Continuity Tested with Closed Sets
The open-set characterization of continuity gives a global test: the preimage of every open set in the target must be open in the domain. There is a complementary test using closed sets. It is often more convenient when the target's closed sets are easier to describe, or when a function is defined by conditions whose solution sets are naturally closed.
Throughout, let \(f:X\to Y\) be a function between topological spaces. Recall that \(f^{-1}(F)=\{x\in X:f(x)\in F\}\) is the preimage of \(F\subseteq Y\). The essential fact is that preimages commute with complements: \(f^{-1}(Y\setminus F)=X\setminus f^{-1}(F)\). Thus taking a complement in the target corresponds exactly to taking a complement in the domain.
Proof. Suppose first that \(f\) is continuous, and let \(F\) be closed in \(Y\). Then \(Y\setminus F\) is open in \(Y\). By the open-set characterization of continuity from “Continuity Through Open Sets,” \(f^{-1}(Y\setminus F)\) is open in \(X\). The complement identity gives \(f^{-1}(F)=X\setminus f^{-1}(Y\setminus F)\), so \(f^{-1}(F)\) is closed in \(X\).
Conversely, suppose that \(f^{-1}(F)\) is closed in \(X\) for every closed \(F\subseteq Y\). Let \(V\) be open in \(Y\). Then \(Y\setminus V\) is closed, so \(f^{-1}(Y\setminus V)\) is closed in \(X\). Using the same complement identity, \(f^{-1}(V)=X\setminus f^{-1}(Y\setminus V)\), which is open in \(X\). The open-set characterization now implies that \(f\) is continuous. \(\square\)
The theorem is a dual formulation, not a different notion of continuity. It changes only which sets are tested: open target sets in one version, closed target sets in the other. In either version, it is the preimage that matters. A single closed set whose preimage is not closed is enough to show that a function is not continuous.
Worked Examples: Using Closed Preimages
Worked Example: The Squaring Map
Let \(f:\mathbb{R}\to\mathbb{R}\) be given by \(f(x)=x^2\), with the usual topology on both spaces. We verify the closed-set condition directly. Let \(F\subseteq\mathbb{R}\) be closed, and take any \(x\notin f^{-1}(F)\). Then \(x^2\notin F\), so \(x^2\) belongs to the open set \(\mathbb{R}\setminus F\). There is an \(\varepsilon>0\) such that \((x^2-\varepsilon,x^2+\varepsilon)\subseteq\mathbb{R}\setminus F\).
Choose \(\delta=\min\{1,\varepsilon/(2|x|+1)\}\), which is positive. If \(|y-x|<\delta\), then \(|y-x|<1\), and hence \(|y+x|\leq |y-x|+2|x|<2|x|+1\). Therefore $$ |y^2-x^2|=|y-x||y+x|<\delta(2|x|+1)\leq\varepsilon. $$ It follows that \(y^2\in(x^2-\varepsilon,x^2+\varepsilon)\), and thus \(y^2\notin F\). So every \(x\notin f^{-1}(F)\) has an open interval around it contained in \(\mathbb{R}\setminus f^{-1}(F)\). This complement is open, \(f^{-1}(F)\) is closed, and the closed-set criterion proves that \(f\) is continuous.
The calculation handles every closed \(F\) at once. It does not depend on whether \(F\) is an interval, a finite set, or a more complicated closed subset of the real line.
Worked Example: A Step Function into a Discrete Space
Give \(Y=\{0,1\}\) the discrete topology, so every subset of \(Y\) is closed. Define \(g:\mathbb{R}\to Y\) by \(g(x)=0\) for \(x<0\) and \(g(x)=1\) for \(x\geq 0\). The closed set \(\{1\}\subseteq Y\) has preimage $$ g^{-1}(\{1\})=[0,\infty). $$ This set is closed in \(\mathbb{R}\), but the closed set \(\{0\}\subseteq Y\) has preimage $$ g^{-1}(\{0\})=(-\infty,0), $$ which is not closed: for example, the sequence \(-1/n\) lies in \((-\infty,0)\) and converges to \(0\), which is not in that set. Thus \(g\) fails the closed-set test and is not continuous.
Checking one target closed set would have been sufficient to detect the failure. Checking both fibers makes clear why the jump at \(0\) matters: the two preimages do not both have the required closedness.
