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Let $\left\{A_{\alpha} | \alpha \in \mathcal{A} \right\}$ be a family of connected subsets of $X$, and assume that there exists a connected set $A$ with $A \cap A_{\alpha} \neq \emptyset$. Show that $ \displaystyle A \cup \bigcup_{\alpha}^{}{A_{\alpha}}$ is connected.

My attempt:

Assume that $\Phi=A \cup \bigcup_{\alpha}^{}{A_{\alpha}}$ is not connected, there exists open sets $V, W \subset X$ such that $$V \cap W =\emptyset, \quad V \cup W= \Phi$$ $A \subset V \cup W$ then $A \subset V$ or $A \subset W$ and $\displaystyle \bigcup_{\alpha}^{}{A_{\alpha}} \subset V \cup W$ then $A_\alpha \subset V \cup W$ for each $\alpha \in \mathcal{A}$ that is $A_\alpha \subset V$ or $A_\alpha \subset W$, thus $$A \cap A_\alpha \subset V \cap W= \emptyset$$ that is $A \cap A_\alpha = \emptyset$ for each $A_\alpha$ which is a contradiction.

Please check, if this proof is correct or not ?

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I don't think the last argument is correct. Yes, $A\subset V$ or $A\subset W$, and for every $\alpha$ we have $A_\alpha\subset V$ or $A_\alpha\subset W$, but you don't show how it implies that $A\cap A_\alpha\subset V\cap W$. In fact, let's assume that, for some $\alpha$, $A_\alpha\subset V$ and also $A\subset V$, then $A\cap A_\alpha\subset V$ and it is not obvious how this intersection suddenly becomes empty (i.e. how $W$ comes into play here).

However, your proof can be fixed. In fact, if (without loss of generality) $A\subset V$, then for every $\alpha$ you must have $A_\alpha\subset V$. (You cannot have $A_\alpha\subset W$ because then you would have the situation as in your proof where $A\cap A_\alpha\subset V\cap W=\emptyset$.) Thus, $A\cup\bigcup_\alpha A_\alpha\subset V$, i.e. $\Phi\subset V$ and then $\Phi=V$ and $W=\emptyset$.

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To be more precise: let $U, V$ be open subsets of $\Phi=A \cup \bigcup_\alpha A_\alpha$ such that $U \cup V = \Phi$ and $U \cap V=\emptyset$. (You have to take open sets of the subspace, not of $X$).

Now $A$ is connected and $A \cap U$ and $A \cap V$ are open subsets of $A$ that are disjoint and cover $A$ (as $A \subseteq \Phi$) so by connectedness one of them is empty, say (WLOG) $A \cap U=\emptyset$ so that $A \cap V =A$, or equivalently $A \subseteq V$.

Now fix any $\alpha$. Then we know again that $U \cap A_\alpha$ and $V \cap A_\alpha$ form an open partition of $A_\alpha$ so by connectedness one of them is empty. But we already know that $$\emptyset \neq A \cap A_\alpha \subseteq A_\alpha \cap V$$

so that $A_\alpha \cap V = V$ (which means that $A_\alpha \subseteq V$ also) and as $\alpha$ was arbitrary, $\Phi \subseteq V$ and $V=\Phi$ and $U=\emptyset$ showing connectedness of $\Phi$.

Note that we start by considering the "glue set" $A$ first as this gives us the candidate for the one non-empty open set in out starting partition.

This proposition is a large ingredient in the proof that $X \times Y$ is connected when $X$ and $Y$ are, suing one "axis" as glue and parallel copies of the other space as the other sets.

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