# Find the infimum and supremum of $\{x\in Q \mid x² < 9\}$

I have to find (and prove) the infimum and supremum of the following set:

$$M_1:=\{x\in\mathbb{Q} \mid x^2 < 9\}$$

On first glance, I would say:

$$\inf M_1=-3$$
$$\sup M_1=3$$

Now I have to prove that these really are the infimum and supremum of the set, and that's the point where I'm having problems. According to the definition of $$\inf$$ and $$\sup$$, this means, that $$-3$$ is the biggest lower bound and 3 is the lowest upper bound:

$$\forall x\in\mathbb(M_1): -3 \leq x \leq 3$$

We can see, that -3 and 3 are not elements of M1, which means:

$$\forall x\in\mathbb(M_1):-3

But how can I show that -3 and 3 are the $$\textbf{biggest / smallest}$$ bound? I mean, for example, what if there is a number bigger than -3 that acts like a lower bound to the set? Obviously there isn't a bigger lower bound, but how can I mathematically show it? Do you guys have any advice? Thanks in advance, and sorry for my English :D

• Assume that $3-p$ is supremum can you find a rational number between $3-p$ and $3$? Oct 25, 2019 at 23:20
• You can to use of Archimedean property. en.m.wikipedia.org/wiki/Archimedean_property Oct 25, 2019 at 23:21

Let $$I= \{ x\in\mathbb{Q} : x^2 <9\}.$$ $$x^2<9\Rightarrow -3 < x < 3.$$ Assume that $$3-p,-3+p\in\mathbb{Q}$$ are $$\sup I$$ and $$\inf I$$ respectively, where $$p>0.$$ By definition, the greatest lower bound $$l$$ of $$I$$ is the value satisfying $$\forall \epsilon >0,\exists x\in I\;(x Similarly, the least upper bound $$u$$ of $$I$$ is the value satisfying $$\forall \epsilon >0,\exists y\in I\; (u-\epsilon By the density of the rationals, we may choose a sequence of rationals $$q_{n_x}\in I$$ such that $$-3 and a sequence of rationals $$q_{n_y}\in I$$ such that for all $$y,3-p This is a contradiction, as by assumption, $$-3+p$$ and $$3-p$$ are $$\inf I$$ and $$\sup I$$ respectively. Since $$p$$ cannot be less than $$0$$ (otherwise $$-3+p$$ and $$3-p$$ would both not be in $$I$$), we thus have that $$p=0$$ and so $$\inf I = -3$$ and $$\sup I = 3.$$
Notice that we have directly satisfied the $$\epsilon$$ definitions of infimum and supremum.