A Set with a Chosen Notion of Openness
The previous tutorial explained why it is useful to study open sets without requiring a distance. A topological space makes that viewpoint precise: we start with a set and specify which of its subsets count as open. The specification must satisfy the topology axioms, but it need not come from a metric. Two spaces can even have the same underlying set and different topologies, and so have different open sets.
We will use \(\mathcal{T}\) to denote a topology on a set \(X\). The pair \((X,\mathcal{T})\) is the space; the set \(X\) alone does not specify which topology is being used. The definition of a topology was introduced in the previous tutorial. In brief, \(\mathcal{T}\) contains \(\varnothing\) and \(X\), is closed under arbitrary unions, and is closed under finite intersections. Its members are the open sets of the space.
The topology is part of the data, not an automatic consequence of the set. For example, on a set with three elements, one topology might declare only \(\varnothing\) and the whole set open, while another might declare every subset open. Both are legitimate, but they give different spaces on the same set. This distinction matters whenever a definition or result refers to openness: its meaning depends on the chosen topology.
Open and Closed Sets
The word “closed” is defined using the topology, rather than by requiring that a set contain its limit points. That latter description is useful in metric spaces, but it is not the general definition. In any topological space, closedness is defined by taking a complement.
A set can be both open and closed, or neither. “Closed” does not mean “not open.” For instance, in every topological space, \(\varnothing\) and \(X\) are both open and closed: each is open by the topology axioms, and each is the complement of the other. In \(\mathbb{R}\) with its usual topology, an interval such as \((0,1)\) is open, while \([0,1]\) is closed; neither example changes the definition, which is always about membership of the complement in the specified topology.
Proof. Since \(X\) and \(\varnothing\) are open, their complements \(\varnothing\) and \(X\) are closed.
Let \(\{F_\lambda:\lambda\in\Lambda\}\) be any family of closed subsets of \(X\). For each \(\lambda\), the complement \(X\setminus F_\lambda\) is open. By De Morgan’s law, $$ X\setminus\bigcap_{\lambda\in\Lambda}F_\lambda =\bigcup_{\lambda\in\Lambda}(X\setminus F_\lambda). $$ The right-hand side is open because arbitrary unions of open sets are open. Therefore the intersection is closed. If \(\Lambda\) is empty, the intersection is understood to be \(X\), which is already closed.
Now let \(F_1,\ldots,F_n\) be finitely many closed sets. Their complements are open, and $$ X\setminus\bigcup_{j=1}^{n}F_j =\bigcap_{j=1}^{n}(X\setminus F_j). $$ The right-hand side is open because finite intersections of open sets are open. Hence the union is closed. For \(n=0\), the union is \(\varnothing\), also closed. This proves all the stated properties. \(\square\)
These properties can also be used in reverse. If a collection \(\mathcal{F}\) of subsets contains \(\varnothing\) and \(X\), is closed under arbitrary intersections, and is closed under finite unions, then the collection of complements \(\{X\setminus F:F\in\mathcal{F}\}\) is a topology. Indeed, taking complements exchanges arbitrary intersections with arbitrary unions and finite unions with finite intersections. Thus open-set and closed-set descriptions encode the same structure.
Worked Example: A Topology on Three Points
Let \(X=\{a,b,c\}\), and consider $$ \mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}. $$ This is a topology. It contains \(\varnothing\) and \(X\). For unions, the only proper nonempty members are \(\{a\}\) and \(\{a,b\}\); a union of members of \(\mathcal{T}\) is therefore one of these sets or \(X\), unless every member is empty, in which case the union is \(\varnothing\). For finite intersections, the intersection of \(\{a\}\) and \(\{a,b\}\) is \(\{a\}\); intersections involving \(\varnothing\) or \(X\) also belong to \(\mathcal{T}\). So the required union and intersection properties hold.
The closed sets are the complements of the members of \(\mathcal{T}\). They are $$ \varnothing,\quad X,\quad \{b,c\},\quad \{c\}. $$ For example, \(\{c\}\) is closed because its complement \(\{a,b\}\) is open, even though \(\{c\}\) itself is not open in this topology. This example also shows that openness is relative to the topology: one cannot infer that every singleton is open just because it is a subset of the space.
Two Extreme Topologies
Every set has two especially simple topologies. The discrete topology declares every subset open. The indiscrete topology declares only the empty set and the whole set open. These names describe the amount of openness allowed: the discrete topology allows as many open sets as possible, while the indiscrete topology allows as few as the axioms permit.
Worked Example: Comparing the Two Extreme Topologies
Take \(X=\{p,q\}\). With the discrete topology, \(\{p\}\) is open, and so is its complement \(\{q\}\). With the indiscrete topology, the only open sets are \(\varnothing\) and \(X\), so neither singleton is open. In both cases, \(\varnothing\) and \(X\) are open, as required.
The closed sets in the discrete topology are also all subsets, because every complement is open. In the indiscrete topology the only closed sets are \(\varnothing\) and \(X\), since a set is closed precisely when its complement is one of those two open sets. Thus the same underlying set supports notably different collections of open and closed sets.
At the other end of the range, a topology need not contain every subset. On an infinite set, the cofinite topology from the previous tutorial is an example: its open sets are \(\varnothing\) and the subsets with finite complement. Such examples make it important to check statements against the actual topology rather than rely on intuition drawn only from the usual topology of the real line.
