# Math Help - Supremums and Infimums

1. ## Supremums and Infimums

If possible find the supremum and infimum of the following sets.

a) (0, infinity)
b) {1/n: n is an element of the natural numbers}
c) rational numbers AND [sqrt(2), sqrt(3)]
d) {x is an element of the real numbers: x+2>x^2}
e) {1/n-1/m: n,m are elements of the natural numbers}

a) supremum: none; infimum: 0
b) supremum: none; infimum: 1
c) supremum: none; infimum: none
d) supremum: none; infimum: 0
e) supremum: none; infimum: 0

I haven't had much practice with these, and I would appreciate any feedback!

2. ## Re: Supremums and Infimums

Originally Posted by lovesmath
If possible find the supremum and infimum of the following sets.

a) (0, infinity)
b) {1/n: n is an element of the natural numbers}
c) rational numbers AND [sqrt(2), sqrt(3)]
d) {x is an element of the real numbers: x+2>x^2}
e) {1/n-1/m: n,m are elements of the natural numbers}

a) supremum: none; infimum: 0
b) supremum: none; infimum: 1
c) supremum: none; infimum: none
d) supremum: none; infimum: 0
e) supremum: none; infimum: 0

I haven't had much practice with these, and I would appreciate any feedback!

For d) note that you have \displaystyle \begin{align*} x + 2 > x^2 \end{align*}, which means

\displaystyle \begin{align*} x^2 &< x + 2 \\ x^2 - x - 2 &< 0 \\ x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 2 &< 0 \\ \left(x - \frac{1}{2}\right)^2 - \frac{9}{4} &< 0 \\ \left(x - \frac{1}{2}\right)^2 &< \frac{9}{4} \\ \left| x - \frac{1}{2}\right| &< \frac{3}{2} \\ -\frac{3}{2} < x - \frac{1}{2} &< \frac{3}{2} \\ -1 < x &< 2 \end{align*}

For e) I would argue that there is no infimum, because if \displaystyle \begin{align*} n > m \end{align*} then \displaystyle \begin{align*} \frac{1}{n} < \frac{1}{m} \end{align*} and therefore \displaystyle \begin{align*} \frac{1}{n} - \frac{1}{m} < 0 \end{align*}.

3. ## Re: Supremums and Infimums

Originally Posted by Prove It
For e) I would argue that there is no infimum, because if \displaystyle \begin{align*} n > m \end{align*} then \displaystyle \begin{align*} \frac{1}{n} < \frac{1}{m} \end{align*} and therefore \displaystyle \begin{align*} \frac{1}{n} - \frac{1}{m} < 0 \end{align*}.
Why does this mean that there is no infimum?

4. ## Re: Supremums and Infimums

Hey lovesmath.

Just a question regarding the infininum: does this refer strictly to the greatest lower bound?

The reason I ask is that if you have an open set and one endpoint, then if that end point is the lower point it means that the infinum won't actually exist you will never actually have a greatest lower bound.

You can think of it with say (0,infinity) where you can't have zero (since it's not included) but if you pick any value (say 0.0001) then you can always show there is a lower value (say 0.0000001) but you repeat this forever and you never actually get a fixed value.

5. ## Re: Supremums and Infimums

Originally Posted by chiro
Just a question regarding the infininum: does this refer strictly to the greatest lower bound?

The reason I ask is that if you have an open set and one endpoint, then if that end point is the lower point it means that the infinum won't actually exist you will never actually have a greatest lower bound.

You can think of it with say (0,infinity) where you can't have zero (since it's not included) but if you pick any value (say 0.0001) then you can always show there is a lower value (say 0.0000001) but you repeat this forever and you never actually get a fixed value.
Is not an "infininum" or an "infinum" but "infimum." A lower bound, including the greatest lower bound, does not have to belong to the set it bounds. If a set has a lower bound, then it has the greatest lower bound — this follows from one of the axioms of real numbers.

