Find the max, min, sup and inf of a sequence 
Let $\{a_n\}=\{x\mid x\in\mathbb {Q},x^2 <2\}$, find the max, min, sup and inf of a sequence


Clearly, sup is $\sqrt {2} $ and inf is $-\sqrt {2} $, so we have $-\sqrt {2}<a_n <\sqrt {2} $. Since $\mathbb {Q} $ is dense of $\mathbb {R} $, max and min both don't exist cause there are many small number between sup and inf.

After I look at the solution, the answer for max and min both are not "DNE", can anyone tell me why cause I don't see it. Thanks.

Okay, I guess my first answer wasn't clear.
If you have a bounded set A one of three things can happen.
1) A has a maximum element x.  If so then x is also the sup of A and the sup of A is a member of the set A.  (sup means least upper bound and if x is maximal it is a least upper bound.)  So max A = sup A = x; x $\in$ A.
The same is true about minimum elements and the inf.
2) A doesn't have a maximum element.  Then if A has a sup, x, the sup is not a member of the set.  If the metric space has the least upper bound property (as R does) the the sup must exist.  So max A does not exist; sup A = x; x $\notin$ A.
The same is true about minimum elements and the inf.
3) If the metric space does not have the least upper bound property (as Q does not) then it is possible (but not always true) that if A doesn't have a maximum element nor does it have a sup.  max A does not exist, sup A does not exist; metric space X does not have least upper bound property.
So if A = {$a_n$} = {$x| x \in Q, x^2 < 2$} is bounded and, I presume, we are viewing it in R which has the least upper bound property, so 3 isn't possible. 
So if A has a max element then max = sup = x and $x^2 < 2$.  Is this possible?
If not, then max does not exist and sup A is the smallest real number that is larger than all of the elements of A.  What real number is that?
