A Subset Becomes a Metric Space
A metric space can be studied from within one of its subsets. The points under consideration change, but the distances between retained points do not: we simply restrict the original metric to pairs of points in the subset. The resulting space is called a subspace. This viewpoint is useful because properties such as openness, closure, and distance must then be interpreted relative to the points the subspace actually contains.
Several facts about restricted metrics and their open and closed sets were established earlier in this course. In particular, the Balls in a Restricted Metric Theorem describes how subspace balls are obtained from ambient balls, while the Open Sets in a Restricted Metric and Closed Sets in a Restricted Metric theorems describe relative openness and closedness. We will use those results and develop a further formula for closure in a subspace.
The restricted metric is a metric by the Balls in a Restricted Metric Theorem (which includes the fact that restriction preserves the metric axioms). The metric space \(X\) is called the ambient space when we want to distinguish it from the subspace \(A\). Notice that only distances between points of \(A\) are part of \(d_A\). If \(x\) is outside \(A\), it is not a point of the metric space \((A,d_A)\).
Subspace Balls and Relative Open Sets
For a point \(a\in A\) and a radius \(r>0\), the ball in \(A\) is $$ B_r^A(a)=\{x\in A:d_A(a,x)<r\}. $$ Since \(d_A(a,x)=d(a,x)\) for \(x\in A\), this ball is exactly the portion of the ambient ball that lies in \(A\): $$ B_r^A(a)=A\cap B_r^X(a). $$ This identity is a direct way to keep track of which points are available when working in the subspace.
Worked Example: A Ball Near the Edge of an Interval
Let \(X=\mathbb{R}\) with its usual metric, \(A=(0,1)\), and \(a=\frac{1}{10}\). For radius \(r=\frac{1}{5}\), the ambient ball is $$ B_{1/5}^{\mathbb{R}}(1/10)=(-1/10,3/10), $$ because a real number \(x\) is in this ball precisely when $$ |x-1/10|<1/5, $$ or equivalently \(-1/10<x<3/10\). The subspace ball is the intersection with \(A\): $$ B_{1/5}^{A}(1/10)=(0,1)\cap(-1/10,3/10)=(0,3/10). $$ The left endpoint \(-1/10\) is excluded by the ball inequality, and every point at or below zero is also excluded from the subspace. The radius and the distances have not changed; only the available points have changed.
A set can be open in \(A\) without being open in \(X\), and it can be closed in \(A\) without being closed in \(X\). This is why one speaks of a set being relatively open or relatively closed when the relevant space is a subspace. The Open Sets in a Restricted Metric Theorem gives an equivalent description: a subset \(U\) of \(A\) is open in \(A\) exactly when there is an open set \(G\) in \(X\) such that \(U=A\cap G\). The corresponding statement for closed sets follows from the Closed Sets in a Restricted Metric Theorem.
Worked Example: Relatively Open but Not Ambiently Open
Take \(A=[0,1]\) as a subspace of \(\mathbb{R}\), and let \(U=[0,\frac{1}{2})\). This is open in \(A\), since $$ U=A\cap(-1,\tfrac{1}{2}). $$ The set \((-1,\frac{1}{2})\) is open in \(\mathbb{R}\), so the characterization of open sets in a restricted metric applies. But \(U\) is not open in \(\mathbb{R}\): every ambient ball centered at \(0\) contains negative numbers, which are not in \(U\). Openness in the subspace only tests whether points of \(A\) sufficiently close to a given point remain in \(U\).
Worked Example: Relatively Closed but Not Ambiently Closed
Let \(A=(0,1)\) and \(F=(0,\frac{1}{2}]\). This set is closed in \(A\), because $$ F=A\cap[0,\tfrac{1}{2}], $$ and \([0,\frac{1}{2}]\) is closed in \(\mathbb{R}\). However, \(F\) is not closed in \(\mathbb{R}\). For example, the sequence \(1/n\), for integers \(n\geq2\), belongs to \(F\) and converges in \(\mathbb{R}\) to \(0\), which is not in \(F\). There is no contradiction: the limit \(0\) is not a point of the subspace \(A\).
Closure in a Subspace
Closure depends on which space is being used. For \(E\subseteq A\), the closure of \(E\) in \(A\), denoted \(\overline{E}^{\,A}\), consists of the points \(a\in A\) for which every subspace ball centered at \(a\) meets \(E\). The ambient closure, denoted \(\overline{E}^{\,X}\), tests every ambient ball instead. The next theorem relates these two tests exactly.
