Find the complement C^c of the set C with respect to the...

Find the complement $C^{c}$ of the set $C$ with respect to the space $\mathcal{C}$ if (a) $\mathcal{C}=\{x: 0<x<1\}, C=\left\{x: \frac{5}{8}<x<1\right\}$. (b) $\mathcal{C}=\left\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 1\right\}, C=\left\{(x, y, z): x^{2}+y^{2}+z^{2}=1\right\}$. (c) $\mathcal{C}=\{(x, y):|x|+|y| \leq 2\}, C=\left\{(x, y): x^{2}+y^{2}<2\right\}$.

Transcript

00:01 So first of all, let's make sure we understand what exactly we're talking about, right? for some set a, we have a plus a complement is equal to the probability space or the space in which you're working, right? here it just says space, but in general, the idea is that whatever space you're in, the set plus its complement cover everything, right? it's like whatever part of the planet is not dark is being lit by the sun, right? so every event has to be in either the original set or its complement and can only be in one.

00:46 So we have three sets here.

00:49 And as i've just produced here, the first space, right, this is my space, x between zero and one in the open set.

00:58 The set c, and there's too many c's in this question, let's be honest, is between 5 eighths and 1.

01:06 So the complement of c here would be x between 0 and 5 .5 over 8, right? this covers together, right, c plus c complement is equal to the space.

01:22 If you jammed those two things together.

01:25 Now for b, so for the second space, we have the set of x, y, z such that the squares of the coordinates are less than equal to one.

01:35 This is what some people might call the unit ball, right, as opposed to the unit circle.

01:40 And we're given that set c is equal to the x, y, and c that satisfy the equality, right? so again, if you think of this space being the unit ball, set c here is giving you the outer shell of the ball, right? it's giving you every point that's as far away from the center of zero zero zero as possible.

02:04 So c complement here is equal to the x, y, and z such that, and, you know, in a perfect world, fine, put these in brackets, it doesn't really matter, is where x squared plus y squared plus z squared is strictly less.

02:21 Than one, right? because now i've comprised the entire set b.

02:25 I've got the space is all the points where the squares are less than or equal to one...

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