A Supremum Proof of the Intermediate Value Theorem

Why Use a Supremum?

The Intermediate Value Theorem guarantees that a continuous function takes every value between its endpoint values. A supremum proof turns that guarantee into a precise argument: it gathers inputs whose function values are below a chosen target, then locates the boundary of that collection.

The key idea is not to assume the function is increasing or that the inputs below the target form a single interval. Instead, the supremum identifies the rightmost boundary of all such inputs. Continuity forces the function value at that boundary to be exactly the target.

Definitions and Assumptions

The least-upper-bound property of the real numbers says that every nonempty subset of \(\mathbb{R}\) that is bounded above has a supremum in \(\mathbb{R}\). This completeness property will provide the point \(c\) used in the proof.

We use continuity in its epsilon-delta form. A function \(f\) is continuous at an interior point \(c\) of its domain if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(|x-c|<\delta\) implies \(|f(x)-f(c)|<\varepsilon\). At an endpoint of a closed interval, the corresponding condition is applied to inputs in the interval on the available side.

The inequalities in the theorem are strict, so neither endpoint already has function value \(r\). The proof will show that the point it constructs lies inside the interval. The version with \(f(b)<r<f(a)\) follows by applying the same argument to \(-f\) and target \(-r\).

Build the Set and Locate Its Boundary

Consider all inputs in the interval where the function value is strictly below \(r\):

Because \(f(a)<r\), the set \(S\) is nonempty: it contains \(a\). It is bounded above because every element of \(S\) lies in \([a,b]\), so \(b\) is an upper bound. The least-upper-bound property therefore gives a real number \(c=\sup S\).

The definition of supremum alone does not yet tell us that \(f(c)=r\). First, we must show that \(c\) is an interior point. Then continuity at \(c\) will rule out both possibilities \(f(c)<r\) and \(f(c)>r\).

Complete Supremum Proof

Here are the details behind those steps. Since \(f(a)<r\), let \(\varepsilon_a=(r-f(a))/2>0\). Continuity at \(a\) gives some \(\delta_a>0\) such that \(x\in[a,b]\) and \(|x-a|<\delta_a\) imply \(|f(x)-f(a)|<\varepsilon_a\). Consequently,

There is therefore an input \(x>a\) in \(S\), chosen close enough to \(a\). As \(c\) is an upper bound for \(S\), this proves \(c>a\).

At the other endpoint, let \(\varepsilon_b=(f(b)-r)/2>0\). By continuity at \(b\), all inputs \(x\in[a,b]\) sufficiently close to \(b\) satisfy

Choose a positive distance \(d\) small enough that every input in \([b-d,b]\) is within this continuity neighborhood. None of those inputs belongs to \(S\), so every element of \(S\) is less than \(b-d\). Thus \(c\le b-d<b\). We have proved \(a<c<b\).

Suppose first that \(f(c)<r\). Set \(\varepsilon=(r-f(c))/2>0\). By continuity at \(c\), all inputs sufficiently close to \(c\) have function values less than

Because \(c<b\), there is an input \(x>c\) close enough to \(c\) for this conclusion to apply. Then \(f(x)<r\), so \(x\in S\). This contradicts the fact that \(c\) is an upper bound for \(S\). Hence \(f(c)\) cannot be less than \(r\).

Now suppose that \(f(c)>r\). Set \(\varepsilon=(f(c)-r)/2>0\). Continuity at \(c\) gives a neighborhood in which

Since \(c=\sup S\), there must be an element \(s\in S\) in this neighborhood. Indeed, if there were no such element, some number strictly less than \(c\) would still be an upper bound for \(S\), contradicting the definition of \(c\). But \(s\in S\) means \(f(s)<r\), whereas the neighborhood condition gives \(f(s)>r\). This is a contradiction. Therefore \(f(c)\) cannot be greater than \(r\).

The value \(f(c)\) is real, so the only remaining possibility is \(f(c)=r\). This proves the theorem.

Worked Examples

Where Continuity Is Essential

Continuity is used at specific points in the proof, not as a general decoration. At \(a\) and \(b\), it ensures that the strict endpoint inequalities persist for nearby inputs. This places the supremum strictly inside the interval. At \(c\), continuity ensures that a strict inequality at \(f(c)\) would persist nearby. Either persistence would contradict the defining boundary property of \(c\).

Without continuity, those nearby function values need not stay on the same side of the target. The supremum still exists for a nonempty bounded set, but existence of that boundary alone says nothing about the function value there. The proof depends on combining the completeness of \(\mathbb{R}\), which supplies \(c\), with continuity, which forces \(f(c)=r\).

Common Pitfall: Using Only Inputs Below the Supremum

A frequent incomplete argument takes elements of \(S\) approaching \(c\), uses continuity, and concludes only that \(f(c)\le r\). For example, with \(f(x)=x\) on \([0,2]\) and \(r=1\), the set is \(S=[0,1)\). Values at inputs in \(S\) approaching \(1\) do show that \(f(1)\le1\), but that one-sided observation alone does not establish equality.

The correction is to rule out both strict inequalities. If \(f(c)<r\), continuity puts points just to the right of \(c\) in \(S\), contradicting that \(c\) is an upper bound. If \(f(c)>r\), continuity excludes elements of \(S\) close to \(c\), contradicting that \(c\) is the least upper bound. Both directions are needed to conclude \(f(c)=r\).