What a Homeomorphism Must Preserve
A homeomorphism is a bijection that respects the topology in both directions. This means it does more than match points: it matches the open sets that describe the spaces. As a result, some features can be used to rule out a homeomorphism before trying to construct one. We will examine three such features: cardinality, isolated points, and the pattern of open sets.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. A map \(f:X\to Y\) is a homeomorphism if it is a bijection, is continuous, and its inverse \(f^{-1}:Y\to X\) is continuous. A bijection matches points one-to-one, while continuity in both directions ensures that open sets correspond.
The Invariance Theorem
- \(X\) and \(Y\) have the same cardinality;
- \(f(\operatorname{Iso}(X))=\operatorname{Iso}(Y)\);
- for every \(U\subseteq X\), \(U\in\mathcal{T}_X\) if and only if \(f(U)\in\mathcal{T}_Y\).
Proof. By definition, \(f\) is a bijection, so it gives a one-to-one correspondence between the points of \(X\) and \(Y\). Thus the spaces have the same cardinality.
Let \(U\) be open in \(X\). Since \(f^{-1}:Y\to X\) is continuous, the preimage of \(U\) under \(f^{-1}\) is open in \(Y\). That preimage is \(f(U)\): a point \(y\in Y\) belongs to it exactly when \(f^{-1}(y)\in U\), or equivalently \(y\in f(U)\). Hence \(f(U)\) is open in \(Y\). Conversely, if \(V\) is open in \(Y\), continuity of \(f\) implies \(f^{-1}(V)\) is open in \(X\). Since \(f\) is onto, \(f(f^{-1}(V))=V\). Therefore \(U\) is open if and only if \(f(U)\) is open.
Apply this open-set correspondence to \(U=\{x\}\). Since \(f(\{x\})=\{f(x)\}\), the singleton \(\{x\}\) is open in \(X\) if and only if \(\{f(x)\}\) is open in \(Y\). Thus \(x\) is isolated if and only if \(f(x)\) is isolated. Because \(f\) is onto, every isolated point of \(Y\) is \(f(x)\) for some \(x\in X\); the equivalence just proved makes that \(x\) isolated. This proves \(f(\operatorname{Iso}(X))=\operatorname{Iso}(Y)\).
Finally, a bijection maps inclusions to inclusions and satisfies \(f(\bigcup_i U_i)=\bigcup_i f(U_i)\) and \(f(U\cap W)=f(U)\cap f(W)\). The latter equality uses injectivity: if a point lies in both images, its two preimages must be the same point. The open-set correspondence therefore preserves inclusion, unions, and intersections. \(\square\)
The theorem gives necessary conditions for two spaces to be homeomorphic. If even one of these conditions fails, no homeomorphism exists. Passing these tests does not by itself construct a homeomorphism; the tests identify obstructions, rather than providing a complete classification.
Using Cardinality and Isolated Points
Worked Example: Different Numbers of Points
Let \(X=\{a,b\}\) and \(Y=\{u,v,w\}\), each with its discrete topology (every subset is open). Could \(X\) and \(Y\) be homeomorphic?
The set \(X\) has two elements and \(Y\) has three. A homeomorphism must be a bijection, but no bijection can match two elements with three: if each point of \(X\) is assigned a distinct point of \(Y\), one point of \(Y\) remains unmatched. Thus the spaces are not homeomorphic. The topologies do not remove this obstruction: the theorem applies regardless of which topologies are placed on the sets.
Check: Both spaces have every point isolated, because every singleton is open in the discrete topology. That invariant agrees, but cardinality does not, so the cardinality obstruction is decisive.
Worked Example: Same Cardinality, Different Isolated Points
Take the same underlying set \(X=\{a,b,c\}\) with two different topologies. Let \(\mathcal{T}_D=\mathcal{P}(X)\) be the discrete topology, and let \(\mathcal{T}_I=\{\varnothing,X\}\) be the indiscrete topology. Are \((X,\mathcal{T}_D)\) and \((X,\mathcal{T}_I)\) homeomorphic?
In \((X,\mathcal{T}_D)\), every subset is open, so \(\{a\},\{b\},\{c\}\) are open and \(\operatorname{Iso}(X)=X\). In \((X,\mathcal{T}_I)\), the only open sets are \(\varnothing\) and \(X\). None of the singletons is open, since each singleton is nonempty and is not all of \(X\). Therefore \(\operatorname{Iso}(X)=\varnothing\) in the indiscrete topology.
