Derived Sets

The Derived Set Records Points Approached by a Set

The closure of a set includes both its original points and any points that can be approached by them. The derived set focuses on the latter: it records points that are approached by points of the set other than the candidate point itself. That distinction matters when a set has isolated points. An isolated point belongs to the set and its closure, but it is not a limit point.

Definition (Derived Set): Let $E\subseteq\mathbb{R}$. The derived set of $E$, denoted by $E'$, is the set of all limit points of $E$: $$ E'=\{x\in\mathbb{R}: \text{for every }r>0,\ (B_r(x)\setminus\{x\})\cap E\neq\varnothing\}. $$

The puncture $B_r(x)\setminus\{x\}$ prevents $x$ itself from being the point that witnesses intersection with $E$. Thus, if $x\in E$, that fact alone does not show that $x\in E'$. Every neighborhood around $x$ must contain another point of $E$. If $x\notin E$, the puncture makes no practical difference, since $x$ could not be a point of $E$ anyway.

Earlier in the course, the Sequential Characterization of Limit Points established that $x\in E'$ if and only if there is a sequence of points of $E\setminus\{x\}$ converging to $x$. We will use that result to calculate derived sets. We will also use the previously established theorem that $E'$ is closed, even when $E$ is not closed. Neither fact says that $E'=E$: for example, an isolated point of $E$ is not in $E'$.

Inclusion and Isolated Points

A first useful property is monotonicity: adding points to a set cannot remove any of its limit points. This follows directly from the punctured-neighborhood definition.

Theorem (Monotonicity of the Derived Set): If $A\subseteq B\subseteq\mathbb{R}$, then $A'\subseteq B'$.

Proof. Let $x\in A'$, and take any $r>0$. Since $x\in A'$, there is a point $y\in (B_r(x)\setminus\{x\})\cap A$. Because $A\subseteq B$, the same point belongs to $(B_r(x)\setminus\{x\})\cap B$. This holds for every $r>0$, so $x\in B'$. Therefore $A'\subseteq B'$. $\square$

Monotonicity concerns the limit points, not the sets themselves: in general, $A'$ need not be a subset of $A$. For example, the points $1/n$ approach $0$, even though $0$ is not one of those points. The theorem does imply that if $A\subseteq B$, then every point approached by $A$ is also approached by $B$.

To describe what the derived set leaves out of the closure, it helps to name the points that are isolated within a set.

Definition (Isolated Point): A point $x\in E$ is isolated in $E$ if there is an $r>0$ such that $B_r(x)\cap E=\{x\}$.

Theorem (Closure Decomposes into Limit Points and Isolated Points): For every $E\subseteq\mathbb{R}$, $$ \overline E=E'\cup\{x\in E:x\text{ is isolated in }E\}, $$ and the two sets on the right are disjoint.

Proof. First, every point of $E'$ belongs to $\overline E$, since every punctured ball around it meets $E$, and hence every ball meets $E$. Every isolated point $x\in E$ also belongs to $\overline E$, because each ball around $x$ contains $x\in E$. Thus the right side is contained in $\overline E$.

For the reverse inclusion, let $x\in\overline E$. The Closure as the Union of a Set and Its Derived Set Theorem, established earlier, gives $\overline E=E\cup E'$. If $x\in E'$, it is already in the first set on the right. Otherwise $x\in E$ and $x\notin E'$. By the definition of $E'$, there is an $r>0$ such that $(B_r(x)\setminus\{x\})\cap E=\varnothing$. Since $x\in E$, this means $B_r(x)\cap E=\{x\}$; hence $x$ is isolated in $E$. This proves the reverse inclusion. Finally, no isolated point of $E$ can belong to $E'$, because its isolating ball has no point of $E$ other than itself. The two sets are disjoint. $\square$

This decomposition clarifies the difference between closure and derived set. The closure retains every point already in $E$, including isolated points. The derived set retains only points that are approached by other points of $E$. In particular, a finite set can have a nonempty closure but an empty derived set.

