Every Subset Is Open
A topology specifies which subsets of a set are to count as open. One particularly direct choice is to declare every subset open. This choice works whether the underlying set has finitely many elements, infinitely many elements, or no elements at all. The resulting topology is called the discrete topology.
To check this definition, recall the topology axioms. A topology \(\mathcal{T}\) on a set \(X\) is a collection of subsets of \(X\) such that \(\varnothing\) and \(X\) belong to \(\mathcal{T}\), arbitrary unions of members of \(\mathcal{T}\) belong to \(\mathcal{T}\), and finite intersections of members of \(\mathcal{T}\) belong to \(\mathcal{T}\). The members of \(\mathcal{T}\) are the open sets.
The condition \(A\subseteq X\) is automatic when \(A\) is already being considered as a subset of \(X\). So in a discrete topology there is no additional test for openness: every subset of the underlying set is open. The topology may have many members, but its rule is simple.
Why This Collection Is a Topology
Proof. We check each topology axiom. First, \(\varnothing\subseteq X\) and \(X\subseteq X\), so \(\varnothing,X\in\mathcal{P}(X)\). Next, let \((A_i)_{i\in J}\) be any family of members of \(\mathcal{P}(X)\), where \(J\) is any index set. Each \(A_i\subseteq X\), so every element of \(\bigcup_{i\in J}A_i\) lies in \(X\). Therefore \(\bigcup_{i\in J}A_i\subseteq X\), and this union belongs to \(\mathcal{P}(X)\). This also covers an empty index set, whose union is \(\varnothing\). Finally, if \(A_1,\ldots,A_n\in\mathcal{P}(X)\) for a positive integer \(n\), then every element of \(\bigcap_{k=1}^n A_k\) lies in \(A_1\), hence in \(X\). Thus \(\bigcap_{k=1}^n A_k\subseteq X\) and belongs to \(\mathcal{P}(X)\). The intersection of zero sets, by convention, is \(X\), which also belongs to \(\mathcal{P}(X)\). All three axioms hold. \(\square\)
The union axiom permits any number of open sets to be united, including infinitely many. The intersection axiom requires only finite intersections. In the discrete topology this distinction causes no difficulty: both the union and intersection of any collection of subsets of \(X\) are still subsets of \(X\), so both are open. The topology axioms require less than this, but the discrete topology satisfies the stronger statement too.
Worked Example: Listing the Topology on a Three-Point Set
Let \(X=\{a,b,c\}\) have the discrete topology. To list \(\mathcal{P}(X)\), choose whether each of \(a,b,c\) is included in a subset. The resulting subsets are
These are all the open sets, because every subset of \(X\) is open. There are \(2\cdot2\cdot2=8\) subsets: each of the three elements has two choices, included or not included. For a direct check of the topology axioms, \(\varnothing\) and \(X\) appear in the list; the union of any members is a subset of \(X\) and so appears in the list; and the intersection of any finite number of members is also a subset of \(X\) and so appears in the list.
Check: for example, \(\{a,b\}\cup\{b,c\}=\{a,b,c\}\), and \(\{a,b\}\cap\{b,c\}=\{b\}\). Both results are in the displayed collection. The list includes the empty set and the whole set, as required.
Finite and Infinite Underlying Sets
If \(X\) has \(n\) elements, each element can independently be either included in or excluded from a subset. Thus the discrete topology has \(2^n\) open sets. In particular, the empty underlying set has \(2^0=1\) subset: \(\varnothing\). Its discrete topology is \(\{\varnothing\}\), which is also \(\{\varnothing,X\}\) because \(X=\varnothing\) in this case.
On an infinite set, the same definition applies without any change. The topology is still the collection of all subsets; there is no restriction that open sets be finite or have finite complements. An infinite underlying set has infinitely many distinct open sets, since its singleton subsets are all open and are distinct for distinct elements. But the complete description remains simpler than a finite list: every subset is open.
Worked Example: Open Sets on the Infinite Set of Positive Integers
Take \(X=\mathbb{N}=\{1,2,3,\ldots\}\) with the discrete topology. Consider the subsets \(E=\{2,4,6,\ldots\}\) and \(O=\{1,3,5,\ldots\}\). Both are subsets of \(X\), so both are open. They are infinite, but finiteness is not required for openness in this topology.
Every positive integer is either even or odd, and no positive integer is both. Therefore \(E\cup O=\mathbb{N}\) and \(E\cap O=\varnothing\). These are open as well. The singleton sets \(\{n\}\) are open for every \(n\in\mathbb{N}\), and
Indeed, \(2k\) is a positive even integer for each \(k\in\mathbb{N}\), and every positive even integer has the form \(2k\) for some \(k\in\mathbb{N}\). This is an infinite union of open sets, and its result is open. The complement of \(E\) in \(X\) is \(O\), also open.
