I'm trying to learn how to find supremum and infimum of a given set as this is essential in my further studying. Here's a problem I want to tackle:

$A=\{\frac{n-k}{n+k}:n,k\in\Bbb{N}\}.$ Find $\sup(A), \inf(A)$.

How to approach such a problem? Any hints? I kinda suspect that $\sup(A)$ will be near 1, but I don't know how to prove it formally and this is really important for me. As for $\inf(A)$ I think -1 might be a good candidate.

PS I can't use derivatives yet, I only learned the definitions of $\sup$ and $\inf$ and I'm allowed to use them in order to prove suspected bounds.

  • $\begingroup$ Are you defining $0 \in \mathbb N$ or not? $\endgroup$ – Thoth19 Oct 18 '14 at 23:57
  • $\begingroup$ No. $0 \notin \Bbb{N}$ $\endgroup$ – qiubit Oct 18 '14 at 23:58

You have come up with good candidates for the supremum and the infimum. The strategy of proof is this. Let the candidate for the supremum be $s$. You first show that $s$ is an upper bound for the set. Then you pick an arbitrary $\epsilon \gt 0$ and show that there exists an element $a \in A$ such that $s \lt a$. For example,

$ \frac{n - k}{n + k} \lt 1$ and hence $1$ is an upper bound for $A$. Let $\epsilon $ be arbitrary. Fix $k$. You can pick $n \in \Bbb N$ such that $\frac{2k}{n + k} \lt \epsilon $. Then, $1 - \epsilon \lt 1 - \frac{2k}{n + k} = \frac{n - k}{n + k} $ and you are done.

For the infimum, you need to show that your candidate $d$ is a lower bound for the set and then if you pick a point which is at an arbitrarily small distance to the right of $d$ on the line, there is an element $a' \in A$ which is less than that point.

First, you can note that $ \frac{|n - k|}{|n + k|} = \frac{|n - k|}{n + k} \lt 1 $ and hence by definition, $-1$ is a lower bound for the set $A$. Now as before let $\epsilon \gt 0$ be arbitrary. We need to show there exists an element $a' \in A$ such that $a' \lt -1 + \epsilon $. Again pick $ \frac{2k}{n + k} \lt \epsilon $ as before which can be done since $n$ can be made arbitrarily large and you will be done.

  • 1
    $\begingroup$ "Then you pick an arbitrary $\epsilon \gt 0$ and show that there exists an element $a \in A$ such that $s \lt a$." Is this for the sake of getting a contradiction? $\endgroup$ – qiubit Oct 19 '14 at 0:16
  • 1
    $\begingroup$ Yes and No. It is a counter-argument to establish the definition. The supremum is the least upper bound of the set. So $S = \sup A \implies $ for any $\epsilon \gt 0$ there exists an element $a \in A, \; a \gt s - \epsilon$. If there is no such element $a$ then $s - \epsilon$ is an upper bound for $A$ which cannot happen since it is less than the supremum which is the least upper bound. $\endgroup$ – Ishfaaq Oct 19 '14 at 0:19
  • 1
    $\begingroup$ The definition I have used is $s = \sup A$ if and only if $\;\;(1),\;s$ is an upper bound for $A$ and $(2),\;$ given any $\epsilon \gt 0$ there exists $a \in A$ such that $a \gt s - \epsilon$. This particular definition can be easily proven with any definition that you use and comes in seriously handy when writing proofs (or so I've found). $\endgroup$ – Ishfaaq Oct 19 '14 at 0:21

So for the sup, the answer is going to be 1. Basically, we shrink k to as small as possible. First we check that 1 is an upper bound. This is obvious because $n-k < n+k$ for positive $k$. Then we want to say 1 is the least upper bound. So we suppose there could be a smaller upper bound. Let this value be $x$. By algebraic manipulation we can get that $n(1-x)/(x+1)<k$ But we can choose n to be large enough such that this is false. You probably have a lemma somewhere that says for a real, 0 a. By contradiction, 1 must be the sup.

We can do similar for the inf. We can shrink n super small so the expression is basically $\frac{-k}{k}$ By the same logic as above, we can show that this is -1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.