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Chapter 1: Problem 2

Find the complement \(C^{c}\) of the set \(C\) with respect to the space \(\mathcal{C}\) if (a) \(\mathcal{C}=\\{x: 0

Short Answer

Expert verified
The complements are (a) \(C^{c} = \left\{ x: 0 < x \leq \frac{5}{8} \right\}\), (b) \(C^{c} = \left\{(x, y, z): x^{2}+y^{2}+z^{2} < 1\right\}\) and (c) \(C^{c} = \left\{(x, y): |x|+|y| = 2, x^{2}+y^{2} \geq 2\right\}\).

Step by step solution

01

Find the complement for (a)

From the problem, \(\mathcal{C}= \{x: 0 < x < 1\}\) and \(C=\left\{x: \frac{5}{8}<x<1\right\}\). The complement \(C^{c}\) will be all those elements in \(\mathcal{C}\) which are not in C. So the complement will be \(C^{c} = \left\{ x: 0 < x \leq \frac{5}{8} \right\}\), because only these elements are in \(\mathcal{C}\) but not in C.
02

Find the complement for (b)

For this part, \(\mathcal{C}=\left\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 1\right\}\) and \(C=\left\{(x, y, z): x^{2}+y^{2}+z^{2}=1\right\}\). The elements \(C^{c}\) are points that lie within the sphere, but not on its surface. So the complement \(C^{c}\) will be \(\left\{(x, y, z): x^{2}+y^{2}+z^{2} < 1\right\}\), since these elements are in \(\mathcal{C}\) but not in C.
03

Find the complement for (c)

Lastly, given \(\mathcal{C}=\{(x, y):|x|+|y| \leq 2\}\) and \(C=\{(x, y): x^{2}+y^{2}<2\}\), the complement \(C^{c}\) will be those elements within the diamond formed by \(\mathcal{C}\) but outside the circle formed by C. So the complement \(C^{c}\) will be \(\left\{(x, y): |x|+|y| = 2, x^{2}+y^{2} \geq 2\right\}\), because only these elements are in \(\mathcal{C}\) but not in C.

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Most popular questions from this chapter

If a sequence of sets \(C_{1}, C_{2}, C_{3}, \ldots\) is such that \(C_{k} \supset C_{k+1}, k=1,2,3, \ldots\), the sequence is said to be a nonincreasing sequence. Give an example of this kind of sequence of sets.

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