A Two-Point Space with One Open Singleton
On a two-point set, the discrete topology makes every subset open, while the indiscrete topology makes only the empty set and the whole space open. The Sierpinski topology lies strictly between these extremes: it makes one singleton open and leaves the other singleton nonopen. That small asymmetry is useful because it gives a concrete example in which the two points have different collections of neighborhoods.
Let \(X=\{a,b\}\), where \(a\ne b\). The chosen point \(a\) will be the point whose singleton is open. Specifying the points matters: the topology below treats \(a\) and \(b\) differently.
To verify that \(\mathcal{T}_{S}\) really is a topology, we use the topology axioms: the empty set and the whole space must belong to the collection; an arbitrary union of its members must belong to it; and the intersection of finitely many of its members must belong to it. The empty union is \(\varnothing\), and the intersection of an empty finite family is \(X\).
Verifying the Topology Axioms
Proof. The two required sets are present: \(\varnothing\in\mathcal{T}_{S}\) and \(X\in\mathcal{T}_{S}\). For arbitrary unions, consider any family of members of \(\mathcal{T}_{S}\). If the family has no members, its union is \(\varnothing\). If at least one member is \(X\), the union is \(X\). If no member is \(X\) but at least one member is \(\{a\}\), the union is \(\{a\}\). If every member is \(\varnothing\), the union is \(\varnothing\). Every possible union is therefore in \(\mathcal{T}_{S}\).
For finite intersections, the intersection of an empty family is \(X\). For a nonempty finite family, if one member is \(\varnothing\), the intersection is \(\varnothing\). Otherwise, all members are either \(\{a\}\) or \(X\). If at least one is \(\{a\}\), the intersection is \(\{a\}\); if all are \(X\), the intersection is \(X\). Each possibility belongs to \(\mathcal{T}_{S}\), so all the topology axioms hold. \(\square\)
The union and intersection checks work because the three sets are ordered by inclusion: \(\varnothing\subseteq\{a\}\subseteq X\). In particular, a union cannot produce \(\{b\}\): none of the allowed sets contains \(b\) without also containing \(a\). An intersection cannot produce \(\{b\}\) either. This explains why the absent singleton does not appear when the axioms are applied.
Worked Example: Checking the Axioms for Specific Families
Let \(X=\{a,b\}\) and \(\mathcal{T}_{S}=\{\varnothing,\{a\},X\}\). Check a union and an intersection directly, including the sets that result.
Take the family \(\{U_1,U_2,U_3\}\) with \(U_1=\{a\}\), \(U_2=X\), and \(U_3=\varnothing\). Then \(U_1\cup U_2\cup U_3=\{a\}\cup\{a,b\}\cup\varnothing=\{a,b\}=X\), which belongs to \(\mathcal{T}_{S}\).
For a finite intersection, take \(V_1=\{a\}\), \(V_2=X\), and \(V_3=\{a\}\). Then \(V_1\cap V_2\cap V_3=\{a\}\cap\{a,b\}\cap\{a\}=\{a\}\), also in \(\mathcal{T}_{S}\). As an additional check on the role of \(\varnothing\), \(U_1\cap U_3=\{a\}\cap\varnothing=\varnothing\), which is allowed. Thus these computations agree with the general axiom verification: unions and finite intersections stay in the collection.
Comparing the Three Topologies
For topologies on the same set, \(\mathcal{T}_1\subseteq\mathcal{T}_2\) means every \(\mathcal{T}_1\)-open set is also \(\mathcal{T}_2\)-open. In that case, \(\mathcal{T}_2\) is finer than \(\mathcal{T}_1\), and \(\mathcal{T}_1\) is coarser than \(\mathcal{T}_2\). Comparing the three collections on \(X=\{a,b\}\) gives an inclusion chain.
Proof. The indiscrete topology is \(\{\varnothing,X\}\), and both of its members belong to \(\mathcal{T}_{S}\). Thus \(\{\varnothing,X\}\subseteq\mathcal{T}_{S}\). The inclusion is strict because \(\{a\}\in\mathcal{T}_{S}\) but \(\{a\}\notin\{\varnothing,X\}\). Next, each member of \(\mathcal{T}_{S}\) is a subset of \(X\), so each belongs to the discrete topology \(\mathcal{P}(X)\). Thus \(\mathcal{T}_{S}\subseteq\mathcal{P}(X)\). This inclusion is strict because \(\{b\}\in\mathcal{P}(X)\) but \(\{b\}\notin\mathcal{T}_{S}\). Therefore both inclusions are strict. \(\square\)
Worked Example: Testing Set Membership in Each Topology
On \(X=\{a,b\}\), decide whether each of \(\varnothing\), \(\{a\}\), \(\{b\}\), and \(X\) is open in the indiscrete, Sierpinski, and discrete topologies.
