Comparing Topologies on One Set
A set can carry many different topologies, so the same identity function can behave differently depending on which topology is used on its domain and codomain. The key is to compare the collections of open sets. In particular, continuity tests open sets in the codomain by taking preimages, while openness tests open sets in the domain by taking images.
Throughout, let \(\mathcal{T}_1\) and \(\mathcal{T}_2\) be topologies on the same set \(X\). The notation \((X,\mathcal{T}_1)\) and \((X,\mathcal{T}_2)\) distinguishes the two spaces even though their underlying set is unchanged. The identity function from the first space to the second is \(\operatorname{id}_X(x)=x\) for every \(x\in X\).
Two topologies are comparable if one is finer than the other. They are incomparable if neither collection of open sets contains the other. Equality counts as comparability: if \(\mathcal{T}_1=\mathcal{T}_2\), each is finer than the other.
The Three Identity-Map Tests
- \(\operatorname{id}_X:(X,\mathcal{T}_1)\to(X,\mathcal{T}_2)\) is continuous if and only if \(\mathcal{T}_2\subseteq\mathcal{T}_1\).
- \(\operatorname{id}_X:(X,\mathcal{T}_1)\to(X,\mathcal{T}_2)\) is open if and only if \(\mathcal{T}_1\subseteq\mathcal{T}_2\).
- \(\operatorname{id}_X:(X,\mathcal{T}_1)\to(X,\mathcal{T}_2)\) is closed if and only if \(\mathcal{T}_1\subseteq\mathcal{T}_2\).
Proof. For continuity, the definition says that the preimage of every \(\mathcal{T}_2\)-open set must be \(\mathcal{T}_1\)-open. For every subset \(U\subseteq X\), the identity has \(\operatorname{id}_X^{-1}(U)=U\). Therefore continuity holds exactly when every member of \(\mathcal{T}_2\) belongs to \(\mathcal{T}_1\), which is \(\mathcal{T}_2\subseteq\mathcal{T}_1\).
For openness, every \(\mathcal{T}_1\)-open set \(U\) must have an open image in the codomain. Its image under the identity is \(U\) itself, so this holds exactly when every member of \(\mathcal{T}_1\) belongs to \(\mathcal{T}_2\). Hence the condition is \(\mathcal{T}_1\subseteq\mathcal{T}_2\).
For closedness, every set closed in \((X,\mathcal{T}_1)\) must be closed in \((X,\mathcal{T}_2)\). If \(\mathcal{T}_1\subseteq\mathcal{T}_2\), then for every \(\mathcal{T}_1\)-closed set \(C\), the complement \(X\setminus C\) belongs to \(\mathcal{T}_1\), hence to \(\mathcal{T}_2\). Thus \(C\) is \(\mathcal{T}_2\)-closed. Conversely, suppose the identity is closed and \(U\in\mathcal{T}_1\). Then \(X\setminus U\) is \(\mathcal{T}_1\)-closed, so it is \(\mathcal{T}_2\)-closed. Its complement \(U\) is therefore \(\mathcal{T}_2\)-open. This proves \(\mathcal{T}_1\subseteq\mathcal{T}_2\). The arguments also apply when \(X=\varnothing\), since both topologies then contain only \(\varnothing\). \(\square\)
The two criteria in the second and third bullets agree: an identity map is open exactly when it is closed. Continuity points in the opposite direction. If the domain topology is finer, continuity holds; if the codomain topology is finer, openness and closedness hold.
Worked Example: Discrete and Coarser Topologies on Three Points
Let \(X=\{a,b,c\}\), let \(\mathcal{T}_1=\mathcal{P}(X)\) be the discrete topology, and let \(\mathcal{T}_2=\{\varnothing,\{a\},X\}\). The second collection is a topology: it contains \(\varnothing\) and \(X\), and its only nontrivial unions and finite intersections remain in the collection. Also \(\mathcal{T}_2\subseteq\mathcal{T}_1\), since all three of its sets are subsets of \(X\).
Consider first the identity from the discrete space to \((X,\mathcal{T}_2)\). The inclusion \(\mathcal{T}_2\subseteq\mathcal{T}_1\) gives continuity. But \(\mathcal{T}_1\not\subseteq\mathcal{T}_2\), because \(\{b\}\in\mathcal{T}_1\) and \(\{b\}\notin\mathcal{T}_2\). The map is neither open nor closed by the theorem. Directly, \(\{b\}\) is a discrete-open set whose image is not \(\mathcal{T}_2\)-open. Also \(\{a\}\) is closed in the discrete space, but not in \((X,\mathcal{T}_2)\), since its complement \(\{b,c\}\) is not \(\mathcal{T}_2\)-open.
