From Intermediate Values to Connected Images
The Intermediate Value Theorem can be read as a statement about the shape of a function’s range. If a continuous function is defined on an interval, then its image has no gaps: whenever it takes two values, it also takes every value between them. Connectedness makes this explanation precise. An interval is connected, a continuous image of a connected set is connected, and connected subsets of the real line are intervals.
The first two facts were established earlier in the course as the Theorem (Intervals Are Connected) and the Theorem (Continuous Images of Connected Sets). The Theorem (Characterization of Connected Subsets of \(\mathbb{R}\)) says that a subset of \(\mathbb{R}\) is connected exactly when it is an interval. Taken together, these results express the Intermediate Value Theorem as a consequence of connectedness. We will also establish a converse: a subset of the real line is connected precisely when every continuous real-valued function on it has the intermediate-value property.
The Range Form of the Intermediate-Value Property
The relevant property can be stated for an arbitrary function, without assuming continuity. This separates the meaning of “takes every intermediate value” from the reason a function might have that property.
The condition includes the endpoint values: if \(y=f(u)\), one may take \(c=u\), and if \(y=f(v)\), one may take \(c=v\). If \(E\) is empty or has only one point, the condition is vacuous or immediate, respectively.
Proof. Suppose first that \(f\) has the intermediate-value property. Take any \(s,t\in f(E)\), and let \(y\) be between \(s\) and \(t\). By the definition of the image, there are \(u,v\in E\) such that \(f(u)=s\) and \(f(v)=t\). The intermediate-value property gives a point \(c\in E\) with \(f(c)=y\). Therefore \(y\in f(E)\). This is exactly the defining property of an interval, so \(f(E)\) is an interval.
Conversely, suppose \(f(E)\) is an interval. Let \(u,v\in E\), and let \(y\) lie between \(f(u)\) and \(f(v)\). Both \(f(u)\) and \(f(v)\) belong to \(f(E)\). Since \(f(E)\) is an interval, every value between them belongs to \(f(E)\), so \(y\in f(E)\). By the definition of the image, there is some \(c\in E\) such that \(f(c)=y\). Thus \(f\) has the intermediate-value property. \(\square\)
This criterion identifies exactly what the Intermediate Value Theorem guarantees about a range: it is an interval. It does not say that the interval must be closed or bounded. It also does not require the function to be continuous once the intermediate-value property itself is known.
Worked Example: A Continuous Function with Range \([0,1]\)
Let \(f:[0,2]\to\mathbb{R}\) be defined by \(f(x)=x(2-x)\). The domain is an interval, so it is connected. The function is a polynomial and hence continuous. To identify its range, rewrite \(f(x)=1-(x-1)^2\). For \(x\in[0,2]\), we have \(-1\leq x-1\leq1\), and therefore \(0\leq(x-1)^2\leq1\). It follows that \(0\leq f(x)\leq1\), so \(f([0,2])\subseteq[0,1]\).
Direct substitution gives \(f(0)=0(2-0)=0\) and \(f(1)=1(2-1)=1\). By the connected-image argument, the range is an interval containing both \(0\) and \(1\), and so it contains every value in \([0,1]\). Combining the two inclusions gives \(f([0,2])=[0,1]\). The interval shape of the range follows from connectedness; the calculation identifies its endpoints.
IVT as a Connectedness Theorem
The range criterion lets us state the connection in a form that also gives a converse. Here “every continuous function” is essential: connectedness is a property of the domain, so its characterization must test the domain independently of a particular choice of function.
Proof. Suppose first that \(E\) is connected, and let \(f:E\to\mathbb{R}\) be continuous. By the Theorem (Continuous Images of Connected Sets), \(f(E)\) is connected. By the Theorem (Characterization of Connected Subsets of \(\mathbb{R}\)), \(f(E)\) is an interval. The Range Criterion for the Intermediate-Value Property now implies that \(f\) has the intermediate-value property on \(E\).
