I'm trying to understand Ronnie Brown's answer here: union of two simply connected open , with open and non empty intersection in $R^2$
Let $X$ be a topological space and $U,V$ be simply connected open subsets of $X$ such that $U\cap V $ is path-conneted.
Fix $x\in U\cap V$.
Let $r$ be a loop in $X$ based at $x$.
By Lebesgue number lemma, there is a partition $\{s_0,\cdots,s_n\}$ of $[0,1]$ such that $r([s_i,s_{i+1}])\subset U$ or $r([s_i,s_{i+1}])\subset V$.
Now, define $r_j(t)=r((1-t)s_j + t\cdot s_{j+1})$ for $1\leq i < n$. ($t\in [0,1]$)
Then, each $r_j$ is a path from $r(s_j)$ to $r(s_{j+1})$.
I completely understand till here.
However, he says that for each $j$, it is possible to choose a path $\alpha$ "in $U\cap V$" such that $\alpha(0)=x$ and $\alpha(1)=r(x_j)$.
How is that possible?
Since $r$ is an arbitrary path, it ranges over all $X$ not restricted in $U\cap V$. Since $U\cap V$ is path connected, if $r(x_j)$ is in $U\cap V$, then that makes sense, but it's possible that $r(s_j)\notin U\cap V$.
Would someone please complete the proof?
Best Answer
You are correct - there is a gap in the proof, but it is easily filled. It is possible that you have chosen the partition so that two consecutive $s_j$'s are in $U$, and then it is possible that $r(s_j)$ is not in $U \cap V$, and no such $\alpha$ exists.
However, all you need to do to fix this is to change your partition. It is always possible to choose a subset $\{t_1,\dots,t_k\} \subseteq \{s_1,\dots,s_n\}$ which satisfies
The second condition ensures that $r(t_i) \in U \cap V$ for each $i$, and then you know you can choose a path as Brown suggested.