6. ## Re: Supremums and Infimums

An infimum requires an inequality of x >= a where a is your so called infimum.

If you are dealing with the real numbers and you have an open end-point, then the infimum doesn't exist.

7. ## Re: Supremums and Infimums

Originally Posted by chiro
If you are dealing with the real numbers and you have an open end-point, then the infimum doesn't exist.
Could you write your claim more precisely, using symbols and formulas instead of words?

9. ## Re: Supremums and Infimums

Originally Posted by chiro
The only two claims I see in the top half of p. 30 are:

(1) A finite set always has a maximum element;
(2) [0, 1] has maximum element 1, but [0, 1) has no maximum element.

These claims are about a maximum element, not a supremum (or infimum). They are not the same.

The rest of the top half of p. 30 consists of definitions, not claims.

10. ## Re: Supremums and Infimums

The infimum is a definition.

On Page 30 it gave a definition for the infimum. Do you agree with it or not?

11. ## Re: Supremums and Infimums

You are saying strange things. In post #7 I asked you to state the following claim more precisely.

Originally Posted by chiro
If you are dealing with the real numbers and you have an open end-point, then the infimum doesn't exist.
You also said the following earlier.

Originally Posted by chiro
The reason I ask is that if you have an open set and one endpoint, then if that end point is the lower point it means that the infinum won't actually exist you will never actually have a greatest lower bound.

You can think of it with say (0,infinity) where you can't have zero (since it's not included) but if you pick any value (say 0.0001) then you can always show there is a lower value (say 0.0000001) but you repeat this forever and you never actually get a fixed value.
It seems that you are saying in these two quotes that $\inf\{x\in\mathbb{R}\mid 0 < x\}$ does not exists, which is incorrect. Therefore I asked you to clarify what your claim is, which you have not done. Instead, you referred me to a definition of infimum, which is indeed standard.

12. ## Re: Supremums and Infimums

So in this example where you have a set corresponding to all x, where x > 0, where x is any real number that even though there is no actual greatest lower bound, the infimum of this set is 0. Is this what you are saying?

13. ## Re: Supremums and Infimums

Originally Posted by chiro
So in this example where you have a set corresponding to all x, where x > 0, where x is any real number that even though there is no actual greatest lower bound, the infimum of this set is 0. Is this what you are saying?
The greatest lower bound and the infimum are the same thing, so one cannot exist without the other. For the set {x | 0 < x} both exist and are 0. The minimum of this set, on the other hand, does not exist because the infimum does not belong to the set.

14. ## Re: Supremums and Infimums

Well this is a little silly IMO if the infimum doesn't actually belong to the set (i.e. it's not an element of the set) in general.

It's not you: I've looked at what they consider this so called infimum to be but the idea of having a greatest lower bound where that element doesn't exist seems a misnomer.

I'd see it as more of an actual limit as opposed to an actual fixed bound and it's a little silly, but that's just me.

15. ## Re: Supremums and Infimums

Originally Posted by chiro
It's not you: I've looked at what they consider this so called infimum to be but the idea of having a greatest lower bound where that element doesn't exist seems a misnomer.
This is precisely the difference between the infimum and the minimum: minimum is the infimum that belongs to the set.

The name "greatest lower bound" does not suggest that it should be an element of the set. After all, a lower bound is just a number that bounds a set from below. For example, -5 is a lower bound of (0, ∞). The greatest lower bound is just the greatest of those bounds, i.e., the maximum of the set of lower bounds. It is not obvious that it exists because not every set, even bounded from above, has a maximum. Indeed, in rational numbers the greatest lower bound of the set $(\sqrt{2},\infty})$ does not exist. It is the completeness property of real numbers that guarantees that this maximum of lower bounds exists.

As an answer to the original question, the supremum of $\{1/n-1/m\mid n,m\in\mathbb{N}\}$ is 1 and the infimum is -1.

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