Compactness of Finite Topological Spaces

Why Finiteness of the Space Is Enough

Compactness asks whether an arbitrary open cover can be reduced to finitely many of its members while still covering the space. The word “arbitrary” matters: a cover need not be finite to begin with. When the underlying set has finitely many points, however, we can make a finite selection directly. For each point, choose one open set from the cover that contains it. The chosen sets cover the whole space, and there can be no more distinct choices than there are points.

Throughout, \((X,\mathcal{T})\) denotes a topological space. An open cover of \(X\) is a family \(\{U_i:i\in J\}\) of open subsets of \(X\) such that every point of \(X\) belongs to at least one \(U_i\). A subcover is a subfamily that still covers \(X\). The space is compact if every open cover has a finite subcover. A finite topological space means a topological space whose underlying set \(X\) is finite; no special condition is imposed on its topology.

Proof. Let \(\{U_i:i\in J\}\) be an open cover of \(X\). First suppose \(X\) is nonempty, and list its distinct points as \(x_1,\ldots,x_n\). Since the family covers \(X\), for each \(k\) there is an index \(i_k\in J\) such that \(x_k\in U_{i_k}\). Select the sets \(U_{i_1},\ldots,U_{i_n}\). Each point \(x_k\) belongs to its selected set \(U_{i_k}\), so their union contains every point of \(X\). All selected sets are members of the original cover, so their union is also a subset of \(X\). Thus they form a subcover. There are \(n\) selections, and repetitions can only reduce the number of distinct selected sets, so this subcover has at most \(n\) members.

If \(X=\varnothing\), the empty subfamily is a finite subcover: its union is \(\varnothing=X\). This handles the case with no points to select. The finite selection in the nonempty case can be made successively through the finite list \(x_1,\ldots,x_n\); it does not require an axiom of choice for an infinite family of choices. In every case, an arbitrary open cover has a finite subcover, so \(X\) is compact. \(\square\)

Choosing Sets Point by Point

The proof does not need a detailed description of the open sets. It uses only two facts: the family covers \(X\), and there are finitely many points to handle. At each point, the cover condition supplies at least one suitable open set. Once one such set is selected for every point, their union covers \(X\) because each point has been accounted for.

The bound is “at most \(n\),” not “exactly \(n\).” Two or more points may be covered by the same selected open set. Also, a selected set may coincide with one already selected. The selection argument guarantees a finite subcover, but it does not promise a subcover with the fewest possible members.

A Consequence for Finite Subspaces

Proof. The subspace topology on \(A\) consists of the sets \(A\cap U\), where \(U\) is open in \(X\). In particular, it is a topology on the finite underlying set \(A\). Let any open cover of \(A\) in this subspace topology be given. The theorem applies to the finite topological space \(A\), so this cover has a finite subcover. If \(A=\varnothing\), the empty subfamily covers it; otherwise the point-by-point proof selects one cover member for each point of \(A\). Therefore \(A\) is compact in its subspace topology. \(\square\)

This conclusion does not say that the topology on \(A\) must be discrete, or that \(A\) must be open or closed in \(X\). Compactness here is a statement about open covers in the topology that \(A\) inherits. The proof depends on the finiteness of the underlying set \(A\), not on how its open subsets sit inside \(X\).

Why the Topology Does Not Change the Argument

Different topologies on the same finite set can have very different open sets. In the indiscrete example, there are only two open sets; in the discrete example, every subset is open. Yet the proof has the same structure in both cases. Given a cover and a point, the definition of cover guarantees at least one member containing that point. Finiteness then allows one selection per point, regardless of how many or how few open sets the topology contains.

There is also a second way to see why finite spaces are manageable. If \(X\) has \(n\) points, it has \(2^n\) subsets, so its topology, being a collection of subsets of \(X\), is finite. Thus a cover considered as a set of distinct open subsets has only finitely many members. The point-by-point proof is still useful: it gives the sharper bound of at most \(n\) selected members, and it applies without confusion when a cover is presented as an indexed family in which the same open set may appear repeatedly.

Common pitfall. A point-by-point selection does not necessarily produce exactly \(n\) distinct open sets. For example, let \(X=\{a,b\}\) with the discrete topology and consider the cover \(U_1=X\), \(U_2=\{a\}\). Choose \(U_1\) for \(a\), and choose \(U_1\) for \(b\). Counting the two selections as two distinct sets would be incorrect: both selections name the same set \(U_1\). The selected family of distinct sets is \(\{U_1\}\), and its union is \(X\). The correct conclusion is that the number of distinct selected sets is at most \(n\), not necessarily equal to \(n\).

Compactness is therefore a property that finite spaces have without needing separation assumptions such as Hausdorffness. For instance, in the Sierpinski space above, the only open neighborhood of \(b\) is \(X\); every open neighborhood of \(a\) is either \(\{a\}\) or \(X\). No choice of one open neighborhood of each point gives disjoint neighborhoods, yet the space is compact by the theorem. The open-cover proof establishes compactness directly and does not require points to be separable by neighborhoods.