Normality in a Space and in Its Subspaces
Normality is a separation property: disjoint closed sets must be placed inside disjoint open sets. The qualification “closed” depends on the space under consideration. A set can be closed in a subspace without being closed in the larger space, so normality of a space and normality of all its subspaces are different requirements.
Throughout, a space is normal if it is \(T_1\) and every two disjoint closed subsets have disjoint open neighborhoods containing them. Thus, for disjoint closed \(C,F\subseteq X\), normality requires open \(U,V\subseteq X\) with \(C\subseteq U\), \(F\subseteq V\), and \(U\cap V=\varnothing\). We use the \(T_4\) convention that includes \(T_1\). A subset \(Y\subseteq X\) has the subspace topology, whose open sets are \(Y\cap U\) for open \(U\subseteq X\).
The universal quantifier over subspaces is the key difference. Ordinary normality concerns closed pairs in \(X\). Hereditary normality concerns closed pairs in every \(Y\), where “closed in \(Y\)” need not mean “closed in \(X\).” Metric spaces satisfy the stronger property because every subspace inherits a metric, and the metric itself provides a separation construction.
Distance to a Set Gives the Separation
Let \((Y,d)\) be a metric space and let \(C\subseteq Y\) be nonempty. For \(y\in Y\), define its distance to \(C\) by \(\operatorname{dist}(y,C)=\inf\{d(y,c):c\in C\}\). This is a nonnegative real number. The crucial facts are that this function is continuous and that it is positive at every point outside a closed set.
Proof. The subspace \(Y\) is a metric space with the restricted metric \(d|_{Y\times Y}\). First suppose \(C\) and \(F\) are both nonempty, and define \(f(y)=\operatorname{dist}(y,C)\) and \(g(y)=\operatorname{dist}(y,F)\) for \(y\in Y\).
For any \(y,z\in Y\) and \(c\in C\), the triangle inequality gives \(d(y,c)\le d(y,z)+d(z,c)\). Taking the infimum over \(c\in C\) gives \(f(y)\le d(y,z)+f(z)\). Interchanging \(y\) and \(z\) gives \(|f(y)-f(z)|\le d(y,z)\). The same argument gives \(|g(y)-g(z)|\le d(y,z)\). Therefore \(h=f-g\) satisfies \(|h(y)-h(z)|\le 2d(y,z)\), so \(h\) is continuous.
Set \(U=\{y\in Y:h(y)<0\}\) and \(V=\{y\in Y:h(y)>0\}\). Continuity makes both sets open, and their defining strict inequalities make them disjoint. If \(c\in C\), then \(f(c)=0\). Since \(F\) is closed in \(Y\) and \(c\notin F\), there is \(r>0\) such that the metric ball \(B_d(c,r)\cap Y\) misses \(F\). Thus \(d(c,b)\ge r\) for every \(b\in F\), so \(g(c)\ge r>0\). Hence \(h(c)<0\), and \(c\in U\). For \(b\in F\), the same argument using that \(C\) is closed gives \(f(b)>0=g(b)\), so \(b\in V\).
If \(C=\varnothing\), take \(U=\varnothing\) and \(V=Y\). If \(F=\varnothing\), take \(U=Y\) and \(V=\varnothing\). These choices also work if both sets are empty. Finally, every metric space is \(T_1\): for a point \(x\), each \(y\ne x\) has a ball about \(y\) of radius \(d(x,y)/2\) that misses \(x\), so \(Y\setminus\{x\}\) is open. Consequently every subspace \(Y\) is normal, proving hereditary normality. \(\square\)
The proof uses closedness in the space being separated. For example, closedness of \(F\) in \(Y\) is exactly what ensures that a point \(c\in C\setminus F\) has a ball in \(Y\) missing \(F\). No assumption that \(Y\) itself is closed in \(X\) is needed.
Worked Example: Separating Closed Sets in a Subspace of the Line
Let \(Y=(0,1)\cup\{2\}\) with the metric inherited from \(\mathbb{R}\). Consider \(C=(0,1/3]\) and \(F=[2/3,1)\cup\{2\}\). These sets are disjoint and closed in \(Y\): their complements in \(Y\) are, respectively, \((1/3,1)\cup\{2\}\) and \((0,2/3)\), both open in \(Y\).
Take \(U=Y\cap(-\infty,1/2)\) and \(V=Y\cap(1/2,\infty)\). Each is open in \(Y\), because the corresponding interval is open in \(\mathbb{R}\). Every point of \(C\) is less than \(1/2\), so \(C\subseteq U\); every point of \(F\), including \(2\), is greater than \(1/2\), so \(F\subseteq V\). No real number is both less than and greater than \(1/2\), hence \(U\cap V=\varnothing\).
Check: The endpoint \(1/2\) is in neither \(U\) nor \(V\). This does not affect the separation: the requirement is that the neighborhoods contain \(C\) and \(F\), not that they cover \(Y\).
Worked Example: Closed Sets in the Rational Subspace
Let \(Y=\mathbb{Q}\) with the metric inherited from \(\mathbb{R}\), and set \(C=\{q\in\mathbb{Q}:q\le 0\}\) and \(F=\{q\in\mathbb{Q}:q\ge 1\}\). The sets are disjoint. They are closed in \(Y\), since their complements are \(\{q\in\mathbb{Q}:q>0\}\) and \(\{q\in\mathbb{Q}:q<1\}\), each open in the subspace topology.