Check: \(2\in E\) because \(2=2\cdot1\); \(5\in O\) because \(5=2\cdot2+1\); and neither \(2\) nor \(5\) belongs to both sets. The union contains every positive integer and the intersection contains none, in agreement with the computations above.
Worked Example: A Discrete Topology on the Real Numbers
Let the underlying set be \(\mathbb{R}\), but equip it with the discrete topology \(\mathcal{P}(\mathbb{R})\). Here \(\mathbb{R}\) means the set of real numbers; the topology in this example is explicitly the discrete topology. The subset \(A=[0,1]\) is open because it is a subset of \(\mathbb{R}\). The singleton \(\{\sqrt{2}\}\) is open for the same reason, and so is the set \(\mathbb{R}\setminus[0,1]\).
The union and intersection of these particular open sets can be computed directly. Since \(\sqrt{2}>1\), the point \(\sqrt{2}\) belongs to \(\mathbb{R}\setminus[0,1]\). Hence
are both open. The union is a subset of \(\mathbb{R}\), and the intersection is the empty set. This example emphasizes that openness depends on the chosen topology, not just on the name or shape of a subset. Here the topology has been specified to be discrete, so every subset is open.
Check: \(0\in[0,1]\), \(1\in[0,1]\), and \(\sqrt{2}\notin[0,1]\) because \(\sqrt{2}>1\). Therefore the intersection contains no points, while adjoining \(\sqrt{2}\) to \([0,1]\) gives exactly \([0,1]\cup\{\sqrt{2}\}\), as stated.
Recognizing the Discrete Topology
The definition can also be recognized from the open singletons. A singleton \(\{x\}\) is the set consisting of one point \(x\). If every singleton is open, then any subset can be assembled as a union of singletons. The arbitrary-union axiom then forces every subset to be open.
Proof. Suppose first that \(\mathcal{T}=\mathcal{P}(X)\). Every singleton \(\{x\}\) is a subset of \(X\), so it belongs to \(\mathcal{P}(X)=\mathcal{T}\). Conversely, suppose every singleton \(\{x\}\), for \(x\in X\), is open. Let \(A\subseteq X\). If \(A=\varnothing\), then \(A\in\mathcal{T}\) by the empty-set axiom. If \(A\ne\varnothing\), then \(A=\bigcup_{x\in A}\{x\}\). Each singleton in this union is open by assumption, so the arbitrary-union axiom gives \(A\in\mathcal{T}\). Thus every subset of \(X\) is in \(\mathcal{T}\), and \(\mathcal{T}=\mathcal{P}(X)\). If \(X=\varnothing\), the first direction has no singleton to check, and the converse follows from \(\varnothing\in\mathcal{T}\); the only subset of \(X\) is \(\varnothing\). \(\square\)
In a discrete topology every subset is also closed: if \(A\subseteq X\), its complement \(X\setminus A\) is a subset of \(X\), hence open. By the definition of closed set, this makes \(A\) closed. Thus in this topology every set is both open and closed. This conclusion follows directly from the definition and does not mean that “open” and “closed” are the same condition in every topological space.
A Common Mistake
A common error is to define a topology on an infinite set using only the empty set, the whole set, and the finite subsets. For \(X=\mathbb{N}\), consider the proposed collection consisting of \(\varnothing\), \(\mathbb{N}\), and all finite subsets of \(\mathbb{N}\). Every singleton belongs to this collection. But the union of the open singletons \(\{2\},\{4\},\{6\},\ldots\) is the infinite set \(E=\{2,4,6,\ldots\}\), which is neither finite nor all of \(\mathbb{N}\). So this union is missing from the proposed collection, violating the arbitrary-union axiom. The correction is to include every subset of \(\mathbb{N}\) when defining the discrete topology.
Check Your Understanding
Use the definition of the discrete topology and the topology axioms to justify your answers.
- List all open sets in the discrete topology on \(X=\{p,q\}\), and state how many there are.
- For \(X=\{0,1,2,3\}\) with the discrete topology, is \(\{0,2\}\) open? Is its complement in \(X\) open?
- On \(\mathbb{N}\) with the discrete topology, explain why the set of odd positive integers is open even though it is infinite.
- Prove that if every singleton in a topology on \(X\) is open, then every subset of \(X\) is open.
- Why does the collection consisting of \(\varnothing\), \(\mathbb{N}\), and all finite subsets of \(\mathbb{N}\) fail the arbitrary-union axiom?