For the converse, suppose every continuous function from \(E\) to \(\mathbb{R}\) has the intermediate-value property. Consider the identity function \(i:E\to\mathbb{R}\), defined by \(i(x)=x\). It is continuous, and \(i(E)=E\). By the assumption, \(i\) has the intermediate-value property. Given any \(u,v\in E\) and any \(y\) between \(u\) and \(v\), apply this property to \(i(u)=u\) and \(i(v)=v\). It gives \(c\in E\) with \(i(c)=y\). Since \(i(c)=c\), this means \(y\in E\). Thus \(E\) is an interval, including the cases where \(E\) is empty or a singleton. The Characterization of Connected Subsets of \(\mathbb{R}\) then implies that \(E\) is connected. \(\square\)
In particular, if \(J\subseteq\mathbb{R}\) is an interval and \(f:J\to\mathbb{R}\) is continuous, then \(J\) is connected. The theorem gives the intermediate-value property for \(f\), which is the Intermediate Value Theorem. The familiar interval version is therefore one consequence of a broader principle: continuous functions preserve connectedness, and in the real line connectedness means having no gaps.
Worked Example: A Gap in the Domain Can Block Intermediate Values
Let \(E=[-2,-1]\cup[1,2]\), and let \(i:E\to\mathbb{R}\) be the identity function \(i(x)=x\). This function is continuous on \(E\). The set \(E\) is not an interval: it contains \(-1\) and \(1\) but does not contain \(0\). By the Characterization of Connected Subsets of \(\mathbb{R}\), it is disconnected.
The failure of the intermediate-value property is explicit. We have \(i(-1)=-1\) and \(i(1)=1\), while \(0\) lies between these values. There is no \(c\in E\) with \(i(c)=0\), because \(0\notin E\). Equivalently, the image \(i(E)=E\) is not an interval. This example shows why continuity alone does not give the Intermediate Value Theorem: the domain must also have the appropriate connectedness.
What a Single Function Can and Cannot Tell Us
The connectedness characterization quantifies over every continuous real-valued function on \(E\). It does not say that a particular continuous function can have an interval image only when its domain is connected. A function may map a disconnected set onto an interval by sending different pieces of the domain to overlapping ranges.
Worked Example: A Disconnected Domain with an Interval as Its Image
Use \(E=[-2,-1]\cup[1,2]\), and define \(g:E\to\mathbb{R}\) by \(g(x)=|x|-1\). The absolute-value function is continuous on \(\mathbb{R}\), so its restriction \(g\) is continuous on \(E\). For every \(x\in E\), \(1\leq|x|\leq2\), and hence \(0\leq g(x)\leq1\). Thus \(g(E)\subseteq[0,1]\).
Conversely, take any \(t\in[0,1]\) and set \(x=1+t\). Then \(1\leq x\leq2\), so \(x\in E\), and \(g(x)=|1+t|-1=1+t-1=t\). Therefore every \(t\in[0,1]\) belongs to \(g(E)\), and \(g(E)=[0,1]\). The image is an interval even though the domain is disconnected. The universal quantifier in the connectedness theorem cannot be replaced by a test of just one function.
There is a related distinction between the interval property and closedness or boundedness. For example, the identity function on \((2,5)\) is continuous and has image \((2,5)\), which is an interval but is neither closed nor bounded. The connectedness argument supplies the absence of gaps; it does not supply endpoint values or bounds. Those require additional hypotheses or separate theorems.
Why This Viewpoint Matters
The connectedness formulation is useful when the domain is not presented as a closed interval. If a domain \(E\) is known to be connected, then every continuous real-valued function on it has an interval as its range, even when \(E\) is open, unbounded, or otherwise not compact. The key question is not whether the domain has endpoints, but whether it is connected.
It also gives a diagnostic for claims about the Intermediate Value Theorem. A proof using connectedness has two distinct steps: continuity makes the image connected, and the real-line characterization turns that connected image into an interval. A missing step in either part leaves a gap in the argument. Conversely, when the domain is disconnected, the identity function exposes the missing values; another function may still have an interval image, as the preceding example shows.
Thus the Intermediate Value Theorem is not merely a rule about function values at the endpoints of a closed interval. It reflects a structural fact about continuous maps: connected input sets have connected images. In \(\mathbb{R}\), that structure is exactly what ensures that no value between two attained values is skipped.
Check Your Understanding
Use the range criterion and connectedness characterization to answer the following questions.
- Why is the intermediate-value property of a function equivalent to its image being an interval?
- Which earlier results connect continuity on a connected set to the interval shape of the image?
- Why does the identity function prove the converse direction of the connectedness characterization?
- In the example with \(g(x)=|x|-1\), how can a disconnected domain still have image \([0,1]\)?
- Does connectedness alone imply that the image of a continuous function is closed or bounded? Explain.