Ambient Sets and Relative Sets
A subset can have different openness or closedness properties depending on the space in which it is considered. For example, an interval that is not open in the real line may be open in a smaller space containing it. The subspace topology explains this difference: it tests openness using intersections with the subspace, rather than requiring the set itself to be open in the ambient space.
Let \((X,\mathcal{T})\) be a topological space and let \(A\subseteq X\). The subspace topology on \(A\) is \(\mathcal{T}_A=\{A\cap U:U\in\mathcal{T}\}\). Thus the open subsets of \(A\) are exactly the intersections of \(A\) with open subsets of \(X\). A subset \(B\subseteq A\) is closed in \(A\) when \(A\setminus B\) is open in \(A\).
- If \(U\) is open in \(X\), then \(A\cap U\) is open in \(A\). Conversely, \(B\subseteq A\) is open in \(A\) if and only if \(B=A\cap U\) for some open \(U\subseteq X\).
- If \(F\) is closed in \(X\), then \(A\cap F\) is closed in \(A\). Conversely, \(B\subseteq A\) is closed in \(A\) if and only if \(B=A\cap F\) for some closed \(F\subseteq X\).
Proof. For the first assertion, if \(U\) is open in \(X\), then \(A\cap U\) is one of the sets defining \(\mathcal{T}_A\), so it is open in \(A\). Conversely, if \(B\) is open in \(A\), the definition of the subspace topology says that there is an open \(U\subseteq X\) such that \(B=A\cap U\). This also proves the stated “if and only if.”
For the closed-set assertion, if \(F\) is closed in \(X\), then \(X\setminus F\) is open in \(X\). The relative complement of \(A\cap F\) in \(A\) is \(A\setminus(A\cap F)=A\cap(X\setminus F)\), which is open in \(A\) by the first assertion. Therefore \(A\cap F\) is closed in \(A\). Conversely, suppose \(B\) is closed in \(A\). Then \(A\setminus B\) is open in \(A\), so there is an open \(U\subseteq X\) such that \(A\setminus B=A\cap U\). Taking complements within \(A\) gives \(B=A\setminus(A\cap U)=A\cap(X\setminus U)\). Since \(X\setminus U\) is closed in \(X\), this is the required representation. \(\square\)
The theorem is about intersections. An ambient open set \(U\) need not be a subset of \(A\), so it is generally \(A\cap U\), not \(U\) itself, that is a subset of \(A\) and is open there. If an ambient open set \(U\) does satisfy \(U\subseteq A\), then \(A\cap U=U\), so \(U\) is open in \(A\). The same observation applies to ambient closed sets.
Worked Applications on the Real Line
Worked Example: An Ambient Open Set Produces a Relative Open Set
Take \(X=\mathbb{R}\) with its standard topology, \(A=[0,1]\) with the subspace topology, and the ambient open interval \(U=(-1,1)\). Compute \(A\cap U\) and decide whether the result is open in \(A\) and in \(\mathbb{R}\).
A real number \(x\) lies in \(A\cap U\) exactly when \(0\leq x\leq1\) and \(-1<x<1\). Together these conditions are \(0\leq x<1\), so \(A\cap U=[0,1)\). Because \(U\) is open in \(\mathbb{R}\), the intersection test gives that \([0,1)\) is open in \(A\).
It is not open in \(\mathbb{R}\). It contains \(0\), but every open interval around \(0\) contains a negative real number, which is not in \([0,1)\). Thus no open neighborhood of \(0\) in \(\mathbb{R}\) is contained in \([0,1)\).
Check: the intersection identity is exact: \(x\in[0,1]\cap(-1,1)\) is equivalent to \(0\leq x<1\), hence \([0,1]\cap(-1,1)=[0,1)\), not \([0,1]\). This calculation confirms both the relative-open conclusion and the distinction from ambient openness.
Worked Example: An Ambient Closed Set Produces a Relative Closed Set
Let \(X=\mathbb{R}\) with its standard topology, \(A=(0,1)\), and \(F=(-\infty,1/2]\). The set \(F\) is closed in \(\mathbb{R}\), since its complement \((1/2,\infty)\) is open. Find \(A\cap F\) and verify relative closedness directly.
A point belongs to \(A\cap F\) when \(0<x<1\) and \(x\leq1/2\). The combined conditions give \(0<x\leq1/2\), so \(A\cap F=(0,1/2]\). By the theorem this intersection is closed in \(A\). Directly, its complement within \(A\) is \(A\setminus(A\cap F)=(1/2,1)\). This is open in \(A\), because \((1/2,1)=A\cap(1/2,2)\), and \((1/2,2)\) is open in \(\mathbb{R}\). This verifies the relative closedness from the definition.
Check: the endpoint \(1/2\) is included because it satisfies \(x\leq1/2\), while \(0\) is excluded because it does not belong to \(A=(0,1)\). Thus the intersection is \((0,1/2]\), and its relative complement is exactly \((1/2,1)\).
Why the Converse Can Fail
The intersection tests have converses in representation form: every relatively open or relatively closed set is an intersection with an ambient set of the corresponding type. But that does not say the relative set itself is open or closed in \(X\). Its representation uses \(A\), which may change the set substantially. The following example checks both failures in one subspace.