Worked Example: A Finite Set Has No Limit Points

Let $E=\{-3,2,7\}$. We show that $E'=\varnothing$. If $x\in E$, choose $r$ smaller than the distance from $x$ to each of the other two elements. The punctured ball $B_r(x)\setminus\{x\}$ then contains no point of $E$. If $x\notin E$, the three distances from $x$ to the elements of $E$ are all positive. Choose $r$ smaller than their minimum; then $B_r(x)\cap E=\varnothing$. In either case, $x$ is not a limit point, so $E'=\varnothing$.

Nevertheless, $E$ is closed, and $\overline E=E$. Its three points are isolated, which is exactly why they occur in the closure but not in the derived set.

Finite Changes Do Not Affect the Derived Set

A limit point depends on points occurring arbitrarily close to the candidate, not on any fixed finite collection of points. Adding or removing finitely many points can create or eliminate isolated points, but it cannot change which points are approached by the set.

Theorem (Invariance Under Finite Changes): If $E,F\subseteq\mathbb{R}$ and their symmetric difference $E\triangle F=(E\setminus F)\cup(F\setminus E)$ is finite, then $E'=F'$.

Proof. Write $D=E\triangle F$, which is finite. We first show $E'\subseteq F'$. Let $x\in E'$. Choose $\rho>0$ so that $B_\rho(x)$ contains no point of $D\setminus\{x\}$. Such a radius exists: if $D\setminus\{x\}$ is nonempty, take $\rho$ smaller than the minimum of the finitely many positive distances $|d-x|$ for $d\in D\setminus\{x\}$; if it is empty, take any $\rho>0$.

Now let $s>0$ and put $t=\min\{s,\rho\}$. Since $x\in E'$, there is a $y\in E$ with $0<|y-x|<t$. This point is not in $D$, because it lies in $B_\rho(x)$ and differs from $x$. Any point in $E\setminus D$ also belongs to $F$, so $y\in F$. Moreover, $0<|y-x|<s$. We have shown that every punctured ball around $x$ meets $F$, and therefore $x\in F'$. This proves $E'\subseteq F'$. Interchanging $E$ and $F$ gives $F'\subseteq E'$, so $E'=F'$. $\square$

Worked Example: Adding an Isolated Point

Let $E=(0,1)$ and $F=(0,1)\cup\{4\}$. The sets differ by just one point, so the Invariance Under Finite Changes Theorem gives $E'=F'$. More explicitly, the points approached by $(0,1)$ are exactly the points of $[0,1]$, so $E'=[0,1]$. The added point $4$ is isolated in $F$: for instance, $B_1(4)\cap F=\{4\}$. It creates no new limit point. Thus $F'=[0,1]$ as well, while $F$ itself contains the additional isolated point.

The finite-change theorem is useful when a set has a complicated but finite collection of exceptional points. Those points can be removed or added without changing the derived set. The finiteness condition is essential: an infinite collection of added points can accumulate and create new limit points.

Derived Sets and Arbitrary Unions

Monotonicity immediately gives one inclusion for unions. Every limit point of one set in a family remains a limit point of the union, since the union contains that set. For a finite union, the earlier Limit Points of a Finite Union Theorem gives equality between the derived set of the union and the union of the derived sets. For arbitrary unions, equality can fail: points from different sets in the family can collectively approach a point even when none of the individual sets has a limit point.

