When Is a Point Close Enough to a Set?
A sequence in a subset may converge in the ambient metric space to a point that is missing from the subset. Closure gives a precise way to collect those missing limit points together with the original set. It turns the question “Can points of this set get arbitrarily close to \(x\)?” into a property of a point and its neighborhoods.
In the previous tutorial, convergence in a subspace was distinguished from convergence in the ambient space. Closure makes that distinction useful: it identifies exactly which ambient points can be approached by points of a given set. The Sequential Characterization of Closed Sets, established earlier in this course, will help us relate this neighborhood-based idea to sequences.
The definition includes points of \(E\): if \(x\in E\), then \(x\in B_r(x)\cap E\) for every \(r>0\). Thus \(E\subseteq\overline{E}\). A point outside \(E\) belongs to \(\overline{E}\) precisely when no positive-radius ball around it can avoid \(E\). The empty set has empty closure, because every ball has empty intersection with it.
Closure Points and Convergent Sequences
In a metric space, the ball condition has an equivalent sequential form. If every ball around \(x\) meets \(E\), we can choose a point of \(E\) within distance \(1/n\) of \(x\) for each positive integer \(n\). Conversely, a sequence from \(E\) converging to \(x\) eventually enters every ball around \(x\).
Proof. Suppose first that \(x\in\overline{E}\). For every positive integer \(n\), the ball \(B_{1/n}(x)\) intersects \(E\). Choose \(e_n\in B_{1/n}(x)\cap E\). Then \(d(e_n,x)<1/n\). Given \(\varepsilon>0\), choose a positive integer \(N\) with \(1/N<\varepsilon\). For \(n\geq N\), $$ d(e_n,x)<\frac{1}{n}\leq\frac{1}{N}<\varepsilon. $$ Therefore \(e_n\to x\).
Conversely, suppose \(e_n\in E\) for every \(n\) and \(e_n\to x\). Let \(r>0\). By convergence, there is an index \(N\) such that \(d(e_n,x)<r\) whenever \(n\geq N\). In particular, \(e_N\in B_r(x)\cap E\), so this intersection is nonempty. Since this holds for every \(r>0\), \(x\in\overline{E}\). If \(E\) is empty, neither a point in its closure nor a sequence with terms in \(E\) exists, so the equivalence also covers that case. \(\square\)
This characterization says that closure records all possible limits of sequences drawn from \(E\), not only limits that are already in \(E\). The sequence may be constant when \(x\in E\); when \(x\notin E\), it witnesses how points of the set approach a missing point.
Worked Example: The Closure of a Reciprocal Sequence
In \(\mathbb{R}\) with its usual metric, let \(E=\{1/n:n\geq1\}\). We claim that $$ \overline{E}=E\cup\{0\}. $$ First, \(E\subseteq\overline{E}\). Also, the sequence \(e_n=1/n\) belongs to \(E\) and converges to \(0\), so the Sequential Characterization of Closure gives \(0\in\overline{E}\).
It remains to show that no other real number is in the closure. If \(x<0\), then for every \(n\geq1\), $$ \left|x-\frac{1}{n}\right|=\frac{1}{n}-x>-x. $$ Thus \(B_{-x/2}(x)\) misses \(E\). Now suppose \(x>0\) and \(x\notin E\). Choose a positive integer \(N\) such that \(1/N<x/2\). For \(n\geq N\), we have $$ \left|x-\frac{1}{n}\right|=x-\frac{1}{n}>\frac{x}{2}. $$ For the finitely many indices \(1\leq n<N\), every distance \(\left|x-1/n\right|\) is positive because \(x\notin E\). If there are any such indices, let \(\delta\) be the minimum of these finitely many positive distances; if there are none, put \(\delta=x/2\). Then \(r=\min\{\delta,x/2\}/2\) is positive, and \(B_r(x)\cap E=\varnothing\). Hence \(x\notin\overline{E}\). This proves the claimed equality.
Closure as the Smallest Closed Superset
The definition of closure uses balls, but it also has a useful set-theoretic description: it is the smallest closed set containing \(E\). We establish this fact directly. The proof that \(\overline{E}\) is closed uses the Sequential Characterization of Closed Sets from earlier in the course.
Proof. We already know that \(E\subseteq\overline{E}\). To prove that \(\overline{E}\) is closed, take any sequence \((x_n)\) in \(\overline{E}\) that converges to \(x\in X\). Let \(\varepsilon>0\). Since \(x_n\to x\), choose \(n\) such that \(d(x_n,x)<\varepsilon/2\). Because \(x_n\in\overline{E}\), the ball \(B_{\varepsilon/2}(x_n)\) meets \(E\). Choose \(e\in B_{\varepsilon/2}(x_n)\cap E\). The triangle inequality gives $$ d(e,x)\leq d(e,x_n)+d(x_n,x)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$ Consequently \(B_\varepsilon(x)\cap E\ne\varnothing\). This holds for every \(\varepsilon>0\), so \(x\in\overline{E}\). The Sequential Characterization of Closed Sets now implies that \(\overline{E}\) is closed.