How Topologies Can Be Combined
Topologies on a fixed set can be compared by inclusion. If \(\mathcal{T}_1\subseteq\mathcal{T}_2\), then every set open for \(\mathcal{T}_1\) is also open for \(\mathcal{T}_2\); \(\mathcal{T}_2\) has at least as many open sets. For example, the indiscrete topology is contained in every topology on \(X\), and every topology on \(X\) is contained in the discrete topology.
An important way to combine topologies is to intersect them. Unlike the union of two topologies, the intersection of a nonempty family of topologies on the same set is always a topology. This gives a systematic way to find the smallest topology that contains some specified open sets.
Proof. Every \(\mathcal{T}_\lambda\) contains \(\varnothing\) and \(X\), so their intersection contains both. Let \(\{U_i:i\in I\}\) be any family of sets in \(\bigcap_{\lambda\in\Lambda}\mathcal{T}_\lambda\). Each \(U_i\) belongs to every \(\mathcal{T}_\lambda\). For any fixed \(\lambda\), the topology axiom gives \(\bigcup_{i\in I}U_i\in\mathcal{T}_\lambda\). Since this is true for every \(\lambda\), the union belongs to \(\bigcap_{\lambda\in\Lambda}\mathcal{T}_\lambda\). The same reasoning applies to a finite intersection: each \(\mathcal{T}_\lambda\) contains that intersection, so it belongs to their common intersection. Therefore the common intersection satisfies the topology axioms. \(\square\)
For a family \(\mathcal{S}\) of subsets of \(X\), there is at least one topology containing every member of \(\mathcal{S}\): the discrete topology contains every subset of \(X\). Intersect all topologies on \(X\) that contain \(\mathcal{S}\). By the theorem, their intersection is itself a topology; it contains \(\mathcal{S}\), and it is contained in every topology that contains \(\mathcal{S}\). It is therefore the smallest topology containing the specified family. This construction is useful when a topology is prescribed by the sets that must be open, rather than by listing every open set in advance.
Worked Example: The Union of Two Topologies Can Fail
Let \(X=\{a,b,c\}\) and define $$ \mathcal{T}_1=\{\varnothing,X,\{a\}\}, \qquad \mathcal{T}_2=\{\varnothing,X,\{b\}\}. $$ Each is a topology: the union or intersection of its members is again one of the three listed sets. Their union as collections is $$ \mathcal{T}_1\cup\mathcal{T}_2 =\{\varnothing,X,\{a\},\{b\}\}. $$ But \(\{a\}\) and \(\{b\}\) both belong to this collection, while their union \(\{a,b\}\) does not. The collection is not closed under unions and is therefore not a topology.
This is why the intersection theorem does not have a matching assertion for unions. When given a collection of subsets and asked whether it is a topology, the topology axioms must be checked; taking the union of known topologies does not automatically preserve them.
A Common Pitfall: Arbitrary Versus Finite Operations
The topology axioms deliberately treat unions and intersections differently: arbitrary unions of open sets are open, but only finite intersections are guaranteed to be open. Dually, arbitrary intersections of closed sets are closed, but only finite unions are guaranteed to be closed. Replacing “finite” by “arbitrary” in either statement is generally false.
Worked Example: An Infinite Intersection of Open Sets
In \(\mathbb{R}\) with its usual topology, each interval \(U_n=(-1/n,1/n)\), for \(n\geq1\), is open. A real number \(x\) belongs to every \(U_n\) exactly when \(|x|<1/n\) for every positive integer \(n\). If \(x\ne0\), choose \(n>1/|x|\); then \(1/n<|x|\), so \(x\notin U_n\). The number \(0\) belongs to every \(U_n\), and hence $$ \bigcap_{n=1}^{\infty}(-1/n,1/n)=\{0\}. $$ The singleton \(\{0\}\) is not open in the usual topology: every open interval around \(0\) contains nonzero real numbers. Thus an infinite intersection of open sets need not be open.
There is no contradiction with the topology axioms. For each finite collection of these intervals, the intersection is the smallest of the intervals in that collection and is still an open interval. The axioms guarantee finite intersections, not intersections of infinitely many sets.
Using the Topological-Space Viewpoint
Thinking of a space as \((X,\mathcal{T})\) keeps two questions separate: which points are present, and which subsets are designated as open? This distinction allows the same set to carry several topologies, and it lets us use open-set definitions in settings where no metric has been supplied. When a metric is available, its open sets give one possible topology, as established earlier in the course; the topology itself records the open-set structure without retaining numerical distances.
A reliable first step in a topological argument is to identify the topology being used. To show a set is open, check that it belongs to \(\mathcal{T}\), or use a description of the open sets already provided. To show a set is closed, check that its complement is open. And when combining many open or closed sets, keep track of whether the operation is finite or arbitrary. These habits make the axioms practical tools rather than a list to memorize.
Check Your Understanding
Use the definitions and results above to answer the following questions.
- What information does the pair \((X,\mathcal{T})\) specify that the set \(X\) alone does not?
- Why is a set closed exactly when its complement belongs to \(\mathcal{T}\)?
- Which closure properties hold for closed sets under arbitrary intersections and finite unions?
- Why does the intersection of a nonempty family of topologies on \(X\) remain a topology?
- In the example with \(U_n=(-1/n,1/n)\), why is the intersection not open even though each \(U_n\) is open?
- What topology axioms fail for the union of \(\mathcal{T}_1\) and \(\mathcal{T}_2\) in the three-point example?