Finding supremum and infimum of a set of rational numbers

Question

Let $f(n,m) = \dfrac{mn}{1+m+n} $. Put $S = \{ f(n,m) : n, m \in \mathbb{N} \} $. Find $\inf S $ and $\sup S$

Attempt:

First, it is clear that $f(n,m) \geq 0$ for all $n,m$. So $0$ is a lower bound. We need show that if $ f(n,m) \geq l$, then $0 \geq l$. If $l > 0$, then by the archimidean property we find and integer $N$ so that $\dfrac{1}{N} < l $ but observe that

$$ \dfrac{1}{N+2} < \dfrac{1}{N} < l $$

This does not seem to help because I can't find a $f(N,M)$ that is less than $l$. Any ideas?

As for the supremum, clearly, $\sup S = \infty$. To prove this: We can find some $f(n_0,m_0)$ greater than $f(n,m)$. For example

$$ \dfrac{n^2 }{1+2n} \geq \dfrac{mn}{1+m+n} $$

does this work?

Answer 1

I'm going to assume, as much as it pains me, that $0\not\in\mathbb{N}$. Otherwise, $f(0,0)=0\le f(n,m)$ for all $n,m$, and we're done.

Note that $f(1,1)=\frac{1}{3}$, so $\frac{1}{3}\ge \inf S$.

Given $n,m\in\mathbb{N}$, $$ f(n,m)=\frac{1}{\frac{1}{nm}+\frac{1}{n}+\frac{1}{m}}\ge \frac{1}{1+1+1}=\frac{1}{3} $$

Answer 2

Actually, $\inf S=\frac13=\frac{1\times1}{1+1+1}$. If $m,n\in\Bbb N$, then $m\geqslant n$ or $n\geqslant m$. If $m\geqslant n$, then$$3mn\geqslant3m=m+m+m\geqslant m+n+1$$which means that$$\frac{mn}{m+n+1}\geqslant\frac13,$$and, of course, if $n\geqslant m$, then you reach the same conclusion.

And, yes, $S$ has no upper bounds and so $\sup S=\infty$ (or it doesn't exist, depending upon the definition).

Answer 3

As for the upper bound, to prove $\sup S=\infty$ it is not enough to show that for every $(n,m)$ there exists $(n_0,m_0)$ such that $f(n_0,m_0)>f(n,m)$ (this criterion means you can always exceed the value of $f(n,m)$ that you have in front of you). For example your criterion holds for $$ f(n,m):= \frac{m+n}{m+n+1}, \qquad m,n\in{\mathbb N} $$ [given $(n,m)$, you could take $(n_0,m_0)=(n+1,m+1)$] but for this $S$ you'll find $\sup S=1$.

To prove $\sup S=\infty$ you need to show that for every $K$ you can find $(n,m)$ such that $f(n,m)>K$. Your choice $(n,n)$ does lead to a solution, since $$ f(n,n)=\frac{n^2}{1+2n}\ge \frac{n^2}{4n}=\frac n4 $$ will exceed $K$ when $n$ is big.