Discrete Topology and Compactness
Compactness is a condition on all open covers: no matter how a space is covered by open sets, some finite selection of those sets must still cover the whole space. In a discrete space, every subset is open, so in particular each singleton is open. For an infinite discrete space, the singleton sets give a direct way to test the finite-subcover condition. The same test also makes the contrast with finite discrete spaces precise.
Let \(X\) be a set. Its discrete topology is the topology \(\mathcal{P}(X)\), the collection of all subsets of \(X\). Thus \((X,\mathcal{P}(X))\) is a topological space in which every subset is open. An open cover of \(X\) is a family of open sets whose union is \(X\). A finite subcover is a finite subfamily that still has union \(X\). By definition, a space is compact if every open cover has a finite subcover.
Finite Discrete Spaces Are Compact
Proof. Let \(X\) be finite with the discrete topology, and let \(\mathcal{U}\) be an open cover of \(X\). If \(X=\varnothing\), the empty subfamily has union \(\varnothing=X\), so it is a finite subcover. Now suppose \(X\ne\varnothing\). For each \(x\in X\), because \(\mathcal{U}\) covers \(X\), there is some \(U_x\in\mathcal{U}\) with \(x\in U_x\). The family of selected sets \(\{U_x:x\in X\}\) is finite because \(X\) is finite. Its union contains every \(x\in X\), and all its members are subsets of \(X\); hence its union is exactly \(X\). It is a finite subcover, so \(X\) is compact. \(\square\)
The proof does not require that the cover itself be finite, or that its members be singletons. It uses only that there are finitely many points to cover. Different points may be covered by the same member of \(\mathcal{U}\), in which case the selected subfamily is smaller still.
Worked Example: A Three-Point Discrete Space
Let \(X=\{a,b,c\}\) have the discrete topology, and consider the open cover \(\mathcal{U}=\{U_1,U_2,U_3\}\), where \(U_1=\{a,b\}\), \(U_2=\{b,c\}\), and \(U_3=\{a,c\}\). Each member is open because every subset of \(X\) is open.
First check that this is a cover: \(a\in U_1\), \(b\in U_1\), and \(c\in U_2\), so \(U_1\cup U_2\cup U_3=X\). In fact, only two members are needed: \(U_1\cup U_2=\{a,b\}\cup\{b,c\}=\{a,b,c\}=X\). Thus \(\{U_1,U_2\}\) is a finite subcover. The proposition proves that this conclusion holds for every open cover of every finite discrete space, not just this particular cover.
Check: the selected sets contain all three points: \(a,b\in U_1\) and \(c\in U_2\). Their union has no points outside \(X\), so it is exactly \(X\).
Why an Infinite Discrete Space Is Not Compact
Proof. If \(X\) is finite, the proposition gives compactness, including when \(X=\varnothing\). For the converse, suppose that \(X\) is infinite and discrete. Consider the family \(\mathcal{U}=\{\{x\}:x\in X\}\). Each member is open by discreteness. Every point \(x\in X\) belongs to its singleton \(\{x\}\), so the union of the family is \(X\): this is an open cover.
Now take any finite subfamily of \(\mathcal{U}\), say the singletons \(\{x_1\},\ldots,\{x_k\}\). Its union is \(\{x_1,\ldots,x_k\}\), which is finite. Since \(X\) is infinite, this finite set cannot equal \(X\); therefore the subfamily does not cover \(X\). This holds for every finite subfamily, so the displayed open cover has no finite subcover. Hence \(X\) is not compact. Together with the finite case, this proves the equivalence. \(\square\)
The key is not merely that the cover has infinitely many members. An infinite cover can have a finite subcover. What matters here is that each member is a singleton: a finite selection of them contains only finitely many points, while the space has infinitely many.
Worked Example: The Natural Numbers with the Discrete Topology
Give \(\mathbb{N}=\{1,2,3,\ldots\}\) the discrete topology. Consider the family \(\mathcal{U}=\{\{n\}:n\in\mathbb{N}\}\). Every \(\{n\}\) is open, and every natural number \(n\) belongs to the family member \(\{n\}\), so \(\bigcup_{n\in\mathbb{N}}\{n\}=\mathbb{N}\).
