0
$\begingroup$

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?

$\endgroup$
5

1 Answer 1

1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ is a path the same thing as connected? $\endgroup$ Apr 17, 2023 at 14:04
  • $\begingroup$ ? You defined path in your question $\endgroup$
    – Keshav
    Apr 26, 2023 at 22:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .