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I have to find the supremum and infimum of $A=\{x\in \mathbb{Q}: 4<x^2\leq7\}$

My proof:

$4<x^2\leq7\implies 2<x\leq\sqrt{7}$

I will prove that $\sup A=\sqrt7$ and $\inf A=2$

Since $\sqrt7\geq x$, $\forall x\in A$ then it's an upper bound for $A$

For $\sqrt7$ to be the supremum $\forall ε>0$ there should be a $x_0\in A$ such that

$\sqrt7-ε<x_0$

Which is true due to the density of rationals in $\mathbb{R}$

So $\sup A=\sqrt7$

Since $2<x, \forall x\in A$ then it's a lower bound

For $2$ to be the infimum $∀ε>0$ there should be a $x_0\in A$ such that

$2+ε>x_0$ again from the density of rationals the statement is true

Is this proof correct/sufficient? Thanks in advance

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  • $\begingroup$ Note that $A$ contains negative numbers. $\endgroup$ – lulu Sep 17 '18 at 11:31
  • $\begingroup$ Since $2<\left | x \right | < \sqrt7$ right? So does that mean that $-\sqrt7$ is the infimum? $\endgroup$ – SlimJim Sep 17 '18 at 11:37
  • $\begingroup$ Right on both counts. $\endgroup$ – lulu Sep 17 '18 at 11:42
  • $\begingroup$ Infimum and supremum in which set? $\endgroup$ – edm Sep 17 '18 at 11:50

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