In R^2, a set S is defined as S = {(x,y)| x=(1/t)cos(t), y=(1/t)sin(t), t ≥ π} which is the spiral with infinitely many turns about the origin.

1) Is S open, closed or neither? Why?

2) Is R^2\S, the complement of S, open, closed or neither? Why?

Printable View

- Sep 12th 2009, 11:15 PMtjkuboOpen vs. closed sets
In R^2, a set S is defined as S = {(x,y)| x=(1/t)cos(t), y=(1/t)sin(t), t ≥ π} which is the spiral with infinitely many turns about the origin.

1) Is S open, closed or neither? Why?

2) Is R^2\S, the complement of S, open, closed or neither? Why? - Sep 13th 2009, 12:25 AMTaluivren
Hi, the set can't be closed because point (0,0) belongs to the boundary of S but doesn't belong to S. To see this, consider this sequence of points from S:

{(x,y)| x=(1/t)cos(t), y=(1/t)sin(t), t = 2kπ, k>0 is a natural number} = {(1/t,0)| t = 2kπ, k>0 is a natural number}. This sequence converges to (0,0).

It is clear that S can't be open (consider arbitrary small epsilon ball centered at some point from S)

Complement of S is neither closed or open too, because to be open, S would have to be closed, and to be closed, S would have to be open. - Sep 13th 2009, 11:04 AMtjkubo
Thank you for the reply. It helped a lot.

How would you prove that the complement of S is not open without using the fact that S is not closed? - Sep 13th 2009, 11:40 AMTaluivren
above we proved first that S is not closed and deduced that complement of S is not open. to prove directly that complement S is not open, we obviously need to use the problematic point (0,0) again, in combination with suitable definition of open sets in R^2.

Try it, recall some definition of open sets in R^2 (or in any metric space), preferably a definition that uses notion of neighborhood of a point, and apply it to (0,0). - Sep 13th 2009, 04:39 PMtjkubo
A set is open if it is a neighborhood of each of its points, so to show that the complement of S is not open, I have to show that it is not a neighborhood of (0,0). In other words, I have to show that an open ball centered at (0,0) cannot exist. I guess this is so because no matter how small you make the radius of the ball, if you make t large enough, you will always find a point of the spiral, which is not part of the complement of S, in the ball, making it not an open ball. Is this correct?

- Sep 14th 2009, 04:33 AMTaluivren
yes, correct. But an open ball $\displaystyle \{(x,y)\in \mathbb{R}^2:|(x,y)-(0,0)|<\varepsilon\}$ is always open ball, no matter if some spiral has its points in it. What doesn't exist is an open ball centered at (0,0) which is fully contained in the complement of S.