G$, where $F,G$ are open wrt $\tau_{A\cup B}$, then either $F$ or $G$ is empty. So that's where you should start: Let $A\cup B=F\dot\cup G$, where $F,G$ are open wrt $\tau_{A\cup B}$. Now you have $A=(F\cap A)\dot\cup (G\cap A)$, and the two disjoint sets in this equation are open wrt $\tau_A$. Since $A$ is connected, one of the two sets is empty. Same goes for $F\cap B$ and $G\cap B$. Now use $A\cap B\neq\emptyset$ to show that this implies that either all of $F$ or all of $G$ is empty.

Vercassivelaunos
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