Finite Unions of Compact Subspaces

Combining Compact Pieces

Compactness is defined by a condition on every open cover: from any such cover, finitely many sets must suffice. If a space is built as a finite union of compact subspaces, a natural strategy is to apply compactness separately to each piece. The key is to restrict the original cover to each piece, then collect the resulting finite subcovers. This requires no assumption that the pieces are disjoint or closed.

Let \((X,\mathcal{T})\) be a topological space. For \(A\subseteq X\), the subspace topology on \(A\) consists of the sets \(A\cap V\), where \(V\in\mathcal{T}\). Thus a set \(U\subseteq A\) is open in \(A\) precisely when \(U=A\cap V\) for some open \(V\) in \(X\). All covers and compactness claims about a subspace below use this topology.

Definition: A topological space is compact if every cover of it by open sets has a finite subcover. A subspace \(A\subseteq X\) is compact if it is compact with its subspace topology. A subcover is a selection of sets from the original cover that still covers the space.

The empty space is compact: the empty subfamily covers it. A finite subcover is allowed to contain no sets when the space being covered is empty. This convention handles empty pieces in a finite union without requiring a separate exception later.

Finite Sets and the Main Theorem

Proposition: Every finite subset \(F\) of a topological space \(X\) is compact in its subspace topology.

Proof. Let \(\mathcal{U}\) be an open cover of \(F\). If \(F=\varnothing\), the empty subfamily is a finite subcover. Otherwise write \(F=\{x_1,\ldots,x_m\}\), where \(m\) is a positive integer. For each \(j\in\{1,\ldots,m\}\), the cover property gives a set \(U_j\in\mathcal{U}\) with \(x_j\in U_j\). The selected sets \(U_1,\ldots,U_m\) cover every point of \(F\), and there are at most \(m\) of them. They form a finite subcover. \(\square\)

Theorem (Finite unions of compact subspaces): If \(K_1,\ldots,K_n\) are compact subspaces of \(X\), where \(n\) is a nonnegative integer, then \(K=\bigcup_{i=1}^{n}K_i\) is compact with its subspace topology.

Proof. If \(n=0\), then \(K=\varnothing\), which is compact. Now suppose \(n\geq 1\), and let \(\mathcal{U}\) be any open cover of \(K\). For each \(i\in\{1,\ldots,n\}\), restrict the cover to \(K_i\) by forming the family \(\{U\cap K_i: U\in\mathcal{U}\}\), omitting empty intersections if desired. Each member is open in \(K_i\): since \(U\) is open in \(K\), it has the form \(U=K\cap V\) for some open \(V\subseteq X\), and then \(U\cap K_i=K_i\cap V\). This restricted family covers \(K_i\), because every point of \(K_i\subseteq K\) lies in some member of \(\mathcal{U}\).

Since \(K_i\) is compact, finitely many members of the restricted family cover \(K_i\). Each selected member comes from a set in the original cover \(\mathcal{U}\); select those original sets and call their finite collection \(\mathcal{U}_i\). The union \(\mathcal{U}_1\cup\cdots\cup\mathcal{U}_n\) is finite, since it is a union of finitely many finite collections. It covers \(K\): if \(x\in K\), then \(x\in K_i\) for at least one index \(i\), and \(\mathcal{U}_i\) covers \(K_i\), hence contains a set covering \(x\). Therefore every open cover of \(K\) has a finite subcover, so \(K\) is compact. \(\square\)

The proof uses only compactness of each piece and the fact that the number of pieces is finite. A set belonging to more than one piece causes no difficulty: it may be covered by sets selected for either or both pieces. There is also no need for the \(K_i\) to be open, closed, or disjoint in \(X\).

Worked Applications

Worked Example: Two Overlapping Finite Subspaces

In \(\mathbb{R}\) with its standard topology, let \(K_1=\{0,2\}\), \(K_2=\{2,5\}\), and \(K=K_1\cup K_2=\{0,2,5\}\). Consider the following open sets of \(K\), obtained by intersecting ambient open intervals with \(K\): \(U_0=K\cap(-1,1)\), \(U_2=K\cap(1,3)\), and \(U_5=K\cap(4,6)\).

Substitution of the points of \(K\) gives \(U_0=\{0\}\), \(U_2=\{2\}\), and \(U_5=\{5\}\): for example, \(0\in(-1,1)\), \(2\notin(-1,1)\), and \(5\notin(-1,1)\). The same membership check gives the other two intersections. Thus \(\{U_0,U_2,U_5\}\) covers \(K\). On \(K_1\), the sets \(U_0,U_2\) cover its two points; on \(K_2\), the sets \(U_2,U_5\) cover its two points. Combining these finite subcovers gives the cover \(U_0,U_2,U_5\) of \(K\). Each \(K_i\) is compact by the proposition on finite subsets, and the theorem guarantees compactness for any open cover of \(K\), not just this one.

Check: \(K_1\cup K_2=\{0,2,5\}\), with the shared point \(2\) counted just once as a point of the union. All three points occur in the selected sets, so the displayed subfamily really does cover \(K\).

