Answer to Question #191554 in Real Analysis for Amara

Question

Find the infimum and supremum in each of the following sets of real numbers: S = {x| − x2 + 6x − 3 > 0

Let a be the supremum of a set of real numbers and let ε > 0 be any real number. there is at least one x ∈ S such that

a−ε<x≤a

where S is the set with the given supremum

Answer 1

First Question:

The quadratic equation defined in the set is

$-x^2 +6x - 3 > 0$

$\implies x^2 -6x +3 <0 \\ \implies x^2 -6x +(-3)^2 +3 <0 + (-3)^2 \\ \implies (x-3)^2 < 6$

$\implies x - 3 < \sqrt{6}$ or $x - 3 < - \sqrt{6}$

i.e $x < 3+\sqrt{6}$ or $x < 3-\sqrt{6}$

This inequality is in this region

$x \in (3- \sqrt{6}, 3+ \sqrt{6})$

Hence the sup S = $3+\sqrt{6}$ and inf S $= 3-\sqrt{6}$

Second question:

Suppose such x does not exist, then we have that

$a - \epsilon \geq x \, \forall x \in A$

Hence $a - \epsilon$ is an upper bound of A. And by definition of supremum of A, $a \leq a - \epsilon$

This is a contradiction. Hence, there is at least one $x \in A \ni a - \epsilon <x \leq a$