Choose \(U=\{q\in\mathbb{Q}:q<1/2\}\) and \(V=\{q\in\mathbb{Q}:q>1/2\}\). These are open in \(Y\), since they are intersections of \(\mathbb{Q}\) with open real intervals. If \(q\in C\), then \(q\le0<1/2\), so \(q\in U\). If \(q\in F\), then \(q\ge1>1/2\), so \(q\in V\). The strict inequalities \(q<1/2\) and \(q>1/2\) cannot hold together, so \(U\cap V=\varnothing\).
Check: The separating point \(1/2\) is rational, but it belongs to neither neighborhood. Its presence in \(Y\) does not cause an intersection.
What Ordinary Normality Does—and Does Not—Say
Ordinary normality is a statement about pairs of sets closed in the ambient space. The closed-subspace implication is useful, but it does not by itself establish hereditary normality.
Proof. The subspace \(Y\) is \(T_1\), since \(X\) is \(T_1\). Let \(C,F\) be disjoint closed subsets of \(Y\). A set closed in \(Y\) is the intersection of \(Y\) with a closed subset of \(X\); because \(Y\) is closed in \(X\), both \(C\) and \(F\) are therefore closed in \(X\). Normality of \(X\) gives disjoint open sets \(U,V\subseteq X\) with \(C\subseteq U\) and \(F\subseteq V\). Then \(Y\cap U\) and \(Y\cap V\) are disjoint open sets in \(Y\) containing \(C\) and \(F\), respectively. Thus \(Y\) is normal. \(\square\)
The hypothesis that \(Y\) is closed matters to this argument: it lets us turn closed sets of \(Y\) into closed sets of \(X\), where normality applies. For an arbitrary subspace that conversion may fail. For instance, take \(Y=(0,1)\subseteq\mathbb{R}\). The set \(C=(0,1/2]\) is closed in \(Y\), because \(Y\setminus C=(1/2,1)\) is open in \(Y\). But \(C\) is not closed in \(\mathbb{R}\): every interval about \(0\) meets \(C\), while \(0\notin C\). So one cannot apply normality of \(\mathbb{R}\) directly to \(C\) as though it were closed in the ambient space.
This example identifies a common gap in an argument, rather than a failure of separation: \(C\) can still be separated from any disjoint closed subset of \(Y\) by the metric-space theorem. The point is that ordinary normality is only given for ambient closed sets, whereas hereditary normality asks for separation of all closed pairs in every subspace. The metric proof establishes the stronger property directly by working with the inherited metric and the relative closed sets.
Worked Example: A Finite Metric Subspace
Let \(Y=\{0,1,3\}\subseteq\mathbb{R}\) with the inherited metric, and let \(C=\{0,3\}\) and \(F=\{1\}\). Every subset of \(Y\) is closed: for each point \(y\in Y\), a ball of radius \(1/2\) about \(y\) meets \(Y\) only in \(y\), so each singleton is open; its complement, a union of open singletons, is also open.
Set \(U=\{0,3\}\) and \(V=\{1\}\). The set \(V\) is the ball \(B_d(1,1/2)\cap Y\). The set \(U\) is the union \((B_d(0,1/2)\cap Y)\cup(B_d(3,1/2)\cap Y)\); the first ball intersects \(Y\) in \(\{0\}\), and the second in \(\{3\}\). Hence \(U\) and \(V\) are open in \(Y\). They contain \(C\) and \(F\), respectively, and their displayed elements show \(U\cap V=\varnothing\).
Check: The radius \(1/2\) is less than every distance between distinct points of \(Y\): those distances are \(1\), \(2\), and \(3\). Thus none of the balls used above accidentally includes another point of \(Y\).
A Common Mistake and the Main Distinction
Common mistake. Given a subspace \(Y\subseteq X\), it is incorrect to assume that every subset closed in \(Y\) is also closed in \(X\), and then apply normality of \(X\) to it. In the interval example, \(C=(0,1/2]\) is closed in \(Y=(0,1)\), but it is not closed in \(\mathbb{R}\), since \(0\) is a limit point of \(C\) outside \(C\). The correction is to check closedness in the correct space. For a general subspace, the metric argument above avoids the invalid ambient-closedness step; for a closed subspace, the proposition explains why ordinary normality does pass down.
| Property | Which closed pairs must be separated? | What follows here? |
|---|---|---|
| Normality of \(X\) | Disjoint closed subsets of \(X\) | Applies to \(X\) itself |
| Normality of a closed subspace \(Y\) | Disjoint subsets closed in \(Y\) | Follows when \(X\) is normal and \(Y\) is closed |
| Hereditary normality of \(X\) | Disjoint closed subsets of every subspace \(Y\subseteq X\) | Follows for every metric space by the inherited-metric proof |
Thus, the conclusion for metric spaces is stronger than simply knowing that a metric space is normal: the same separation method works inside every subspace, whether or not that subspace is closed in the original space. The essential ingredients are the restricted metric, closedness in the subspace, and the distance functions that produce disjoint open neighborhoods.
Check Your Understanding
For each question, identify the space in which openness or closedness is being tested.
- State the definition of hereditary normality and explain why it implies ordinary normality.
- In the metric proof, why is \(\operatorname{dist}(c,F)>0\) when \(c\in C\) and \(F\) is closed in \(Y\)?
- Why does the proof define \(U\) using \(h<0\) and \(V\) using \(h>0\), rather than using non-strict inequalities?
- Give an example of a set closed in a subspace of \(\mathbb{R}\) that is not closed in \(\mathbb{R}\), and verify both claims.
- Which extra hypothesis on \(Y\subseteq X\) allows normality of \(X\) to imply normality of \(Y\) by the closed-subspace argument?