# Supremum and infimum of specific sets

• Bunny-chan
In summary: I do not understand what you are trying to say. Can you explain it more clearly?Sorry for my late reply!In summary, Homework Equations, In D, the [ ] denotes the integral part of \frac{n}{3}. C a little harder. Please post some attempt at each of those.
Bunny-chan

## Homework Statement

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 $\frac{n}{3}$.

No equations.

## 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.

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

FactChecker said:
Yes, the supremum must be ≥ (n-1)/2 for every n≥2 ∈ N.
Can I do something similar for set B?

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.

FactChecker said:
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.
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.

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.

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

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

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?

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.

Bunny-chan said:
rational numbers can never be a set's upper bound
Not true.
Bunny-chan said:
which means none of those are upper bounds.
I do not follow the logic of that at all.
Bunny-chan said:
For D, I guess we can also say the supremum is 1, because
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.

haruspex said:
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.
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?

Bunny-chan said:
the supremum for DDD would be 0.6666
yes.

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

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

haruspex said:
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.
$\frac{1}{10}?$

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Bunny-chan said:
$\frac{1}{10}?$
Can you not find an element in the set greater than 0.10000...?

haruspex said:
Can you not find an element in the set greater than 0.10000...?
I'll always be able to find such element, that's why it's an infinite set, isn't it?

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

haruspex said:
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?
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.

Bunny-chan said:
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.
(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.

haruspex said:
(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.
Oh, we can assume that even though the world only isn't there?

Bunny-chan said:
Oh, we can assume that even though the world only isn't there?
I'm fairly sure that's what the question intends. It doesn't make sense otherwise.

haruspex said:
I'm fairly sure that's what the question intends. It doesn't make sense otherwise.
OK, I understand. In that case, since there is no element greater than $\frac{1}{10}$, it is an upper bound for the set.

Bunny-chan said:
OK, I understand. In that case, since there is no element greater than $\frac{1}{10}$, it is an upper bound for the set.
No, there are lots of elements greater than that. You have 0. followed by any sequence of 0s and 1s.

haruspex said:
No, there are lots of elements greater than that. You have 0. followed by any sequence of 0s and 1s.
Yes, but the one element that none of those sequences is greater than is 1, right?

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Bunny-chan said:
Yes, but the one element that none of those sequences is greater than is 1, right?
1 is not in the set. All of the numbers start either 0.0 or 0.1. 0.2 is an upper bound, so the supremum must be between 0.1 and 0.2.

haruspex said:
1 is not in the set. All of the numbers start either 0.0 or 0.1. 0.2 is an upper bound, so the supremum must be between 0.1 and 0.2.
Would $\frac{1}{9} = 0,1111...$ be a valid option to consider then? D:

Bunny-chan said:
Would $\frac{1}{9} = 0,1111...$ be a valid option to consider then? D:
Yes!

haruspex said:
Yes!
Oh, wow, finally!

haruspex said:
D and E are fairly easy. C a little harder. Please post some attempt at each of those.
For $C$, I am noticing that setting $n = 1$, and increasing the values of $m$, the numbers start to approach $0.9999...$ Is that right?

Bunny-chan said:
For $C$, I am noticing that setting $n = 1$, and increasing the values of $m$, the numbers start to approach $0.9999...$ Is that right?
That is certainly true. Can you prove it gets arbitrarily close to 1, but no combination of m and n exceeds it?

haruspex said:
That is certainly true. Can you prove it gets arbitrarily close to 1, but no combination of m and n exceeds it?
I don't know how to effectively prove it, but I can determine the possible values for $m$ and $n$ to see if it could possibly exceed $1$:

For $m = 0$, the values will be obviously $0$;
For $m \lt 0$, the values will be negative, which is inferior to $1$;
For $m > 0, m \geq n$, the denominator can't be less than the numerator (which implies in values smaller than $1$), seeing that $|m| \geq m \Rightarrow |m| + n \gt m$. The same can be said if $m \lt n$.

Is that right?

Bunny-chan said:
I don't know how to effectively prove it, but I can determine the possible values for $m$ and $n$ to see if it could possibly exceed $1$:

For $m = 0$, the values will be obviously $0$;
For $m \lt 0$, the values will be negative, which is inferior to $1$;
For $m > 0, m \geq n$, the denominator can't be less than the numerator (which implies in values smaller than $1$, seeing that $|m| \geq m \Rightarrow |m| + n \gt m$. The same can be said if $m \lt n$.

Is that right?
Yes. To complete the proof, can you find a sequence of elements converging to 1?

haruspex said:
Yes. To complete the proof, can you find a sequence of elements converging to 1?
Do I have to use limits for that?

Bunny-chan said:
Do I have to use limits for that?
Yes, convergence is all to do with limits. Is that a problem? For the purposes of the question, I don't think you actually need to prove the convergence. Just show a sequence which pretty obviously converges to the extremum in your answer.

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