Question

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

(ii) Let a be the supremum of a set of real numbers and let ε > 0 be any real number.Show that there is at least one x ∈ S such that a−ε<x≤a where S is the set with the given supremum.

Answer 1

(i) Given set is-

$S = x: − x^2 + 6x − 3 > 0$

$\Rightarrow x^2-6x+3<0\\ \Rightarrow (x-(3+\sqrt{6})(x-(3-\sqrt{6})<0$

This will only possible when-

$(x-(3+\sqrt{6}))<0 \text{ and }(x-(3-\sqrt{6})>0$

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

Hence inf{S}$=3-\sqrt{6}$

sup{S}$=3+\sqrt{6}$

(ii) As sup{S}=a, where S is set of real numbers.

As $\epsilon >0$ , Implies that There always exist a supremum between the $a-\epsilon$ and $a$ , As a contain the supremum before the $\epsilon$ came into picture.

Hence there is at least one x ∈ S such that $a−\epsilon<x≤a$ where S is the set with the given supremum.