For the identity in the reverse direction, continuity fails: the codomain-open set \(\{b\}\) has preimage \(\{b\}\), which is not \(\mathcal{T}_2\)-open. This reverse identity is open and closed, because \(\mathcal{T}_2\subseteq\mathcal{T}_1\). For a direct check, every set in \(\mathcal{T}_2\) is discrete-open, and every \(\mathcal{T}_2\)-closed set is discrete-closed.
Check: the discrete topology is finer than \(\mathcal{T}_2\). So the identity from the finer topology to the coarser one is continuous, while the identity the other way is open and closed.
Comparing the Standard and Lower Limit Topologies
On \(\mathbb{R}\), the standard topology has open intervals as a basis. The lower limit topology, denoted \(\mathbb{R}_\ell\), has the half-open intervals \([a,b)\), where \(a<b\), as a basis. A basis means that open sets are unions of basis elements. We can compare these topologies by checking whether each standard open set is lower-limit open and whether each lower-limit basis element is standard open.
Worked Example: The Identity Between \(\mathbb{R}_\ell\) and \(\mathbb{R}\)
First, every standard-open set \(U\) is lower-limit open. If \(x\in U\), the definition of the standard topology gives some \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). The lower-limit basis interval \([x,x+\varepsilon)\) contains \(x\) and is contained in \(U\). Thus each point of \(U\) has a lower-limit basis neighborhood inside \(U\). This proves that the standard topology is contained in the lower limit topology.
The containment is strict. The set \([0,1)\) is a lower-limit basis element, so it is lower-limit open. It is not standard-open: any standard-open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and that interval contains negative numbers, which are not in \([0,1)\). Therefore the lower limit topology is strictly finer than the standard topology.
It follows that the identity \(\mathbb{R}_\ell\to\mathbb{R}\) is continuous. It is neither open nor closed, because the lower limit topology is not contained in the standard topology. The set \([0,1)\) witnesses failure of openness. For closedness, \(C=\mathbb{R}\setminus[0,1)\) is lower-limit closed but not standard-closed: its complement \([0,1)\) is not standard-open.
In the reverse direction, the identity \(\mathbb{R}\to\mathbb{R}_\ell\) is not continuous. The lower-limit-open set \([0,1)\) has itself as its preimage, which is not standard-open. This reverse identity is both open and closed, because every standard-open set is lower-limit open, and the complement argument shows every standard-closed set is lower-limit closed.
Check: the standard topology is a proper subset of the lower limit topology. The continuous direction is from the lower limit topology to the standard topology; the open and closed direction is from the standard topology to the lower limit topology.
Finite-Complement and Standard Topologies
The finite-complement topology on \(\mathbb{R}\) consists of \(\varnothing\) and all subsets whose complement in \(\mathbb{R}\) is finite. Its open sets are all standard-open as well. Indeed, if \(F\subseteq\mathbb{R}\) is finite and \(x\notin F\), then, when \(F\ne\varnothing\), the finitely many positive distances \(|x-y|\) for \(y\in F\) have a positive minimum. An interval about \(x\) of radius less than that minimum avoids \(F\). Thus \(\mathbb{R}\setminus F\) is standard-open. When \(F=\varnothing\), the complement is \(\mathbb{R}\), which is standard-open.
Worked Example: Identity Maps Involving the Finite-Complement Topology
Let \(\mathcal{T}_{\mathrm{fc}}\) be the finite-complement topology on \(\mathbb{R}\), and let \(\mathcal{T}_{\mathrm{std}}\) be the standard topology. The preceding argument proves \(\mathcal{T}_{\mathrm{fc}}\subseteq\mathcal{T}_{\mathrm{std}}\). The inclusion is strict: \((0,1)\) is standard-open, but its complement is infinite, so it is not finite-complement open.
Consequently, the identity \((\mathbb{R},\mathcal{T}_{\mathrm{std}})\to(\mathbb{R},\mathcal{T}_{\mathrm{fc}})\) is continuous. It is neither open nor closed. The standard-open set \((0,1)\) is not finite-complement open, so it witnesses failure of openness. The set \([0,1]\) is standard-closed, but it is not finite-complement closed: closed sets in the finite-complement topology are \(\mathbb{R}\) and the finite sets, whereas \([0,1]\) is infinite and is not all of \(\mathbb{R}\).
