This is question 3 on page 152 in Chapter 23 of James R. Munkers’ Topology 2nd Edition textbook.

Question

Let \({A_\alpha}\) be a collection of connected subspaces of \(X\); Let \(A\) be a connected subspace of \(X\). Show that if \(A \cap A_\alpha = \emptyset\) for all \(\alpha\), then \(A \cup (\cup A_\alpha)\) is connected.

Proof

Suppose \(A \cap A_\alpha \ne \emptyset\) for every \(\alpha\). Thus \(\cup A_\alpha\) and \(A\) have some point in common. By Theorem 23.3, the collection of \(A\) and \(\cup A_\alpha\) must be connected. Thus \(A \cup (\cup A_\alpha)\) is connected.

Old Question

Let \({A_\alpha}\) be a collection of connected subspaces of \(X\); Let \(A\) be a connected subspace of \(X\). Show that if \(A \cap A_\alpha = \emptyset\) for all \(\alpha\), then \(A \cup (\cup A_\alpha)\) is connected.

Old Proof

Suppose \(A \cap A_\alpha = \emptyset\) for all \(\alpha\). To show that \(A \cup (\cup A_\alpha)\) is connected, suppose, for the sake of contradiction, that \(A \cup (\cup A_\alpha)\) is separated. Viewing \(A \cup (\cup A_\alpha)\) as a collection of connected subspaces of X and using Theorem 23.3, the sets \(A\) and \(\cup A_\alpha\) must have an element in common. However, we know that since \(A \cap A_\alpha = \emptyset\) for every \(\alpha\), the sets \(A\) and \(\cup A_\alpha\) cannot have any elements in common, reaching a contradiction. Thus, \(A \cup (\cup A_\alpha)\) is connected.

\[\tag*{$\blacksquare$}\]

Extras

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