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?

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- 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