Why List the Open Sets?
For a finite space, the topology is a finite collection of subsets, so continuity can be checked by a finite calculation. This makes finite examples useful for seeing exactly what a homeomorphism must preserve: not distances or labels, but the pattern of open sets. The reliable procedure is to list the open sets in both spaces, compute preimages for continuity, and check the inverse map as well.
Recall that a map \(f:X\to Y\) is continuous if \(f^{-1}(U)\) is open in \(X\) for every open set \(U\) in \(Y\). A homeomorphism is a bijection \(f:X\to Y\) for which both \(f\) and \(f^{-1}:Y\to X\) are continuous. All the spaces below are finite, and their topologies are given explicitly.
A Criterion Using Open Sets
Proof. Suppose first that \(f\) is a homeomorphism. If \(V\in\mathcal{T}_X\), continuity of \(f^{-1}:Y\to X\) says that \((f^{-1})^{-1}(V)\) is open in \(Y\). Since \(f\) is bijective, \((f^{-1})^{-1}(V)=f(V)\). Thus \(f(V)\in\mathcal{T}_Y\), so \(\{f(V):V\in\mathcal{T}_X\}\subseteq\mathcal{T}_Y\). Conversely, if \(U\in\mathcal{T}_Y\), continuity of \(f\) says \(f^{-1}(U)\in\mathcal{T}_X\), and bijectivity gives \(f(f^{-1}(U))=U\). Hence \(U\) belongs to \(\{f(V):V\in\mathcal{T}_X\}\). This proves equality.
Now suppose the displayed equality holds. For every \(U\in\mathcal{T}_Y\), equality supplies \(V\in\mathcal{T}_X\) with \(U=f(V)\). Bijectivity then gives \(f^{-1}(U)=V\), which is open in \(X\). Therefore \(f\) is continuous. For every \(V\in\mathcal{T}_X\), the set \(f(V)\) is open in \(Y\), and \((f^{-1})^{-1}(V)=f(V)\). Thus \(f^{-1}\) is continuous as well, so \(f\) is a homeomorphism. \(\square\)
This criterion converts the two continuity checks into one comparison of open-set lists, provided the proposed map is bijective. In an actual calculation, it is often clearest to show both lists and match each open set under the map.
Worked Example: Matching Two Nested Open-Set Lists
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}_X=\{\varnothing,\{a\},\{a,b\},X\}\), and let \(Y=\{1,2,3\}\) have topology \(\mathcal{T}_Y=\{\varnothing,\{2\},\{2,3\},Y\}\). These are topologies: in each list, the nonempty proper open sets are nested, so unions and finite intersections of listed sets remain listed. Define \(f(a)=2\), \(f(b)=3\), and \(f(c)=1\). Determine whether \(f\) is a homeomorphism.
The three images \(2,3,1\) are distinct and exhaust \(Y\), so \(f\) is bijective. The images of the four open sets of \(X\) are \[ f(\varnothing)=\varnothing,\qquad f(\{a\})=\{2\},\qquad f(\{a,b\})=\{2,3\},\qquad f(X)=Y. \] These are exactly the four open sets of \(Y\). By the theorem, \(f\) is a homeomorphism.
We can also check continuity directly. The preimages of the open sets in \(Y\) are \[ f^{-1}(\varnothing)=\varnothing,\qquad f^{-1}(\{2\})=\{a\},\qquad f^{-1}(\{2,3\})=\{a,b\},\qquad f^{-1}(Y)=X. \] Every one is open in \(X\). For the inverse, the images of the open sets in \(X\), listed above, are all open in \(Y\); equivalently, the preimage under \(f^{-1}\) of each open set in \(X\) is open in \(Y\). Thus both continuity requirements are checked.
Check: the open-set lists correspond term by term, and the four preimages just computed are precisely \(\varnothing,\{a\},\{a,b\},X\), the full topology on \(X\).
When Every Bijection Works
Some finite topologies distinguish no points by their open sets. The indiscrete topology on a set \(X\) is \(\{\varnothing,X\}\). Between two indiscrete spaces, the only continuity checks are the preimages of the empty set and the whole target. For a bijection, these preimages are again the empty set and the whole domain.
Worked Example: All Bijections Between Indiscrete Spaces
Let \(X=\{a,b,c\}\) and \(Y=\{1,2,3\}\), each with its indiscrete topology. Consider the bijection \(g:X\to Y\) defined by \(g(a)=3\), \(g(b)=1\), and \(g(c)=2\). The target open sets are \(\varnothing\) and \(Y\), and \[ g^{-1}(\varnothing)=\varnothing,\qquad g^{-1}(Y)=X. \] Both are open in \(X\), so \(g\) is continuous.
