Closed Sets Relative to an Ambient Space

Closedness Depends on the Ambient Space

Whether a set is closed is not determined by its points alone: it also depends on the topology in which the set is being considered. A subset can be closed within a subspace without being closed in the larger space. The useful test is precise: a set \(C\) is closed in a subspace \(A\) exactly when there is a closed set \(F\) in the ambient space \(X\) such that \(C=A\cap F\).

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 a set is open in \(A\) when it is the intersection of \(A\) with some open set of \(X\). Closedness in \(A\) is defined by taking complements within \(A\), not within all of \(X\).

Definition: A subset \(C\subseteq A\) is closed in \(A\) (or relatively closed in \(A\)) if \(A\setminus C\) is open in the subspace topology on \(A\). A subset \(F\subseteq X\) is closed in \(X\) if \(X\setminus F\) is open in \(X\).

The distinction between \(A\setminus C\) and \(X\setminus C\) matters. The first tests closedness within \(A\); the second tests closedness in \(X\). We now prove that the two tests are related by intersection with \(A\).

The Ambient-Closed-Set Criterion

Theorem: Let \(A\subseteq X\) and \(C\subseteq A\). Then \(C\) is closed in \(A\) if and only if there is a closed set \(F\subseteq X\) such that \(C=A\cap F\).

Proof. Suppose first that \(C\) is closed in \(A\). By the definition of relative closedness, \(A\setminus C\) is open in \(A\). By the definition of the subspace topology, there is an open set \(U\subseteq X\) with \(A\setminus C=A\cap U\). Let \(F=X\setminus U\). Since \(U\) is open, \(F\) is closed in \(X\). Also, because \(C\subseteq A\), the sets \(C\) and \(A\setminus C\) partition \(A\). Therefore \(C=A\setminus(A\setminus C)=A\setminus(A\cap U)=A\cap(X\setminus U)=A\cap F\).

Conversely, suppose \(F\) is closed in \(X\) and \(C=A\cap F\). Then \(A\setminus C=A\setminus(A\cap F)=A\cap(X\setminus F)\). The set \(X\setminus F\) is open in \(X\), so its intersection with \(A\) is open in the subspace topology on \(A\). Hence \(A\setminus C\) is open in \(A\), and \(C\) is closed in \(A\). Both directions also apply when \(A=\varnothing\): then \(C=\varnothing\), and \(C=A\cap\varnothing\), with \(\varnothing\) closed in \(X\). \(\square\)

The theorem supplies an ambient witness \(F\) for relative closedness. That witness need not equal \(C\): it only needs to agree with \(C\) on the points of \(A\). In fact, the same relative closed set can have many witnesses, because changing \(F\) outside \(A\) does not change \(A\cap F\).

Worked Example: Relatively Closed but Not Closed in \(\mathbb{R}\)

Give a set that is closed in the subspace \(A=(0,1)\) of \(\mathbb{R}\), but is not closed in \(\mathbb{R}\). Throughout this example, \(\mathbb{R}\) has its standard topology.

Take \(C=(0,\tfrac12]\), which is a subset of \(A\), and choose the ambient set \(F=(-\infty,\tfrac12]\). Its complement in \(\mathbb{R}\) is \((\tfrac12,\infty)\), an open set, so \(F\) is closed in \(\mathbb{R}\). Compute the intersection: \(A\cap F=(0,1)\cap(-\infty,\tfrac12]=(0,\tfrac12]=C\). The endpoint \(\tfrac12\) is included because it belongs to both sets, while \(0\) is excluded because \(0\notin A\). The theorem shows that \(C\) is closed in \(A\).

To check the relative definition directly, \(A\setminus C=(\tfrac12,1)\). This equals \(A\cap(\tfrac12,2)\), and \((\tfrac12,2)\) is open in \(\mathbb{R}\). Hence \(A\setminus C\) is open in \(A\).

But \(C\) is not closed in \(\mathbb{R}\). If it were, its complement \(\mathbb{R}\setminus C=(-\infty,0]\cup(\tfrac12,\infty)\) would be open. This complement contains \(0\), but every open interval around \(0\) contains positive numbers less than \(\tfrac12\), and those numbers belong to \(C\). Thus no open interval around \(0\) is contained in \(\mathbb{R}\setminus C\), so that complement is not open. The two closedness claims differ because they use different ambient spaces.

Check: the witness calculation gives \(A\cap F=C\), while the complement test in \(\mathbb{R}\) fails at \(0\). These checks confirm that \(C\) is relatively closed but not ambiently closed.

Intersections and the Direction of Implication

The criterion gives one implication without any extra assumption on \(A\): if \(C\subseteq A\) is closed in \(X\), then \(C\) is closed in \(A\). Indeed, use \(F=C\), since \(A\cap C=C\). The reverse implication requires care. If \(C=A\cap F\), the fact that \(F\) is closed does not by itself make \(A\cap F\) closed in \(X\); the other intersected set \(A\) may fail to be closed.

Proposition: If \(C\subseteq A\subseteq X\), then (a) if \(C\) is closed in \(X\), it is closed in \(A\); and (b) if \(A\) is closed in \(X\) and \(C\) is closed in \(A\), then \(C\) is closed in \(X\).

