# prove that there is a continuous path between any two points in a connected, open subset

Let $$U$$ be a connected, open subset of the plane $$\mathbb R^2$$ . Prove that any two points $$p_0$$ and $$p_1$$ in $$U$$ can be joined by a path, that is, there is a continuous map $$f : [0, 1] \rightarrow U$$ such that $$f(0) = p_0$$ and $$f(1) = p_1$$ .

EDIT

I started by assuming U is disconnected, but there is a path between two points in different components in U. Since U is disconnected, there is a separation A, B. THe path must intersect both A and B. Let f be the path and C be the range of f, then C intersects both A and B.

Since A and B are disjoint and closed, their complements, U-A and U-B are open and cover U. Since C intersects both A and B, it must also intersect U-A and U-B. Thus C intersects bothe closed sets A and U-A and also both closed sets B and U-B.

This implies C is not connected which contradicts that U is disconnected but there is a path between two points in different components of U.Therefor U is connected and any two points in U can be joined by a path.

I would love some feedback on this proof, is it the right approach? Is it a rigorous solution?

• what have you tried? Apr 16, 2023 at 20:22
• I am struggling with creating an epsilon ball contained in U, and how that is related to connectedness. Apr 16, 2023 at 20:32
• I suggest: assume that $U$ is disconnected but there is a path between two points in different components of $U$. Since the set is disconnected, there is a separation $A,B$. Consider the intersection of $A$ and $B$ with the range of the path. Then…?
– MJD
Apr 16, 2023 at 20:36
• Does this answer your question? Showing an open connected subset of $\mathbb{R}^n$ is path-connected Apr 18, 2023 at 17:07
• Apr 18, 2023 at 17:08

Hint: Try defining the set $$\{p\in U,~ \text{there is a path joining p_0 to p}\}$$