Closed Maps: Follow Closed Sets Forward
Continuity is tested by taking an open set in the target and pulling it back: its preimage must be open in the domain. A closed map asks a different question. It takes a closed set in the domain and pushes it forward, and the resulting image must be closed in the target. These operations involve different sets and different topologies, so continuity alone does not settle whether a map is closed.
Throughout, a map \(f:X\to Y\) is considered with specified topologies on \(X\) and \(Y\). A subset is closed when its complement in its own space is open. In particular, being closed in a subspace is not the same as being closed in the larger space.
For comparison, the closed-set test for continuity says that \(f\) is continuous if and only if \(f^{-1}(F)\) is closed in \(X\) for every closed \(F\subseteq Y\). The difference is the direction of the operation: a closed map concerns \(f(C)\), whereas continuity concerns \(f^{-1}(F)\). The definitions do not make either property imply the other.
A Closed Map on the Real Line
Worked Example: The Squaring Function
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), with the standard topology on both copies of \(\mathbb{R}\). We show that \(f\) is closed.
Take any closed set \(C\subseteq\mathbb{R}\). If \(C=\varnothing\), then \(f(C)=\varnothing\), which is closed. Suppose \(C\ne\varnothing\), and take any convergent sequence \((y_n)\) in \(f(C)\), with limit \(y\in\mathbb{R}\). For each \(n\), choose \(x_n\in C\) such that \(x_n^2=y_n\). Since \(y_n\geq 0\) for every \(n\), its limit satisfies \(y\geq 0\): if \(y<0\), then all sufficiently late terms of a sequence converging to \(y\) would be negative, which is impossible.
Each chosen \(x_n\) is either nonnegative or negative. At least one of these two classes occurs for infinitely many indices. Restrict to those indices. If \(x_n\geq0\) on this subsequence, then \(x_n=\sqrt{y_n}\), so along it \(x_n\to\sqrt y\). If \(x_n<0\) on the subsequence, then \(x_n=-\sqrt{y_n}\), so along it \(x_n\to-\sqrt y\). These limits also apply when \(y=0\). In either case, a subsequence of points of \(C\) converges to a point \(x\) with \(x^2=y\). Since \(C\) is closed, \(x\in C\), and therefore \(y=f(x)\in f(C)\).
We have shown that every convergent sequence in \(f(C)\) has its limit in \(f(C)\). In \(\mathbb{R}\), this implies that \(f(C)\) is closed: if a set were not closed, a point in its closure but outside it could be approximated by a sequence from the set, by choosing a point within distance \(1/n\) for each \(n\). Thus \(f(C)\) is closed, as required.
Check: for \(C=\{-2,3\}\), the image is \(f(C)=\{4,9\}\), a closed finite subset of \(\mathbb{R}\). In the general argument, the selected limit has square \(y\), so it verifies membership in the image rather than merely nonnegativity of \(y\).
The example establishes closedness by considering an arbitrary closed set in the domain. It is not enough to test just one especially simple set. Conversely, finding even one closed set whose image is not closed proves that a map is not closed.
A Projection That Is Not Closed
Worked Example: Projecting a Hyperbola
Give \(\mathbb{R}^2\) its Euclidean topology and \(\mathbb{R}\) its standard topology. Consider the projection \(\pi_1:\mathbb{R}^2\to\mathbb{R}\), defined by \(\pi_1(x,y)=x\), and the set \(F=\{(x,y)\in\mathbb{R}^2:xy=1\}\).
First, \(F\) is closed. To check this directly, take \((a,b)\notin F\), so \(ab\ne1\). For \(|x-a|<\delta\), \(|y-b|<\delta\), and \(\delta\leq1\), the triangle inequality gives \(|xy-ab|\leq |x||y-b|+|b||x-a|\leq (|a|+|b|+1)\delta\). Choose \(\delta>0\) less than both \(1\) and \(|ab-1|/[2(|a|+|b|+1)]\). Then \(|xy-1|\geq |ab-1|-|xy-ab|>|ab-1|/2>0\). The open rectangle specified by these two inequalities is a neighborhood of \((a,b)\) disjoint from \(F\). Thus the complement of \(F\) is open, so \(F\) is closed.
Now compute the image. If \((x,y)\in F\), then \(xy=1\), so \(x\ne0\). Conversely, for each \(x\ne0\), the point \((x,1/x)\) belongs to \(F\) and projects to \(x\). Hence \(\pi_1(F)=\mathbb{R}\setminus\{0\}\). This image is not closed: \(1/n\in\pi_1(F)\) for every \(n\in\mathbb{N}\), while \(1/n\to0\) and \(0\notin\pi_1(F)\). Therefore \(\pi_1\) is not a closed map.
