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

Last updated Nov 1, 2022

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

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

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

Let $y, z \in D$. Then there exists $P_{y}, P_{z} \in \mathcal{P}$ s.t. $y \in P_{y}$ and $z \in P_{z}$. Since both are Path-Connected, there exist Continuous Paths $$\begin{align*} f: [0, 1] \to D \text{ s.t. } f(0)=y,f(1)=x\\ g: [0, 1] \to D \text{ s.t. } g(0)=x,g(1)=z \end{align*}$$ Then we can take $h := f*g$. Since $f(1) = g(0)$, Continuous Path Concatenation is valid here so $h$ is a Continuous Path. Then $h(0) = f(0) = y$ and $h(1) = g(1) = z$. Since $y,z \in D$ were arbitrary, $D$ is Path-Connected. $\blacksquare$