The inverse is given by \(g^{-1}(1)=b\), \(g^{-1}(2)=c\), and \(g^{-1}(3)=a\). The open sets in its target \(X\) are \(\varnothing\) and \(X\), whose preimages under \(g^{-1}\) are \(\varnothing\) and \(Y\). Both are open in \(Y\); hence \(g^{-1}\) is continuous and \(g\) is a homeomorphism.
In fact, this argument applies to every bijection between these two indiscrete spaces: the preimage of the empty set is empty, and the preimage of the whole target is the whole domain, regardless of how the points are matched. There are \(3\cdot2\cdot1=6\) bijections here, and all six are homeomorphisms. The calculation depends on the listed topologies, not on a special choice of labels or formula.
Check: the six possible bijections arise by choosing the image of \(a\) in \(3\) ways, the image of \(b\) from the \(2\) remaining points, and then the image of \(c\) from the \(1\) remaining point. Each has the same two continuity checks above.
Open-Set Invariants Can Rule Out a Homeomorphism
A homeomorphism sends open sets bijectively to open sets. Consequently, homeomorphic spaces must have the same number of open sets. More precisely, a homeomorphism preserves the number of points in each open set: if \(f\) is a bijection, then \(|f(V)|=|V|\) for every \(V\subseteq X\).
Proof. By the theorem, \(f\) sends every open set of \(X\) to an open set of \(Y\), and every open set of \(Y\) is the image of an open set of \(X\). This correspondence is one-to-one: if \(f(V_1)=f(V_2)\), injectivity of \(f\) implies \(V_1=V_2\). Therefore the two topologies have the same number of open sets. Since \(f\) is a bijection of points, its restriction to \(V\) is a bijection from \(V\) onto \(f(V)\), so \(|V|=|f(V)|\). The correspondence therefore preserves the sizes of open sets. \(\square\)
Worked Example: The Same Number of Open Sets Does Not Suffice
Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}_X=\{\varnothing,\{a\},X\}\), and let \(Y=\{1,2,3\}\) have topology \(\mathcal{T}_Y=\{\varnothing,\{1,2\},Y\}\). Each list is a topology: it contains the empty set and the whole space, and unions and finite intersections of its listed sets remain listed. Both spaces have exactly three open sets.
Suppose there were a homeomorphism \(f:X\to Y\). The open set \(\{a\}\) would have to map to an open set of \(Y\) containing exactly one point, because \(f\) is a bijection and \(|\{a\}|=1\). But the only open sets of \(Y\) are \(\varnothing\), \(\{1,2\}\), and \(Y\), of sizes \(0,2,3\) respectively. None has size \(1\). This contradicts preservation of open sets and their sizes, so the spaces are not homeomorphic.
Thus matching the number of open sets is a necessary condition, not a sufficient one. Here both topologies have three open sets, but their open sets have different size patterns: \(0,1,3\) for \(X\) and \(0,2,3\) for \(Y\).
Check: a bijection would send a one-point subset to a one-point subset, since it sends its sole element to exactly one element. The list for \(Y\) contains no one-point open set, confirming the obstruction.
A Practical Finite-Space Procedure
For a proposed homeomorphism between finite spaces, use the following sequence of checks. First verify that the proposed map is a bijection. Next list all open sets in the domain and target. Compute the image of each domain open set, or compute the preimage of each target open set. If the image list is exactly the target topology, the criterion proves that the map and its inverse are continuous. If it is not, the map is not a homeomorphism. When looking for any homeomorphism rather than checking a given one, compare simple invariants first: the number of points, the number of open sets, and the sizes of open sets. These can rule out a match quickly, although matching them does not by itself construct a homeomorphism.
Common pitfall. It is tempting to conclude that two finite spaces are homeomorphic as soon as their topologies contain the same number of open sets. In the preceding example, both topologies contain three open sets, but one has a one-point open set and the other does not. The correction is to treat the count only as a necessary test, then check the actual open-set pattern under a proposed bijection.
Check Your Understanding
For each question, use the listed open sets rather than relying on a picture or on the names of the points.
- Let \(X=\{p,q\}\) and \(Y=\{r,s\}\) both have the discrete topology, whose open sets are all subsets. Show that every bijection \(X\to Y\) is a homeomorphism by checking the open sets.
- For the spaces in the first worked example, compute \(f^{-1}(\{2,3\})\) and \(f(\{b,c\})\). Which of these sets is open in its respective space?
- Let \(X=\{a,b,c\}\) have topology \(\{\varnothing,\{a\},X\}\). Can it be homeomorphic to a three-point space with topology \(\{\varnothing,\{1,3\},Y\}\)? Explain using the sizes of open sets.
- Why does equality of the numbers of open sets not, by itself, prove that two finite spaces are homeomorphic?