The two spaces have the same three points, but their numbers of isolated points differ: three versus zero. A homeomorphism would have to carry isolated points exactly to isolated points, so no homeomorphism exists.
Check: The open sets themselves also differ. The discrete topology has \(2^3=8\) open sets, one for each subset of \(X\); the indiscrete topology has \(2\). This is consistent with the theorem, which requires a homeomorphism to give a bijection between the open sets.
Open-Set Patterns as an Invariant
The open-set correspondence is stronger than a comparison of how many open sets there are. Any pattern expressed entirely in terms of open sets and their intersections, inclusions, or unions must be shared by homeomorphic spaces. For example, consider the property that every two nonempty open sets intersect. If \(f\) is a homeomorphism and \(U,V\) are nonempty open sets in \(X\), then \(f(U)\) and \(f(V)\) are nonempty open sets in \(Y\). Also, injectivity gives \(f(U\cap V)=f(U)\cap f(V)\). Thus \(U\cap V\) is empty exactly when \(f(U)\cap f(V)\) is empty. The property is preserved in either direction.
Worked Example: The Standard and Finite-Complement Topologies
Compare \(\mathbb{R}\) with its standard topology and \(\mathbb{R}\) with the finite-complement topology, in which the open sets are \(\varnothing\) and all subsets with finite complement. These spaces have the same underlying set, so they have the same cardinality. Neither has isolated points, but they are not homeomorphic.
First, neither space has isolated points. In the standard topology, every open neighborhood of a real number \(x\) contains an open interval around \(x\), and that interval contains points other than \(x\). Thus \(\{x\}\) is not open. In the finite-complement topology, \(\{x\}\) is nonempty but its complement \(\mathbb{R}\setminus\{x\}\) is infinite, so it is not open either.
Now examine intersections of nonempty open sets. In the finite-complement topology, let \(U\) and \(V\) be nonempty open sets. Their complements are finite, so \(\mathbb{R}\setminus(U\cap V)=(\mathbb{R}\setminus U)\cup(\mathbb{R}\setminus V)\) is finite. Since \(\mathbb{R}\) is infinite, \(U\cap V\) cannot be empty: if it were empty, its complement would be all of \(\mathbb{R}\), which is not finite. Thus every two nonempty open sets in this topology intersect.
In the standard topology, \((0,1)\) and \((2,3)\) are nonempty open sets and are disjoint, since every point of the first is less than \(1\), while every point of the second is greater than \(2\). The two spaces therefore have different open-set patterns. By the invariance theorem, no homeomorphism between them exists.
Check: The obstruction is not cardinality or the number of isolated points; those invariants agree. It is the existence of two disjoint nonempty open sets, a property the finite-complement topology on \(\mathbb{R}\) does not have.
A Common Pitfall: A Bijection Need Not Preserve Open Sets
It is tempting to think that a bijection automatically carries open sets to open sets. The following calculation shows why that is wrong. Use the standard topology on \(\mathbb{R}\), and define \(f:\mathbb{R}\to\mathbb{R}\) by swapping \(0\) and \(1\) and fixing every other point. This map is a bijection: applying it twice returns every point to its starting value.
Consider the open interval \(U=(-1/2,1/2)\). It contains \(0\) but not \(1\). Every point of \(U\) except \(0\) is fixed, and \(0\) maps to \(1\). Therefore \(f(U)=((-1/2,1/2)\setminus\{0\})\cup\{1\}\). This set is not open: it contains \(1\), but for any \(r>0\), the interval \((1-r,1+r)\) contains points greater than \(1/2\) and sufficiently close to \(1\), and those points are not in \(f(U)\). Hence \(f(U)\) is not a neighborhood of \(1\). Since a homeomorphism must take open sets to open sets, this bijection is not a homeomorphism.
The correct test is not whether a proposed map is bijective, but whether it is a bijection that preserves the topology in both directions. The invariants in this tutorial are useful because they can rule out a homeomorphism without needing to test a proposed formula: compare point counts, isolated points, or specific features of the open-set structure.
Check Your Understanding
For each question, identify which invariant applies and state the reasoning.
- Can a space with four points be homeomorphic to a space with five points? Why?
- In a space with the discrete topology, which points are isolated? What does this imply about any space homeomorphic to it?
- Why can the discrete and indiscrete topologies on a three-point set not be homeomorphic?
- Suppose one space has two disjoint nonempty open sets, while every two nonempty open sets in another space intersect. What does this tell you?
- Must a bijection between two topological spaces carry every open set to an open set? Give the reason your answer follows from the definition of homeomorphism.