Let's say $P$ is a collection of path collected parts of $\mathbb{R^{p}}$, with the standard metric. We know that $\cap P\neq 0$. I want to prove that $\cup P$ is also path connected.
This is what I already have but I'm a little stuck with writing the rest of my idea down.
Take $x,y \in \cup P$ random. I'm going to prove that there exist a function $f\mapsto \cup P$ so that $f(0)=x$ and $f(1)=y$. Define $P_{1},P_{2}\subset \cup P$ with $x \in P_{1}$ and $y\in P_{2}$. Because $\cap P$ isn't empty there is an $z\in \cap P$. This implies that $z\in P_{1}$ and $z \in P_{2}$. Because $P_{1}$ and $P_{2}$ are path connected parts (given) there are functions $h:[0,1] \mapsto P_{1}$ and $g:[0,1]\mapsto P_{2}$ so that $h(0)=x$ and $h(1)=z$ and $g(0)=z$ and $g(1)=y$.
My idea for the rest is that i can write $f$ as $g$ and $h$ but I'm not sure how to do that. I wanted to change $h$ into $h:[0,1/2] \mapsto P_{1}$ and $g$ to $g:[1/2,1] \mapsto P_{2}$. Can someone help me to write the rest of my solution down.
EDIT: I think i can defini f so that $f:[0,1]\mapsto\cup P$ with f(x)=g(2x) for $x \in [0,1/2[$ and f(x)=h(2x-1) for $x \in [1/2,1]$
Then I only needed to verify that $f$ is a continuous bijection. But because f is a composition of two continuous bijections it's okej. But does this prove that $\cup P$ is also pad connected?
EDIT 2: Now i prove that f is continuous in 1/2
Let's take the right limit of x->1/2 lim(f(x))=g(1)=z let's take the left limit of x->1/2 lim(f(x))=h(0)=z
so continuous in 1/2