Connected Domains Constrain the Range
The fixed-point results concerned points where a function’s output agrees with its input. A broader question is what values a continuous function can take at all. If the domain is connected, continuity places a strong restriction on the range: it cannot have a gap between two values that the function attains.
Recall the Theorem (Continuous Images of Connected Sets) from earlier in this course: if \(E\subseteq\mathbb{R}\) is connected and \(f:E\to\mathbb{R}\) is continuous, then \(f(E)\) is connected. Also recall the Characterization of Connected Subsets of \(\mathbb{R}\): a subset of the real line is connected exactly when it is an interval. Together, these results say that \(f(E)\) is an interval. We will use that conclusion to investigate the shape and endpoints of a bounded range, without repeating the proof of the continuous-image theorem.
Here, \(f(E)=\{f(x):x\in E\}\) denotes the image, or range, of \(f\). The assertion that this range is an interval means that whenever \(u\) and \(v\) are attained values and \(u<w<v\), the value \(w\) is also attained. The endpoints of the interval need not themselves be attained. That depends on the function and its domain.
Classifying a Bounded Range
For a nonempty bounded range, its infimum and supremum identify its lower and upper endpoints, even if the function does not attain either one. Whether an endpoint belongs to the range is a separate question. The next theorem gives an exact classification.
Proof. Since \(E\) is nonempty, so is \(f(E)\). Its boundedness and the completeness of \(\mathbb{R}\) give the real numbers \(\alpha=\inf f(E)\) and \(\beta=\sup f(E)\), with \(\alpha\leq\beta\).
If \(\alpha=\beta\), every \(y\in f(E)\) satisfies \(\alpha\leq y\leq\beta\), so \(y=\alpha\). Nonemptiness then gives \(f(E)=\{\alpha\}\).
Now suppose \(\alpha<\beta\). Every \(y\in f(E)\) lies in \([\alpha,\beta]\). To see that every \(w\) with \(\alpha<w<\beta\) belongs to \(f(E)\), the definition of infimum gives some \(u\in f(E)\) with \(u<w\). Otherwise \(w\) would be a lower bound for \(f(E)\) larger than \(\alpha\). Similarly, the definition of supremum gives some \(v\in f(E)\) with \(v>w\). The Continuous Images of Connected Sets Theorem says \(f(E)\) is connected, and hence an interval. Since \(u,w,v\) are ordered with \(u<w<v\) and \(u,v\in f(E)\), the interval property implies \(w\in f(E)\).
Thus every point strictly between \(\alpha\) and \(\beta\) belongs to the image, and no point outside \([\alpha,\beta]\) does. The point \(\alpha\) belongs to \(f(E)\) precisely when the infimum is attained; the point \(\beta\) belongs precisely when the supremum is attained. These possibilities give exactly the four listed intervals. \(\square\)
This classification separates two facts that are easy to conflate. Connectedness determines that there are no missing interior values. It does not determine whether the boundary values are attained. In particular, continuity on a connected domain does not by itself guarantee a minimum or maximum; compactness is one familiar condition that can provide those guarantees.
Worked Example: An Image with Neither Endpoint
Let \(E=(0,3)\) and define \(f(x)=x^2\). The domain is an interval, hence connected, and \(f\) is continuous on \(E\). For every \(x\in E\), \(0<x<3\), so \(0<x^2<9\). Thus \(f(E)\subseteq(0,9)\).
Conversely, let \(y\in(0,9)\) and take \(x=\sqrt{y}\). Then \(0<y<9\) implies \(0<\sqrt y<3\), so \(x\in E\), and \(f(x)=x^2=(\sqrt y)^2=y\). Therefore \(f(E)=(0,9)\). The infimum is \(0\) and the supremum is \(9\), but neither endpoint is attained because neither \(0\) nor \(3\) lies in the domain.
Worked Example: Both Endpoints Are Attained
Let \(E=[-2,1]\) and \(f(x)=x^2\). The domain is connected and \(f\) is continuous. For \(x\in[-2,1]\), we have \(0\leq x^2\leq4\): the lower bound follows because squares are nonnegative, and the upper bound follows from \(|x|\leq2\).
Every \(y\in[0,4]\) is attained. If \(0\leq y\leq1\), then \(x=\sqrt y\) lies in \([0,1]\subseteq E\) and \(f(x)=y\). If \(1<y\leq4\), then \(x=-\sqrt y\) lies in \([-2,-1)\subseteq E\) and \(f(x)=y\). Hence \(f(E)=[0,4]\). The minimum \(0\) is attained at \(x=0\), and the maximum \(4\) is attained at \(x=-2\).