Finally, let \(F\) be any closed subset of \(X\) with \(E\subseteq F\). If \(x\in\overline{E}\), the Sequential Characterization of Closure gives a sequence \((e_n)\) in \(E\) converging to \(x\). Since every \(e_n\) belongs to \(F\), the Sequential Characterization of Closed Sets implies \(x\in F\). Therefore \(\overline{E}\subseteq F\). The same conclusion is immediate when \(E\) is empty, because then \(\overline{E}=\varnothing\). This proves the theorem. \(\square\)
The smallest-closed-superset description is particularly helpful when a set is given by a simple condition but its closure is not immediately visible from the ball definition. It also clarifies why closure is a natural operation: it adds only the points that are forced by closedness.
Worked Example: Closing an Open Interval
Let \(E=(2,5)\subseteq\mathbb{R}\). Every point of \(E\) is in \(\overline{E}\). The endpoints belong to the closure as well: for every \(r>0\), the point \(2+\min\{r/2,1/2\}\) lies in \(E\) and is at distance less than \(r\) from \(2\); similarly, \(5-\min\{r/2,1/2\}\) lies in \(E\) and is at distance less than \(r\) from \(5\).
If \(x<2\), then every \(y\in E\) satisfies \(|x-y|>2-x\), so \(B_{(2-x)/2}(x)\) misses \(E\). If \(x>5\), every \(y\in E\) satisfies \(|x-y|>x-5\), so \(B_{(x-5)/2}(x)\) misses \(E\). Thus no point outside \([2,5]\) is in the closure, and $$ \overline{(2,5)}=[2,5]. $$ Indeed, \([2,5]\) is closed and contains \((2,5)\); the theorem also says it is the smallest such closed set.
Distance to a Set
There is another way to measure whether a point is in the closure: take the infimum of its distances to points of the set. For a nonempty set \(E\), define the distance from \(x\) to \(E\) by $$ d(x,E)=\inf\{d(x,e):e\in E\}. $$ This number is nonnegative. It need not be the distance from \(x\) to some particular point of \(E\), since the infimum may not be attained.
Proof. Suppose \(x\in\overline{E}\). For every \(r>0\), some \(e\in E\) satisfies \(d(x,e)<r\). Hence \(0\leq d(x,E)\leq d(x,e)<r\). If \(d(x,E)\) were positive, taking \(r=d(x,E)\) would contradict this strict inequality. Therefore \(d(x,E)=0\).
Conversely, suppose \(d(x,E)=0\). Let \(r>0\). If \(B_r(x)\cap E\) were empty, every \(e\in E\) would satisfy \(d(x,e)\geq r\), implying \(d(x,E)\geq r\), contrary to \(d(x,E)=0\). Thus every such ball meets \(E\), and \(x\in\overline{E}\). \(\square\)
Worked Example: A Set That Is Closed in the Discrete Metric
Let \(X\) be a nonempty set with the discrete metric \(\delta\), where \(\delta(x,y)=0\) if \(x=y\) and \(\delta(x,y)=1\) if \(x\ne y\). Take any \(E\subseteq X\). If \(x\in E\), every ball centered at \(x\) meets \(E\), so \(x\in\overline{E}\). If \(x\notin E\), then \(\delta(x,e)=1\) for every \(e\in E\). When \(E\) is nonempty, this gives \(d(x,E)=1\), and the Distance Characterization of Closure shows \(x\notin\overline{E}\). If \(E\) is empty, its closure is empty by definition. In either case, \(\overline{E}=E\).
The metric matters: unlike the reciprocal set in the usual metric on \(\mathbb{R}\), no distinct points in a discrete metric space can approach one another arbitrarily closely. This example also shows why a closure must always be understood relative to a specified metric space.
Using Closure Carefully
The ball definition, sequential characterization, smallest-closed-superset theorem, and distance characterization are different tools for the same set. The ball definition is often best for checking a particular point directly. Sequences are useful when a natural approximating sequence is available. The smallest-closed-superset theorem is effective when a candidate closed set is easy to identify, and distance is useful when an infimum can be calculated.
A common pitfall is to treat closure as independent of the ambient space. It is not: the available points and the metric determine which balls meet the set, and therefore which points are closure points. In particular, the closure of a set viewed as a subspace can differ from its closure in a larger space. The Closure in a Subspace result from earlier in this course gives the precise relation between these two viewpoints.
Another pitfall is to assume that a point belongs to a closure only when the distance to the set is attained. The distance characterization requires the infimum to be zero, not the existence of a nearest point. For example, \(0\) is in the closure of \(\{1/n:n\geq1\}\), even though no member of that set has distance zero from \(0\).
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why does a sequence converging to \(x\) from within \(E\) show that every ball centered at \(x\) meets \(E\)?
- How can one construct a sequence in \(E\) converging to a point known to lie in \(\overline{E}\)?
- What makes \(\overline{E}\) the smallest closed superset of \(E\), rather than merely a closed superset?
- Why does \(d(x,E)=0\) not require a point \(e\in E\) with \(d(x,e)=0\)?
- What is the closure of a subset \(E\) of a discrete metric space, and which feature of the metric explains the answer?