Worked Example: A Cover of Two Compact Intervals

In \(\mathbb{R}\) with its standard topology, let \(K_1=[0,2]\), \(K_2=[1,3]\), and \(K=[0,3]\). The intervals \(K_1\) and \(K_2\) are compact by the open-cover result for closed intervals. Consider the open cover of \(K\) given by \(U_1=K\cap(-1,1.5)\), \(U_2=K\cap(1.4,2.5)\), and \(U_3=K\cap(2.4,4)\).

To check that these sets cover \(K\), points in \([0,1.5)\) lie in \(U_1\), points in \([1.5,2.4]\) lie in \(U_2\), and points in \((2.4,3]\) lie in \(U_3\). The middle range is contained in \((1.4,2.5)\), including both \(1.5\) and \(2.4\), so no endpoint is missed. On \(K_1\), \(U_1\) covers \([0,1.5)\) and \(U_2\) covers \([1.5,2]\); together they cover \(K_1\). On \(K_2\), \(U_2\) covers \([1,2.5)\) and \(U_3\) covers \([2.5,3]\); together they cover \(K_2\). Combining the selections gives the finite subcover \(U_1,U_2,U_3\) of \(K\).

Check: The endpoints are included: \(0\in U_1\), \(2\in U_2\), and \(3\in U_3\). The intervals overlap, so the choices on each piece cover through the points where one selected set stops containing points of the interval.

Worked Example: Infinite Pieces in an Indiscrete Space

Let \(X=\mathbb{N}\) with the indiscrete topology \(\{\varnothing,X\}\), and let \(K_1=\{2,4,6,\ldots\}\) and \(K_2=\{1,3,5,\ldots\}\). Each subspace \(K_i\) has the indiscrete topology: its only open sets are \(\varnothing\) and \(K_i\), by intersecting the two open sets of \(X\) with \(K_i\).

Each \(K_i\) is compact. Indeed, an open cover of the nonempty space \(K_i\) must include a nonempty open set, and the only such open set is \(K_i\) itself. That one set is a finite subcover. Their union is \(X\). An open cover of \(X\) must likewise include \(X\), so \(X\) is compact as well.

Check: \(K_1\cap K_2=\varnothing\), and every positive integer is either even or odd, so \(K_1\cup K_2=X\). This example has infinite pieces and an infinite union, but still only two pieces; the theorem applies because the number of pieces, rather than the size of each piece, is what must be finite.

Why the Finite Condition Matters

A common mistake is to extend the theorem to an arbitrary union: “Each piece is compact, so their union is compact.” The proof identifies the problem with that reasoning. For finitely many pieces, the finite subcovers can be combined into a finite family. For infinitely many pieces, combining one finite subcover for each piece can produce infinitely many selected sets.

Worked Example: An Infinite Union That Is Not Compact

In \(\mathbb{R}\) with its standard topology, each singleton \(K_n=\{1/n\}\), for \(n\in\mathbb{N}\), is compact by the finite-subset proposition. Their union is \(A=\{1/n:n\in\mathbb{N}\}\). Define open intervals \(V_1=(1/2,3/2)\) and, for each integer \(n\geq2\), \(V_n=(1/(n+1),1/(n-1))\).

These intervals cover \(A\). The interval \(V_1\) contains \(1\). For \(n\geq2\), taking reciprocals of positive numbers shows \(1/(n+1)<1/n<1/(n-1)\), so \(1/n\in V_n\). In fact \(V_n\cap A=\{1/n\}\): for an integer \(k\geq1\), membership \(1/(n+1)<1/k<1/(n-1)\) is equivalent to \(n-1<k<n+1\), which forces \(k=n\). Also \(V_1\cap A=\{1\}\), since \(1/2<1/k<3/2\) for a positive integer \(k\) holds exactly when \(k=1\).

Consequently, any finite selection from the cover \(\{V_n:n\in\mathbb{N}\}\) covers only finitely many points of \(A\), and \(A\) is infinite. The cover has no finite subcover, so \(A\) is not compact. The incorrect argument treats “one finite subcover for each \(K_n\)” as though the combined selection must be finite. Here each piece needs just one set, but infinitely many pieces require infinitely many sets.

Check: For every \(n\geq2\), \(V_n\cap A\) contains exactly the point \(1/n\), while \(V_1\cap A\) contains exactly \(1\). A finite selection therefore misses \(1/k\) for any index \(k\) not selected.

The useful method is specific and reliable: start with one arbitrary open cover of the union, restrict it to each compact piece, take a finite subcover on each piece, and combine the selections. The finite number of pieces is what turns those separate finite selections into a single finite subcover.

Check Your Understanding

Use the open-cover definition in each question.

Why is the restriction \(U\cap K_i\) open in \(K_i\) when \(U\) is open in \(K=\bigcup_{i=1}^{n}K_i\)?
In the theorem’s proof, where is the assumption that the number of compact pieces is finite used?
Does the finite-union result require the compact subspaces to be disjoint? Explain why or why not.
Give an open cover of \(\{1/n:n\in\mathbb{N}\}\) in the standard topology of \(\mathbb{R}\) that has no finite subcover.
Why is an arbitrary union of compact subspaces not covered by the finite-union theorem?