- #1

Bunny-chan

- 105

- 4

## Homework Statement

I'm in need of some help to be able to determine the supremum and infimum of the following sets:[tex]A = \left\{ {mn\over 1+ m+n} \mid m, n \in \mathbb N \right\}[/tex][tex]B = \left\{ {mn\over 4m^2+m+n^2} \mid m, n \in \mathbb N \right\}[/tex][tex]C = \left\{ {m\over \vert m\vert +n} \mid m \in \mathbb Z, n \in \mathbb N \right\}[/tex][tex]D = \left\{ {n\over 3} - \left[{n\over 3}\right] \mid m, n \in \mathbb N \right\}[/tex][tex]E = \{x \mid x \text{ is decimal fraction between } 0 \text{ and } 1 \text{ that has the digits } 0 \text{ and } 1\}[/tex]P.S: In [itex]D[/itex], the [itex][ \ ][/itex] denotes the integral part of [itex]\frac{n}{3}[/itex].

## Homework Equations

No equations.

## The Attempt at a Solution

In my progress so far I was able to verify that [itex]\text{sup}\{x \in A\} = +∞[/itex]:

Setting [itex]m = (n-1)[/itex] to get[tex]\frac{mn}{1+m+n}= \frac{n(n-1)}{1 + n + n-1} = \frac{n(n-1)}{2n} = \frac{n-1}{2}[/tex]That way [itex]\frac{n-1}{2}\in A[/itex] for every [itex]n\geq 2[/itex]. Thus, the supremum of [itex]A[/itex] is greater than [itex]\frac{n-1}{2}[/itex] for every integer [itex]n[/itex]. Is that correct?

I would really appreciate some guidance on these. I struggle at proving and verifying these conditions... x_x.