4.2 The Intermediate Value Theorem
In this section we will prove the important Intermediate value theorem. The following Lemma will be
required.
Lemma 4.2.1. Let \(f:X\rightarrow Y\) be a function between topological spaces. If \(X\) is connected, then \(f(X)\) is connected
Proof. Let \(A=f(X)\), so that \(f:X\rightarrow A\). If \(A\) is not connected, there exists a non-empty proper subset \(B\) of \(A\) such that
\(B\) is both open and closed in \(A\). Since \(f\) is continuous from \(X\) to \(A\), and \(A=f(x)\subseteq Y\) means \(f^{-1}(B)\) is a non-empty proper
subset of \(X\) that is both open and closed in \(X\). This contradicts the fact that \(X\) is connected. Hence \(A\)
is connected.
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Proof.
An interval is connected
Suppose an interval \(I\) were disconnected, so \(I = A\cup B\) with \(A,B\) non-empty, open in \(I\) and disjoint. Pick \(a\in A\) and \(b\in B\) with \(a<b\); since \(I\) is an interval, \([a,b]\subseteq I\). Let \[c = \sup \left \{x\in [a,b] : x\in A\right \}.\] If \(c\in A\) then, \(A\) being open, some interval \((c-\delta ,c+\delta )\) lies in \(A\), and as \(c<b\) this puts points above \(c\) in \(A\), contradicting the supremum. If \(c\in B\) then, \(B\) being open, some \((c-\delta ,c+\delta )\) lies in \(B\), so no point of \(A\) exceeds \(c-\delta \), again contradicting the supremum. As \(c\) lies in \(A\cup B\), both cases are impossible and \(I\) is connected.
A connected set is an interval
If \(S\subseteq \mathbb {R}\) is not an interval there are \(a<c<b\) with \(a,b\in S\) and \(c\notin S\). Then \[S = \left (S\cap (-\infty ,c)\right )\ \cup \ \left (S\cap (c,\infty )\right )\] splits \(S\) into two non-empty disjoint relatively open sets, so \(S\) is disconnected. □
Theorem 4.2.3 (Intermediate Value Theorem). Let \(f\) be a continuous real-valued function on the closed bounded interval \([a,b]\). Then \(f\) takes every value between \(f(a)\) and \(f(b)\): for any \(C\) strictly between \(f(a)\) and \(f(b)\) there exists \(c\in (a,b)\) with \(f(c)=C\).
Proof. Suppose \(f(a)<C<f(b)\); the other case follows by applying this one to \(-f\). The interval \([a,b]\) is connected, and the continuous image of a connected set is connected, so \(f\left ([a,b]\right )\) is a connected subset of \(\mathbb {R}\). By the corollary above, a connected subset of \(\mathbb {R}\) is an interval. Since \(f(a)\) and \(f(b)\) both lie in \(f\left ([a,b]\right )\) and an interval contains every point between any two of its points, \(C \in f\left ([a,b]\right )\). So \(C=f(c)\) for some \(c\in [a,b]\), and \(c\) is neither \(a\) nor \(b\) because \(f(a)\neq C \neq f(b)\). □
Remark. The theorem is usually proved in a first analysis course by bisection, which requires the completeness of \(\mathbb {R}\) directly. The proof above uses no estimates at all: it needs only that \([a,b]\) is connected, that continuity preserves connectedness, and that connected subsets of \(\mathbb {R}\) are intervals. That is the case for doing topology — a theorem about real numbers turns out to be a theorem about connectedness, and the same argument proves the corresponding statement on any connected domain.
The following result, known as the fixed point theorem, follows, from the intermediate value
theorem.
Corollary 4.2.4 (Fixed Point theorem). If \(f:[a,b]\rightarrow [a,b]\) is a continuous function, then there exists a point
\(x_0\in [a,b]\) such that \(f(x_0)=x_0\).
Proof. Let \(f\) be continuous on \([a,b]\) and \(f(x)\in [a,b]\) for every \(x\in [a,b]\).
If \(f(a)=a\) or \(f(b)=b\), then f has a finite point. Suppose \(f(a)>a\) and \(f(b)<b\). Define \(g(x)=f(x)-X\) for every \(x\in [a,b]\). By definition, g is continuous
on \([a,b]\). Also, \(g(a)>0\) and \(g(b)<0\). So 0 is an intermediate value for \(g\) on \([a,b]\). Hence by the intermediate value theorem,
there exist an \(x_0\in [a,b]\) such that \(g(x_0)=0\). Hence \(f(x_0)=x_0\).
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