# 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.

• Is that exactly how the problem is stated? Because your answers are correct. But the problem statement is weird, so I am wondering if maybe you are misinterpreting something about the problem and have passed that misinterpretation on to us. Commented Oct 13, 2015 at 16:18
• That is the question statement Commented Oct 13, 2015 at 16:22
• Can you see what the answers for max and min are? or does it just reject "DNE"? Commented Oct 13, 2015 at 16:34
• The answer says" max and min both exist" Commented Oct 13, 2015 at 18:55
• Well then, either your course is using a different definition of max and min than the rest of the world, or the answer is wrong, and you are right. Commented Oct 13, 2015 at 18:57

The max (and, similarly, the min) does not exist because there is no rational $r$ such that $r^2 = 2$ and, for any rational $r < \sqrt{2}$ there is another rational $s$ such that $r < s < \sqrt{2}$.

• How do you know that if r < $\sqrt 2$ then there is an s such that r < s < $\sqrt 2$? Commented Oct 13, 2015 at 16:22
• @fleablood - because there is a rational number between any two distinct real numbers. Commented Oct 13, 2015 at 16:25
• @PaulSinclair How do you know that $\sqrt 2$ is a well defined real number? Commented Oct 13, 2015 at 16:41
• @fleablood: See Soba noodles's answer. And, after you read it, it would be nice to upvote it. Commented Oct 13, 2015 at 19:05
• Because there is no rational number $r$ such that $r^2 = 2$. That is why Dedekind's method of cuts for defining the reals in terms of partitions of the rationals is so brilliant. Look at Landau's "Foundations of Analysis" for all the gory details. Commented Oct 14, 2015 at 4:00

This is a classic example of $\mathbb{Q}$ not being a complete field while being dense in $\mathbb{R}$.

Let's denote the set containing the sequence $\{a_n\}$ as $S$. Then, for any $q\in S$, using Archimedes's axiom for the numbers $2-q^2>0$ and $2q+1>0$ there exists an $n\in \mathbb{N}$ such that $n(2-q^2)>2q+1$.

Thus we have: $n^2(2-q^2)>n(2q+1)=2nq+n\geq 2nq+1 \Rightarrow (q+\frac{1}{n})^2<2$, so there always exists a $q'=q+\frac{1}{n} \in S, q'>q$, ergo there is no maximum of this sequence in $\mathbb{Q}$. The reasoning for the nonexistance of a minimum is similar.

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?

• From the statement, the sequence is an enumeration of the rationals with square < 2. Commented Oct 13, 2015 at 16:23
• @PaulSinclair I don't see your point. Commented Oct 13, 2015 at 16:25
• Your first sentence. The sequence is given as a sequence of rational numbers (which are necessarily in the reals). And because any set of rational numbers is countable, and this one is infinite, such a sequence necessarily exists. Commented Oct 13, 2015 at 16:30
• @PaulSinclair: The sequence has no inf or sup in $\mathbb{Q}$, but it has both in $\mathbb{R}$. So the context matters; what space are we working in?
– user169852
Commented Oct 13, 2015 at 16:35
• @Bungo That was precisely my point. Commented Oct 13, 2015 at 16:39