If Connected Sets share a point, then their Union is Connected

Last updated Nov 1, 2022

Let $X$ be a Topological Space, $x \in X$, and suppose $\mathcal{C} \subset \mathcal{P}(X)$ is such that for all $C \in \mathcal{C}$

1. $C$ is Connected
2. $x \in C$

Then $D := \bigcup\limits_{C \in \mathcal{C}} C$ is Connected.

Suppose there were some disconnecting Open $U, V \subset X$. Then because $x \in D$, only one of $x \in U$ or $x \in V$ holds. Without Loss of Generality, let $x \in V$. Because all $C \in \mathcal{C}$ are Connected and $C \cap V \neq \emptyset$, we must have $C \cap V = C$. Thus $C \cap U = \emptyset$. But since this holds for all $C \in \mathcal{C}$: $$D \cap U = \bigcup\limits_{C \in \mathcal{C}} C \cap U = \bigcup\limits_{C \in \mathcal{C}} \emptyset = \emptyset.$$ Thus $D \cap U = \emptyset$ and $D \cap V = D$, so $D$ is Connected. $\blacksquare$