The same example also shows that a connected image need not come from an injective function: \(f(-1)=f(1)=1\), even though \(f(E)\) is the interval \([0,4]\).
Worked Example: A Bounded Image of an Unbounded Domain
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x/(1+|x|)\). The denominator is positive everywhere, and the formula defines a continuous function. For \(x\geq0\), \(f(x)=x/(1+x)\), which satisfies \(0\leq f(x)<1\). For \(x<0\), \(f(x)=x/(1-x)\), and \(-1<x/(1-x)<0\). Thus \(f(\mathbb{R})\subseteq(-1,1)\).
To check that every value in this interval is attained, let \(y\in(-1,1)\). If \(0\leq y<1\), choose \(x=y/(1-y)\), which is nonnegative. Then \(1+x=1/(1-y)\), so \(x/(1+x)=y\), and therefore \(f(x)=y\). If \(-1<y<0\), choose \(x=y/(1+y)\), which is negative. In this case, \(1-x=1/(1+y)\), so \(x/(1-x)=y\), and again \(f(x)=y\). Therefore \(f(\mathbb{R})=(-1,1)\).
The domain \(\mathbb{R}\) is unbounded, but the image is bounded. Its infimum and supremum are \(-1\) and \(1\), neither of which is attained. The classification concerns the image’s shape and does not require the domain itself to be bounded.
Finite Range Forces Constancy
An interval containing two distinct points contains every real number between them, so it cannot be finite. This observation gives a useful consequence for functions that are known to take only finitely many values.
Proof. If \(E\) is empty, the assertion that \(f\) is constant is vacuous. Otherwise the Continuous Images of Connected Sets Theorem implies that \(f(E)\) is connected. By the earlier theorem on finite connected subsets of \(\mathbb{R}\), a finite connected set is either empty or a singleton. Since \(E\) is nonempty, \(f(E)\) is nonempty, so it must be a singleton, say \(f(E)=\{c\}\). Hence \(f(x)=c\) for every \(x\in E\), as required. \(\square\)
Worked Example: Why a Jump Prevents Continuity
Define \(h:[0,2]\to\mathbb{R}\) by \(h(x)=0\) when \(0\leq x\leq1\), and \(h(x)=1\) when \(1<x\leq2\). Its range is the finite set \(\{0,1\}\), but \(h\) is not continuous at \(1\). Indeed, for \(x_n=1+1/n\), where \(n\geq2\), we have \(x_n\in[0,2]\), \(x_n\to1\), and \(h(x_n)=1\), while \(h(1)=0\). Continuity at \(1\) would require \(h(x_n)\to h(1)\), which fails.
The theorem therefore does not say that every function on an interval with finite range is constant. It says that a function with finite range on a connected domain cannot also be continuous unless it is constant. Here the two different values require a discontinuity.
What Connectedness Does—and Does Not—Guarantee
The main conclusion is a restriction on the range, not a guarantee that the range is open, closed, or nontrivial. The set \((0,9)\) in the first example omits both endpoints, whereas \([0,4]\) in the second includes both. A constant function on any nonempty connected domain has a singleton image. These are all intervals, as the continuous-image theorem requires.
It is also important not to reverse the theorem. A disconnected domain can have a connected image. For example, take \(E=\{-2,2\}\) and \(f(x)=x^2\). The domain is disconnected, but the image is the singleton \(\{4\}\), which is connected. Continuity alone does not make every image reveal whether its domain was connected; the theorem gives a forward implication.
When applying the result, check the hypotheses before drawing conclusions: the function must be continuous on the domain, and the domain must be connected. Then use interval structure to rule out gaps. If the image is bounded, its infimum and supremum locate the possible endpoints, while direct verification determines whether either endpoint is attained.
Check Your Understanding
Use the connected-image theorem and the results in this tutorial to answer the following questions.
- Why must every value strictly between two attained values belong to the image of a continuous function on a connected domain?
- For a nonempty bounded image, what determines whether its infimum and supremum belong to the image?
- What is the image of \(f(x)=x^2\) on \((0,3)\), and why are its endpoints omitted?
- Why must a continuous function on a connected domain be constant if its range is finite?
- Does a connected image imply that the domain is connected? Give the example from this tutorial that supports your answer.