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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.

Michael Hardy
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Simple
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  • 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. – Paul Sinclair Oct 13 '15 at 16:18
  • That is the question statement – Simple Oct 13 '15 at 16:22
  • Can you see what the answers for max and min are? or does it just reject "DNE"? – Paul Sinclair Oct 13 '15 at 16:34
  • The answer says" max and min both exist" – Simple Oct 13 '15 at 18:55
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    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. – Paul Sinclair Oct 13 '15 at 18:57
  • Actually, with that answer, it is clear that the author dropped "do not" from careless typing. Otherwise they would have given you the values for max and min. – Paul Sinclair Oct 13 '15 at 19:11

3 Answers3

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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 $r < s < \sqrt{2}$.

marty cohen
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  • How do you know that if r < $\sqrt 2$ then there is an s such that r < s < $\sqrt 2$? – fleablood Oct 13 '15 at 16:22
  • @fleablood - because there is a rational number between any two distinct real numbers. – Paul Sinclair Oct 13 '15 at 16:25
  • @PaulSinclair How do you know that $\sqrt 2$ is a well defined real number? – fleablood Oct 13 '15 at 16:41
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    @fleablood: See Soba noodles's answer. And, after you read it, it would be nice to upvote it. – marty cohen Oct 13 '15 at 19:05
  • @fleablood - are you serious? It is evident from the OP that this is not a "start from scratch, assume you know nothing" problem. It is also evident from the work shown, and that the sup and inf answers were accepted as correct that this is a problem in the reals. If you really don't know the answer to your question, then I can answer it. But otherwise, why are you asking it? – Paul Sinclair Oct 13 '15 at 19:05
  • I ask because the question asked if there were min and max. If there are min and max then there is a greatest rational less than 2. Than that is the sup of the rationals whose squares are less than 2. And therefore the definition of $\sqrt(2)$ as the sup of the rationals whose squares are less than 2 is no longer correct and there is no well defined number to be called the $\sqrt 2$. So, in other words entire existence of $\sqrt 2$ relies upon the answer to this question being there is no max or min. – fleablood Oct 13 '15 at 21:02
  • So basically you're saying that $\sqrt{2}$ exists because it's supposedly defined as the sup of this sequence? What's wrong with defining it as $(\sqrt{2})^2=2$? You avoid the mess with sequences altogether, and Pythagoras is much more pleased. – Soba noodles Oct 13 '15 at 21:46
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    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. – marty cohen Oct 14 '15 at 04:00
  • In that case, I certainly will. Thank you for the reference – Soba noodles Oct 14 '15 at 18:42
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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.

Soba noodles
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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?

fleablood
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  • From the statement, the sequence is an enumeration of the rationals with square < 2. – Paul Sinclair Oct 13 '15 at 16:23
  • @PaulSinclair I don't see your point. – fleablood Oct 13 '15 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. – Paul Sinclair Oct 13 '15 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? –  Oct 13 '15 at 16:35
  • @Bungo That was precisely my point. – fleablood Oct 13 '15 at 16:39