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:
$\forall x\in\mathbb(M_1):-3<x<3$
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
 A: One must resort to the definitions when proving whether something is the supremum or infinimum of a set.
1. Show that 3 is an upper bound of the set.
2 show that 3 is the smallest upper bound. In other words suppose there existed some a such that a is an upper bound but a < 3. By density of rationals we can choose a q in Q such that a < q < 3. Then q is in your set, condtradicting the fact that a is an upper bound.
Alternatively, we can do the epsilon definition of supremum as suggested by @kingW3.
A: 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 $\forall \epsilon >0,\exists x\in I\;(x<l+\epsilon).$ Similarly, the least upper bound $u$ of $I$ is the value satisfying $\forall \epsilon >0,\exists y\in I\; (u-\epsilon<y).$ By the density of the rationals, we may choose a sequence of rationals $q_{n_x}\in I$ such that $-3<q_{n_x}<-3+p\;\forall x$ and a sequence of rationals $q_{n_y}\in I$ such that for all $y,3-p<q_{n_y}<3.$ 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.
