# What is 'inf' & 'sup' ?

1. Mar 3, 2004

### Jules

I'm trying to understand a text, 'Dynamics on Time Scales'.
Some of you may be familiar with this subject.
Unfortunately, the first page has already confused me. So it appears that I'm either a math retard or it's been so long since my university courses that new things have arisen that I'm not aware of.
I'm going to be optimistic and choose the latter option.
So, have the following new functions (like cos, sin, etc.) been defined during my absence from the field:

'inf' and 'sup' ? ,
and if so, what the hell do they mean ?

or am I so mathematically inept that I cannot grasp the basic definitions required to discuss the topic?

2. Mar 3, 2004

### NateTG

Inf and sup stand for infimum and supremum respectively. They're typically used on sets of real numbers.

http://mathworld.wolfram.com/Infimum.html
http://mathworld.wolfram.com/Supremum.html

The infimum or greatest lower bound of a set $$S$$ is the largest number $$x$$ so that $$s \in S$$ implies that $$s \geq x$$.

Similarly the supremum is the least upper bound of the set.

3. Mar 3, 2004

### suyver

Just a simple example: take the open interval A=[0,1). What is the largest number in A? That doesn't exist, so max(A) is undefined. That's why the supremum is so useful, the least upper bound is well defined: sup(A)=1.

In general (I believe), for any finite-valued interval X, sup(X) always exists but max(X) does not always exist.

Similar arguments for min(X) and inf(X).

4. Mar 3, 2004

### Jules

Oh dear, more questions...

NateTG & suyver,
Thank you, both for the Wolfram link which I'm finding quite useful, and the example you provided.

Although I'm considerably demystified there is still some confusion for which I hope you have patience:

The open interval A=[0,1) is all numbers between 0 and 1, including 0 but excluding 1, no ?
The maximum number in A will be 0.999999999... , a number which cannot be fixed since you can always add a positive quantity q such that 0.999999999...+ q > 0.999999999... but still less than 1. And this makes it undefined ?
Is sup A therefore just the limit of A as it tends to its upper bound ?
Is inf A = 0 ?

[I'm now thinking that not only has time made me rusty but the university math courses I did were for an engineering degree and therefore didn't examine things such as inf & sup & intervals etc.]

5. Mar 3, 2004

### matt grime

.9999999... is equal to 1, a well defined number. to see that just realize that .999999.... is the sum of a geometric progression, and if you don't like that one, then it is 3 times .333333..... which everyone calls a third, so three times one third is 1.

A has no maximum element.

6. Mar 3, 2004

### Integral

Staff Emeritus
This is where you are wrong. .999... + 10-N where N is ANY integer is > 1, You cannot find a real number q such as you have claimed.

Read this it has 2 proofs, the first is simply using the known forumula for the sum of a geometric series. The second elaborates on my above comment.

Last edited: Mar 3, 2004
7. Mar 3, 2004

### NateTG

Re: Oh dear, more questions...

Consider, for example that 100, &pi; and 1 are all upper bounds for the set $$(0,1)$$, so it makes little sense to talk about a set having a unique upper bound. I expect that you mean 'least upper bound' when you write 'its upper bound' in the statement, since the statement implies a unique upper bound for A.

Moreover, talking about the limit of A (a set) as A approaches some bound doesn't make sense because:
1. It doesn't really make sense for the set A to approach a number.
2. Since A is invariant in the limit, it should still be A, and not some numerical result.

Regarding the 'largest' number in $$(0,1)$$. If you give me any number $$x \in (0,1)$$, then $$\frac{x+1}{2} > x$$ and $$\frac{x+1}{2} \in (0,1)$$, so $$x$$ cannot be the largest number in $$(0,1)$$.

That said, if A has a largest number &alpha;, then &alpha; is the lub (supremum) of A.

Last edited: Mar 3, 2004
8. Mar 3, 2004

### curiousbystander

Since all the technical aspects have been aptly explained, here's a nontechnical way to think about sup and inf that seemed to help a few of my students (please take it with a grain of salt-- it's only to help with the concept so you can allow yourself to accept the actual technical details):

When the world is fair and just, and your set A has a maximum, then the sup of A is precisely the maximum of A, but sometimes fate is not that kind.

