$$ \def\R{\mathbb R} \def\N{\mathbb N} \def\Q{\mathbb Q} \def\Z{\mathbb Z} \newcommand{\st}{~:~} \newcommand{\cee}{\mathsf{C}} \newcommand{\bee}{\mathrm{B}} \def\mathhlyellow#1{\bbox[##fff3bf,2px,border:1px solid ##b7791f]{#1}} \def\mathhlgold#1{\bbox[##ffe8a3,2px,border:1px solid ##b7791f]{#1}} \def\mathhlorange#1{\bbox[##ffddb3,2px,border:1px solid ##c05621]{#1}} \def\mathhlgreen#1{\bbox[##d9f7d7,2px,border:1px solid ##3f8f46]{#1}} \def\mathhlmint#1{\bbox[##d4f5e9,2px,border:1px solid ##2c7a7b]{#1}} \def\mathhlblue#1{\bbox[##dceeff,2px,border:1px solid ##2b6cb0]{#1}} \def\mathhlpurple#1{\bbox[##eadffd,2px,border:1px solid ##6b46c1]{#1}} \def\mathhlpink#1{\bbox[##ffdbe8,2px,border:1px solid ##b83280]{#1}} \def\mathhlred#1{\bbox[##ffd6d6,2px,border:1px solid ##c53030]{#1}} \def\mathhlgray#1{\bbox[##e8ecef,2px,border:1px solid ##64748b]{#1}} \def\mathhlgrey#1{\bbox[##e8ecef,2px,border:1px solid ##64748b]{#1}} $$

11  The Min-Max Theorem

Highlights

  • Special properties of how continuous functions interact with closed and bounded subsets of \(\R.\)

11.1 Preparation

The goal of this section is to prove the Min-Max Theorem, which says that for any nonempty, closed, and bounded subset \(C\subseteq\mathbb R\) and any continuous function \(f:C\rightarrow\mathbb R,\) \(f\) achieves a global max and a global min on \(C.\) That is, there are points \(a,b\in C,\) called the global min and global max respectively, such that for all \(x\in C,\) \[f(a)\leq f(x)\leq f(b).\]

Exercise 11.1 Show by example that both closedness and boundedness are necessary conditions for guaranteeing a global max and global min. That is:

  1. Find a closed subset \(C\) of \(\mathbb R\) (which is not bounded) and a continuous function \(f:C\rightarrow\mathbb R\) so that \(f\) has no global max and/or no global min on \(C.\)
  2. Find a bounded set \(C\) of \(\mathbb R\) (which is not closed) and a continuous function \(f:C\rightarrow\mathbb R\) so that \(f\) has no global max and/or no global min on \(C.\)

As we’ll see, closed and bounded subsets of \(\mathbb R\) play an important role, so here are a couple more theorems about such subsets.

Theorem 11.2 Let \(C\) be a closed and bounded subset of \(\mathbb R,\) and let \((a_n)\) be a sequence in \(C.\) Then \((a_n)\) has a subsequence that converges to a point in \(C.\)

Exercise 11.3 Prove Theorem 11.2. Note that you not only need to show that \((a_n)\) has a convergent subsequence but also that the limit of that convergent subsequence is in \(C.\)

11.2 Closed and Bounded Subsets of \(\R\)

The main theoretical result of this section will be Theorem 11.4, which tells us that if \(C\subseteq\R\) is closed and bounded and \(f:C\rightarrow\R\) is continuous, then the image \(f(C)\) is also closed and bounded. As you saw in Exercise 11.3, Bolzano-Weierstrass will serve as the foundation for much of what we do, so we should expect results that hold in \(\R\) but not in \(\Q.\)

Theorem 11.4 Let \(C\) be a closed and bounded subset of \(\mathbb R,\) and suppose \(f:C\rightarrow\mathbb R\) is continuous. Then the image \(f(C)\) is also closed and bounded.

Before we see the proof, it’s worth noting that closed and bounded work together here. It’s not true that the continuous image of any closed set is closed, nor that the continuous image of any bounded set is bounded, as the exercise below shows.

Exercise 11.5 The following examples show that Theorem 11.4 can fail if either condition on \(C\) (closed or bounded) is violated.

  1. Let \(C=[1,\infty)\) and \(f:C\rightarrow\mathbb R\) be defined by \(f(x)=\arctan(x).\) What is \(f(C)?\) Explain how this example shows that the image of a closed subset of \(\mathbb R\) need not be closed.
  2. Let \(C=(0,1]\) and \(f:C\rightarrow\mathbb R\) be defined by \(f(x)=\frac{1}{x}.\) What is \(f(C)?\) Explain how this example shows that the image of a bounded subset of \(\mathbb R\) need not be bounded.

Proof. Proof of Theorem 11.4.

Suppose \(C\) is closed and bounded and that \(f:C\rightarrow \R\) is continuous. First we will prove that \(f(C)\) is closed by showing that the limit of every convergent sequence in \(f(C)\) is in \(f(C).\) So suppose \((y_n)\) is a convergent sequence in \(f(C).\) Let’s call its limit \(b.\) Since \(y_n\in f(C),\) there exists \(x_n\in C\) such that \(f(x_n)=y_n\) for every \(n\in\N.\)1 Then by Theorem 11.2, \((x_n)\) has a subsequence \((x_{i_n})\) that converges to a point in \(C;\) we’ll call that point \(a.\) Since \(f\) is continuous and \(x_{i_n}\rightarrow a,\) we have \(f(x_{i_n})\rightarrow f(a);\) that is, \(y_{i_n}\rightarrow f(a).\) But since \(y_n\rightarrow b,\) we know \(y_{i_n}\rightarrow b,\) so by the uniqueness of sequence limits, we have \(b=f(a).\) Since \(a\in C,\) this proves \(b\in f(C)\) as desired.

