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**okor** Problem: Let $\displaystyle A_{1}$, $\displaystyle A_{2}$, and $\displaystyle A_{3}$ be connected subsets of X, such that $\displaystyle A_{1} \cap A_{2} \neq \emptyset$ and $\displaystyle A_{2} \cap A_{3} \neq \emptyset$. Prove that $\displaystyle A_{1} \cup A_{2} \cup A_{3}$ is connected.

So far this is what I have:

Assume $\displaystyle A_{1} \cup A_{2} \cup A_{3}$ is disconnected, then $\displaystyle (A_{1} \cup A_{2} \cup A_{3}) \cap U \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap V \neq \emptyset, (A_{1} \cup A_{2} \cup A_{3}) \cap U \cap V = \emptyset$.

Therefore $\displaystyle (A_{1} \cap U \cap V) \cup (A_{2} \cap U \cap V) \cup (A_{3} \cap U \cap V) = \emptyset$, $\displaystyle (A_{1} \cap U) \cup (A_{2} \cap U) \cup (A_{3} \cap U) \neq \emptyset$, $\displaystyle (A_{1} \cap V) \cup (A_{2} \cap V) \cup (A_{3} \cap V) \neq \emptyset$

Thus $\displaystyle (A_{1} \cap U \cap V) = (A_{2} \cap U \cap V) = (A_{3} \cap U \cap V) = \emptyset$

Now, I'm not 100% sure if I have been approaching this correctly, but I think from here I have to show for one of the 3 connected subsets its intersection with U and Y is nonempty and thus it is disconnected meaning there is a contradiction, but I am not sure how to prove this part.