Why Closed Subspaces Are the Right Setting
Normality concerns separating two disjoint closed sets by disjoint open sets. In a subspace, however, “closed” and “open” refer to the subspace topology, not necessarily to the ambient space. The key fact is that a relatively closed subset of a closed subspace is itself closed in the ambient space. This lets us apply normality in the larger space, then restrict the separating open sets back to the subspace.
Throughout, a subspace \(Y\subseteq X\) has the subspace topology \(\{Y\cap U: U\text{ is open in }X\}\).
The requirement that the separating open sets be disjoint is essential: normality is stronger than merely finding an open neighborhood of each closed set. The theorem below preserves the separation property, as well as the \(T_1\) condition when normality is understood as \(T_4\).
Relative Closedness and Ambient Closedness
Proof. Suppose first that \(F\) is closed in \(Y\). Then \(Y\setminus F\) is open in \(Y\). By the definition of the subspace topology, there is an open set \(G\subseteq X\) such that \(Y\setminus F=Y\cap G\). Let \(C=X\setminus G\), which is closed in \(X\). For a point \(y\in Y\), membership in \(C\) is equivalent to nonmembership in \(G\), so \(y\in C\) if and only if \(y\notin Y\cap G=Y\setminus F\), which is equivalent to \(y\in F\). Thus \(F=Y\cap C\).
Conversely, suppose \(F=Y\cap C\) for a closed set \(C\subseteq X\). Then \(Y\setminus F=Y\cap(X\setminus C)\). Since \(X\setminus C\) is open in \(X\), this intersection is open in \(Y\) by the definition of the subspace topology. Therefore \(F\) is closed in \(Y\). \(\square\)
In particular, if \(Y\) is closed in \(X\), then every relatively closed \(F\subseteq Y\) is ambient closed: the proposition gives \(F=Y\cap C\), and the intersection of the two closed sets \(Y\) and \(C\) is closed in \(X\). This implication is exactly where the hypothesis that \(Y\) is closed enters.
Worked Example: A Closed Interval in the Real Line
Let \(X=\mathbb{R}\) with its standard topology and \(Y=[0,1]\). Take \(F=[0,1/2]\). First, \(F\) is closed in \(Y\): its relative complement is \(Y\setminus F=(1/2,1]\), and \((1/2,1]=Y\cap(1/2,2)\), an open set of \(Y\).
For an ambient closed set representing \(F\), take \(C=(-\infty,1/2]\). This is closed in \(\mathbb{R}\), and direct intersection gives \(Y\cap C=[0,1]\cap(-\infty,1/2]=[0,1/2]=F\). Also \(Y\) itself is closed in \(\mathbb{R}\), since its complement \((-\infty,0)\cup(1,\infty)\) is open. Thus this is an instance where a relatively closed set is ambient closed, as the proposition predicts.
Check: The endpoints are included: \(0\in Y\cap C\) and \(1/2\in Y\cap C\). The point \(3/4\) belongs to \(Y\) but not \(C\), so it is not in \(F\).
The Inheritance Theorem
Proof. Let \(A\) and \(B\) be disjoint closed subsets of \(Y\). By the proposition, there are closed sets \(C,D\subseteq X\) such that \(A=Y\cap C\) and \(B=Y\cap D\). Since \(Y\) is closed in \(X\), both \(A=Y\cap C\) and \(B=Y\cap D\) are closed in \(X\). They are disjoint because they were disjoint as subsets of \(Y\). Normality of \(X\) therefore gives disjoint open sets \(U,V\subseteq X\) with \(A\subseteq U\) and \(B\subseteq V\). The sets \(Y\cap U\) and \(Y\cap V\) are open in \(Y\), contain \(A\) and \(B\), respectively, and are disjoint, since \(U\cap V=\varnothing\). This proves that \(Y\) is normal.
If \(X\) is \(T_4\), it is also \(T_1\). For each \(y\in Y\), the singleton \(\{y\}\) is closed in \(X\), hence closed in \(Y\) by the subspace topology. Thus \(Y\) is \(T_1\) as well as normal, so \(Y\) is \(T_4\). \(\square\)
The argument includes the cases \(A=\varnothing\) or \(B=\varnothing\): the definition of normality supplies disjoint open sets containing the two closed sets, and the restrictions remain disjoint. If \(Y=\varnothing\), there are no points whose singletons need to be checked, and the normality condition holds because the only subset of \(Y\) is empty.
Worked Example: Opposite Edges of a Square
Let \(X=\mathbb{R}^2\) with the Euclidean topology, and let \(Y=[0,1]\times[0,1]\). The set \(Y\) is closed in \(X\), because it is the intersection of the closed sets \([0,1]\times\mathbb{R}\) and \(\mathbb{R}\times[0,1]\). The earlier result that every metric space is normal applies to \(\mathbb{R}^2\). Consider the bottom and top edges \(A=[0,1]\times\{0\}\) and \(B=[0,1]\times\{1\}\).
