Open Sets in a Subspace Come from the Ambient Space
A subset can have its own topology even when it sits inside a larger space. For compactness, this raises a practical question: how can we work with a cover whose sets are open in the subset, rather than open in the whole space? The answer is to translate each such set into an ambient open set whose intersection with the subset gives the original set. This translation lets us test compactness using either kind of cover.
Let \((X,\mathcal{T})\) be a topological space, and let \(A\subseteq X\). The subspace topology on \(A\) is the collection \(\mathcal{T}_A=\{A\cap U:U\in\mathcal{T}\}\). Thus a set \(V\subseteq A\) is open in \(A\) precisely when there is an open set \(U\) in \(X\) such that \(V=A\cap U\). The set \(U\) is an ambient representative of \(V\). It need not be unique: changing \(U\) outside \(A\) may leave \(A\cap U\) unchanged.
An ambient open family covering \(A\) need not cover all of \(X\). The requirement is only \(A\subseteq\bigcup_{U\in\mathcal{U}}U\). Intersecting each member with \(A\) then gives a cover by sets open in the subspace. Conversely, each member of a subspace-open cover has at least one ambient open representative. The next result makes this correspondence precise without claiming the representatives are unique.
Proof. For the first assertion, each \(V_i\) is open in \(A\), so by the definition of the subspace topology there is an ambient open set \(U_i\) with \(V_i=A\cap U_i\). Since the \(V_i\) cover \(A\), each \(a\in A\) belongs to some \(V_i\), and hence to \(U_i\). Thus the \(U_i\) cover \(A\). For the converse, \(A\cap U_i\) is open in \(A\) by the definition of the subspace topology. Each point \(a\in A\) belongs to some \(U_i\), so it belongs to \(A\cap U_i\). Therefore these traces cover \(A\). \(\square\)
The Ambient-Open-Cover Criterion
Proof. Suppose first that \(A\) is compact, and let \(\{U_i:i\in J\}\) be ambient open sets covering \(A\). By Theorem 1, the sets \(A\cap U_i\) form an open cover of the subspace. Compactness gives indices \(i_1,\ldots,i_k\) such that \(A\subseteq\bigcup_{j=1}^k(A\cap U_{i_j})\). Since \(A\cap U_{i_j}\subseteq U_{i_j}\), the finite family \(U_{i_1},\ldots,U_{i_k}\) covers \(A\).
Conversely, suppose every ambient open family covering \(A\) has a finite subfamily covering \(A\), and let \(\mathcal{V}\) be an open cover of \(A\) in its subspace topology. Consider the family \(\mathcal{W}=\{U\in\mathcal{T}:A\cap U\in\mathcal{V}\}\). This is an ambient open family covering \(A\): for each \(a\in A\), some \(V\in\mathcal{V}\) contains \(a\), and since \(V\) is subspace-open, \(V=A\cap U\) for some ambient open \(U\). By the assumed property, finitely many members \(U_1,\ldots,U_k\) of \(\mathcal{W}\) cover \(A\). Their traces \(A\cap U_1,\ldots,A\cap U_k\) belong to \(\mathcal{V}\) and cover \(A\). Thus \(A\) is compact. If \(A=\varnothing\), the empty family already covers it and is a finite subcover, so the equivalence includes the empty subspace. \(\square\)
The criterion changes only the language of the cover: it does not require the ambient open sets to cover points outside \(A\). In practice, one can use whichever description makes the cover easiest to express. A relative cover may be simple to write down, while an ambient family may make it easier to see that no finite subcover exists.
Worked Example: Translating a Cover of a Closed Interval
Take \(X=\mathbb{R}\) with its standard topology and \(A=[0,1]\) with the subspace topology. Consider the two ambient open sets \(U_1=(-1,\frac{3}{5})\cup(\frac{4}{5},2)\) and \(U_2=(\frac{1}{2},\frac{9}{10})\).
Their traces on \(A\) are \(V_1=A\cap U_1=[0,\frac{3}{5})\cup(\frac{4}{5},1]\) and \(V_2=A\cap U_2=(\frac{1}{2},\frac{9}{10})\). These sets cover \(A\). Indeed, \(0\le x<\frac{3}{5}\) implies \(x\in V_1\); \(\frac{3}{5}\le x\le\frac{4}{5}\) implies \(\frac{1}{2}<x<\frac{9}{10}\), so \(x\in V_2\); and \(\frac{4}{5}<x\le1\) implies \(x\in V_1\). The boundary values are included correctly: \(\frac{3}{5}\) and \(\frac{4}{5}\) belong to \(V_2\), while \(0\) and \(1\) belong to \(V_1\).
In the other direction, starting with the relative cover \(\{V_1,V_2\}\), the displayed \(U_1,U_2\) are ambient representatives because \(A\cap U_i=V_i\) for both indices. They cover \(A\), although they do not cover all of \(\mathbb{R}\). This example also shows why ambient representatives are not unique: \(U_1\cup(3,4)\) is another ambient open representative of \(V_1\), since \((3,4)\cap A=\varnothing\).
