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The question: If $\{A_j : j \in S\}$ is a family of connected sets and if one set of the family, $A$ intersects all the others, prove that $X = \cup_{j \in S} A_j$ is connected.

My attempt at the proof:

Suppose for contradiction that $X = B \cup C$, $B,C$ nonempty and separate.

claim: Then, for any $A_j$, $A_j \subset B$ or $A_j \subset C$, but not both. This is because $A_j \subset X$, and $A_j = (B \cap A_j) \cup (C \cap A_j)$. Given that $B, C$ separate, we know that $A_j$ is disconnected, a contradiction.

So now we have $B = \cup_{A_j \in B} A_j$, $C = \cup_{A_j \in C} A_j$.

Suppose WLOG that $A \subset B$, then $A \cap C = A \cap \cup_{A_j \in C} A_j \neq \emptyset$ (since $C \neq \emptyset$).

Then, $A \cap C \neq 0 \implies B \cap C \neq 0$, which is a contradiction.

Is the above proof correct?


marked as duplicate by José Carlos Santos general-topology Dec 4 '18 at 10:29

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