Why Regularity Is Inherited
Regularity asks whether a point can be separated from a closed set that does not contain it. Passing to a subspace changes which sets count as open and closed, so it is not enough simply to reuse the ambient neighborhoods without checking their relative versions. The key is that a closed set in a subspace is the intersection of that subspace with an ambient closed set. This lets the separation property in the original space produce the separation needed in the subspace.
For this tutorial, a space is regular if it is \(T_1\) and, whenever \(x\) is outside a closed set \(C\), there are disjoint open sets containing \(x\) and \(C\), respectively. The \(T_1\) condition means every singleton is closed. This is the \(T_3\) convention: both the \(T_1\) condition and the separation condition are required.
In particular, open neighborhoods in a subspace need not themselves be open in the ambient space. They are intersections with the subspace. The closed-set description is just as important: when applying regularity in \(X\), we will separate a point from an ambient closed set representing the relatively closed set.
Two Subspace Facts
Proof. The set \(A\cap\overline{B}^{\,X}\) is closed in \(A\), because \(\overline{B}^{\,X}\) is closed in \(X\). It contains \(B\), since \(B\subseteq A\) and \(B\subseteq\overline{B}^{\,X}\). Now let \(F\) be any closed subset of \(A\) containing \(B\). By the definition of the subspace topology, \(F=A\cap C\) for some closed \(C\subseteq X\). Since \(B\subseteq F\subseteq C\), the ambient closure \(\overline{B}^{\,X}\) is contained in \(C\). Therefore \(A\cap\overline{B}^{\,X}\subseteq A\cap C=F\). So \(A\cap\overline{B}^{\,X}\) is the smallest relatively closed set containing \(B\), which proves the formula. This argument also applies when \(B=\varnothing\). \(\square\)
The same reasoning describes neighborhoods: if \(a\in A\), every open neighborhood of \(a\) in \(A\) has the form \(A\cap U\), where \(U\) is an ambient open neighborhood of \(a\). Conversely, every such intersection is a neighborhood in \(A\). We will use both facts in the examples.
Proof. First suppose \(A\ne\varnothing\). Since \(X\) is \(T_1\), every singleton \(\{a\}\), for \(a\in A\), is closed in \(X\). Hence \(\{a\}=A\cap\{a\}\) is closed in \(A\), so \(A\) is \(T_1\).
Next let \(a\in A\) and let \(F\) be closed in \(A\) with \(a\notin F\). Write \(F=A\cap C\) for some closed \(C\subseteq X\). Because \(a\in A\) but \(a\notin A\cap C\), we have \(a\notin C\). Regularity of \(X\) gives disjoint open sets \(U,V\subseteq X\) with \(a\in U\) and \(C\subseteq V\). Then \(A\cap U\) and \(A\cap V\) are open in \(A\), contain \(a\) and \(F\) respectively, and are disjoint because \(U\cap V=\varnothing\). Thus \(A\) satisfies the required separation condition. If \(A=\varnothing\), both the \(T_1\) and separation conditions hold vacuously. Therefore every subspace of \(X\) is regular. \(\square\)
Notice that the argument did not require \(A\) to be closed or open in \(X\). It used only the subspace descriptions of open and closed sets and the regularity of the ambient space.
Relative Neighborhoods and Closures in Intervals
Worked Example: An Endpoint in the Closed Unit Interval
Let \(A=[0,1]\) have the subspace topology inherited from \(\mathbb{R}\), and let \(B=(0,1/2)\subseteq A\). Find \(\overline{B}^{\,A}\) and describe neighborhoods of the endpoints of \(A\).
The ambient closure of \(B\) in \(\mathbb{R}\) is \([0,1/2]\). Indeed, every open interval about a point of \([0,1/2]\) meets \((0,1/2)\), while a point outside \([0,1/2]\) has an open interval disjoint from \((0,1/2)\). The closure formula now gives \(\overline{B}^{\,A}=A\cap[0,1/2]=[0,1/2]\).
At \(0\), an open neighborhood in \(A\) can be obtained by intersecting an ambient interval \((-\varepsilon,\varepsilon)\), with \(\varepsilon>0\), with \(A\). The result is \([0,\varepsilon)\) if \(0<\varepsilon\leq 1\), and it contains points of \(B\) for every such \(\varepsilon\). At \(1\), the corresponding intersection with \((1-\varepsilon,1+\varepsilon)\) is a neighborhood of \(1\), but for \(0<\varepsilon\leq 1/2\) it misses \(B\). Thus \(0\) lies in the relative closure of \(B\), while \(1\) does not.
Check: The answer \([0,1/2]\) is closed in \(A\), since its complement in \(A\) is \((1/2,1]\), open in \(A\). It contains \(B\), and the neighborhood calculation shows that its endpoint \(0\) cannot be removed from the closure.
Worked Example: A Relatively Closed Set That Is Not Ambient Closed
Let \(A=(0,1)\subseteq\mathbb{R}\) and \(B=(0,1/2]\subseteq A\). Find the closure of \(B\) in \(A\), and compare closedness in the two spaces.
