Is it possible to create a region in the $xy$ plane that has infinite area but contains all of its boundary points? In other words, if I am introducing a theorem that begins: "Let R be a closed region in the $xy$ plane", do I need to also specify that R is finite?
Asked
Active
Viewed 223 times
1
-
Closed can mean multiple things, including: topologically closed (contains all its boundary points), or enclosed by one or more loops (a Jordan region), or (in the theory of surfaces) compact and boundaryless. If one convention isn't clear from context, your readers will thank you for being explicit. – Andrew D. Hwang Jun 09 '17 at 15:53
-
Thanks. I've edited the question to hopefully clarify – Michael Stachowsky Jun 09 '17 at 15:54
-
Trivially yes, $\mathbb R^2$ as it is open and closed. But maybe you are thinking in a proper subset. – Rafa Budría Jun 09 '17 at 15:58
-
I believe so. I think I meant that it is strictly closed, not clopen. Does that change things? – Michael Stachowsky Jun 09 '17 at 16:00
3 Answers
2
What about the region $$ D = \left\{(x,y)\mid x \geq 1, 0 \leq y \leq 1/x\right\} $$ A few more examples came to mind, but then I wasn't sure on the definition of region.
Matthew Leingang
- 26,112
-
I believe I meant "simply connected domain". intuitively, in the finite case, a "region" in my definition is a "shape" that can be drawn in the plane that does not contain any pinch points...forgive my lack of rigour :-) – Michael Stachowsky Jun 09 '17 at 16:32
2
To address your concern in the comments, it's a theorem that a topological space contains nontrivial clopen sets (sets other than $\emptyset$ and the space itself) $\iff$ the space is disconnected. Because $\mathbb{R}^2$ is connected, any nonempty, proper, closed subset of $\mathbb{R}^2$ is not also open.
Since closed sets are defined to be the complements of open sets, your question will have an affirmative answer if we can establish the existence of open sets of finite area. Such open sets exist; in particular, the metric topology has a basis consisting of the collection of open balls of finite radius.
Kaj Hansen
- 33,011
-
1Is it totally equivalent? The complement of an closed set with infinite area is an open set, but not necessarily with finite area. – Matthew Leingang Jun 09 '17 at 16:16
-
-
Do I need to specify that my region is simply connected in order for the answer to be affirmative? – Michael Stachowsky Jun 09 '17 at 16:31
-
1Though path-connected, the complement of any open ball won't be simply connected, and yet it is a closed set with infinite area. – Kaj Hansen Jun 09 '17 at 16:40
