Compactness and Subspaces
A space is compact when every open cover has a finite subcover. It is tempting to expect this property to pass to every subset: surely a smaller space should be easier to cover. The key point is that a cover of a subspace need only cover the points of that subspace. It may leave out points of the ambient space, and compactness of the ambient space applies only to covers that cover all of it.
Let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). The subspace topology on \(A\) consists of the sets \(A\cap U\), where \(U\in\mathcal{T}\). Thus an open set in \(A\) need not be open in \(X\), although it is always the intersection of \(A\) with an open set of \(X\). A family of sets covers \(A\) if its union contains every point of \(A\).
Here a finite subcover means a finite selection from the given cover whose union still contains every point of the space being covered. The choice of space matters: a family that covers \(A\) may not cover \(X\).
Translating a Subspace Cover
Proof. Suppose first that \(\{G_i:i\in J\}\) covers \(A\). By the definition of the subspace topology, for each \(i\) there is an open set \(U_i\in\mathcal{T}\) with \(G_i=A\cap U_i\). Every \(a\in A\) belongs to some \(G_i\), so \(a\in U_i\). Therefore the \(U_i\) cover \(A\). Conversely, suppose \(G_i=A\cap U_i\) and the \(U_i\) cover \(A\). For any \(a\in A\), some \(U_i\) contains \(a\), so \(a\in A\cap U_i=G_i\). Thus the \(G_i\) cover \(A\). \(\square\)
This translation preserves the finite-subcover question for the subspace: a finite selection of the \(G_i\) covers \(A\) exactly when the corresponding finite selection of the \(U_i\) covers \(A\). It does not say that the \(U_i\) cover all of \(X\). That distinction is central to the example below.
The Open Interval Inside the Compact Interval
Worked Example: A Cover of (0,1) With No Finite Subcover
Give \([0,1]\) its subspace topology from \(\mathbb{R}\) with the standard topology, and give \(A=(0,1)\) the subspace topology inherited from \([0,1]\). For each integer \(n\ge 2\), let \(U_n=(1/n,1)\). Each \(U_n\) is open in \(\mathbb{R}\), so \(U_n=A\cap U_n\) is open in \(A\).
It covers \(A\). Fix \(x\in(0,1)\). Since \(1/x\) is a real number, choose an integer \(n\ge2\) with \(n>1/x\). Taking reciprocals of positive numbers gives \(1/n<x\), and \(x<1\) by the choice of \(x\). Hence \(x\in(1/n,1)=U_n\). This proves that \(\{U_n:n\ge2\}\) covers \(A\).
No finite selection covers \(A\). Choose any finite nonempty selection \(U_{n_1},\ldots,U_{n_k}\), and let \(N\) be the largest of the selected indices. If \(n_i\le N\), then \(1/n_i\ge1/N\), so \(U_{n_i}\subseteq U_N\). Since \(U_N\) itself is selected, the union of the selected sets is \(U_N=(1/N,1)\). But \(1/(2N)\in(0,1)\) and \(1/(2N)\le1/N\), so \(1/(2N)\notin U_N\). The finite selection therefore misses a point of \(A\). The empty selection also fails to cover \(A\), since \(A\) is nonempty.
Conclusion. The displayed family is an open cover of \(A\) with no finite subcover, so \(A=(0,1)\) is not compact.
On the other hand, \([0,1]\) is compact in its subspace topology from \(\mathbb{R}\): this is the compactness of closed intervals established by the open-cover proof for closed intervals. Thus a compact space can have a subspace that is not compact. In particular, compactness is not hereditary.
Proof. The interval \([0,1]\) is compact by the open-cover result for closed intervals. The set \((0,1)\) is a subspace of it, with the subspace topology. The family \(U_n=(1/n,1)\), \(n\ge2\), consists of open sets in that subspace and covers it, as verified above. Every finite selection misses \(1/(2N)\), where \(N\) is the largest selected index; the empty selection also fails because \((0,1)\ne\varnothing\). Thus \((0,1)\) is not compact, which disproves hereditary compactness. \(\square\)
Why Ambient Compactness Does Not Fix the Cover
The sets \(U_n\) cover \((0,1)\), but they do not cover \([0,1]\): none contains either endpoint. Therefore they are not an open cover of the compact ambient interval, and the finite-subcover property of \([0,1]\) cannot be applied to them as they stand.