In a cruel and capricious world where A does not have a maximum, the sup of A ([size=small]legaleese about boundedness omitted[/size]) is the number that wants to be the maximum-- the number that WOULD be the maximum if some cruel twist of fate hadn't seen fit to keep it from being in A-- like the 1 in (0,1).

Similarly, the inf of a set ([size=small]same legaleese[/size]) is the minimum of that set, unless you're back in that heartless world where the set doesn't have a minimum. Then the inf is the number that WOULD be the minimum if it could.

9. Mar 4, 2004

### Jules

Sorry people, when I wrote 0.9999.. I meant a finite number of 9s and then zeros, not 9s infinitely recurring. Poorly specified it was. But I think I get the point about the existence of a maximum.

Now when NateTG says 100, n and 1 are all upper bounds for the set (0,1) I'm confused once again, for does not (0,1) imply all numbers between 0 and 1, excluding 0 and 1, thereby not including 100 ?

Thanks curiousbystander, the world needs more teachers like you.

Okay, just to check,
if A = Z (all integers), then
inf A = -infinity
& sup A = +infinity ?
Are the max & min considered unique ?

and say, A = N (natural numbers) intersection [1,2) then,
inf A = min A = 0 (assuming N contains zero)
& sup A = 2, max A non existant ?

(Sorry, haven't yet figured out how to include the proper symbolic fonts)

10. Mar 4, 2004

### suyver

Yes, the set A=(0,1) contains all numbers between 0 and 1, excluding 0 and 1, thereby not including 100. However, he did not state that 100 is an element of A. He just said that 100 is an upper bound for this set A. It's just not a very precise upper bound in the sense that we can think of a better (lower valued) one: 1. This upper bound is the best we can do, and it's unique. Therefore, we call this upper bound the supremum of the set: sup(A)=1.

In general: max (if it exists!) and sup (if it exists!) are unique and they are also well-defined. If for some set X, x=max(X) then also x=sup(X). However, the other way around does not work, since max(X) is not always defined when sup(X) is.

Assuming you ment A=[0,2) (typo?) then this seems correct. You got it!

11. Mar 4, 2004

### Jules

"100 is an upper bound for this set A"
How can it be an upper bound if it is not even in the set ?

Yes there was sort of a typo, I actually meant

A = [0] $$\cap$$ [1,2)

12. Mar 4, 2004

### suyver

The upper bound is simply a number that is at least as large as the largest number in the set.

Consider again my set A=(0,1). We already determined that max(A) does not exist and that sup(A)=1.
What is an upper bound for A? Any number >=1 is an upper bound for A! This is because any number >=1 is at least as big as the largest number in the set. So, 100 is an upper bound for the set A. The number 1 is the smallest upper bound imaginable. That's why it's the supremum. I can't explain it simpler than that, but maybe someone else can?

Maybe an example outside mathematics will help: Suppose you have $6.83 in your wallet. I'm sure you agree that you then have less than$100 in your wallet. So, $100 is an upper bound for the amount of money in your wallet. Also$10 is an upper bound. Since $10 is less than$100, we say that $10 is a better upper bound.$6.84 is also an upper bound and it is an even better one.

Hope this helps.

13. Mar 4, 2004

### matt grime

Definition

if S is a subset of R, then U is an upper bound if, for all s in S, s<=U.

where does one require U to be in the set S?

Definition.

Sup(S) is an upper bound, U, with the further property that for all e>0, there is an s in S with U-e <x<=U

again no requirement that U actually be in S.

1 is the sup of[0,1) since it is AN unpper bound and, for all e>0, there exists an n with 1/n <e, hence

1-e<1-1/n<1, and 1-1/n is in [0,1)

14. Mar 4, 2004

### suyver

Don't you mean: ... there is an s in S with U-e <s<=U ?

15. Mar 4, 2004

### matt grime

that would certainly make more sense - slight user error between brain and keyboard. again.

16. Mar 5, 2004

### Jules

Aah thanks for the help gentlemen (hmm I'm assuming you're all male...).

I was not previously aware of the exact definition of a 'bound'.
I will return to my book this weekend & hopefully make some progress past the first page.