The Proof Strategy: Pull a Separation Back
The previous tutorial used the fact that a continuous image of a connected set is connected to describe possible ranges. Here we prove that result. The central idea is to work backward from a hypothetical separation of the image: continuity makes the pieces of that separation pull back to relatively open pieces of the domain.
Recall that a separation of a set \(Y\subseteq\mathbb{R}\) is a representation \(Y=A\cup B\), where \(A\) and \(B\) are nonempty, disjoint, and relatively open in \(Y\). A set is connected if it has no separation. Relative openness is important here: \(A\) and \(B\) need not be open in all of \(\mathbb{R}\). But if \(A\) is relatively open in \(Y\), there is an open set \(G\subseteq\mathbb{R}\) such that \(A=Y\cap G\). This lets us use continuity, which guarantees that the preimage of an open set is relatively open in the domain.
Proof. Put \(U=f^{-1}(A)\) and \(V=f^{-1}(B)\). Since \(A\) and \(B\) are disjoint, \(U\cap V=\varnothing\). Since every \(f(x)\), for \(x\in E\), lies in \(Y=A\cup B\), each \(x\in E\) belongs to \(U\) or \(V\). Thus \(E=U\cup V\).
Because \(A\) is relatively open in \(Y\), there is an open set \(G\subseteq\mathbb{R}\) with \(A=Y\cap G\). For \(x\in E\), the value \(f(x)\) already belongs to \(Y\). Consequently, \(x\in U\) exactly when \(f(x)\in G\), so \(U=f^{-1}(G)\). Continuity of \(f\) makes \(U\) relatively open in \(E\). Likewise, \(B=Y\cap H\) for some open \(H\subseteq\mathbb{R}\), and \(V=f^{-1}(H)\) is relatively open in \(E\).
Finally, \(U\) and \(V\) are both nonempty. Since \(A\subseteq f(E)\), each point of \(A\) has a preimage in \(E\), and hence \(U\ne\varnothing\); the same reasoning gives \(V\ne\varnothing\). We have shown that \(U\) and \(V\) are disjoint, nonempty, relatively open, and have union \(E\). They form a separation of \(E\). \(\square\)
Proof of the Continuous-Image Theorem
Proof. If \(E\) is empty, then \(f(E)\) is empty and has no separation, so the conclusion holds. Suppose \(E\) is nonempty. Assume, for contradiction, that \(f(E)\) is disconnected. By the definition of disconnectedness, there are nonempty disjoint sets \(A\) and \(B\), relatively open in \(f(E)\), such that \(f(E)=A\cup B\). Thus \(A\) and \(B\) form a separation of the image.
By the Separation of the Image Pulls Back Lemma, \(f^{-1}(A)\) and \(f^{-1}(B)\) form a separation of \(E\). This contradicts the assumption that \(E\) is connected. Therefore \(f(E)\) is connected. \(\square\)
The proof depends on both hypotheses in distinct ways. Connectedness of \(E\) rules out the separation produced by the lemma. Continuity ensures that the preimages of relatively open pieces of the image are relatively open in \(E\). Without continuity, the pullback may still partition the domain, but it need not be a separation.
Choose a separation \(f(E)=A\cup B\).
The sets \(f^{-1}(A)\) and \(f^{-1}(B)\) are nonempty, disjoint, and cover \(E\).
Relative openness of \(A\) and \(B\) in the image, together with continuity, makes their preimages relatively open in \(E\).
The preimages would separate \(E\), contrary to its connectedness.
Worked Applications
Worked Example: The Range of a Cubic on a Closed Interval
Let \(f:[-2,1]\to\mathbb{R}\) be given by \(f(x)=x^3\). The interval \([-2,1]\) is connected, and the polynomial \(f\) is continuous. We first verify the range directly. For \(-2\leq x\leq1\), the cube function is increasing, so \(-8=(-2)^3\leq x^3\leq1^3=1\).
Conversely, take any \(y\in[-8,1]\), and let \(x=\sqrt[3]{y}\), the real cube root. Since the cube function is increasing, \(-8\leq y\leq1\) implies \(-2\leq x\leq1\). Also \(f(x)=x^3=(\sqrt[3]{y})^3=y\). Therefore \(f([-2,1])=[-8,1]\), a connected set. The theorem ensures connectedness without requiring this direct range calculation in every application.
