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This question already has an answer here:

Let G1 and G2 be two regions. Suppose that G1 ∩ G2 ≠ 0. Show that G1 U G2 is connected.

I know I have to first show that G1 U G2 is open and then show it is connected but I have no idea where to start or how to even prove this.

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marked as duplicate by José Carlos Santos, Namaste, Xander Henderson, Lord Shark the Unknown, Deepesh Meena Sep 25 '18 at 4:58

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  • $\begingroup$ This really doesn’t have anything to do with complex analysis:) $\endgroup$ – MathIsLife12 Sep 24 '18 at 22:04
  • $\begingroup$ it's for my complex analysis course $\endgroup$ – Zainab Husain Sep 24 '18 at 22:05
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Let $G_1$ and $G_2$ be connected subspaces of $\mathbb{C}$. We want to show that $G:=G_1 \cup G_2$ is connected. Let $A$ and $B$ be two disjoint open subsets of $G$ such that $G = A \cup B$. Since $G_1$ is connected, we must have either $G_1 \subseteq A$ or $G_1 \subseteq B$. Without loss of generality, assume $G_1 \subseteq A$. Since $G_2$ is also connected, we either have $G_2 \subseteq A$ or $G_2 \subseteq B$. Let $z \in G_1 \cap G_2$ and observe that $z \in G_1 \subseteq A$. Because $A \cap B = \varnothing$, the only possibility is $G_2 \subseteq A$.

It follows that $G_1 \cup G_2 \subseteq A \subseteq G = G_1 \cup G_2$. We conclude that $A = G_1 \cup G_2$ and that $B = \varnothing$. In other words, $G_1 \cup G_2$ does not admit a separation and must be connected.

To show that $G_1 \cup G_2$ is open, simply use that the union of open sets is again open.

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