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I am to find supremum and infimum of given set, and prove they are in fact supremum and infimum. $$A=\lbrace x\in \Bbb{R}:x^2-5|x|+4\lt0\rbrace $$ $$\sup A = 4, \inf A = -4$$ I tried to prove it in the following way, let $\varepsilon\gt0, a\in A$, then if $a\gt 4-\varepsilon$, $4-\varepsilon$ is no upper bound, so I can set $a$ to be equal to $4-\frac{\varepsilon}{2}$ and therefore $4-\varepsilon$ is not upper bound what proves that 4 is least upper bound. Proof for greatest lower bound is similar. My question is, is my proof correct?

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  • $\begingroup$ How do you know $\exists a \in A$ such that $a > 4 - \epsilon?$ $\endgroup$ – dhk628 Dec 13 '15 at 10:44
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Equivalently, $A = \{x \in \mathbf {R} | 1 < |x| < 4 \}$. This proves that $$\inf A = -4 \qquad \sup A = 4.$$

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