For the reverse identity, continuity fails because the standard-open set \((0,1)\) has a preimage that is not finite-complement open. The map is open and closed: every finite-complement-open set is standard-open, and every finite-complement-closed set (a finite set or \(\mathbb{R}\)) is standard-closed.
Check: \((0,1)\) has infinitely many points and its complement has infinitely many points, so it is not finite-complement open; \([0,1]\) is infinite and proper, so it is not finite-complement closed. These witnesses agree with the topology-inclusion criteria.
When the Topologies Are Incomparable
If neither topology contains the other, none of the three properties holds for either direction of the identity. The continuity test needs one inclusion, while the open and closed tests need the reverse inclusion. Incomparability rules out both.
Worked Example: Two Incomparable Topologies on Three Points
Let \(X=\{a,b,c\}\), \(\mathcal{T}_a=\{\varnothing,\{a\},X\}\), and \(\mathcal{T}_b=\{\varnothing,\{b\},X\}\). Each collection is a topology: it contains the empty set and \(X\), and unions and finite intersections of its listed sets remain listed. Neither topology contains the other, since \(\{a\}\in\mathcal{T}_a\setminus\mathcal{T}_b\) and \(\{b\}\in\mathcal{T}_b\setminus\mathcal{T}_a\).
For the identity from \((X,\mathcal{T}_a)\) to \((X,\mathcal{T}_b)\), continuity fails because the target-open set \(\{b\}\) has preimage \(\{b\}\), which is not \(\mathcal{T}_a\)-open. Openness fails because the domain-open set \(\{a\}\) has image \(\{a\}\), which is not \(\mathcal{T}_b\)-open. To check closedness directly, \(\{b,c\}\) is closed in \((X,\mathcal{T}_a)\), since its complement \(\{a\}\) is open. It is not closed in \((X,\mathcal{T}_b)\), since its complement \(\{a\}\) is not \(\mathcal{T}_b\)-open.
For the reverse identity, continuity fails because the target-open set \(\{a\}\) has preimage \(\{a\}\), not open in \(\mathcal{T}_b\). Openness fails using the domain-open set \(\{b\}\), which is not \(\mathcal{T}_a\)-open. Closedness fails because \(\{a,c\}\) is closed in \((X,\mathcal{T}_b)\), but not in \((X,\mathcal{T}_a)\): its complement \(\{b\}\) is not \(\mathcal{T}_a\)-open.
Check: each failure is witnessed by a set whose membership in one topology does not carry over to the other. The two topologies are incomparable, so there is no identity direction with any of the three properties.
A Common Direction Error
Common pitfall. One might guess that a finer codomain topology makes the identity continuous because it has more open sets. That reverses the preimage test. For example, let \(X=\{a,b\}\), with the discrete topology \(\mathcal{T}_d=\mathcal{P}(X)\) and the indiscrete topology \(\mathcal{T}_i=\{\varnothing,X\}\). The identity \((X,\mathcal{T}_d)\to(X,\mathcal{T}_i)\) is continuous: the preimages of the only target-open sets, \(\varnothing\) and \(X\), are open in the discrete domain. But the reverse identity is not continuous, because the codomain-open set \(\{a\}\) has preimage \(\{a\}\), which is not open in the indiscrete domain.
A reliable procedure is to write down the domain topology first and the codomain topology second. For continuity, check whether every codomain-open set is domain-open. For openness, check whether every domain-open set is codomain-open. For closedness, use the same inclusion as for openness, or apply the complement definition directly.
Check Your Understanding
For each question, identify the domain and codomain topology before applying an inclusion test.
- If \(\mathcal{T}_2\subseteq\mathcal{T}_1\), which identity direction is continuous?
- For the identity \((X,\mathcal{T}_1)\to(X,\mathcal{T}_2)\), state the inclusion condition for openness and for closedness.
- In the standard and lower limit topologies on \(\mathbb{R}\), which topology is finer? Give a set witnessing strictness.
- For the finite-complement and standard topologies on \(\mathbb{R}\), which identity direction is continuous?
- If \(\mathcal{T}_1\) and \(\mathcal{T}_2\) are incomparable, can either identity direction be open? Explain using the inclusion criterion.