Now by way of contradiction, suppose \(f(C)\) is not bounded. Then there exists a sequence \((y_n)\) in \(f(C)\) with no convergent subsequence. As before, since \(y_n\in f(C),\) there exists \(x_n\in C\) such that \(f(x_n)=y_n\) for all \(n\in\N.\) Again by Theorem 11.2, \((x_n)\) has a subsequence \((x_{i_n})\) that converges to some point \(a\) in \(C.\) Since \(f\) is continuous and \(x_{i_n}\rightarrow a,\) we have \(y_{i_n}\rightarrow f(a),\) contradicting the assumption that \((y_n)\) has no convergent subsequence.

Exercise 11.6 This exercise is meant to help you see what’s going on in the first part of the proof of Theorem 11.4.

Let \(C=[-2,2]\) and define \(f:C\rightarrow\R\) by \(f(x)=x^2.\) Let \((y_n)\) be the sequence \(3,3.9,3.99,3.999\ldots.\)

  1. What is \(f(C)\) in this case? Is \((y_n)\) in \(f(C)?\)
  2. Find a sequence \((x_n)\) in \(C\) so that (a) \(f(x_n)=y_n,\) and (b) \((x_n)\) does not converge. (Hint: you’ll need to take advantage of the fact that \(f\) is not injective.)
  3. Does the sequence \((x_n)\) you chose have a convergent subsequence? If so, what does it converge to?

Lemma 11.7 Let \(C\subseteq\R\) be non-empty, closed, and bounded. Then \(\sup C\in C\) and \(\inf C\in C.\)

The proof is part of Exercise 11.9 below.

Theorem 11.8 (Min-Max Theorem) Let \(C\) be a non-empty, closed, and bounded subset of \(\R\) and suppose \(f:C\rightarrow \R\) is continuous. Then there exist points \(a,b\in C,\) which we call the global min and global max respectively, such that for all \(x\in C\) \[f(a)\leq f(x)\leq f(b).\]

Exercise 11.9  

  1. Prove Lemma 11.7.
  2. Use Lemma 11.7 to help you prove Theorem 11.8.

We conclude with another consequence of Theorem 11.2 that we’ll need for later on in the course.

Theorem 11.10 Let \(C\subseteq\R\) be non-empty, closed, and bounded and let \(f:C\rightarrow\R\) be a function. Then \(f\) is uniformly continuous if and only if \(f\) is continuous.

Proof. We already know that uniform continuity implies continuity, so we just need to prove the backwards implication.

By way of contradiction, suppose \(f:C\rightarrow\R\) is continuous but not uniformly continuous. Then there exists \(\epsilon>0\) and sequences \((x_n)\) and \((y_n)\) in \(C\) such that \(|x_n-y_n|\rightarrow0\) and yet for all \(n\in\N,\)

\[ |f(x_n)-f(y_n)|\geq\epsilon . \tag{11.1}\]

Then by Theorem 11.2, the sequence \((x_n)\) has a subsequence \((x_{i_n})\) that converges to some point \(a\in C.\) Now consider the corresponding subsequence \((y_{i_n})\) of \((y_n).\) Since \(|x_n-y_n|\rightarrow0,\) the subsequence \(|x_{i_n}-y_{i_n}|\rightarrow0\) and therefore \(x_{i_n}-y_{i_n}\rightarrow0.\) Then \(y_{i_n}=x_{i_n}-(x_{i_n}-y_{i_n}),\) so \(y_{i_n}\rightarrow a-0=a.\)

So since \(f\) is continuous and \(x_{i_n}\rightarrow a,\) we know \(f(x_{i_n})\rightarrow f(a).\) Similarly, since \(y_{i_n}\rightarrow a,\) we know \(f(y_{i_n})\rightarrow f(a).\) Therefore \(f(x_{i_n})-f(y_{i_n})\rightarrow0,\) so there exists \(M\in\N\) such that for all \(m>M,\) \(|f(x_{i_m})-f(y_{i_m})|<\epsilon,\) contradicting Equation 11.1.

Exercise 11.11 We proved earlier that if \(C\subset\R\) is nonempty, closed, and bounded, and \(f:C\rightarrow \R\) is continuous and injective, then \(f^{-1}:f(C)\rightarrow C\) is continuous. Give a new proof that uses Theorem 11.4 (and what we know about inverse images of open sets under continuous maps).

11.3 Review Questions

  1. Give examples to show that the conditions of boundedness and closedness “work together” in Theorem 11.4 and Theorem 11.8 by showing that neither property on its own is sufficient to imply the conclusions of the theorems. Include pictures.
  2. Note that many of the theorems above are marked with maroon bars, indicating that they rely (one way or another) on the Least Upper Bound Axiom. Give examples to show that these would be false if our universe were \(\Q\) instead of \(\R.\)
  3. Define a subset \(U=\{1,\frac{1}{2},\frac{1}{3},\ldots\}\cup\{0\}.\) Suppose \(f:U\rightarrow\R\) is continuous. Prove that there exists \(c\in U\) such that \(f(c)\geq f(x)\) for all \(x\in U.\)
  4. Since we are seeing the previous problem in this section, it’s clear what tools to use. But out of context, it would be harder. Write another example of a problem that tests this content that would be tricky if you saw it out of context.

  1. We aren’t assuming \(f\) is one-to-one, so \(x_n\) need not be unique.↩︎