Worked Example: Relatively Closed but Not Ambiently Closed
Let \(X=\mathbb{R}\) with its standard topology, \(A=(0,1)\), and \(B=(0,1/2]\). First verify that \(B\) is closed in \(A\), and then show that \(B\) is not closed in \(X\).
Since \(B\subseteq A\), its complement within \(A\) is \(A\setminus B=(1/2,1)\). The equality holds because the points of \(A\) not in \(B\) are precisely those strictly greater than \(1/2\) and strictly less than \(1\). Also \((1/2,1)=A\cap(1/2,2)\), where \((1/2,2)\) is open in \(\mathbb{R}\). Hence \(A\setminus B\) is open in \(A\), and \(B\) is closed in \(A\).
However, \(B\) is not closed in \(\mathbb{R}\). Its complement contains \(0\), and every open interval around \(0\) contains a positive real number less than \(1/2\), hence a point of \(B\). Therefore the complement of \(B\) is not open in \(\mathbb{R}\), so \(B\) is not closed there.
Check: for any radius \(r>0\), the number \(x=\min(r,1)/4\) satisfies \(0<x<r\) and \(x\leq1/4<1/2\). Thus \(x\in B\cap(-r,r)\). This verifies that every open interval around \(0\) meets \(B\), as used to show that \(X\setminus B\) is not open.
A parallel failure occurs for relative openness. In the first example, \([0,1)\) is open in \(A=[0,1]\) but not open in \(\mathbb{R}\). The reason is visible in its representation: \([0,1)=A\cap(-1,1)\). Intersecting with \(A\) removes all points outside \(A\), including the negative points that would prevent this set from being an ambient neighborhood of \(0\).
Proof. If \(B\) is open in \(A\), write \(B=A\cap U\) for an open \(U\subseteq X\). When \(A\) is open in \(X\), the intersection of the two open sets \(A\) and \(U\) is open in \(X\), so \(B\) is ambiently open. If \(B\) is closed in \(A\), write \(B=A\cap F\) for a closed \(F\subseteq X\). When \(A\) is closed in \(X\), the intersection of the two closed sets \(A\) and \(F\) is closed in \(X\). \(\square\)
The hypotheses of the corollary matter. In the relative-open example, \(A=[0,1]\) is not open in \(\mathbb{R}\), and the relatively open set \([0,1)\) is not ambiently open. In the relative-closed example, \(A=(0,1)\) is not closed in \(\mathbb{R}\), and the relatively closed set \((0,1/2]\) is not ambiently closed. These examples show why one cannot simply discard the condition on \(A\).
A Reliable Test and a Common Pitfall
When a problem asks whether a set is open or closed in a subspace, first check that the set is actually a subset of the subspace. Then use the relevant test: express a relatively open set as \(A\cap U\) with \(U\) ambiently open, or express a relatively closed set as \(A\cap F\) with \(F\) ambiently closed. A proposed representation is useful only if its intersection has been computed correctly.
Common pitfall. For \(A=[0,1]\subseteq\mathbb{R}\), one might mistakenly claim \([0,1]=A\cap(-1,1)\), then use that equality to discuss relative openness. The calculation fails at \(1\): although \(1\in A\), it is not in \((-1,1)\). In fact, \(A\cap(-1,1)=[0,1)\), since membership requires both \(0\leq x\leq1\) and \(-1<x<1\), which together give \(0\leq x<1\). The corrected identity proves that \([0,1)\) is open in \(A\); it does not prove that \([0,1]\) is open in \(A\).
| Ambient information | Conclusion in the subspace \(A\) | What is not automatic |
|---|---|---|
| \(U\) is open in \(X\) | \(A\cap U\) is open in \(A\) | \(A\cap U\) need not be open in \(X\) |
| \(F\) is closed in \(X\) | \(A\cap F\) is closed in \(A\) | \(A\cap F\) need not be closed in \(X\) |
| \(A\) is open in \(X\), and \(B\) is open in \(A\) | \(B\) is open in \(X\) | Without \(A\) open, the conclusion can fail |
| \(A\) is closed in \(X\), and \(B\) is closed in \(A\) | \(B\) is closed in \(X\) | Without \(A\) closed, the conclusion can fail |
Check Your Understanding
For each question, keep track of which space supplies the topology and compute the intersections before drawing a conclusion.
- Let \(A=[-1,2]\subseteq\mathbb{R}\) and \(U=(0,3)\). Compute \(A\cap U\) and state why it is open in \(A\).
- Let \(A=(0,2)\subseteq\mathbb{R}\) and \(F=(-\infty,1]\). Compute \(A\cap F\) and verify that it is closed in \(A\) by finding its relative complement.
- Give a relatively open subset of \(A=[0,1]\) that is not open in \(\mathbb{R}\), and justify both claims.
- For \(A=(0,1)\) and \(B=(0,1/2]\), show that \(B\) is closed in \(A\) but not closed in \(\mathbb{R}\).
- State the additional condition on \(A\) that ensures every relatively open subset of \(A\) is open in \(X\). Prove the implication using the intersection characterization.