Closed Subspaces of Compact Spaces

From a Relative Cover to an Ambient Cover

Compactness is defined by open covers, but a cover of a subspace consists of sets open in that subspace, not necessarily open in the entire space. The key step is to express those sets as intersections with ambient open sets. When the subspace is closed, its complement is open; adjoining that complement then gives an open cover of the whole compact space.

Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). The subspace topology on \(A\) consists of all sets \(A\cap U\), where \(U\in\mathcal{T}\). A family \(\{V_i:i\in J\}\) of open sets in a space is an open cover of that space if their union is the whole space. The space is compact if every open cover has a finite subcover. A subset \(A\) is closed in \(X\) if \(X\setminus A\) is open in \(X\); it is compact as a subspace if it is compact with the subspace topology.

Theorem: If \(X\) is compact and \(A\) is closed in \(X\), then \(A\), with its subspace topology, is compact.

Proof. Let \(\{V_i:i\in J\}\) be any open cover of \(A\) in the subspace topology. For each \(i\in J\), the definition of that topology gives an open set \(U_i\in\mathcal{T}\) such that \(V_i=A\cap U_i\). Since the \(V_i\) cover \(A\), the family consisting of all the \(U_i\), together with \(X\setminus A\), covers \(X\): points of \(A\) lie in some \(U_i\), and points outside \(A\) lie in \(X\setminus A\). The set \(X\setminus A\) is open because \(A\) is closed. Thus this is an open cover of \(X\), so compactness of \(X\) gives a finite subcover. Keep the finitely many selected sets of the form \(U_i\), and discard \(X\setminus A\) if it was selected. Every point of \(A\) must belong to one of the retained \(U_i\), since no point of \(A\) lies in \(X\setminus A\). Therefore the corresponding finitely many sets \(A\cap U_i=V_i\) cover \(A\). If \(A=\varnothing\), the empty family is already a finite subcover. In all cases every open cover of \(A\) has a finite subcover, so \(A\) is compact. \(\square\)

The proof uses closedness at one specific point: it makes \(X\setminus A\) an open set that can be adjoined to the cover. The finite-subcover property of \(X\) then supplies finitely many ambient sets, and restricting those sets back to \(A\) supplies finitely many members of the original relative cover.

Working Through the Cover Construction

Worked Example: A Closed Interval Inside a Compact Interval

Take \(X=[0,1]\) with the subspace topology inherited from \(\mathbb{R}\) with its standard topology, and let \(A=[0,1/2]\). The interval \(X\) is compact by the open-cover theorem for closed intervals. The set \(A\) is closed in \(X\), since \(X\setminus A=(1/2,1]\), which is open in \(X\): it is \(X\cap(1/2,2)\). The theorem therefore says \(A\) is compact.

Here is the cover construction for a particular relative open cover. Let \(V_1=A\cap(-1,2/5)\) and \(V_2=A\cap(3/10,1)\). These are open in \(A\). They cover \(A\): a point \(x\in A\) with \(x<2/5\) belongs to \(V_1\); if \(x\geq2/5\), then \(x>3/10\) and \(x\leq1/2<1\), so \(x\in V_2\).

Choose the ambient open sets \(U_1=(-1,2/5)\) and \(U_2=(3/10,1)\). Adjoin the open complement \(X\setminus A=(1/2,1]\). The three sets cover \(X\): points in \([0,2/5)\) lie in \(U_1\); points in \([2/5,1/2]\) lie in \(U_2\); points in \((1/2,1]\) lie in the complement. Restricting \(U_1\) and \(U_2\) to \(A\) returns \(V_1\) and \(V_2\), which cover \(A\). In this example the displayed relative cover is already finite; the theorem’s argument works in exactly the same way when the starting cover has infinitely many members.

Check: the endpoints are handled by the stated strict inequalities. The point \(2/5\) is not in \(U_1\), but it is in \(U_2\), since \(2/5>3/10\). The point \(1/2\) is not in \(X\setminus A\), but it is in \(U_2\), since \(1/2<1\). Thus no endpoint is lost in the cover.

Worked Example: A Closed Subspace of a Finite Space

Let \(X=\{a,b,c\}\) with the discrete topology, in which every subset is open. Every subset is also closed, because its complement is open. The space \(X\) is compact: from any open cover, choose one covering set for each of the three points; the chosen sets form a subcover with at most three members. In particular \(A=\{a,c\}\) is closed and compact.

For an explicit relative cover, take \(V_1=\{a\}\), \(V_2=\{c\}\), and \(V_3=\{a,c\}\). These sets are open in \(A\) and cover \(A\). They are also ambient open sets, and the complement \(X\setminus A=\{b\}\) is open. The family \(\{V_1,V_2,V_3,\{b\}\}\) covers \(X\). One finite subcover is \(\{V_3,\{b\}\}\); discarding the complement leaves \(V_3\), which covers \(A\).

