Three Different Questions About a Map
Continuity, openness, and closedness all describe how a map interacts with the topology, but they ask different questions. Continuity tests preimages of open sets; openness and closedness test images of open and closed sets. A map can have one of these properties without having the others. The distinction matters when deciding whether a continuous bijection is a homeomorphism, or when studying how a particular map changes the sets in its domain.
Throughout, a map \(f:X\to Y\) is between topological spaces. A set is closed when its complement is open. The image of \(A\subseteq X\) is \(f(A)=\{f(x):x\in A\}\), and the preimage of \(B\subseteq Y\) is \(f^{-1}(B)=\{x\in X:f(x)\in B\}\).
The direction of the set operation is a useful reminder: continuity concerns preimages, while openness and closedness concern images. In particular, a continuous map need not send open sets to open sets, or closed sets to closed sets.
A Projection from a Square
Worked Example: Projecting a Closed Square
Let \(Q=[0,1]\times[0,1]\subseteq\mathbb{R}^2\) and give \(Q\) the subspace topology from the usual Euclidean topology on \(\mathbb{R}^2\). Give \([0,1]\) the subspace topology from the usual topology on \(\mathbb{R}\). Consider the coordinate projection \(p:Q\to[0,1]\), \(p(x,y)=x\). We check continuity, openness, and closedness separately.
Continuity. Let \(V\) be open in \([0,1]\). By the definition of the subspace topology, \(V=G\cap[0,1]\) for some open \(G\subseteq\mathbb{R}\). Then \(p^{-1}(V)=Q\cap(G\times\mathbb{R})\). The set \(G\times\mathbb{R}\) is open in the Euclidean plane, so this preimage is open in \(Q\). Thus \(p\) is continuous.
Openness. Let \(U\) be open in \(Q\), and take any \(x\in p(U)\). There is \(y\in[0,1]\) with \((x,y)\in U\). Since \(U\) is relatively open, there is an open Euclidean ball of radius \(r>0\) around \((x,y)\) whose intersection with \(Q\) lies in \(U\). For every \(s\in[0,1]\) with \(|s-x|<r\), the point \((s,y)\) lies in that ball and in \(Q\), so \(s\in p(U)\). Therefore \(p(U)\) contains a relative neighborhood of each of its points and is open in \([0,1]\).
Closedness. The square \(Q\) is compact by the Heine–Borel theorem. A closed subset \(C\) of \(Q\) is compact: add the open set \(Q\setminus C\) to any open cover of \(C\), take a finite subcover of \(Q\), and discard \(Q\setminus C\) if it is not needed. A continuous image of a compact set is compact, since an open cover of the image pulls back to an open cover of the original set and a finite subcover pulls forward to a finite subcover. Finally, compact subsets of \(\mathbb{R}\) are closed. Hence \(p(C)\) is compact and closed in \(\mathbb{R}\), so it is also closed in \([0,1]\). Thus \(p\) is closed.
Check: the whole square \(Q\) is closed in itself and \(p(Q)=[0,1]\), which is closed in \([0,1]\). Also, the open set \(Q\) maps to \([0,1]\), open in the relative topology of \([0,1]\). These tests agree with the conclusions; the arguments above establish them for every open or closed subset, not just \(Q\).
This example has all three properties, but that combination is not automatic for a projection. Even for familiar subsets of the plane, the domain and codomain must be specified and each property checked.
Worked Example: Projecting an Open Disk
Let \(D=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}\), with its subspace topology, and let \(p:D\to\mathbb{R}\) be \(p(x,y)=x\). This projection is continuous and open, but it is not closed.
For continuity, if \(V\subseteq\mathbb{R}\) is open, then \(p^{-1}(V)=D\cap(V\times\mathbb{R})\), which is open in \(D\). To see openness, take an open \(U\subseteq D\) and \((x,y)\in U\). Since \(D\) is open in \(\mathbb{R}^2\), the subspace-open set \(U\) is itself open in \(\mathbb{R}^2\). Choose \(r>0\) so that the ball of radius \(r\) about \((x,y)\) lies in \(U\). For every real \(s\) with \(|s-x|<r\), \((s,y)\) lies in that ball, so \(s\in p(U)\). Thus \(p(U)\) is open in \(\mathbb{R}\).
Now let \(F=\{(x,0):-1<x<1\}\). This is closed in \(D\), because it is the intersection of \(D\) with the closed horizontal axis in \(\mathbb{R}^2\). But \(p(F)=(-1,1)\), which is not closed in \(\mathbb{R}\): its complement is not open at \(1\), since every interval around \(1\) contains points of \((-1,1)\). Therefore \(p\) is not a closed map.
Inclusions and Identity Maps
For an inclusion, the domain is a subset with its subspace topology. Its behavior can be characterized exactly.