In \(\mathbb{R}\), the closure of \(B\) is \([0,1/2]\): intervals about \(0\) meet \(B\), points in \(B\) are already in the closure, and any point outside \([0,1/2]\) has a neighborhood missing \(B\). Therefore \(\overline{B}^{\,A}=A\cap[0,1/2]=(0,1/2]\).
The set \(B\) is closed in \(A\), because its complement in \(A\) is \((1/2,1)\), which is open in \(A\). But \(B\) is not closed in \(\mathbb{R}\): it does not contain \(0\), even though every open neighborhood of \(0\) intersects \(B\). The relative closure formula explains the difference: the ambient closure adds \(0\), but \(0\notin A\), so it does not appear in the closure within \(A\).
Check: Intersecting the ambient closure with the subspace gives exactly the proposed relative closure: \((0,1)\cap[0,1/2]=(0,1/2]\).
Euclidean Subspaces, Including the Rational Line
For a useful application, every metric space is regular. Here is the separation argument. Let \((X,d)\) be a metric space, let \(C\subseteq X\) be closed, and let \(x\notin C\). Since \(X\setminus C\) is open, there is \(r>0\) such that \(B_d(x,r)\subseteq X\setminus C\). The open sets \(U=B_d(x,r/2)\) and \(V=\{y\in X:d(x,y)>r/2\}\) are disjoint, \(U\) contains \(x\), and \(C\subseteq V\): each \(c\in C\) lies outside \(B_d(x,r)\), so \(d(x,c)\geq r>r/2\). The set \(V\) is open because if \(d(x,y)>r/2\), then every point sufficiently close to \(y\) also has distance from \(x\) greater than \(r/2\), by the triangle inequality. Metric spaces are \(T_1\), since for \(x\ne y\), the ball \(B_d(x,d(x,y)/2)\) misses \(y\). This proves regularity. In particular, \(\mathbb{R}\) with its standard topology is regular, and the theorem shows that each of its subspaces is regular.
Worked Example: Closure in the Rational Line
Give \(\mathbb{Q}\) the subspace topology inherited from \(\mathbb{R}\), and take \(B=\mathbb{Q}\cap(0,1)\). Find the closure of \(B\) in \(\mathbb{Q}\).
The ambient closure of \((0,1)\) in \(\mathbb{R}\) is \([0,1]\). Applying the subspace closure formula gives \(\overline{B}^{\,\mathbb{Q}}=\mathbb{Q}\cap[0,1]\).
To see the endpoint behavior directly, every neighborhood of \(0\) in \(\mathbb{Q}\) contains a set of the form \(\mathbb{Q}\cap(-\varepsilon,\varepsilon)\), where \(\varepsilon>0\). This contains a rational number strictly between \(0\) and \(\min(\varepsilon,1)\), so it meets \(B\). Every neighborhood of \(1\) likewise contains a rational number strictly between \(\max(0,1-\varepsilon)\) and \(1\). Points of \(\mathbb{Q}\cap(0,1)\) already belong to \(B\). If \(q\in\mathbb{Q}\) is outside \([0,1]\), there is an open interval around \(q\) disjoint from \([0,1]\); its intersection with \(\mathbb{Q}\) is a neighborhood missing \(B\).
Check: The endpoint rationals \(0\) and \(1\) are included, but no irrational numbers are points of this subspace at all. Thus the relative closure is precisely \(\mathbb{Q}\cap[0,1]\), not the full real interval \([0,1]\). Regularity of \(\mathbb{Q}\) follows from the theorem, even though \(\mathbb{Q}\) is not a closed subset of \(\mathbb{R}\).
Common Pitfall: Confusing Relative and Ambient Closedness
A common error is to assume that if \(F\) is closed in \(A\), then \(F\) is closed in \(X\), and to apply regularity in \(X\) directly to \(F\). The second worked example disproves that assumption: \((0,1/2]\) is closed in \(A=(0,1)\), but it is not closed in \(\mathbb{R}\).
For instance, take \(a=3/4\in A\) and \(F=(0,1/2]\). The incorrect approach would call \(F\) an ambient closed set and separate \(a\) from it on that basis. The correction is to use an ambient closed representative: \(F=A\cap[0,1/2]\), where \([0,1/2]\) is closed in \(\mathbb{R}\) and does not contain \(3/4\). Ambient regularity separates \(3/4\) from \([0,1/2]\); intersecting the resulting open sets with \(A\) gives the required relative separation. This is exactly the step that makes the general proof work.
Check Your Understanding
In each question, keep track of which space supplies the topology.
- Let \(A=[-1,1]\subseteq\mathbb{R}\) and \(B=(-1,0)\). Find \(\overline{B}^{\,A}\) and describe a neighborhood of \(-1\) in \(A\).
- Give an ambient closed set \(C\subseteq\mathbb{R}\) such that \((0,1)\cap C=(0,1/2]\).
- Explain why every subspace of a \(T_1\) space is \(T_1\).
- Let \(A\subseteq X\), where \(X\) is regular, and let \(F\) be closed in \(A\). Explain why a point \(a\in A\setminus F\) cannot belong to an ambient closed representative \(C\) with \(F=A\cap C\).
- For \(B=\mathbb{Q}\cap(2,3)\) in \(\mathbb{Q}\), calculate the closure in \(\mathbb{Q}\) using the ambient closure formula.