# Supremum and infimum of specific sets

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1. Apr 22, 2017

### Bunny-chan

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

2. Relevant equations
No equations.

3. The attempt at a solution
In my progress so far I was able to verify that $\text{sup}\{x \in A\} = +∞$:

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 greater than $\frac{n-1}{2}$ for every integer $n$. Is that correct?

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

2. Apr 23, 2017

### FactChecker

Yes, the supremum must be ≥ (n-1)/2 for every n≥2 ∈ N.

3. Apr 23, 2017

### Bunny-chan

Can I do something similiar for set B?

4. Apr 23, 2017

### FactChecker

You can probably do something similar. It's not clear to me what techniques you are expected to use in your class. Each problem may take a different technique.

5. Apr 23, 2017

### Bunny-chan

I'm not required to use any specific technique, I just need to find these parameters. And besides, I don't know many of those anyway.

6. Apr 23, 2017

### FactChecker

The answer for the supremum of A was simplified by it being infinity. If it had been something finite, your approach may not have gotten the right answer because the values with m=(n-1) may not be the highest. So you will have to be careful.
There are ways to find extreme points of continuous functions of two variables that may help in some of the problems. You will still need to be careful when you are restricted to integers or natural numbers.

7. Apr 24, 2017

### haruspex

D and E are fairly easy. C a little harder. Please post some attempt at each of those.

8. Apr 25, 2017

### Bunny-chan

Anyway, let me see... For $E$, I know that rational numbers can never be a set's upper bound, so if we take a number $a$ such that $0 \lt a \lt 1$, there will always be an $a + q$ such that $a \lt a + q \lt 1$, which means none of those are upper bounds. Thus, $\sup{(E)} = 1$. The same can be said for $0$; seeing we can infinitely take rational numbers greater than $0$, none of them is the greatest lower bound, and therefore $\inf{(E)} = 0$

For $D$, I guess we can also say the supremum is $1$, because $$n = 2 \Rightarrow \frac{2}{3} - \left[\frac{2}{3}\right] = 0.6666... - 0 = 0.6666... \\ n = 3 \Rightarrow \frac{3}{3} - \left[\frac{3}{3}\right] = 1 - 1 = 0 \\ n = 4 \Rightarrow \frac{4}{3} - \left[\frac{4}{3}\right] = 1.3333... - 1 = 0.3333...$$ And since all values are positive, $0$ is the infimum.

Well, is that right?

9. Apr 25, 2017

### FactChecker

I think you may be concentrating on the wrong thing. A set can have an element that is equal to its supremum and it can have an element that is equal to its infimum. the set {1} has 1=maximum=supremum=infimum=minimum.

For a finite supremum, s, of set X you must show:
1) ∀x∈X, x≤s.
2) ∀ε>0, ∃x∈X such that s-ε<x.
For a finite infimum, i, of set X you must show:
1) ∀x∈X, i≤x.
2) ∀ε>0, ∃x∈X such that x<i+ε.​

Make obvious adjustments for infinite supremum and infimum.

10. Apr 25, 2017

### haruspex

Not true.
I do not follow the logic of that at all.
No, as you showed D only has three elements.

I think you do not understand the concept of a supremum. If a subset of ℝ is bounded above then its supremum is the least element of ℝ (not necessarily a member of the set) which is an upper bound for the set.
For a finite set, this is simply the greatest member of the set.

11. Apr 25, 2017

### Bunny-chan

When I said that none of those were upper bounds, I actually meant least upper bound, in the sense that we can infinitely place numbers between them. At least that's what I thought! D:

But anyway, I think that is more clear to me now. So the supremum for $D$ would be $0.6666...$, considering the set is finite?

12. Apr 25, 2017

### haruspex

yes.

13. Apr 25, 2017

### Bunny-chan

Is there a way I could show that conclusion (both for sets $D$ and $E$) in a little bit more mathematically rigorous way?

14. Apr 25, 2017

### haruspex

It is easy to show, rigorously, that the greatest element in a finite ordered set is its supremum. Show that the two conditions FactChecker posted are met (post #9).

E is not a finite set. Do you have an answer for that now? The supremum is between 0 and 1.

15. Apr 25, 2017

### Bunny-chan

$\frac{1}{10}?$

Last edited: Apr 25, 2017
16. Apr 25, 2017

### haruspex

Can you not find an element in the set greater than 0.10000....?

17. Apr 25, 2017

### Bunny-chan

I'll always be able to find such element, that's why it's an infinite set, isn't it?

18. Apr 25, 2017

### haruspex

I don't understand your reasoning. How does its being an infinite set imply you can find an element > 0.10000....? The set {0.1, 0.01, 0.001, ...} is infinite but contains no element >0.1.

If you can find an element greater than that (i.e. greater than 1/10) then 1/10 cannot be a supremum for the set. Can you find a specific element in the set > 0.1?

19. Apr 25, 2017

### Bunny-chan

Nevermind, I understand what you mean. I got a bit confused by the numbers.

It's because, for instance $\frac{91}{100} = 0,91$ is a decimal fraction between $0$ and $1$ that has the digits $0$ and $1$. And so we can go along increasing the number as much as we like. But maybe I interpreted it wrong.

20. Apr 25, 2017

### haruspex

(Previous reply deleted because it crossed with an edit to post #19)

0,91 is not in the set. Only digits 0 and 1 are allowed.