When Nearby Points Keep Appearing
The previous tutorial studied sets formed by taking limits of sequences. We now focus on a particular kind of limit: a point that can be approached by points of a set different from the point itself. Such a point is called an accumulation point. The distinction matters because a point may belong to a set without nearby points of that set gathering around it, and an accumulation point need not belong to the set at all.
For a set \(E\subseteq\mathbb{R}\), this notion is also called a limit point, and the set of its accumulation points is the derived set \(E'\), as defined in “Limit Points.” The definition below makes explicit the role of neighborhoods and the requirement that the nearby point differ from the point being tested.
The strict inequality \(0<|y-x|\) excludes \(x\) itself. Without this exclusion, every point of \(E\) would automatically pass the test using \(y=x\), even if no other point of \(E\) were nearby. An accumulation point can be in \(E\), as happens for interior points of an interval, or outside \(E\), as happens for the point \(0\) and the set \(\{1/n:n\geq1\}\).
The Neighborhood Characterization of Closure from “Arbitrary Intersections of Closed Sets” says that \(x\in\overline E\) precisely when every ball around \(x\) meets \(E\). The accumulation-point condition adds the requirement that the meeting point differ from \(x\). Accordingly, the established Closure as the Union of a Set and Its Derived Set gives \(\overline E=E\cup E'\). We will use these earlier results as context, not re-prove them here.
Examples of Accumulation Points
Worked Example: The Accumulation Points of an Open Interval
Let \(E=(3,7)\). We show that every point of \([3,7]\) is an accumulation point of \(E\), while no point outside \([3,7]\) is one.
First take \(x\in(3,7)\). For any \(r>0\), choose $$ y=x+\min\left\{\frac r2,\frac{7-x}{2}\right\}. $$ The added quantity is positive, so \(y>x\). It is less than \(r\), and it is at most \((7-x)/2\), so \(y<7\). Since \(x>3\), we have \(y>3\) as well. Thus \(y\in E\) and \(0<|y-x|<r\).
At \(x=3\), choose \(y=3+\min\{r/2,1\}\). This point is greater than \(3\) and at most \(4\), hence belongs to \((3,7)\). Its distance from \(3\) is less than \(r\): if the minimum is \(r/2\), the distance is \(r/2<r\); if the minimum is \(1\), then \(r\geq2\), so \(1<r\). At \(x=7\), the same reasoning with \(y=7-\min\{r/2,1\}\) gives a point in \((3,7)\) within distance \(r\) of \(7\).
Now suppose \(x<3\). Set \(r=(3-x)/2>0\). If \(y\in B_r(x)\), then \(y<x+r=(x+3)/2<3\), so \(B_r(x)\cap E=\varnothing\). If \(x>7\), choose \(r=(x-7)/2\); every \(y\in B_r(x)\) satisfies \(y>x-r=(x+7)/2>7\), again leaving the ball disjoint from \(E\). Therefore the accumulation points are exactly \([3,7]\).
Worked Example: A Convergent Set with One Accumulation Point
Let \(E=\{4+1/n:n\geq1\}\). The points \(4+1/n\) are distinct: if \(m>n\), then \(1/m<1/n\), so \(4+1/m<4+1/n\). Also, $$ \left|(4+1/n)-4\right|=\frac1n\longrightarrow 0. $$ Thus for every \(r>0\), all sufficiently large \(n\) give a point of \(E\) different from \(4\) and within distance \(r\) of \(4\). Hence \(4\) is an accumulation point.
To see that there are no others, we will use the sequential characterization proved below: an accumulation point is the limit of a sequence of distinct points of the set. Suppose \(z\) is an accumulation point and choose distinct \(z_k\in E\) with \(z_k\to z\). Write \(z_k=4+1/n_k\). Since the \(z_k\) are distinct, the positive integer indices \(n_k\) are distinct. For any positive integer \(M\), at most \(M\) of these indices can be at most \(M\); therefore, for all sufficiently large \(k\), \(n_k>M\). This proves \(n_k\to\infty\), and hence \(z_k=4+1/n_k\to4\). Uniqueness of limits gives \(z=4\). The sole accumulation point is \(4\), which is not itself in \(E\).
Worked Example: Every Real Number Is an Accumulation Point of the Rationals
Fix any \(x\in\mathbb{R}\). By density of the rationals, for each positive integer \(n\) choose $$ q_n\in\mathbb{Q}\cap\left(x+2^{-(n+1)},\,x+2^{-n}\right). $$ These intervals are pairwise disjoint: the upper endpoint of the interval for \(n+1\) equals the lower endpoint of the interval for \(n\), and all intervals with larger indices lie still farther to the left. Since the intervals are open, even consecutive intervals have no point in common. Thus the selected rationals \(q_n\) are distinct. Moreover, \(q_n>x\) and $$ 0<|q_n-x|<2^{-n}. $$ As \(2^{-n}\to0\), we have \(q_n\to x\). For every \(r>0\), a sufficiently large \(n\) satisfies \(2^{-n}<r\), so \(q_n\) is a rational different from \(x\) within distance \(r\) of \(x\). Therefore \(x\) is an accumulation point of \(\mathbb{Q}\). Since \(x\) was arbitrary, every real number is an accumulation point of the rationals.
A Sequential Characterization
The examples suggest a useful way to detect accumulation points: instead of checking every radius separately, construct one sequence of distinct set points that converges to the candidate point. The requirement of distinctness is important. A sequence that repeats one point forever could converge to a member of \(E\) without showing that other points of \(E\) gather nearby.
