Home Math Teaching About Me

2. The \(\sup\) and \(\inf\)

≪ 1. On Writing Proofs | Table of Contents

During lecture, you have seen a progression of number systems throughout the centuries, starting with the natural numbers \(\mathbb{N}\), to the integers \(\mathbb{Z}\), to the rationals \(\mathbb{Q}\), and finally to the real numbers \(\mathbb{R}\). The key feature of the real numbers that allows for real analysis to actually happen is the completeness axiom — the fact that least upper bounds and greatest lower bounds exist and are real numbers.

While we have defined both \(\sup\) and \(\inf\) in lecture, today we’ll be focusing on the \(\sup\). Almost everything you can say about the \(\sup\) can also bo said about the \(\inf\), perhaps with some inequalities flipped around. If we have time though, we’ll describe how to reformulate everything we’ve said about the \(\sup\) for the \(\inf\) as well.

Recall that the supremum or \(\sup\) of a nonempty set of real numbers is its “least upper bound”. More precisely,

Something to note about the supremum of a set \(S\) is that it’s unique, i.e. a set \(S\) cannot have more than one supremum! More specifically, if \(M_1\) and \(M_2\) both satisfy the property in the definition above, then \(M_1 \leq M_2\) is an upper bound) and also \(M_2 \leq M_1\)! Thus \(M_1 = M_2\). So saying the supremum makes sense (whereas “the” upper bound of a set wouldn’t make much sense).

We can often intuit the \(\sup\) of a set of real numbers, but proving that something satisfies the rather bulky definition above can be somewhat cumbersome. So, this theorem may be helpful by providing an alternative but still equivalent definition of the \(\sup\) of a set of numbers:

What this is essentially saying is that the real numbers can be split into two pieces: the right piece is the set of all numbers that are upper bounds of \(S\), and the left piece is the set of all real numbers that aren’t. The supremum is then the point at which the two pieces meet. Draw a picture!

Proof

First, we need to prove the “forward direction”: if \(M = \sup S\), then for all \(\epsilon > 0\), there is an element \(x\in S\) such that \(x > M - \epsilon \).

We’ll prove the contrapositive instead. The negation of the conclusion is: “there exists an \(\epsilon > 0\) such that there is no \(x\in S\) such that \(x > M - \epsilon \)”. This means that every \(x\in S\) satisfie \(x \leq M - \epsilon \). But then \(M - \epsilon \) is an upper bound for \(S\), and clearly \( M - \epsilon < M\)! Thus \(M\) cannot be the supremum by definition.

Next, we need to prove the “reverse direction”: if for all \(\epsilon > 0\), there is an element \(x\in S\) such that \(x > M - \epsilon \), then \(M = \sup S\).

For this, let \(M’\) be any upper bound \(S\). For any \(\epsilon > 0\), there is an \(x\in S\) such that \(x > M - \epsilon \). But \(M’ \geq x\) since it’s an upper bound, so \(M’ > M - \epsilon \). This is true for any \(\epsilon > 0\), so \(M’ \geq M\). This is true for any upper bound \(M’\), hence \(M = \sup S\) by definition. \(\square \)

It’s important to keep this in mind because sometimes, we can intuit the supremum of a set of real numbers, but proving it straight from the definitions can be kind of hard. Instead, one often ends up using some version of the above alternative definition to rigorously justify a supremum.

Solution

One can perhaps intuit that \(1\) should be the least upper bound. Clearly \(1\) is an upper bound.

For any \(\epsilon > 0\), there exists some \(n \in \mathbb{N} \) such that \(\frac{1}{n} < \epsilon \), hence \(1 - \frac{1}{n} > 1- \epsilon \). Thus, \(1 - \epsilon \) is not an upper bound for any \(\epsilon > 0\)! It follows from the preceeding proposition that \(1\) is the least upper bound for \(S\). \(\square\)

I don’t think that there is a proof straight from our original definition. If \(M’\) is an upper bound of \(S\), if one were to show that \(M’ \geq 1\), one would most likely proceed by showing that \(M’\) is not less than \(1\) by the above argument!

Solution

First, we claim that \(\sup \left( S_1 \right) + \sup \left( S_2 \right)\) is an upper bound of \(S_1 + S_2\). This is because if \(x_1 \in S_1\) and \(x_2 \in S_2\), we have \(x_1\leq \sup \left( S_1 \right)\) and \(x_2 \leq \sup \left( S_2 \right)\), hence \[x_1 + x_2 \leq \sup \left( S_1 \right) + \sup \left( S_2 \right).\]

Next, we’ll show that for all \(\epsilon > 0\), the quantity \(\sup \left( S_1 \right) + \sup \left( S_2 \right) - \epsilon \) is not an upper bound for \(S_1 + S_2\). By the earler proposition, this will show that indeed \(\sup \left( S_1 \right) + \sup \left( S_2 \right) = \sup \left( S_1 + S_2 \right)\).

Let \(\epsilon > 0\). Then, by the proposition from before, there exist elements \(x_1 \in S_1\) and \(x_2\in S_2\) such that \[ \begin{align*} x_1 > \sup \left( S_1 \right) - \frac{\epsilon }{2} && \textrm{and} && x_2 > \sup \left( S_2 \right) - \frac{\epsilon }{2}. \end{align*} \] Together, this means that \[x_1 + x_2 > \sup \left( S_1 \right) + \sup \left( S_2 \right) - \epsilon, \] concluding the proof. \(\square \)

Hint

Try recreating the proof from the preceeding exercise. Try \(f_1(x) = \cos x\) and \(f_2(x) = \sin x\) for the strict inequality.

Just to be safe, let’s describe the infimum a bit. You can imagine that the infimum is the mirror image of the supremum — whereas the latter is the least upper bound, the infimum of a (bounded) set is the greatest lower bound. More specifically,

Really, if you hold up the entire real number line to a mirror, (\sup)’s become (\inf)’s and (\leq )’s become (\geq)’s (and vice-versa). So, it’s hopefully unsurprising to see that our proposition from before becomes:

I’ll leave the proof as an exercise. You should probably try to write and prove analogues of the previous exercises using \(\inf\)’s instead of \(\sup\)’s!