By definition, the indiscrete topology contains exactly \(\varnothing\) and \(X\). Therefore \(\varnothing\) and \(X\) are open in it, while neither singleton is open. The Sierpinski topology contains exactly \(\varnothing\), \(\{a\}\), and \(X\). Therefore \(\{a\}\) is open and \(\{b\}\) is not. The discrete topology is \(\mathcal{P}(X)\), whose members are all four subsets, so both singletons as well as \(\varnothing\) and \(X\) are open.
Check: the displayed lists have respectively two, three, and four members. The inclusions are strict for the same witnesses as before: \(\{a\}\) is in the Sierpinski topology but not the indiscrete one, and \(\{b\}\) is in the discrete topology but not the Sierpinski one.
Here is the comparison in one place. “Not open” means the set is absent from that topology; it does not mean the set is not a subset of \(X\).
| Subset of \(X\) | Indiscrete | Sierpinski | Discrete |
|---|---|---|---|
| \(\varnothing\) | Open | Open | Open |
| \(\{a\}\) | Not open | Open | Open |
| \(\{b\}\) | Not open | Not open | Open |
| \(X\) | Open | Open | Open |
Neighborhoods and Closed Sets
An open neighborhood of a point is an open set containing that point. The open neighborhoods in the Sierpinski space can be read directly from its three open sets. The point \(a\) belongs to \(\{a\}\) and \(X\), while \(b\) belongs only to \(X\). Thus \(a\) has two open neighborhoods and \(b\) has one.
Proof. The open sets are exactly \(\varnothing,\{a\},X\). The sets among these containing \(a\) are \(\{a\}\) and \(X\), and the only one containing \(b\) is \(X\). This proves the neighborhood statements. A set is closed exactly when its complement in \(X\) is open. The complements of \(\varnothing,\{a\},X\) are respectively \(X,\{b\},\varnothing\), so the closed sets are exactly \(X,\{b\},\varnothing\). \(\square\)
Worked Example: Finding Neighborhoods and Closed Sets
For \(X=\{a,b\}\) with \(\mathcal{T}_{S}=\{\varnothing,\{a\},X\}\), find all open neighborhoods of each point and determine whether \(\{b\}\) is open or closed.
To find the open neighborhoods of \(a\), test each open set for membership of \(a\): \(\varnothing\) does not contain \(a\), \(\{a\}\) does, and \(X\) does. The answer is \(\{a\}\) and \(X\). For \(b\), \(\varnothing\) and \(\{a\}\) do not contain \(b\), while \(X\) does, so \(X\) is its only open neighborhood.
The set \(\{b\}\) is not open because it is absent from \(\mathcal{T}_{S}\). Its complement is \(X\setminus\{b\}=\{a\}\), which is open, so \(\{b\}\) is closed. Check: this agrees with the full closed-set list \(\{\varnothing,\{b\},X\}\).
These neighborhood lists make the asymmetry precise: \(a\) has a neighborhood that excludes \(b\), namely \(\{a\}\), whereas every open neighborhood of \(b\) also contains \(a\). No such distinction occurs in the discrete topology, where each point has its own singleton as an open neighborhood. In the indiscrete topology, each point has only \(X\) as an open neighborhood.
A Common Mistake
Common pitfall. Seeing that \(\{a\}\) is open and concluding that the Sierpinski topology is discrete confuses “one singleton is open” with “every subset is open.” The discrete topology on \(X=\{a,b\}\) must contain \(\{b\}\) as well as \(\{a\}\). But \(\{b\}\notin\mathcal{T}_{S}\), so \(\mathcal{T}_{S}\ne\mathcal{P}(X)\). The corrected comparison is \(\{\varnothing,X\}\subsetneq\{\varnothing,\{a\},X\}\subsetneq\mathcal{P}(X)\). The first inclusion adds \(\{a\}\); the second adds \(\{b\}\).
The Sierpinski topology is therefore a simple test case for reasoning directly from a list of open sets. Its topology axioms can be checked without geometric intuition, its position between the two extreme topologies is witnessed by specific subsets, and its different neighborhood lists show exactly where the asymmetry lies.
Check Your Understanding
For each question, use \(X=\{a,b\}\) and \(\mathcal{T}_{S}=\{\varnothing,\{a\},X\}\).
- Compute \(\{a\}\cup\{b\}\) and \(\{a\}\cap X\). Which results belong to \(\mathcal{T}_{S}\)?
- Which subsets of \(X\) are open in the Sierpinski topology but not in the indiscrete topology?
- Which subsets of \(X\) are open in the discrete topology but not in the Sierpinski topology?
- List the open neighborhoods of \(a\) and of \(b\). Why are the lists different?
- Is \(\{a\}\) closed in the Sierpinski topology? Use its complement to justify your answer.