Proof. For (a), let \(F=C\). It is closed in \(X\), and \(A\cap F=A\cap C=C\), since \(C\subseteq A\). The theorem implies that \(C\) is closed in \(A\). For (b), the theorem gives a closed set \(F\subseteq X\) with \(C=A\cap F\). Both \(A\) and \(F\) are closed in \(X\), and a finite intersection of closed sets is closed: its complement is the union of the two open complements. Thus \(C=A\cap F\) is closed in \(X\). \(\square\)

So if \(A\) is closed in \(X\), relative closedness of \(C\subseteq A\) does imply ambient closedness. When \(A\) is not closed, this conclusion can fail, as the preceding example demonstrates: \(A=(0,1)\) is not closed in \(\mathbb{R}\), and \(C=(0,\tfrac12]\) is closed in \(A\) but not in \(\mathbb{R}\).

Worked Example: A Finite Space and Its Subspace

Let \(X=\{a,b,c\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\). This is a topology: arbitrary unions and finite intersections of its listed sets are again among the listed sets. Let \(A=\{a,b\}\) and \(C=\{b\}\).

First list the closed sets of \(X\), by taking complements of the four open sets. They are \(X\), \(\{b,c\}\), \(\{c\}\), and \(\varnothing\). In particular, \(\{b,c\}\) is closed. Its intersection with \(A\) is \(A\cap\{b,c\}=\{a,b\}\cap\{b,c\}=\{b\}=C\). The theorem therefore verifies that \(C\) is closed in \(A\).

For a direct check, the subspace open sets are the intersections of \(A\) with the open sets of \(X\). These are \(\varnothing\), \(\{a\}\), and \(A\). The complement of \(C\) within \(A\) is \(A\setminus C=\{a\}\), which is open in \(A\). In contrast, \(C\) is not closed in \(X\): its complement \(X\setminus C=\{a,c\}\) is not one of the four open sets in \(\mathcal{T}\). This finite example makes the role of the ambient space explicit: the witness \(\{b,c\}\) intersects \(A\) in \(C\), but it is not the same set as \(C\).

Check: relative closedness follows because \(A\setminus C=\{a\}\) is open in the subspace. Ambient closedness fails because \(\{a,c\}\notin\mathcal{T}\), so \(X\setminus C\) is not open.

Worked Example: When a Closed Subspace Makes the Difference

In \(\mathbb{R}\) with the standard topology, let \(A=[0,2]\) and \(C=[0,1]\). Verify relative closedness, and explain why it also gives ambient closedness in this case.

The set \(A\) is closed in \(\mathbb{R}\), since its complement \((-\infty,0)\cup(2,\infty)\) is open. Within \(A\), the complement of \(C\) is \(A\setminus C=(1,2]\). It is relatively open because \(A\cap(1,3)=[0,2]\cap(1,3)=(1,2]\), and \((1,3)\) is open in \(\mathbb{R}\). Thus \(C\) is closed in \(A\).

The ambient-closed-set criterion also gives the witness \(F=(-\infty,1]\), which is closed because its complement \((1,\infty)\) is open. Computing the intersection gives \(A\cap F=[0,2]\cap(-\infty,1]=[0,1]=C\). Moreover, since \(A\) and \(F\) are both closed in \(\mathbb{R}\), their intersection \(C\) is closed in \(\mathbb{R}\). This outcome is guaranteed by part (b) of the proposition, not by relative closedness alone.

Check: the relative complement \((1,2]\) is \(A\cap(1,3)\), and the ambient witness intersection is \([0,2]\cap(-\infty,1]=[0,1]\). The endpoint \(1\) belongs to \(C\); the endpoint \(2\) belongs to \(A\setminus C\).

A Common Mistake

Common pitfall. From \(C=A\cap F\) with \(F\) closed in \(X\), it is incorrect to conclude that \(C\) must be closed in \(X\). The calculation only expresses \(C\) as the intersection of \(F\) with \(A\); if \(A\) is not closed, this intersection need not be closed in \(X\).

For a concrete correction, take \(A=(0,1)\), \(F=(-\infty,\tfrac12]\), and \(C=A\cap F=(0,\tfrac12]\) in standard \(\mathbb{R}\). The set \(F\) is closed and \(C\) is closed in \(A\), but \(C\) is not closed in \(\mathbb{R}\), as shown by the failure of its complement to be open at \(0\). The valid conclusion is exactly the theorem's one: \(C\) is closed in the subspace \(A\). If one additionally knows that \(A\) is closed in \(X\), then the proposition does imply that \(C\) is closed in \(X\).

When checking a claim about closedness, name the space first. Then use the complement in that space: \(A\setminus C\) for relative closedness, or \(X\setminus C\) for ambient closedness. The intersection criterion translates between these tests without conflating them.

Check Your Understanding

For each question, keep track of which space supplies the complement or the closed-set witness.

Let \(A=(0,2)\) and \(C=(0,1]\) in standard \(\mathbb{R}\). Find a closed \(F\subseteq\mathbb{R}\) such that \(A\cap F=C\).
If \(C\subseteq A\subseteq X\) and \(C\) is closed in \(X\), must \(C\) be closed in \(A\)? Give the reason.
In the finite-space example, list the closed sets of the subspace \(A=\{a,b\}\) and identify the relative complement of \(\{b\}\).
Suppose \(A\) is closed in \(X\) and \(C\) is closed in \(A\). Which theorem and proposition together show that \(C\) is closed in \(X\)?
Give an example of a set closed in a subspace but not in the ambient space, and identify both the subspace and the ambient space.