If a finite selection uses the singletons \(\{n_1\},\ldots,\{n_k\}\), its union is \(\{n_1,\ldots,n_k\}\), which has at most \(k\) elements. It cannot contain every natural number. For a concrete selection, the first four singletons have union \(\{1\}\cup\{2\}\cup\{3\}\cup\{4\}=\{1,2,3,4\}\), which misses \(5\), among other natural numbers. The same reasoning applies to any finite selection: if its union is a finite set \(F\subseteq\mathbb{N}\), there is a natural number not in \(F\).
Check: the family covers all of \(\mathbb{N}\), but each finite subfamily has a finite union. Since \(\mathbb{N}\) is infinite, no such union is \(\mathbb{N}\).
Worked Example: An Uncountable Discrete Space
Let the underlying set be \(\mathbb{R}\), but give it the discrete topology \(\mathcal{P}(\mathbb{R})\), rather than the standard topology. The family \(\mathcal{V}=\{\{x\}:x\in\mathbb{R}\}\) consists of open sets, and its union is \(\mathbb{R}\), since each real number \(x\) belongs to \(\{x\}\).
A finite subfamily, with selected points \(x_1,\ldots,x_k\), has union \(\{x_1,\ldots,x_k\}\). This union is finite and cannot equal \(\mathbb{R}\). For example, if the selected points are \(0,1,-1\), their union as singletons is \(\{0,1,-1\}\), which misses \(2\). For any finite selection, the same obstruction follows because \(\mathbb{R}\) is infinite. Thus this open cover has no finite subcover, and the discrete space on \(\mathbb{R}\) is not compact.
Check: the proof uses only that the underlying set is infinite, not that it is countable. The topology matters too: these singletons are open because the topology here is discrete.
A Different Cover and a Common Mistake
The singleton cover is canonical, but other covers can also expose the failure of compactness. The next example shows how an open cover may have members that overlap and still have no finite subcover.
Worked Example: Overlapping Open Sets on an Infinite Set
Let \(X=\{p\}\cup\{q_n:n\in\mathbb{N}\}\), where all the points \(p,q_1,q_2,\ldots\) are distinct, and give \(X\) the discrete topology. For each \(n\in\mathbb{N}\), define \(V_n=\{p,q_n\}\). Every \(V_n\) is open. The family \(\{V_n:n\in\mathbb{N}\}\) covers \(X\): it contains \(p\) in every member, and each \(q_n\) belongs to \(V_n\).
For a finite set \(F\subseteq\mathbb{N}\) of selected indices, the union of the corresponding members is \(\bigcup_{n\in F}V_n=\{p\}\cup\{q_n:n\in F\}\). Because \(\mathbb{N}\) is infinite and \(F\) is finite, choose \(m\in\mathbb{N}\setminus F\). Then \(q_m\in X\), but \(q_m\notin\bigcup_{n\in F}V_n\). Therefore no finite subfamily covers \(X\).
Check: every point of \(X\) is covered by the full family, while each finite selection misses at least one point \(q_m\). The overlap at \(p\) does not help cover the infinitely many distinct points \(q_n\).
Common pitfall. It is incorrect to argue that a singleton cover has a finite subcover simply because every cover member is open. For the discrete space \(\mathbb{N}\), selecting \(\{1\},\{2\},\{3\}\) gives union \(\{1,2,3\}\), not \(\mathbb{N}\). More generally, selecting \(k\) singleton members gives a union with at most \(k\) points; an infinite space has points outside that union. Openness ensures the family is an open cover, but compactness demands the additional finite-subcover property.
This criterion is useful because it completely classifies compact discrete spaces: there are no infinite exceptions, whether the underlying set is countable or uncountable. For finite discrete spaces, finiteness of the point set supplies a finite selection from any cover. For infinite discrete spaces, the open singletons form one specific cover that defeats every finite selection.
Check Your Understanding
For each question, identify the topology and distinguish an open cover from a finite subcover.
- Why is the family of all singletons an open cover of a nonempty discrete space?
- If \(X\) is infinite and discrete, what is the union of a finite selection of distinct members of its singleton cover?
- Explain why the empty space is compact under the finite-subcover definition.
- In the discrete space \(X=\{p\}\cup\{q_n:n\in\mathbb{N}\}\), why does every finite selection from \(\{V_n:n\in\mathbb{N}\}\), where \(V_n=\{p,q_n\}\), miss a point?
- Can a cover with infinitely many members have a finite subcover? Explain why the number of members of a cover alone does not decide compactness.