Theorem (Derived Set of a Union): For any family $\{E_\alpha:\alpha\in I\}$ of subsets of $\mathbb{R}$, $$ \bigcup_{\alpha\in I}E_\alpha'\subseteq\left(\bigcup_{\alpha\in I}E_\alpha\right)'. $$

Proof. Let $x\in\bigcup_{\alpha\in I}E_\alpha'$. Then $x\in E_{\alpha_0}'$ for some index $\alpha_0\in I$. Every punctured ball around $x$ meets $E_{\alpha_0}$. Since $E_{\alpha_0}\subseteq\bigcup_{\alpha\in I}E_\alpha$, the same punctured ball meets the union. This is true for every radius, so $x\in(\bigcup_{\alpha\in I}E_\alpha)'$. $\square$

Worked Example: An Infinite Union Creates a Limit Point

For each positive integer $n$, let $E_n=\{1/n\}$. Each $E_n$ is a singleton, so the finite-set argument shows that $E_n'=\varnothing$. Consequently, $$ \bigcup_{n=1}^{\infty}E_n'=\varnothing. $$ But the union of the sets is $S=\{1/n:n\geq1\}$, and $0\in S'$. Indeed, given $r>0$, choose a positive integer $n>1/r$. Then $1/n\in S$, $1/n\neq0$, and $$ \left|\frac1n-0\right|=\frac1n<r. $$ Thus every punctured ball around $0$ meets $S$, and $0$ belongs to $S'$. Therefore $$ \bigcup_{n=1}^{\infty}E_n'\subsetneq\left(\bigcup_{n=1}^{\infty}E_n\right)'. $$ The new limit point arises from points supplied by infinitely many different members of the family.

Calculating a Derived Set from a Sequence

For a set formed from the terms of a convergent sequence, the sequence limit is often a candidate limit point, but it is important to check whether other points qualify. The sequential characterization of limit points provides an efficient way to do so: a candidate is a limit point exactly when points of the set other than the candidate can be chosen in a sequence converging to it.

Worked Example: The Derived Set of the Reciprocal Integers

Let $S=\{1/n:n\geq1\}$. Since $1/n\to0$ and every $1/n$ is different from $0$, the Sequential Characterization of Limit Points gives $0\in S'$.

Now suppose $x\in S'$. By that same characterization, there is a sequence $y_k\in S\setminus\{x\}$ with $y_k\to x$. Write $y_k=1/n_k$, where each $n_k$ is a positive integer. For every fixed positive integer $M$, only the $M$ values $1,1/2,\ldots,1/M$ have indices at most $M$. Each such value can occur only finitely often in $(y_k)$: if one occurred infinitely often, a constant subsequence would converge to that value, contradicting $y_k\to x$ and $y_k\neq x$. Hence eventually $n_k>M$. It follows that $n_k\to\infty$, and therefore $y_k=1/n_k\to0$. Since also $y_k\to x$, uniqueness of limits gives $x=0$. We conclude that $$ S'=\{0\}. $$ The terms of $S$ are isolated, while their common accumulation point is not in $S$.

A frequent error is to use membership in $E$ as evidence that a point belongs to $E'$. The definition requires other points of $E$ arbitrarily close; an isolated point fails that test. Another is to assume that derived sets always distribute over unions. The inclusion for arbitrary unions is valid, but the reciprocal-integers example shows that it can be strict when infinitely many sets are involved.

Test the punctured neighborhoods.
A point is in $E'$ only if every positive-radius ball around it contains a point of $E$ different from the center.

Use sequences to identify candidates.
A sequence of points of $E\setminus\{x\}$ converging to $x$ proves that $x$ is a limit point.

Check for isolated points and infinite unions.
Isolated points belong to the closure but not the derived set; an infinite union may create limit points absent from each individual derived set.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

What condition on punctured balls defines membership in $E'$?
Can a point of $E$ fail to belong to $E'$? Describe the relevant property of that point.
If $A\subseteq B$, what inclusion relates $A'$ and $B'$?
What does the Invariance Under Finite Changes Theorem say about sets whose symmetric difference is finite?
For an arbitrary family of sets, which inclusion always holds between the union of their derived sets and the derived set of their union?
Why is $0$ a limit point of $\{1/n:n\geq1\}$ even though it is not a member of that set?

What You'll Learn