One can add open sets in \([0,1]\) that cover the endpoints, but the finite-subcover conclusion then concerns the enlarged family, not the original cover of \((0,1)\). For example, relative to \([0,1]\), the sets \([0,1/2)\) and \((1/2,1]\) are open. Together with \(U_3=(1/3,1)\), they cover \([0,1]\): the first covers 0 through points below \(1/2\), the second covers points above \(1/2\) through 1, and \(U_3\) contains \(1/2\). This finite cover of \([0,1]\) uses the added endpoint sets. It does not give a finite selection from the original family \(\{U_n:n\ge2\}\) that covers \((0,1)\).
Worked Example: A Finite Subspace Is Compact
Take \(F=\{1/3,2/3\}\subseteq[0,1]\), with the subspace topology inherited from the standard topology on \(\mathbb{R}\). Let \(\mathcal{G}\) be any open cover of \(F\). Since \(1/3\in F\), some member \(G_1\in\mathcal{G}\) contains \(1/3\). Since \(2/3\in F\), some member \(G_2\in\mathcal{G}\) contains \(2/3\). These two selected members cover every point of \(F\). If \(G_1=G_2\), one member already suffices. Thus every open cover has a finite subcover and \(F\) is compact.
Check: this argument uses only that \(F\) has two points and that each point belongs to some member of the cover. It applies regardless of which open sets of the subspace topology make up the cover.
Worked Example: A Compact Subspace With Infinitely Many Points
Let \(A=\{0\}\cup\{1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\), with the subspace topology inherited from the standard topology. We prove directly from the open-cover definition that \(A\) is compact.
Let \(\mathcal{G}\) be an arbitrary open cover of \(A\), and choose \(G\in\mathcal{G}\) with \(0\in G\). Because \(G\) is open in \(A\), there is an open set \(O\subseteq\mathbb{R}\) such that \(G=A\cap O\). Since \(0\in O\) and \(O\) is open, there is an \(\varepsilon>0\) with \((-\varepsilon,\varepsilon)\subseteq O\). Choose \(N\in\mathbb{N}\) so large that \(1/N<\varepsilon\). For every \(n\ge N\), positivity and \(n\ge N\) give \(0<1/n\le1/N<\varepsilon\), so \(1/n\in O\), and hence \(1/n\in G\). Thus \(G\) contains 0 and all points \(1/n\) with \(n\ge N\).
The only points of \(A\) not yet accounted for are among the finite list \(1,1/2,\ldots,1/(N-1)\). For each point on that list, choose one member of \(\mathcal{G}\) that contains it. Together with \(G\), these choices form a finite subcover. If \(N=1\), the list is empty and \(G\) alone covers \(A\). Since the argument applies to every open cover, \(A\) is compact.
Check: the proof does not assume that every infinite subspace is noncompact. The finite-subcover property depends on the particular space and its topology, not simply on whether it has infinitely many points.
A Common Mistake
Common pitfall. One might argue: “\([0,1]\) is compact and \((0,1)\subseteq[0,1]\), so \((0,1)\) is compact.” The conclusion does not follow from the definition. Compactness of \([0,1]\) gives a finite subcover only for a cover of all of \([0,1]\); the cover \(U_n=(1/n,1)\) covers only \((0,1)\). The explicit check shows that no finite selection covers that subspace. The correct conclusion is that compactness can fail to pass to a subspace, not that it always fails: the finite subspace and the infinite subspace above are both compact.
When testing compactness of a subspace, first identify its own open sets and then test covers of that subspace. Translating relative open sets to ambient open sets is useful, but it does not turn a cover of the subspace into a cover of the whole ambient space. The endpoints in the interval example explain precisely why the ambient compactness argument does not apply.
Check Your Understanding
For each question, keep track of which space the proposed family is required to cover.
- For \(n\ge2\), explain why \(U_n=(1/n,1)\) is open in the subspace \((0,1)\) and why the family covers every point of that subspace.
- Given finitely many members \(U_{n_1},\ldots,U_{n_k}\), identify the largest index \(N\) and a point of \((0,1)\) that the selected family misses.
- Why can the compactness of \([0,1]\) not be applied directly to the cover \(\{U_n:n\ge2\}\)?
- Prove directly that every finite topological space is compact, using an arbitrary open cover.
- In the proof for \(A=\{0\}\cup\{1/n:n\in\mathbb{N}\}\), why does an open set containing 0 contain all \(1/n\) for sufficiently large \(n\)?