Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$

Question

In $\Bbb{R}^2$ and $\Bbb{R}^3$ are all open sets in the form of an open ball of some positive radius? In other words, do open sets look like anything else except balls/spheres?

I know any union of open sets is again open, but that doesn't really help me picture the possibility of anything else, just a union of open balls.

The interior of any subset of $\mathbb{R}^n$ set is open in $\mathbb{R}^n$. — mez, Jun 17 '13 at 23:04

score 4 · Accepted Answer · answered Jun 17 '13 at 23:00

4

The open rectangle $\{(x,y)\mid 0<x<1,0<y<1\}$ is open. In general, open sets can look very very different than just balls or spheres. You can draw any smooth closed and simple curve in the plane, and consider the points inside it. That's an open set. Similarly, you can consider any closed simple surface in $\mathbb R^3$, and consider the points it bounds. That will be an open set.

More generally, If you remove any closed set from either $\mathbb R^2$ or $\mathbb R^3$, the remaining set is open.

answered Jun 17 '13 at 23:00

Ittay Weiss

79,840
7
141
236

So the unit intveral on the x-axis wouldn't be open in $\Bbb{R}^2$ right? – TheRealFakeNews Jun 17 '13 at 23:04
No: a set is open iff it is equal to its interior, and the interior of a line is empty in $\mathbb R^2$ – Clement C. Jun 17 '13 at 23:09
no, it's not open. – Ittay Weiss Jun 17 '13 at 23:11

score 4 · Answer 2 · answered Jun 17 '13 at 23:23

Clarifying Berci's idea a bit, suppose that $f:{\mathbb R}^n\rightarrow\mathbb R$ is any continuous function. Then $$\{{\mathbf x}: f({\mathbf x})>0\}$$ is an open set. This can be quite crazy. If $n=2$ and $$f(x,y)=\left(x^2+y^2-5\right) \cos (x \, y),$$ we get the following:

enter image description here

score 1 · Answer 3 · answered Jun 17 '13 at 22:59

1

Take any shape, and consider it without its border. E.g. an open square or cube, or anything..

answered Jun 17 '13 at 22:59

Berci

90,745

Stahl · Answer 4 · 2013-06-17T23:11:25.403

0

Take the union of two open balls in $\Bbb{R}^2$, one centered at $1$ and one centered at $-1$, both of radius $1$. This set is open, but it should be clear (just draw the picture!) that it is not an open ball.

Edit: To give an example of a set that doesn't look like the union of open balls, consider $\Bbb{R}^n$ without the origin. Since points are closed in $\Bbb{R}^n$, this set is open.

edited Jun 17 '13 at 23:11

answered Jun 17 '13 at 22:57

Stahl

23,212

Yes, but it is still a union of open balls. So in some sense, are open balls the basis of all open sets in $\Bbb{R}^2$? – TheRealFakeNews Jun 17 '13 at 22:58
2

Exactly. They are a basis. – Berci Jun 17 '13 at 23:00
1

Any open set can be written as a union of open balls! If $U$ is open, then by definition, each point $p\in U$ has some open ball $B_p$ around it completely contained in $U$. Then $U = \cup_{p\in U} B_p$. – Stahl Jun 17 '13 at 23:01

score 0 · Answer 5 · edited Apr 13 '17 at 12:21

Despite all of the complicated open subsets of $\mathbb{R}^n$ being bandied about, the usual topology has the special property that it is second countable. This means that every open set can be written as the union of sets from a countable collection. One particular collection for $\mathbb{R}^2$ is the countable set of all balls $B(p,q;r)$ of rational radius $r$ with center point $(p,q)$ having rational coordinates. So, yes, every open set may be thought of as "built up" from open balls.

Can you think of another countable basis for $\mathbb{R}^2$?

Here is a related Math.SE discussion.

Open sets in $\Bbb{R}^2$ and $\Bbb{R}^3$

5 Answers5