Worked Example: An Absolute-Value Function
Define \(f:[0,4]\to\mathbb{R}\) by \(f(x)=|x-1|\). The function is continuous and its domain is an interval, hence connected. If \(0\leq x\leq1\), then \(f(x)=1-x\), which lies in \([0,1]\). If \(1\leq x\leq4\), then \(f(x)=x-1\), which lies in \([0,3]\). Thus every value of the function belongs to \([0,3]\).
For any \(y\in[0,3]\), choose \(x=1+y\). Then \(1\leq x\leq4\), and \(f(x)=|(1+y)-1|=|y|=y\), because \(y\geq0\). Hence every point of \([0,3]\) is attained and \(f([0,4])=[0,3]\). In particular, the range has no gap, as the theorem predicts.
Worked Example: A Bounded Range on the Real Line
Consider \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x)=x/(1+x^2)\). Its denominator is positive for every real \(x\), so this rational function is continuous on \(\mathbb{R}\). For all real \(x\), \[ \frac12-\frac{x}{1+x^2} =\frac{(x-1)^2}{2(1+x^2)}\geq0, \qquad \frac12+\frac{x}{1+x^2} =\frac{(x+1)^2}{2(1+x^2)}\geq0. \] The first inequality gives \(f(x)\leq\frac12\), and the second gives \(f(x)\geq-\frac12\). Equality occurs in the first at \(x=1\), since \((1-1)^2=0\), and in the second at \(x=-1\), since \((-1+1)^2=0\).
Thus the image contains both \(-\frac12\) and \(\frac12\), and is contained in the interval \([-\frac12,\frac12]\). The real line is connected, so the theorem says \(f(\mathbb{R})\) is connected; by the characterization of connected subsets of \(\mathbb{R}\), it is an interval. An interval containing both endpoints contains every value between them. Therefore \(f(\mathbb{R})=[-\frac12,\frac12]\).
Why Continuity Cannot Be Dropped
To see where the proof can fail, define \(g:[-1,1]\to\mathbb{R}\) by \(g(x)=0\) for \(x<0\) and \(g(x)=2\) for \(x\geq0\). The domain is connected, but the range is \(\{0,2\}\), which is disconnected: its singleton subsets \(\{0\}\) and \(\{2\}\) form a separation.
The cut at \(1\) pulls back to \(U=g^{-1}((-\infty,1))=[-1,0)\) and \(V=g^{-1}((1,\infty))=[0,1]\). These sets are disjoint, nonempty, and cover \([-1,1]\). The first is relatively open, since \([-1,0)=(-2,0)\cap[-1,1]\). But the second is not relatively open: every relative neighborhood of \(0\) in \([-1,1]\) contains negative points, which are not in \(V\). So these sets do not form a separation.
This failure is exactly what the theorem’s proof predicts. The preimage of an open set need not be relatively open when the function is not continuous. Indeed, if \(x_n=-1/n\), then \(x_n\to0\), but \(g(x_n)=0\) for every \(n\), whereas \(g(0)=2\). Thus \(g\) is not continuous at \(0\).
The theorem also has only a forward implication: connected domain plus continuity forces connected image. A connected image does not imply that the domain is connected. For example, \(E=\{-3,-1,1,3\}\) is disconnected, but the continuous function \(f(x)=0\) on \(E\) has the connected image \(\{0\}\). The theorem constrains what continuity can do to a connected domain; it does not diagnose connectedness of the domain from the image alone.
Check Your Understanding
Use the pullback lemma and the proof to answer the following questions.
- If \(f(E)=A\cup B\) is a separation, why are \(f^{-1}(A)\) and \(f^{-1}(B)\) both nonempty?
- Why does relative openness of \(A\) in \(f(E)\) allow continuity to show that \(f^{-1}(A)\) is relatively open in \(E\)?
- Which contradiction proves that the image of a connected set under a continuous function is connected?
- In the example \(g(x)=0\) for \(x<0\) and \(g(x)=2\) for \(x\geq0\), which preimage fails to be relatively open, and at what point?
- Does a connected image imply that its domain is connected? Explain using the constant function on \(\{-3,-1,1,3\}\).