Proof. For any open \(V\subseteq X\), \(i^{-1}(V)=A\cap V\), which is open in the subspace \(A\). Thus \(i\) is continuous. If \(A\) is open in \(X\), every open subset of \(A\) has the form \(A\cap V\) for an open \(V\subseteq X\), and this intersection is open in \(X\); hence \(i\) is open. Conversely, \(A\) is open in itself, so if \(i\) is open, its image \(i(A)=A\) must be open in \(X\). If \(A\) is closed in \(X\), every closed subset of \(A\) has the form \(A\cap C\) for a closed \(C\subseteq X\), and this intersection is closed in \(X\); hence \(i\) is closed. Conversely, \(A\) is closed in itself, so if \(i\) is closed, its image \(A\) must be closed in \(X\). \(\square\)
Worked Example: A Closed Inclusion That Is Not Open
Give \(A=\{0\}\) the subspace topology inherited from the usual real line, and consider \(i:A\to\mathbb{R}\). The only subsets of \(A\) are \(\varnothing\) and \(A\), and both are open and closed in \(A\). Their images are \(\varnothing\) and \(\{0\}\), both closed in \(\mathbb{R}\), so \(i\) is closed. It is continuous by the proposition. It is not open: \(A\) is open in itself, but its image \(\{0\}\) is not open in \(\mathbb{R}\), since every interval around \(0\) contains nonzero real numbers.
Identity maps between different topologies give another precise comparison. Let \(\mathcal{T}_d\) be a topology on a set \(X\), used on the domain, and let \(\mathcal{T}_c\) be a topology on the same set, used on the codomain. Write \(\operatorname{id}_X:(X,\mathcal{T}_d)\to(X,\mathcal{T}_c)\) for the identity map.
Proof. For any \(V\subseteq X\), \(\operatorname{id}_X^{-1}(V)=V\). Therefore continuity says exactly that every set open in \(\mathcal{T}_c\) is open in \(\mathcal{T}_d\), which is \(\mathcal{T}_c\subseteq\mathcal{T}_d\). The image of any set under the identity is the set itself, so openness says exactly that every \(\mathcal{T}_d\)-open set is \(\mathcal{T}_c\)-open. For closedness, the same inclusion of open sets is equivalent, by taking complements in \(X\), to every \(\mathcal{T}_d\)-closed set being \(\mathcal{T}_c\)-closed. This is precisely the condition that the identity be closed. \(\square\)
Worked Example: A Continuous Identity That Is Neither Open Nor Closed
Put the discrete topology on \(\mathbb{R}\) as the domain and the usual topology on \(\mathbb{R}\) as the codomain. Every usual-open set is open in the discrete topology, so the identity is continuous. It is not open: the domain-open set \(\{0\}\) has image \(\{0\}\), which is not usual-open. It is not closed either: \((0,1)\) is closed in the discrete domain, since every subset is closed there, but its image is not closed in the usual real line. For instance, every interval around \(0\) meets \((0,1)\), while \(0\notin(0,1)\). Thus continuity alone guarantees neither openness nor closedness.
Keep the Tests Separate
A useful way to avoid confusion is to write down the relevant test before examining a map. For continuity, start with an arbitrary open set in the codomain and take its preimage. For openness, start with an arbitrary open set in the domain and take its image. For closedness, start with an arbitrary closed set in the domain and take its image. A single successful example of an open set or a closed set can disprove the corresponding mapping property; establishing the property requires checking all such sets, or applying a proved criterion such as the inclusion proposition.
Common mistake. It is incorrect to argue that a continuous map must be open because “open sets stay open.” That reverses the definition: continuity says preimages of codomain-open sets are domain-open. In the discrete-to-usual identity example, the preimage condition holds for every usual-open set, but the image of the domain-open singleton \(\{0\}\) is not open in the codomain. The correction is to test the required direction of the map on sets, rather than infer a property from the word “continuous.”
The examples also show that no pair of these properties should be silently identified: a projection can be continuous and open but not closed; an inclusion can be continuous and closed but not open; and an identity can be continuous without being either open or closed. The specific hypotheses and topologies determine the answer.
Check Your Understanding
For each question, identify whether the test concerns preimages or images, and use the stated topology.
- For the inclusion \(i:(0,1)\to\mathbb{R}\) with the usual topologies, determine whether it is continuous, open, and closed.
- For the identity from the usual topology on \(\mathbb{R}\) to the discrete topology on \(\mathbb{R}\), determine which of the three properties hold.
- Explain why the horizontal diameter used in the open-disk example is closed in the disk even though its image is not closed in \(\mathbb{R}\).
- State the exact condition on \(\mathcal{T}_d\) and \(\mathcal{T}_c\) for the identity from \((X,\mathcal{T}_d)\) to \((X,\mathcal{T}_c)\) to be continuous.
- Give a reason that proving a map sends one open set to an open set does not establish that it is an open map.