I have a problem about a supremum or an infimum

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This page contains advice about dealing with the supremum or infimum of a set. Clicking on answers to questions below will reveal further advice and/or further questions.

General discussion

The first thing to say is that a typical problem involving sups and infs is usually soluble by the kinds of methods described in I have a problem to solve in real analysis and I do not believe that a fundamental idea is needed. That is, if you understand properly what a supremum or an infimum is, then your problem is likely to be fairly easy.

If you do not understand what a supremum or infimum is, what can you do about it? One thing is to use some simple translation rules. If $A$ is a set, then the statement $x=\sup A$ may conjure up some confused mental picture such as "the biggest element of $A$ except that $A$ may be slightly fuzzy and not actually contain its maximum". That is confused because if $A$ has a maximum then obviously it contains its maximum, and if it does not have a maximum then you can't say that it doesn't contain its maximum as there isn't one. It's fine to have a picture of the supremum of $A$ as something like "the element that would be the maximum if it had one," but for the purposes of writing out proofs one needs to abandon this picture and translate statements that involve "sup" or "inf" into statements that do not involve them. The one complication is that you need more than one statement to do the job, basically because the phrase "least upper bound" contains the two conditions "upper bound" and "least". But that isn't a huge problem. Here is the basic translation rule.

The statement $x=\sup A$ is equivalent to the following pair of statements.

Every element of $A$ is less than or equal to $x.$
For every $y<x$ there exists an element of $A$ that is greater than $y.$

To see how this can be used, let us see how to prove a result that gives us another useful translation of the statement $x=\sup A.$

Theorem 1 Let $A$ be a non-empty bounded set of real numbers and let $x=\sup A.$ Then there is a sequence $(a_n)$ of elements of $a$ such that $a_n\rightarrow x.$

Proof. In order to find such a sequence, we just build it up term by term. In order to ensure that $a_n\rightarrow x$ we attempt to ensure that $a_n$ gets closer and closer to $x$ : an obvious way of doing this is to insist that $|a_n-x|\leq 1/n.$ (This particular choice is not essential: we could just as well ask for $|a_n-x|\leq 1/(n^2+13),$ but we have chosen to use the simplest sequence that tends to $0$ .) Can we find $a_n$ with this property? To answer this we use the translation above. The second statement implies that there exists $a\in A$ such that $a\geq x-1/n,$ and the first statement implies that this $a$ is at most $x.$ Therefore, we can set $a_n$ to be this $a$ and we know that $|a_n-x|\leq 1/n,$ as required.□

Once we have this result, we can give a different translation of $x=\sup A.$

The statement $x=\sup A$ is equivalent to the following pair of statements.

Every element of $A$ is less than or equal to $x.$
There is a sequence $(a_n)$ of elements of $A$ such that $a_n\rightarrow x.$

Let us see exactly why the two are equivalent, as this gives a further illustration about how to reason with the notion of a supremum. We have shown that the first translation (which is basically the definition of the supremum) implies the second. Suppose now that we know that the two sentences in the second translation are true. To deduce that $x=\sup A,$ all we need to do is deduce that for every $y<x$ there exists $a\in A$ such that $a>y.$ We are given a sequence $(a_n)$ such that $a_n\rightarrow x,$ from which it follows that for every $\epsilon>0$ there exists $n$ such that $a_n>x-\epsilon.$ Choosing $\epsilon$ to be $x-y,$ we find that there exists $n$ such that $a_n>y.$ Since we know that $a_n\in A,$ we are done.