# Connected Topological Space

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 ?

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 $$W=\emptyset$$.

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.