Finding Complement of a set with respect to space U

[RJLiberator]

Homework Statement

Find the complement C' of the set C with respect to the space U if:

1. U = {(x,y,z) : x^2+y^2+z^2 ≤1}, C = {(x,y,z) : x^2+y^2+z^2 = 1}
2. U = {(x,y) : |x| + |y| ≤ 2}, C = {(x,y) : x^2 + y^2 < 2}

Homework Equations

Definition of complement: The elements of the space that do not exist inside the set.

The Attempt at a Solution

I wanted to reassure myself that I am doing things right.

For 1: My answer is C' = {(x,y,z): x^2+y^2+z^2 < 1}
However, why is this right? Couldn't I have (sqrt(1/3), sqrt(1/3), sqrt(1/3)) as (x,y,z) for the set C = 1, and then have C' = (sqrt(1/3), sqrt(1/3), 0) and then we have x and y equal in both C and C' ? Is my answer wrong or what am I missing in the understanding here?

For 2: x^2+y^2 < 2 is a circle with radius sqrt(2) where as |x| + |y| ≤ 2 is a square with corners on each x = - 2, x = 2, y= - 2, y = 2. So the area i am looking for is |x| + |y| - x^2 + y^2 ≤ 2.
But I am not sure how to put this into a final answer.

I can visualize it via the area, but it's hard to put this into an answer.

 

[Orodruin]

and then we have x and y equal

It does not matter if x and y are equal if not also z is equal. The elements of your set are combinations of three numbers. If those numbers are not all the same, the elements are not the same.

For 2: x^2+y^2 < 2 is a circle with radius sqrt(2) where as |x| + |y| ≤ 2 is a square with corners on each x = - 2, x = 2, y= - 2, y = 2. So the area i am looking for is |x| + |y| - x^2 + y^2 ≤ 2.

This is simply incorrect. You need to find out what points in the square are not points in the disk.

 

[RJLiberator]

Thanks for your help. For part 1, I understand now and see why I was confused. Now I understand that I have the correct answer.

For part 2, after thinking about it, I also understand why my area-area part did not work out.
It's more to say:
(|x|+|y| ≤ 2) - (x^2+y^2 < 2) = the complement, but that's not a good answer. I need to find the points in the square that are not points in the disk as you stated.

I see that |x|+|y| = 2 is definitely in the complement.

What I am thinking here is

Answer: C' = {(x,y): sqrt(2) ≤ |x|+|y| ≤ 2}

This takes out the middle part and leaves the good ends. However, their seems to be an error in that sqrt(2) is just another square, but we want a circle.

 

[Orodruin]

Hint: You can put several conditions into the definition of the set.

 

[Mark44]

For 2: x^2+y^2 < 2 is a circle with radius sqrt(2) where as |x| + |y| ≤ 2 is a square with corners on each x = - 2, x = 2, y= - 2, y = 2. So the area i am looking for is |x| + |y| - x^2 + y^2 ≤ 2.

This is simply incorrect. You need to find out what points in the square are not points in the disk.

@RJLiberator, if you haven't already done so, sketch a graph of the two sets. If you have done this, then disregard this advice.

 

[RJLiberator]

Hm. I think I was making this harder then I needed to make it:

C' = {(x,y) : x^2+y^2 ≥ 2 and |x|+|y| ≤ 2 } This seems to get the correct area.