LinkBack URL
About LinkBacksIn 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?
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?
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.
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.Originally Posted by tjkubo
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).
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?
  |