Using the Criterion to Test Compactness
Worked Example: An Open Interval Is Not Compact
Let \(A=(0,1)\) with the subspace topology inherited from standard \(\mathbb{R}\). For each integer \(n\ge2\), set \(U_n=(\frac{1}{n},2)\), an ambient open set. Its trace is \(A\cap U_n=(\frac{1}{n},1)\).
These traces cover \(A\). Given any \(x\in(0,1)\), choose an integer \(n\ge2\) with \(n>1/x\), which is possible by the Archimedean property of \(\mathbb{R}\). Then \(1/n<x<1<2\), so \(x\in U_n\), and hence \(x\in A\cap U_n\).
No finite subfamily covers \(A\). For any nonempty finite selection of indices, let \(N\) be the largest selected index. Since \(n\le N\) implies \(1/n\ge1/N\), the selected ambient sets are all contained in \(U_N\). Their traces therefore have union contained in \((\frac{1}{N},1)\). The point \(x=\frac{1}{2N}\) satisfies \(0<\frac{1}{2N}<\frac{1}{N}<1\), so it lies in \(A\) but not in any selected trace. The empty selection does not cover the nonempty set \(A\), either. By Theorem 2, \(A\) is not compact.
Check: for a selected largest index \(N\), the point \(\frac{1}{2N}\) is strictly below the excluded endpoint \(\frac{1}{N}\). Thus the failure to cover is not an endpoint-convention issue.
Worked Example: Applying the Criterion to a Compact Interval
Again let \(A=[0,1]\subseteq\mathbb{R}\). Take any family \(\{U_i:i\in J\}\) of standard-open subsets of \(\mathbb{R}\) that covers \(A\). The traces \(A\cap U_i\) are open in the subspace and cover it, by Theorem 1. The earlier open-cover proof for closed intervals establishes that \([0,1]\) is compact. Therefore there are finitely many indices \(i_1,\ldots,i_k\) such that \([0,1]\subseteq\bigcup_{j=1}^k(A\cap U_{i_j})\). Since each trace is contained in its ambient set, the same finite collection \(U_{i_1},\ldots,U_{i_k}\) covers \([0,1]\). This is precisely the ambient-cover criterion.
The conclusion concerns covers of \([0,1]\), not covers of the whole real line. For example, a single interval \((-1,2)\) covers \([0,1]\), even though it does not cover \(\mathbb{R}\). The finite-subcover question asks whether finitely many members cover the subspace under consideration.
Worked Example: A Finite Subspace
Let \(A=\{2,5\}\subseteq\mathbb{R}\), with its subspace topology, and let \(\{U_i:i\in J\}\) be any family of ambient open sets covering \(A\). Since the family covers \(A\), there is an index \(i_2\) with \(2\in U_{i_2}\), and an index \(i_5\) with \(5\in U_{i_5}\). The at most two selected sets \(U_{i_2},U_{i_5}\) cover \(A\). If a selected set covers both points, only one set is needed. Thus every ambient open cover has a finite subcover, and Theorem 2 shows that \(A\) is compact.
The argument uses the number of points in \(A\), not any special feature of the ambient topology: for a finite subset with \(m\) points, choose one cover member containing each point, obtaining at most \(m\) sets. When \(A\) is empty, zero sets suffice. The translation criterion is useful even in such simple cases because it identifies exactly which topology determines openness: the traces are open in \(A\), regardless of whether they are open in \(\mathbb{R}\).
A Common Pitfall: Relative Open Does Not Mean Ambient Open
For \(A=[0,1]\subseteq\mathbb{R}\), the set \(V=[0,\frac{1}{2})\) is open in \(A\), because \(V=A\cap(-1,\frac{1}{2})\) and \((-1,\frac{1}{2})\) is open in \(\mathbb{R}\). But \(V\) is not open in \(\mathbb{R}\): every real open interval around \(0\) contains negative numbers, which are not in \(V\). A mistaken argument might claim that a subspace-open cover is already a cover by ambient-open sets and apply an ambient compactness statement directly. The correction is to replace each relative-open member \(V_i\) by an ambient representative \(U_i\) satisfying \(V_i=A\cap U_i\), or to use the ambient-open-cover criterion proved above.
The distinction is particularly useful when a cover is given in one form but compactness is easier to apply in the other. For proving compactness, translate a relative cover to ambient representatives and seek a finite family. For disproving compactness, it is enough to exhibit ambient open sets covering the subspace such that every finite selection misses some point of it.
Check Your Understanding
For each question, keep track of whether the sets are open in the subspace or in the ambient space.
- Let \(A=[0,2]\subseteq\mathbb{R}\). Write \([0,1)\) as the intersection of \(A\) with an ambient open set.
- Explain why an ambient open family covering \(A\) need not cover all of \(X\).
- For \(A=(0,1)\), verify directly that the sets \(A\cap(\frac{1}{n},2)\), \(n\ge2\), cover \(A\).
- In the same cover, if the largest selected index is \(N\), give a point of \(A\) missed by all selected sets and verify both strict inequalities.
- State the ambient-open-cover criterion for compactness, including what the ambient open sets are required to cover.