Proof. Suppose first that \(x\) is an accumulation point of \(E\). Theorem (Every Neighborhood of a Limit Point Contains Infinitely Many Set Points), from “Closed Sets and Limit Points,” states that every neighborhood of \(x\) contains infinitely many points of \(E\). We choose points recursively. Select \(x_1\in E\cap B_1(x)\) with \(x_1\ne x\). Once \(x_1,\ldots,x_{n-1}\) have been selected, the ball \(B_{1/n}(x)\) contains infinitely many points of \(E\). Removing the finite set \(\{x,x_1,\ldots,x_{n-1}\}\) cannot remove all of them, so choose \(x_n\) in that ball and outside the removed set. The resulting points are in \(E\), are different from \(x\), and are pairwise distinct. They satisfy $$ |x_n-x|<\frac1n. $$ Given \(\varepsilon>0\), choose \(N\) so that \(1/N<\varepsilon\). If \(n\geq N\), then \(|x_n-x|<1/n\leq1/N<\varepsilon\), so \(x_n\to x\).
Conversely, suppose \(x_n\in E\) are pairwise distinct and \(x_n\to x\). Fix any \(r>0\). Convergence gives an \(N\) such that \(n\geq N\) implies \(|x_n-x|<r\). Since the sequence is pairwise distinct, at most one of its terms can equal \(x\). Consequently, there is some \(n\geq N\) for which \(x_n\ne x\). For this index, \(0<|x_n-x|<r\), with \(x_n\in E\). This verifies the accumulation-point definition for every \(r>0\). \(\square\)
The proof in the forward direction uses the infinitude of set points in every neighborhood, not merely the fact that each neighborhood contains at least one point other than \(x\). Infinitude lets us avoid all earlier choices and produce a genuinely distinct sequence. The converse also explains why a convergent sequence without distinctness would not suffice: its terms might all be the same point of \(E\).
Approach from the Left or the Right
On the real line, an accumulation point must be approached from at least one direction. This gives a more precise description of how the set sits near a point. The one-sided definitions use intervals that exclude \(x\), just as the accumulation-point definition uses punctured neighborhoods.
Proof. If \(x\) is a left accumulation point, then for each \(r>0\) there is \(y\in E\) with \(x-r<y<x\). Thus \(0<|y-x|<r\), so \(x\) is an accumulation point. If \(x\) is a right accumulation point, the inequalities \(x<y<x+r\) give the same conclusion.
For the reverse implication, suppose \(x\) is an accumulation point but is neither a left nor a right accumulation point. Failure of the left condition means there is \(r_L>0\) for which no \(y\in E\) satisfies \(x-r_L<y<x\). Failure of the right condition means there is \(r_R>0\) for which no \(y\in E\) satisfies \(x<y<x+r_R\). Put \(r=\min\{r_L,r_R\}>0\). Any point \(y\) with \(0<|y-x|<r\) lies either to the left of \(x\), in \((x-r_L,x)\), or to its right, in \((x,x+r_R)\). Neither interval contains a point of \(E\) by the choices of \(r_L\) and \(r_R\). This contradicts that \(x\) is an accumulation point. Hence at least one of the two one-sided conditions holds. \(\square\)
Worked Example: Accumulation from Only One Side
Let \(E=\{-1/n:n\geq1\}\). Every point of \(E\) is negative, and \(-1/n\to0\). For any \(r>0\), choose \(n\) large enough that \(1/n<r\). Then \(-r<-1/n<0\), so \(0\) is a left accumulation point of \(E\).
It is not a right accumulation point: for every \(r>0\), the interval \((0,r)\) contains no point of \(E\), because every member of \(E\) is negative. Thus \(0\) is approached from the left only. This example shows why the one-sided theorem says “left or right,” not necessarily “both.”
What the Definition Does—and Does Not—Say
Accumulation points describe repeated proximity, not membership. For example, the rational-number example shows that a point can be an accumulation point of a set even when it does not belong to that set. Conversely, membership alone does not imply accumulation: for the singleton \(E=\{6\}\), the point \(6\) is not an accumulation point, since every ball around \(6\) contains no other point of \(E\). These distinctions are reflected in the earlier identity \(\overline E=E\cup E'\): closure includes points of \(E\) whether or not they are accumulation points, while \(E'\) records only accumulation points.
A common pitfall is to use a sequence from \(E\) that converges to \(x\) but has no distinct terms. The constant sequence \(x_n=6\), for instance, converges to the point \(6\) in \(\{6\}\), yet it gives no evidence of nearby set points different from \(6\). The Sequential Characterization avoids this problem by requiring pairwise distinct terms. A different pitfall is to assume that an accumulation point must be approached from both directions. The one-sided example shows that one direction alone is enough.
In practice, the neighborhood definition is often best for proving that a point is an accumulation point: given an arbitrary radius, produce a different set point inside it. The sequential characterization is often better for ruling out possible accumulation points, especially when distinct elements of a set can be indexed and their limiting behavior is known. The one-sided characterization then refines the conclusion by identifying where the nearby points lie on the real line.
Check Your Understanding
Use the definitions and the two characterizations to answer the following questions.
- Why does the definition of accumulation point require a set point different from \(x\)?
- What additional condition does the sequential characterization impose beyond convergence of a sequence in \(E\)?
- Why can the recursive selection in the sequential theorem avoid all previously chosen points?
- Can a point outside \(E\) be an accumulation point of \(E\)? Give an example from this tutorial.
- Why does a left accumulation point necessarily count as an accumulation point?
- Does every accumulation point have to be approached from both sides? Explain using the one-sided example.