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Suppose, $S= \{ x \in \mathbb R\mid x^2 > 2\}$. Then $\sup S = +\infty$. What is $\inf S$?

I'm guessing that $\inf S \in (-\sqrt 2, \sqrt 2)$. Is that true?

Faithhhhhh
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1 Answers1

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No. What if $x = -2?\quad -4? \quad -333337899999 ?$

Can you see that $\inf S = -\infty$

The interval you suggest, actually the interval $[-\sqrt 2, \sqrt 2]$ is excluded from $S$, so that $S = \mathbb R \setminus [-\sqrt 2, \sqrt 2]$.

amWhy
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  • im sorry im abit slow. but x=-4 would give x^2 =16 so it isnt inf s? ??? – Faithhhhhh Sep 20 '14 at 14:54
  • Right, and $16 \gt 2$, correct? The elements of $S$ here are the $x$'s that makes $x^2 >2$. The members of $S$ are not the values $x^2$. Remember, $S$ consists of all real numbers $x$ for which $x^2 >2$ is true. $;x = -4,$ is one such $,x\in S,$ because $,x^2 = (-4)^2 =16 > 2$. – amWhy Sep 20 '14 at 14:55
  • Oh i get it now since x<-2 or x>2 because x^2>2. so sup S = plus infinity and inf would be the lowest point of x<-2 which is minus infinity! thank you so much! i really appreciate your help~!! – Faithhhhhh Sep 20 '14 at 15:08
  • You are very welcome, faith! – amWhy Sep 20 '14 at 15:12