The projection is nevertheless continuous. If \(U\subseteq\mathbb{R}\) is open, then \(\pi_1^{-1}(U)=U\times\mathbb{R}\). For any \((x,y)\in U\times\mathbb{R}\), choose \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq U\). The Euclidean ball of radius \(\varepsilon\) about \((x,y)\) lies in \(U\times\mathbb{R}\), so this preimage is open. The example is a continuous map that is not closed.
Continuity and Closedness Are Independent
Worked Example: Continuous but Not Closed
Let \(X=[0,1)\) have the subspace topology inherited from the standard real line, let \(Y=\mathbb{R}\) have its standard topology, and define \(i:X\to Y\) by \(i(x)=x\). For every open \(U\subseteq Y\), \(i^{-1}(U)=U\cap[0,1)\), which is open in the subspace topology by its definition. Thus \(i\) is continuous.
The set \(C=X\) is closed in \(X\), because the complement \(X\setminus C=\varnothing\) is open in \(X\). But \(i(C)=[0,1)\) is not closed in \(Y\): every open interval around \(1\) contains points of \([0,1)\), while \(1\notin[0,1)\). Hence \(i\) is not closed.
Check: closedness of \(C\) was tested using the complement in \(X\), which is empty. Closedness of \(i(C)\) was tested in \(Y=\mathbb{R}\); the point \(1\) shows that this image is not closed there.
Worked Example: Closed but Not Continuous
Let \(X\) be the set \(\mathbb{R}\) with the indiscrete topology \(\{\varnothing,\mathbb{R}\}\), and let \(Y=\mathbb{R}\) have the standard topology. Define \(j:X\to Y\) by \(j(x)=x\). The only closed subsets of \(X\) are \(\varnothing\) and \(\mathbb{R}\), because their complements are the only open subsets of \(X\). Their images under \(j\) are \(\varnothing\) and \(\mathbb{R}\), both closed in the standard real line. Therefore \(j\) is a closed map.
However, the standard open interval \((0,1)\subseteq Y\) has preimage \(j^{-1}((0,1))=(0,1)\). This is neither \(\varnothing\) nor \(\mathbb{R}\), so it is not open in the indiscrete topology on \(X\). By the open-preimage definition, \(j\) is not continuous.
The topology on each space matters: the same underlying set map can behave differently when the domain or target topology changes. The example shows that even a closed map need not be continuous.
A Basic Rule for Composing Closed Maps
Proof. Let \(C\) be any closed subset of \(X\). Since \(f\) is closed, \(f(C)\) is closed in \(Y\). Since \(g\) is closed, the image under \(g\) of this closed subset of \(Y\), namely \(g(f(C))\), is closed in \(Z\). The composition satisfies \((g\circ f)(C)=g(f(C))\), so it sends every closed subset of \(X\) to a closed subset of \(Z\). This is precisely the definition of a closed map. \(\square\)
This rule uses the forward-image condition twice, with the intermediate space \(Y\) providing the set to which the second map is applied. It does not say that either map is continuous, and the examples above show why closedness should be checked independently.
A Common Pitfall: Mixing Up the Two Tests
A common incorrect argument is: “The inclusion \(i:[0,1)\to\mathbb{R}\) is closed because \([0,1)\) is closed.” This confuses a set with its image and ignores which space supplies the topology. The whole domain \(X=[0,1)\) is indeed closed in itself, but the definition asks whether its image \(i(X)=[0,1)\) is closed in the target \(\mathbb{R}\). It is not, since every neighborhood of \(1\) meets it.
A second related error is to use continuity as though it tested images of closed sets. Continuity tests \(f^{-1}(F)\) for closed \(F\) in the target; closedness of a map tests \(f(C)\) for closed \(C\) in the domain. The projection example makes the distinction concrete: its continuity does not prevent the closed hyperbola from having a nonclosed image.
| Property | Set being tested | Where it must be closed or open |
|---|---|---|
| Continuity | \(f^{-1}(U)\), for each open \(U\subseteq Y\) | Open in \(X\) |
| Equivalent closed-set test for continuity | \(f^{-1}(F)\), for each closed \(F\subseteq Y\) | Closed in \(X\) |
| Closed map | \(f(C)\), for each closed \(C\subseteq X\) | Closed in \(Y\) |
Check Your Understanding
For each question, keep track of whether the set is being mapped forward or pulled back, and name the topology in which closedness is tested.
- Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\). What is the image of the closed set \(\{-1,2\}\), and why is this consistent with \(f\) being closed?
- For the projection example, verify directly that \(0\notin\pi_1(F)\) but \(1/n\in\pi_1(F)\) for every \(n\in\mathbb{N}\).
- In the inclusion \(i:[0,1)\to\mathbb{R}\), identify a point that prevents \(i([0,1))\) from being closed.
- For the identity map from the indiscrete real line to the standard real line, which open target set has a preimage that is not open in the domain?
- Explain in one sentence the difference between the set tested in continuity and the set tested for a closed map.