Points Approached by a Set
The boundary test from the previous tutorial asks whether every ball around a point meets both a set and its complement. A related question focuses on just one set: does every ball around a point contain a point of the set other than the center itself? Such points are limit points. They describe where a set accumulates, even when the point being approached does not belong to the set.
Throughout, \((X,d)\) is a metric space and \(E\subseteq X\). Balls, closures, and limit points are taken in \(X\), unless another ambient space is specified. As with boundary and closure, the ambient space matters because it determines which points are available in each ball.
Equivalently, every ball around \(x\) contains a point of \(E\) other than \(x\). The strict inequality \(d(x,y)>0\) is essential: the center itself cannot be the point that makes the ball meet \(E\). There is no requirement that \(x\in E\). A point outside \(E\) can be a limit point if points of \(E\) occur arbitrarily close to it.
A point \(x\in E\) is called an isolated point of \(E\) if some ball around \(x\) contains no points of \(E\) other than \(x\). Thus a point of \(E\) is isolated precisely when it is not a limit point of \(E\). A point outside \(E\), by contrast, is not isolated in \(E\) under this terminology; it may or may not be a limit point.
Worked Examples with the Definition
Worked Example: The Limit Points of a Reciprocal Sequence
Let \(E=\{1/n:n\in\mathbb{N}\}\) in \(\mathbb{R}\). First, \(0\) is a limit point. Given \(r>0\), choose an integer \(n>1/r\). Then \(1/n\in E\), \(1/n\ne0\), and \[ |1/n-0|=1/n<r. \]
Now fix \(x<0\). Every \(y\in E\) is positive, so \(|x-y|=|x|+y>|x|\). The ball of radius \(|x|/2\) around \(x\) therefore contains no point of \(E\), and \(x\notin E'\).
Suppose instead that \(x>0\). Choose \(N\) large enough that \(1/N<x/2\). For \(n\geq N\), we have \(1/n\leq1/N<x/2\), and hence \[ |x-1/n|=x-1/n>x/2. \] Among the finitely many terms \(1/n\) with \(n<N\), discard any term equal to \(x\). If any terms remain, their distances from \(x\) have a positive minimum \(m\); if none remain, put \(m=x/2\). Set \(r=\min\{x/2,m\}/2\). Every term of \(E\) other than \(x\) is at distance greater than \(r\) from \(x\): the tail terms are farther than \(x/2\), and the remaining initial terms are at least \(m\) away. Thus \(x\) is not a limit point. We have proved \[ E'=\{0\}. \] In particular, \(0\) is a limit point even though \(0\notin E\).
Worked Example: A Finite Set in a Metric Space
Let \(E=\{a_1,\ldots,a_k\}\) be a finite subset of a metric space, with its elements distinct. We show that \(E'=\varnothing\). Fix any \(x\in X\). If \(x\notin E\), each of the finitely many numbers \(d(x,a_j)\) is positive. Their minimum \(m\) is positive, and the ball \(B_{m/2}(x)\) misses \(E\).
If \(x\in E\), consider the distances from \(x\) to the other points of \(E\). When there are other points, their finitely many distances have a positive minimum \(m\), and \(B_{m/2}(x)\) contains no point of \(E\) other than \(x\). When \(E=\{x\}\), every ball around \(x\) already has that property. In either case, there is a ball with no point of \(E\) distinct from \(x\). Therefore \(x\notin E'\). Since \(x\) was arbitrary, \(E'=\varnothing\).
This example includes the empty set as a special case: it has no limit points. It also shows why a point belonging to a set is not, by itself, evidence that it is a limit point.
Limit Points and Sequences
In metric spaces, the ball definition has an exact sequential counterpart. The requirement that the sequence use distinct points prevents a constant sequence at \(x\) from falsely making \(x\) appear to be a limit point. In fact, the sequence can be chosen with no repeated terms.
Proof. Suppose first that \(x\in E'\). We construct distinct points \(y_n\in E\setminus\{x\}\). Choose \(y_1\in E\) with \(0<d(x,y_1)<1\). Such a point exists by the definition of limit point. Once \(y_1,\ldots,y_{n-1}\) have been chosen, define \[ \delta_n=\min\left(\{1/n\}\cup\{d(x,y_j)/2:1\leq j<n\}\right). \] This is a minimum of a nonempty finite collection of positive numbers, so \(\delta_n>0\). Since \(x\) is a limit point, choose \(y_n\in E\) such that \(0<d(x,y_n)<\delta_n\). For each \(j<n\), this gives \[ d(x,y_n)<d(x,y_j)/2<d(x,y_j), \] so \(y_n\ne y_j\). Also, \(d(x,y_n)<1/n\), and therefore \(y_n\to x\).
Conversely, suppose distinct points \(y_n\in E\setminus\{x\}\) satisfy \(y_n\to x\). Given \(r>0\), convergence gives an index \(N\) such that \(d(y_n,x)<r\) whenever \(n\geq N\). In particular, \(y_N\in E\), \(y_N\ne x\), and \(0<d(x,y_N)<r\). This is the definition of \(x\in E'\). \(\square\)
Worked Example: A Sequence Reveals a Limit Point
Let \(E=\{2+1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\). The points \(y_n=2+1/n\) are distinct, belong to \(E\), and differ from \(2\). Moreover, \[ |y_n-2|=1/n\longrightarrow0. \] The Sequential Characterization of Limit Points therefore shows that \(2\in E'\), although \(2\notin E\).