Worked Example: Inclusion of a Subspace
Let \(A\subseteq X\) be any subspace, and consider the inclusion map \(i:A\to X\), defined by \(i(a)=a\). If \(F\) is closed in \(X\), then \(i^{-1}(F)=A\cap F\). This is closed in the subspace \(A\): its complement relative to \(A\) is \(A\setminus(A\cap F)=A\cap(X\setminus F)\), which is open in \(A\) by the definition of the subspace topology. The closed-set criterion proves that \(i\) is continuous.
For a concrete instance, take \(X=\mathbb{R}\) and \(A=[2,5]\), with their usual and subspace topologies, respectively. For every closed \(F\subseteq\mathbb{R}\), the preimage under the inclusion is \(F\cap[2,5]\), closed in \([2,5]\). The test works without requiring that \(A\) itself be open in \(X\).
Continuity and the Closure of a Set
The closed-set criterion leads to another way to recognize continuity. Instead of testing preimages, we can compare the closure of a set in the domain with the closure of its image in the target. Recall that \(\overline{A}\) denotes the closure of \(A\): it is the smallest closed set containing \(A\).
Proof. Suppose \(f\) is continuous and fix \(A\subseteq X\). The set \(\overline{f[A]}\) is closed in \(Y\). By the closed-set criterion, \(f^{-1}(\overline{f[A]})\) is closed in \(X\). It contains \(A\), because \(f(a)\in f[A]\subseteq\overline{f[A]}\) for every \(a\in A\). Since \(\overline{A}\) is the smallest closed set containing \(A\), we have \(\overline{A}\subseteq f^{-1}(\overline{f[A]})\). Applying \(f\) to this inclusion yields \(f[\overline{A}]\subseteq\overline{f[A]}\).
Conversely, suppose the displayed inclusion holds for every \(A\subseteq X\). Let \(F\) be closed in \(Y\), and put \(A=f^{-1}(F)\). Since \(F\) is closed, \(f[A]\subseteq F\) implies \(\overline{f[A]}\subseteq F\). The assumed inclusion gives \(f[\overline{A}]\subseteq\overline{f[A]}\subseteq F\). Therefore every point of \(\overline{A}\) lies in \(f^{-1}(F)=A\), so \(\overline{A}\subseteq A\). The reverse inclusion \(A\subseteq\overline{A}\) always holds, and hence \(A\) is closed. We have shown that the preimage of every closed \(F\subseteq Y\) is closed. The closed-set criterion proves that \(f\) is continuous. \(\square\)
This formulation expresses a useful constraint: a continuous map cannot send a point in the closure of \(A\) outside the closure of the image of \(A\). The inclusion may be strict. For example, for the inclusion \(j:(0,1)\to\mathbb{R}\) and \(A=(0,1)\), we have \(j[\overline{A}]=(0,1)\) while \(\overline{j[A]}=[0,1]\).
Continuity Is Not the Closed-Map Property
A frequent confusion is to reverse the direction of the map in the closed-set criterion. Continuity says that preimages of closed target sets are closed in the domain. It does not say that images of closed domain sets are closed in the target. A function that does send closed sets to closed sets is called a closed map; this is an additional property, not part of the definition of continuity.
For a simple example, give \((0,1)\) the subspace topology inherited from \(\mathbb{R}\), and let \(j:(0,1)\to\mathbb{R}\) be the inclusion. The subspace example above shows that \(j\) is continuous. But \((0,1)\) is closed as a subset of its own domain, while its image under \(j\) is \((0,1)\), which is not closed in \(\mathbb{R}\). Thus \(j\) is continuous but is not a closed map.
When testing continuity through closed sets, always identify a closed subset of the target first, then take its preimage. The closedness required is in the domain. Keeping track of these roles prevents the most common misuse of the criterion.
Check Your Understanding
Use the closed-set criterion and the closure formulation to answer the following questions.
- State the condition on preimages of closed sets that is equivalent to continuity.
- Why does the closed-set criterion follow from the open-set characterization?
- For the step function into the discrete space \(\{0,1\}\), which closed target set has a nonclosed preimage?
- State the closure-preservation condition equivalent to continuity.
- In the inclusion map \(j:(0,1)\to\mathbb{R}\), why does continuity not imply that the image of every closed domain set is closed in \(\mathbb{R}\)?