Both edges are closed in \(Y\): they are intersections of \(Y\) with the ambient closed sets \(\mathbb{R}\times\{0\}\) and \(\mathbb{R}\times\{1\}\). They are disjoint because a point in their intersection would have second coordinate both \(0\) and \(1\), which is impossible.
Here we can exhibit the separating neighborhoods explicitly. Let \(U=\mathbb{R}\times(-1/3,1/3)\) and \(V=\mathbb{R}\times(2/3,4/3)\). These are open in \(\mathbb{R}^2\), since each is the inverse image of an open interval under the continuous second-coordinate projection. The first-coordinate restrictions are unnecessary: both sets contain the entire corresponding edge. Their second-coordinate intervals are disjoint, so \(U\cap V=\varnothing\). The relative open sets \(Y\cap U\) and \(Y\cap V\) therefore separate \(A\) and \(B\) in \(Y\).
Check: Every point of \(A\) has second coordinate \(0\), and \(-1/3<0<1/3\), so \(A\subseteq Y\cap U\). Every point of \(B\) has second coordinate \(1\), and \(2/3<1<4/3\), so \(B\subseteq Y\cap V\). No second coordinate can belong to both intervals, verifying disjointness.
Worked Example: A Finite Discrete Subspace
Let \(X=\{a,b,c,d\}\) with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), and let \(Y=\{a,b,c\}\). In a discrete topology every subset is both open and closed, so \(X\) is normal: for any disjoint \(A,B\subseteq X\), the choices \(U=A\) and \(V=B\) are disjoint open sets containing them. The set \(Y\) is closed in \(X\).
Take \(A=\{a\}\) and \(B=\{c\}\). They are disjoint and closed in \(Y\). As ambient closed sets, they can be written \(A=Y\cap\{a\}\) and \(B=Y\cap\{c\}\). They are already closed in \(X\), so choose the ambient open sets \(U=\{a\}\) and \(V=\{c\}\). Their restrictions to \(Y\) are \(Y\cap U=\{a\}\) and \(Y\cap V=\{c\}\), which are disjoint open neighborhoods in \(Y\). This verifies the conclusion directly for this pair.
More generally, if \(A,B\) are any disjoint relatively closed subsets of this \(Y\), they are subsets of a discrete space and hence open in \(Y\). Taking those same sets as their neighborhoods separates them. This includes an empty set: if \(A=\varnothing\), take \(Y\cap U=\varnothing\) and \(Y\cap V=B\); these are disjoint relative open sets.
A Common Pitfall: Relative Does Not Mean Ambient
A frequent error is to start with a set closed in \(Y\) and immediately apply normality of \(X\) to it. That is not justified unless the set is also closed in \(X\). The proposition gives a reliable repair: represent a relatively closed set as \(Y\cap C\) for an ambient closed \(C\). When \(Y\) is closed, that intersection is ambient closed.
For a concrete check of the distinction, take \(X=\mathbb{R}\), \(Y=(0,1)\), and \(F=(0,1/2]\). The set \(F\) is closed in \(Y\), because \(Y\setminus F=(1/2,1)=Y\cap(1/2,2)\) is open in \(Y\). But \(F\) is not closed in \(\mathbb{R}\): every open neighborhood of \(0\) contains a positive point less than \(1/2\), hence meets \(F\), while \(0\notin F\). Thus \(0\) lies in the closure of \(F\) in \(\mathbb{R}\), so \(F\) is not ambient closed.
The proposition still represents this \(F\) as an intersection: \(F=Y\cap(-\infty,1/2]\). The intersection is not ambient closed because \(Y\) is not closed. This example shows precisely why the proof for a closed subspace works and why the step “relatively closed, therefore ambient closed” is invalid without the closedness hypothesis. It does not assert that every nonclosed subspace is nonnormal; closedness is a sufficient condition for this inheritance theorem.
Check Your Understanding
For each question, keep ambient and relative open or closed sets distinct.
- Let \(Y\subseteq X\). If \(F\) is closed in \(Y\), what ambient-set representation does the proposition provide?
- In the inheritance proof, at which step is the assumption that \(Y\) is closed used?
- Suppose \(X\) is \(T_4\) and \(Y\) is closed in \(X\). Why is each singleton in \(Y\) closed in the subspace topology?
- For \(Y=(0,1)\subseteq\mathbb{R}\), verify that \((0,1/2]\) is closed in \(Y\) but not closed in \(\mathbb{R}\).
- If \(A\) and \(B\) are disjoint closed subsets of a closed subspace \(Y\), explain why their ambient open separators restrict to disjoint open separators in \(Y\).