The distinctness condition also clarifies a common error. If one merely found a sequence in \(E\) converging to \(x\), that would not suffice: when \(x\in E\), the constant sequence \(y_n=x\) converges to \(x\) but supplies no points other than \(x\). The sequence in the theorem must consist of points in \(E\setminus\{x\}\).
How Limit Points Determine the Closure
Closure asks whether every ball around a point meets \(E\), while the limit-point condition asks whether every ball meets \(E\) away from its center. These conditions differ only when the center itself is the only available point of \(E\) in some ball. This gives a useful description of closure in terms of the set and its derived set.
Proof. Suppose \(x\in\overline E\). If \(x\in E\), then \(x\in E\cup E'\). If \(x\notin E\), every ball centered at \(x\) meets \(E\), and every point of \(E\) is necessarily different from \(x\). Thus every ball contains a point \(y\in E\) with \(0<d(x,y)\), so \(x\in E'\). This proves \(\overline E\subseteq E\cup E'\).
For the reverse inclusion, \(E\subseteq\overline E\) by the definition of closure. If \(x\in E'\), every ball around \(x\) contains a point of \(E\), so \(x\in\overline E\). Hence \(E\cup E'\subseteq\overline E\), proving the equality. \(\square\)
Combining this identity with the Closure Is the Smallest Closed Superset result from earlier in this course gives another characterization: \(E\) is closed if and only if \(E'\subseteq E\). Indeed, \(E\) is closed exactly when \(\overline E=E\); by the theorem, that equality holds exactly when every limit point of \(E\) already belongs to \(E\). A closed set can have many limit points, but it cannot omit any of them.
The Derived Set Is Closed
The set \(E'\) itself has a useful regularity property: it is closed, whether or not \(E\) is closed. The proof uses the definition directly and carefully separates a point from all possible limit points nearby.
Proof. We show that if \(x\notin E'\), then some ball around \(x\) misses \(E'\). Since \(x\) is not a limit point of \(E\), there is \(r>0\) such that \[ B_r(x)\cap(E\setminus\{x\})=\varnothing. \] We claim that \(B_{r/2}(x)\cap E'=\varnothing\). Let \(y\in B_{r/2}(x)\). If \(y=x\), then \(y\notin E'\) by assumption. If \(y\ne x\), choose \[ s=\tfrac12\min\{d(x,y),\,r-d(x,y)\}. \] Both quantities in the minimum are positive: \(d(x,y)>0\), and \(d(x,y)<r/2<r\). For any \(z\in B_s(y)\), the triangle inequality and its reverse give \[ d(x,z)\leq d(x,y)+d(y,z)<d(x,y)+s<r \] and \[ d(x,z)\geq d(x,y)-d(y,z)>d(x,y)-s>0. \] Thus \(z\in B_r(x)\setminus\{x\}\). The choice of \(r\) implies that this ball contains no point of \(E\), so \(B_s(y)\cap E=\varnothing\). In particular, \(y\) is not a limit point of \(E\). We have shown that every \(y\in B_{r/2}(x)\) lies outside \(E'\). Hence the complement of \(E'\) is open, and \(E'\) is closed. \(\square\)
The Ambient Space Can Change the Answer
Worked Example: An Interval in Two Ambient Spaces
Let \(E=(2,4)\). First view it as a subset of \(\mathbb{R}\). Every \(x\in(2,4)\) is a limit point: given \(r>0\), choose \[ t=\tfrac12\min\{r,4-x\}. \] Then \(t>0\), \(x+t<4\), and \(x+t\in E\setminus\{x\}\) with \(|(x+t)-x|=t<r\). The endpoint \(2\) is also a limit point, since for any \(r>0\), the point \(2+\min\{r/2,1\}\) lies in \(E\) and within distance \(r\) of \(2\). Similarly, \(4-\min\{r/2,1\}\) verifies that \(4\) is a limit point. A point \(x<2\) has a ball around it that misses \(E\), for example one of radius \((2-x)/2\); a point \(x>4\) has the corresponding property using radius \((x-4)/2\). Therefore \(E'=[2,4]\) in \(\mathbb{R}\).
Now take \(X=(2,4)\) as the ambient space and again let \(E=(2,4)\), so that \(E=X\). Each point \(x\in X\) is still a limit point: the same choice of \(t\) gives a distinct point \(x+t\in X\) within distance \(r\). Thus \(E'=X=(2,4)\) in this ambient space. The endpoints are no longer limit points because they are not elements of \(X\). The set under consideration is unchanged as a collection of real numbers, but its limit points differ because the ambient space differs.
Limit points connect several ideas developed earlier in this course. The Sequential Characterization of Closure gives sequences approaching points of the closure; the limit-point version sharpens this by requiring the approximating points to differ from the point approached. The closure identity \(\overline E=E\cup E'\) makes the distinction precise, while the closedness criterion \(E'\subseteq E\) gives a practical test for closed sets.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What condition must hold in every ball around a limit point?
- Why does the constant sequence at \(x\) not establish that \(x\) is a limit point of a set containing \(x\)?
- How does the Sequential Characterization of Limit Points produce distinct terms converging to the point?
- Explain why \(\overline E=E\cup E'\) implies that \(E\) is closed exactly when \(E'\subseteq E\).
- Why is the derived set \(E'\) closed even when \(E\) is not closed?
- For \(E=\{1/n:n\in\mathbb{N}\}\) in \(\mathbb{R}\), why is \(0\) a limit point despite not belonging to \(E\)?