This is a topology problem.

Suppose $A \subset \mathbb{R}^n$ is an open connected set, we want to prove that given a point $p \in A$ for any $x \in A$ there is a broken-line path from $x$ to $p$ in A.

Note: A broken-line path consists of a sequence of points and line segments between points adjacent to each other in the sequence.

I propose a solution below. Please let me know if you find any mistakes.

Question

This is a topology problem.

Suppose $A \subset \mathbb{R}^n$ is an open connected set, we want to prove that given a point $p \in A$ for any $x \in A$ there is a broken-line path from $x$ to $p$ in A.

Note: A broken-line path consists of a sequence of points and line segments between points adjacent to each other in the sequence.

I propose a solution below. Please let me know if you find any mistakes.

anonymous · Answer

Step 1)
First we show that given $p \in A$ the set, $P$, of all $x \in A$ such that there is a broken-line path between $p$ and $x$ is open. So we need to show that if for some $x$ there is a broken-line path between $x$ and $p$, then there exists some epsilon ball $B_{\varepsilon}(x) \subset P$. This is easy to see. Since $A$ is open there is certainly a ball $B_{\varepsilon}(x) \subset A$. So we must show that every point in the ball is also broken-line connected to $p$. Since we already know there is a broken-line path from $p$ to $x$ we can simply add a line segment from $x$ to any point in the sphere. So if $B_{\varepsilon}(x) \subset A$ then $B_{\varepsilon}(x) \subset P$. Therefore $P$ is open.

Step 2)
We prove the contrapositive. Let $A$ be open, take some $p \in A$ and consider the set $P$ consisting of all  $x \in A$ such that there is a broken-line path between $p$ and $x$. By step 1, we have that $P$ is open. We show that if there exists some $y \in A$ where $y \not\in P$ then $A$ is disconnected. To prove this we show that $Q = A - P$ is open, since then there would be open disjoint sets $P \cup Q = A$. So consider some $y \in Q$ we show there is a ball $B_{\varepsilon}(y) \subset Q$. Suppose there wasn't such a ball. That would mean for every ball $B_{\varepsilon}(y)$ there is some point in the ball that is broken-line path connected to $p$ but then there would be line segment from that point to $y$. So then $y$ would be broken-line path connected to $p$ contradicting our initial assumption. Since there is such a ball $Q$ is open. So we have disjoint open $P, Q$ such that $P \cup Q = A$. So $A$ is disconnected.

Step 3)
We have shown that if there exists a point for which a broken-line path does not exist then $A$ is disconnected. By contrapositive we have shown that if $A$ is connected there is a broken-line path between all pairs of points.

usukidoll · Answer

I admire your mathematical abilities :) I'm not in that level yet