# Is my understanding of the Supremum correct?

I wanted to know if someone could please tell me if my understanding of the Supremum is correct?

My Understanding

In essence, the "Supremum" is the least upper bound of a subset that is within a partially ordered or ordered set E.g. $\mathbb{R}$. Therefore

$$\sup\{1,2,3\}$$

The least upper bound of this set or the $\sup$ would be $3$. If I had a more complicated example:

$$\sup\{(-1)^n-\frac{1}{n} : n \in \mathbb{N^*}\}$$

Then the $\sup$ would be $1$ because $\lim_{n \rightarrow \infty} (-1)^n = 1$ and $\lim_{n \rightarrow \infty^+} \frac{1}{n} = 0$. Therefore $1-0$ = $1$. Since $\sup$ means least upper bound, can I say for $A = \{1,2,3\}$, that the upper bound is $4$, or $5$ or any $n$? And if not then why does this wikipedia say that for $A$ the upper bound can be $4$?

Also, is finding the $\sup$, in essence, finding the $\lim$ of that "function"? And if there is more than one function in the set then finding the $\lim$'s separately and doing the operation between them?

Let $f(x)$ be some function and let $g(x)$ be another function.

E.g. $A=\{f(x) + g(x) : x \in \mathbb{R}\}$ Then the $\sup$ would be $\sup \{\lim_{x \to \infty} f(x) + \lim_{x \to \infty} g(x)\}$

Thanks a lot for your time!

EDIT:

I just realized that $\lim_{n \to \infty} (-1)^n \not= 1$ and it is indefinite, so I guess that throws off the $\lim$ argument for calculating $\sup$ or does it?

EDIT 2:

It seems as mentioned by one of the answered, that I have gone way ahead of myself by using limits, in the wrong context, to find the $\sup$ but then if that is not the case then how is it possible to know the $\sup$ of a set of functions like the second example?

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no limit as n goes to infinity (-1)^n is not 1... – Lost1 Apr 12 '13 at 10:51
@Lost1 Yes, you are correct, I just checked that it was indefinite. I guess that throws off my $\lim$ argument for find $\sup$ or does it? – Jeel Shah Apr 12 '13 at 10:54
it still has a sup or inf, what is it? – Lost1 Apr 12 '13 at 10:56
@Lost1 The $\sup$ is $1$ but I can't verify that because I got the answer from wiki. I thought I verified it by the $\lim$ argument. – Jeel Shah Apr 12 '13 at 10:58
limit has nothing to do with sup and inf... – Lost1 Apr 12 '13 at 10:58

## 4 Answers

The supreme of a set need not belong to the set! I think this is your point of doubt. The supreme of a set is a number that is associated to the set by setting the supreme.

Another thing you must understand precisely what is a real number $c$ is an upper bounds in a set if $A$ is greater than or equal to all the numbers $a$ living in set $A$.

So we have to talk in set of upper bounds of a set. We fix $$\mathcal{U}(A)=\{c\in\mathbb{R} : c\geq a,\;\forall a\in A\} \quad \mbox{ the set of upper bounds } \\ s=\sup A \Leftrightarrow s\leq c,\;\forall c\in \mathcal{U}(A) \quad \mbox{ supreme as the smallest of all upper bounds }$$

It can happen that one of the upper bounds in all living $A$ as is the case when the number 3 $A = \{1,2,3 \}$. On the other hand if $A =\{(-1)^n-\frac{1}{n} : n \in \mathbb{N^*}\}$ not have any upper bound of the set $A$ living in the set $A$.

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You said yourself that $\sup\{1,2,3\}=3$, and Wikipedia says that $4$ is an upper bound of $\{1,2,3\}$, but not the least upper bound!

Note that $\lim_{n\to\infty} a_n$ need not exist. Moreover, $\sup a_n$ is a bound for all terms, not just for essentially all late terms (the $\lim$ does not depend on early terms). Thus with $$A=\left\{ \frac{(-1)^n}n\Biggm| n\in\mathbb N\right\},$$ we have $\sup A=\frac12$ and $\inf A=-1$, whereas $\lim_{n\to\infty}\frac{(-1)^n}n=0$. There is also the concept of $\limsup$ and $\liminf$ (and they have the property that tehy do coincide with the limit if it exists), but that's a different story and you should grasp $\sup$ and $\inf$ before considering these.

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Okay, so I guess I way ahead of myself with the limit thing and used it in an totally different way then it was meant to be used but how do you know that $\sup A = \frac{1}{2}$? – Jeel Shah Apr 12 '13 at 11:02
Let n=1 and you get infA. Let n=2 and you get supA. All other possible values are in between -1 and 1/2. – Haikal Yeo Apr 13 '13 at 5:57

the set is

{-1-1, 1-1/2, -1-1/3, 1-1/4, ...} = {-2,1/2,-4/3,3/4,...}

So this appears that the smallest value is -2. If the smallest value exist, then it must be the inf )similiarly if the max value exists, it must be the sup)

sup f(x) + g(x) = sup f(x) + sup g(x)

sup and inf are in a sense near 'max' and near 'min' of a set.

max = sup if a maximum element exist, if not, it is the smallest number which beats every element in the set.

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The supremum of a subset of $\mathbb R$ (other posets are possible, but it seems from your question you are interested in $\mathbb R$) is, by definition, the smallest among all upper bounds of the given subset. If the subset has one upper bound, say $M$, then it has infinitely many upper bounds (e.g., $M+r$, $r>0$). Of all the upper bounds of a given subset, the smallest one (which exists since $\mathbb R$ is complete) is called the least upper bound.

So, a subset doesn't a 'the upper bound', but rather 'an upper bound', and if it has one, then it has many. The set $\{1,2,3\}$ certainly has $4$ as upper bound, but also $5$ and also $\pi$ as upper bounds.

Finding the supremum of a set is not (essentially or non-essentially) the same as finding the limit of anything. The supremum is the least upper bound. A limit is something else. A limit of a sequence depends on the order in which the elements of the sequence appear and is insensitive to any amount of finite changes in the sequence. However, the supremum of a set depends, potentially, on all elements in the set and can change drastically by changing the value of just one element in the set.

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