I've recently started to study supremum and infimum, and I'm having trouble determining the supremum and infimum of the following sets:

$ A = \left\{ {mn\over 1+ m+n} \mid m, n \in \mathbb N \right\} $ $ B = \left\{ {mn\over 4m^2+m+n^2} \mid m, n \in \mathbb N \right\} $ $ C = \left\{ {m\over \vert m\vert +n} \mid m \in \mathbb Z, n \in \mathbb N \right\} $ $ D = \left\{ {n\over 3} - \left[{n\over 3}\right] \mid m, n \in \mathbb N \right\} $

  • In $D$, the $[\; ]$ denotes the integral part of ${n\over 3}$.

$E = \{x \mid x \text{ is decimal fraction between } 0 \text{ and } 1 \text{ that has the digits } 0 \text{ and } 1\}$.

In my progress so far I was able to verify that sup $\{x \in A\} = +∞$

Setting $m = (n-1)$ to get $$ \frac{mn}{1+m+n}= \frac{n(n-1)}{1 + n + n-1} = \frac{n(n-1)}{2n} = \frac{n-1}{2} $$

That way $\frac{n-1}{2}\in A$ for every $n\geq 2$. Thus, the supremum of $A$ is bigger than $\frac{n-1}{2}$ for every integer $n$.

I don't know how to go past this. x_x

Any help would be greatly appreciated!

  • $\begingroup$ For A consider: $m = n$ then we read: $\frac{n^2}{1+2n}$ which goes to infinity as n goes to infinity. If you take $m = - n$ you will get $\frac{-n^2}{1} = -n^2$ which goes to minus infinity. Why do you think the supremum can't be inifinite? $\endgroup$ – Piotr Benedysiuk Apr 16 '17 at 17:19
  • $\begingroup$ I don't know. I just presumed that by definition a supremum had to be a specific number. I wasn't aware that infinity could be regarded as it. $\endgroup$ – Bunny-chan Apr 16 '17 at 17:22
  • $\begingroup$ Maybe this is helpful $\endgroup$ – kingW3 Apr 16 '17 at 17:29
  • $\begingroup$ I see. Thank you. I'll edit my post. $\endgroup$ – Bunny-chan Apr 16 '17 at 17:38
  • $\begingroup$ It is often convenient to extend $\mathbb R$ to include $\infty$ or $\pm \infty.$ We just can't do ordinary arithmetic with them. $\endgroup$ – DanielWainfleet Apr 16 '17 at 18:37

For A, since $m\geq 1$ and $n \geq 1$ we have $(m-1)(n-1)\geq 0,$ so $mn\geq m+n-1.$

So $mn/(m+n+1)\geq (m+n-1)/(m+n+1)=1-2/(m+n+1)\geq 1-2/3$ (because $m+n+1\geq 3$).

And we have $1-2/(m+n+1)=1-2/3 \iff m=n=1.$

Also all of the other inequalities are equalities when $m=n=1 .$ So $\min A=1-2/3=1/3.$

  • $\begingroup$ Greatly appreciated input. ^^ $\endgroup$ – Bunny-chan Apr 16 '17 at 18:50

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