Let the set T be given by T = { x in the reals | x^2 < 2}. Describe maximum, minimum, supremum (least upper bound), and infimum (greatest lower bound) for T (they may not exist).
Accepted Solution
A:
Answer:The supremum is [tex]\sqrt{2}[/tex].The infimum is [tex]-\sqrt{2}[/tex].There is no maximum.There is no minimum.Step-by-step explanation:We have the set [tex]T=\{x\in\mathbb{R}:x^2<2\}[/tex]. Now, let us recall that [tex]\sqrt{x^2}=|x|[/tex], and the inequality [tex]x^2<2[/tex] is equivalent to [tex]|x|<\sqrt{2}[/tex], so our set can be written as[tex]T=\{x\in\mathbb{R}:|x|<\sqrt{2}\}[/tex].The inequality [tex]|x|<\sqrt{2}[/tex] is equivalent to [tex]-\sqrt{2}<x<\sqrt{2}[/tex]. So, [tex]T=\{x\in\mathbb{R}:-\sqrt{2}<x<\sqrt{2}\}[/tex]. Thus, [tex]T=(-2,2)[/tex].Now, we have that [tex]T[/tex] is the open interval (-2,2). From this we can extract all the information we need:The supremum is [tex]\sqrt{2}[/tex].The infimum is [tex]-\sqrt{2}[/tex].There is no maximum, because the interval is open.There is no minimum, because the interval is open.