The Extreme Value Theorem For Metric Spaces

Why Compactness Guarantees Extreme Values

A continuous function can preserve the closeness of nearby points without having any value that is largest or smallest. The domain matters: on a compact space, continuity forces the set of output values to have enough structure for both extremes to be attained. The proof has two parts. First, continuity carries compactness from the domain to the image. Second, a nonempty compact subset of the real line contains its supremum and infimum.

Throughout, \((X,d)\) is a metric space with its metric topology, and \(\mathbb{R}\) has its standard topology. A space is compact if every open cover has a finite subcover. A function \(f:X\to\mathbb{R}\) is continuous if the preimage of every open set in \(\mathbb{R}\) is open in \(X\). For a nonempty subset \(K\subseteq\mathbb{R}\), a maximum is a value \(M\in K\) such that \(t\le M\) for every \(t\in K\); a minimum is a value \(m\in K\) such that \(m\le t\) for every \(t\in K\). A supremum or infimum need not belong to the set.

Proof. Let \(\{V_\alpha\}_{\alpha\in J}\) be an open cover of \(f(X)\), where each \(V_\alpha\) is open in the subspace \(f(X)\). By the definition of the subspace topology, for each \(\alpha\) there is an open set \(O_\alpha\subseteq Y\) such that \(V_\alpha=O_\alpha\cap f(X)\). Continuity makes \(f^{-1}(O_\alpha)\) open in \(X\). These preimages cover \(X\): for each \(x\in X\), \(f(x)\in f(X)\), and some \(V_\alpha\) contains \(f(x)\), so \(x\in f^{-1}(O_\alpha)\). Compactness of \(X\) gives finitely many indices \(\alpha_1,\ldots,\alpha_n\) whose preimages cover \(X\). For every \(y\in f(X)\), write \(y=f(x)\) for some \(x\in X\). That \(x\) belongs to one of the selected preimages, so \(y\in V_{\alpha_i}\) for some \(i\). Thus the selected \(V_{\alpha_i}\) cover \(f(X)\), proving it compact. If \(X=\varnothing\), then \(f(X)=\varnothing\), which is compact because every open cover of the empty set has the empty finite subcover. \(\square\)

Compact Subsets of the Real Line Have Extremes

Proof. First we show that \(K\) is bounded. The open intervals \((-n,n)\), for \(n\in\mathbb{N}\), cover \(K\), since every real number lies in one of them. Compactness gives a finite subcover. If \(N\) is the largest index in that finite subcover, then all its intervals lie in \((-N,N)\), so \(K\subseteq(-N,N)\). Since \(K\) is nonempty and bounded above, completeness of \(\mathbb{R}\) gives \(M=\sup K\).

We next show that \(K\) is closed. Fix \(x\in\mathbb{R}\setminus K\). For each \(y\in K\), let \(r_y=|x-y|/3\), which is positive because \(x\ne y\). The intervals \(V_y=(y-r_y,y+r_y)\) cover \(K\). Compactness gives a finite subcover \(V_{y_1},\ldots,V_{y_n}\). Let \(U=\bigcap_{i=1}^n(x-r_{y_i},x+r_{y_i})\), an open neighborhood of \(x\). Each interval \((x-r_{y_i},x+r_{y_i})\) is disjoint from \(V_{y_i}\), since the distance between their centers is \(3r_{y_i}\), while the sum of their radii is \(2r_{y_i}\). If a point of \(U\) belonged to \(K\), it would belong to some \(V_{y_i}\) and also to \((x-r_{y_i},x+r_{y_i})\), which is impossible. Therefore \(U\cap K=\varnothing\). Every point outside \(K\) has an open neighborhood disjoint from \(K\), so \(\mathbb{R}\setminus K\) is open and \(K\) is closed.

For each \(n\in\mathbb{N}\), \(M-1/n\) is not an upper bound for \(K\), because \(M\) is its least upper bound. Thus there is \(k_n\in K\) with \(M-1/n<k_n\le M\). These inequalities imply \(0\le M-k_n<1/n\), so \(k_n\to M\). A closed subset of \(\mathbb{R}\) contains the limit of each convergent sequence of its points: if \(M\notin K\), the open complement would contain an interval around \(M\), contradicting \(k_n\to M\). Hence \(M\in K\), and \(M\) is the maximum of \(K\). Applying the same argument to \(-K=\{-t:t\in K\}\), which is nonempty and compact because \(t\mapsto-t\) is continuous by Theorem 1, shows that \(-K\) has a maximum. Its negative is the minimum of \(K\). \(\square\)

Proof. By Theorem 1, \(f(X)\) is compact in \(\mathbb{R}\). It is nonempty because \(X\) is nonempty. Theorem 2 gives a minimum \(m\in f(X)\) and a maximum \(M\in f(X)\). By the definition of image, there are \(x_{\min},x_{\max}\in X\) with \(f(x_{\min})=m\) and \(f(x_{\max})=M\). Since \(m\le t\le M\) for every \(t\in f(X)\), substituting \(t=f(x)\) proves the required inequalities for every \(x\in X\). \(\square\)

Applying the Theorem in Euclidean Sets

Common Pitfall: A Supremum Is Not Always a Maximum

The Extreme Value Theorem guarantees values actually taken by the function. It does not say that every continuous function on every bounded set attains its supremum and infimum. For example, on \(X=(0,1)\), the function \(f(x)=x\) is continuous and bounded, and its supremum is \(1\) and infimum is \(0\). Neither is a maximum or minimum, because \(1\notin X\) and \(0\notin X\).

A mistaken calculation might say, “\(x<1\) for every \(x\in(0,1)\), so the maximum is \(1\).” The inequality proves only that \(1\) is an upper bound, not that \(1\) is a value of \(f\). Indeed, \(f(x)=1\) would require \(x=1\), which is outside the domain. The domain is bounded but not compact: the open sets \(U_n=(1/n,1)\), for \(n\in\mathbb{N}\), cover \((0,1)\), but no finite selection covers it. If \(N\) is the largest selected index, their union is \((1/N,1)\), which misses points of \((0,1)\). Compactness is the hypothesis that rules out this failure.