Proof. Fix any \(a\in A\). By the metric-ball characterization of closure, \(a\in\overline{E}^{\,A}\) if and only if every subspace ball \(B_r^A(a)\), for \(r>0\), meets \(E\). Using the subspace-ball identity, $$ B_r^A(a)\cap E=(A\cap B_r^X(a))\cap E=B_r^X(a)\cap E, $$ where the last equality holds because \(E\subseteq A\). Thus every subspace ball centered at \(a\) meets \(E\) if and only if every ambient ball centered at \(a\) meets \(E\), which is equivalent to \(a\in\overline{E}^{\,X}\). We have proved, for each \(a\in A\), that \(a\in\overline{E}^{\,A}\) if and only if \(a\in\overline{E}^{\,X}\). Since \(\overline{E}^{\,A}\) contains only points of \(A\), this gives $$ \overline{E}^{\,A}=A\cap\overline{E}^{\,X}. $$ \(\square\)
The intersection with \(A\) is essential: ambient limit points outside the subspace cannot belong to its closure. A useful consequence is that \(E\) is dense in \(A\) precisely when \(A\subseteq\overline{E}^{\,X}\). Indeed, density in \(A\) means \(\overline{E}^{\,A}=A\), and the theorem says this is equivalent to \(A\cap\overline{E}^{\,X}=A\), or \(A\subseteq\overline{E}^{\,X}\).
Worked Example: A Sequence Accumulating at a Missing Endpoint
Let \(A=(0,1)\) be a subspace of \(\mathbb{R}\), and let $$ E=\{1/n:n\geq2\}. $$ In \(\mathbb{R}\), the closure of \(E\) is \(E\cup\{0\}\): each listed point is in \(E\), the sequence \(1/n\) approaches \(0\), and there are no other accumulation points. Applying the Closure in a Subspace Theorem gives $$ \overline{E}^{\,A}=A\cap(E\cup\{0\})=E. $$ The point \(0\) is an ambient limit point but not a point of \(A\), so it is not in the subspace closure. In particular, \(E\) is closed as a subset of \(A\), even though it is not closed as a subset of \(\mathbb{R}\).
Distance to a Set Is Preserved
There is another quantity that can be computed either in the subspace or in the ambient space. If \(E\) is nonempty and \(x\) is a point of the space being considered, the distance from \(x\) to \(E\) is the infimum of the distances from \(x\) to points of \(E\). When \(x\in A\) and \(E\subseteq A\), the available points of \(E\) and all their distances from \(x\) are exactly the same in both spaces.
Proof. For each \(y\in E\), both \(x\) and \(y\) belong to \(A\), so the definition of the restricted metric gives \(d_A(x,y)=d(x,y)\). Therefore the two sets of real numbers whose infima are taken are identical: $$ \{d_A(x,y):y\in E\}=\{d(x,y):y\in E\}. $$ Their infima are equal, as claimed. \(\square\)
Worked Example: Distance in an Interval Subspace
Let \(A=[0,3]\subseteq\mathbb{R}\), take \(x=1\), and set \(E=\{0,3\}\). In the ambient space, $$ d_{\mathbb{R}}(1,E)=\inf\{|1-0|,|1-3|\}=\inf\{1,2\}=1. $$ In the subspace, the distances are unchanged because both points and the metric are restricted from \(\mathbb{R}\): $$ d_A(1,E)=\inf\{d_A(1,0),d_A(1,3)\}=\inf\{1,2\}=1. $$ Although \(A\) contains only points from the interval, both candidates in \(E\) are retained, so the distance computation is the same.
Keep the Ambient Space in View
A subspace uses the same distance formula between its points, but it can have different open sets, closed sets, and closures because it has fewer points. For \(E\subseteq A\subseteq X\), the closure formula makes the distinction explicit: $$ \overline{E}^{\,A}=A\cap\overline{E}^{\,X}. $$ It also explains why an ambient limit point outside \(A\) does not count as a limit point in \(A\). Similarly, distances to a nonempty subset \(E\subseteq A\) agree when measured from \(x\in A\), but this statement does not define a subspace distance from a point \(x\notin A\): such a point is outside the metric space \((A,d_A)\).
When working with a subset, first identify the space in which each claim is made. For balls, intersect the ambient ball with the subspace. For open or closed sets, use the relative characterizations from earlier tutorials. For closure, intersect the ambient closure with the subspace. These checks prevent a common mistake: treating a set as though it has the same boundary points regardless of which space contains it.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- How is the restricted metric on a nonempty subset \(A\) defined?
- Express the ball \(B_r^A(a)\) using an ambient ball in \(X\).
- For \(E\subseteq A\), why might \(\overline{E}^{\,A}\) omit a point of \(\overline{E}^{\,X}\)?
- State the formula relating closure in \(A\) to closure in \(X\).
- Why do the distances from \(x\in A\) to points of a nonempty \(E\subseteq A\) have the same infimum in both spaces?