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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.

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  • $\begingroup$ Are you defining $0 \in \mathbb N$ or not? $\endgroup$
    – Thoth19
    Oct 18, 2014 at 23:57
  • $\begingroup$ No. $0 \notin \Bbb{N}$ $\endgroup$
    – qiubit
    Oct 18, 2014 at 23:58

2 Answers 2

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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.

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    $\begingroup$ "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, 2014 at 0:16
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    $\begingroup$ 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, 2014 at 0:19
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    $\begingroup$ 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, 2014 at 0:21
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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.

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