Check: \(V_3\cup\{b\}=\{a,c\}\cup\{b\}=\{a,b,c\}=X\), and \(V_3=A\cap V_3=A\). This verifies both the ambient cover and its restriction back to \(A\).

Closed Inside a Compact Subspace

Corollary: Let \(X\) be a topological space, let \(A\subseteq X\) be compact as a subspace, and let \(F\subseteq A\) be closed in \(A\). Then \(F\) is compact as a subspace of \(X\).

Proof. The subspace \(A\), with its own topology, is a compact topological space by assumption. The set \(F\) is closed in that space. Applying the theorem to the space \(A\) and its closed subset \(F\) shows that \(F\) is compact with the subspace topology inherited from \(A\). This is the same topology on \(F\) as the one inherited directly from \(X\): in either case its open sets are precisely the sets \(F\cap U\) for \(U\) open in \(X\). To verify the only potentially non-immediate direction, a set open in \(A\) has the form \(A\cap U\) for \(U\) open in \(X\); intersecting it with \(F\subseteq A\) gives \(F\cap U\). Hence compactness here is compactness as a subspace of \(X\), as claimed. \(\square\)

This corollary distinguishes two roles. The ambient space \(X\) need not be compact: it is enough that \(A\) is compact. The subset \(F\) must be closed in \(A\), the space in which compactness is being applied. Closedness in some larger space is not the condition used by the theorem unless it also makes \(F\) closed in \(A\).

Worked Example: A Nonclosed Subspace Can Fail to Be Compact

Let \(X=[0,1]\) with the standard subspace topology from \(\mathbb{R}\), and let \(A=(0,1)\), also with its subspace topology. The space \(X\) is compact, but \(A\) is not closed in \(X\), since \(X\setminus A=\{0,1\}\) is not open in \(X\). Indeed, no relatively open neighborhood of \(0\) in \(X\) is contained in \(\{0,1\}\): every such neighborhood contains points of \(X\) strictly between \(0\) and \(1\).

For each \(n\in\mathbb{N}\), define \(V_n=(1/n,1)\), viewed as a subset of \(A\). It is open in \(A\), since \(V_n=A\cap(1/n,2)\). These sets cover \(A\). If \(x\in(0,1)\), choose \(n\in\mathbb{N}\) with \(n>1/x\). Then \(1/n<x<1\), so \(x\in V_n\).

No finite subfamily covers \(A\). In fact, for any nonempty finite set of indices \(n_1,\dots,n_k\), let \(N\) be the largest of those indices. Since \(1/n_j\geq1/N\) for each \(j\), the union of the selected sets is \(V_N=(1/N,1)\). The point \(1/(2N)\) belongs to \(A\), because \(0<1/(2N)<1\), but it does not belong to \(V_N\), because \(1/(2N)<1/N\). The empty subfamily also fails to cover the nonempty set \(A\). Thus this open cover has no finite subcover, and \(A\) is not compact.

Check: the strict inequality in \(V_N=(1/N,1)\) matters: the point \(1/(2N)\) is below its left endpoint. The failure does not contradict the theorem, because \(A\) is not closed in the compact space \(X\).

Why the Complement Matters

The proof is a practical translation procedure. Start with a cover open in the subspace, represent each member as an intersection with an ambient open set, and adjoin the complement of the subspace. Compactness applies only after this produces a cover of the whole ambient space. Once a finite ambient subcover is obtained, the complement can be discarded because it contains no points of the subspace.

A common mistake is to say that every subspace of a compact space is compact. The example \(A=(0,1)\subseteq[0,1]\) gives a concrete correction: the ambient interval is compact, but the relative open cover \(\{(1/n,1):n\in\mathbb{N}\}\) has no finite subcover. Another mistake is to adjoin \(X\setminus A\) without checking that it is open. The cover definition requires open sets, and this is exactly where the hypothesis that \(A\) is closed enters. The theorem asserts compactness of closed subspaces; it does not say that every compact subspace must be closed.

Check Your Understanding

For each question, identify the space whose open cover is being considered and state where closedness is used.

In the theorem’s proof, why does a relative open set \(V_i\) have the form \(A\cap U_i\) for some open \(U_i\subseteq X\)?
If \(A=\varnothing\), why is it compact under the finite-subcover definition?
In the interval example, verify that \(X\setminus A=(1/2,1]\) is open in the subspace \(X=[0,1]\).
For the cover \(V_n=(1/n,1)\) of \((0,1)\), why does a finite selection with largest index \(N\) have union \((1/N,1)\)?
Can the theorem be applied when \(A\) is compact but not known to be closed in \(X\)? State which hypothesis is missing.