Perfect Sets

Closed Sets Without Isolated Points

The previous tutorial examined points that can be separated from the rest of a set by a small neighborhood. A set with no such points has a very different local structure: every neighborhood of each member contains another member of the set. When this property is combined with closedness, the resulting set is called perfect.

The two requirements play different roles. Having no isolated points is a local condition on the members of the set. Closedness ensures that limits of points in the set do not lie outside it. Together, they give a useful identity involving the derived set $E'$, the set of accumulation points of $E$.

Definition: A set $P\subseteq\mathbb{R}$ is perfect if it is closed and has no isolated points. Equivalently, $P$ is closed and every $x\in P$ is an accumulation point of $P$. The empty set is perfect under this definition.

The equivalence in the definition follows from the Isolation and Accumulation theorem: for a point of $P$, being isolated is exactly the failure to be an accumulation point. The empty set satisfies both requirements vacuously: it is closed, and it has no points that could be isolated.

A Derived-Set Test for Perfectness

A set is perfect precisely when its points are exactly its accumulation points. This gives a concise test that brings together closedness and the absence of isolated points.

Theorem (Derived-Set Characterization of Perfect Sets): A set $P\subseteq\mathbb{R}$ is perfect if and only if $P=P'$.

Proof. Suppose $P$ is perfect. Since $P$ has no isolated points, every $x\in P$ is an accumulation point of $P$. Hence $P\subseteq P'$. Every accumulation point of $P$ belongs to its closure, and $P$ is closed, so $P'\subseteq P$. Therefore $P=P'$.

Conversely, suppose $P=P'$. Every point of $P$ is then an accumulation point of $P$, so no point of $P$ is isolated. By the earlier Closure Decomposes into Limit Points and Isolated Points theorem, the closure of $P$ is the union of $P'$ and the isolated points of $P$. There are no isolated points, and $P'=P$, so $\overline{P}=P$. Thus $P$ is closed and has no isolated points; it is perfect. $\square$

The identity $P=P'$ is especially useful when accumulation points are easier to describe than neighborhoods that isolate points. It also highlights why closedness matters: without it, a set can have no isolated points and still omit some of its accumulation points.

Worked Example: A Closed Interval Is Perfect

Let $P=[a,b]$, where $a<b$. This interval is closed. We check that none of its points is isolated by finding, for every $x\in P$ and $r>0$, a different point of $P$ within distance $r$ of $x$.

If $x<b$, choose $$ \delta=\min\left\{\frac r2,\frac{b-x}{2}\right\} \qquad\text{and}\qquad y=x+\delta. $$ Both entries in the minimum are positive, so $\delta>0$. Also, $\delta<r$, and $y>x$. Since $\delta\leq(b-x)/2$, $$ y\leq x+\frac{b-x}{2}=\frac{x+b}{2}<b. $$ Thus $y\in[a,b]$, $y\ne x$, and $|y-x|=\delta<r$.

The remaining case is $x=b$. Set $$ \delta=\min\left\{\frac r2,\frac{b-a}{2}\right\} \qquad\text{and}\qquad y=b-\delta. $$ Here $\delta>0$, $\delta<r$, and $y<b$. Since $\delta\leq(b-a)/2$, we have $y\geq(b+a)/2>a$. Consequently $y\in[a,b]$, $y\ne b$, and $|y-b|=\delta<r$. Every point of $[a,b]$ is therefore an accumulation point, so $[a,b]$ is perfect.

A Perfect Set That Is Not an Interval

Perfect sets need not contain an interval around any of their points. The Cantor set is a standard example: it is built by repeatedly removing open middle thirds from intervals, yet every point that remains is approached by other points that remain.

Worked Example: Verifying That the Cantor Set Is Perfect

Start with $C_0=[0,1]$. To form $C_{n+1}$ from $C_n$, remove the open middle third of each component interval of $C_n$. Thus $C_1=[0,1/3]\cup[2/3,1]$, and the process continues. Define the Cantor set by $$ C=\bigcap_{n=0}^{\infty} C_n. $$

Each $C_n$ is a finite union of closed intervals and is therefore closed. The Arbitrary Intersections of Closed Sets theorem implies that $C$ is closed. It is nonempty, since the endpoints $0$ and $1$ remain in every $C_n$.

At level $n$, the component intervals have length $3^{-n}$, and distinct components are separated by gaps. Given $x\in C$, there is therefore a component interval $J$ of $C_n$ containing $x$. Both endpoints of $J$ belong to $C$: the construction removes open middle thirds, so it never removes an endpoint of a retained interval. Choose an endpoint $y$ of $J$ different from $x$. Such an endpoint exists because $J$ has positive length. Then $y\in C$, $y\ne x$, and $$ 0<|y-x|\leq 3^{-n}. $$ For any $r>0$, choose $n$ large enough that $3^{-n}<r$. The point $y$ then lies in $C$, differs from $x$, and satisfies $|y-x|<r$. No point of $C$ is isolated. Since $C$ is also closed, it is perfect.

This example separates two ideas that are easy to conflate. A perfect set has no isolated points, but it need not contain an open interval. The Cantor-set argument establishes perfectness directly; it does not rely on a claim that the set fills any neighborhood.

Worked Example: A Closed Set That Is Not Perfect

Let $E=\{-1,3\}$. It is closed because every finite subset of $\mathbb{R}$ is closed. But both its points are isolated. For example, $B_1(-1)=(-2,0)$ meets $E$ only at $-1$, and $B_1(3)=(2,4)$ meets $E$ only at $3$. Thus $E$ is closed but not perfect.

This example shows why closedness alone is insufficient. In the derived-set test, $E'=\varnothing$, whereas $E\ne\varnothing$, so $E\ne E'$.

Every Nonempty Perfect Set Is Uncountable

Perfectness imposes a strong size condition. If a perfect set contains even one point, its lack of isolated points lets us repeatedly find two distinct points in any sufficiently small neighborhood of a point already chosen. Keeping the resulting neighborhoods separate creates a distinct point of the set for every infinite binary sequence.

Theorem (Every Nonempty Perfect Set Is Uncountable): If $P\subseteq\mathbb{R}$ is nonempty and perfect, then $P$ is uncountable.

Proof. Choose $p_{\varnothing}\in P$, and let $I_{\varnothing}=[p_{\varnothing}-1,p_{\varnothing}+1]$. We construct closed intervals $I_s$, with centers $p_s\in P$, for each finite binary word $s$. The empty word is denoted by $\varnothing$; the two children of a word $s$ are $s0$ and $s1$.

Suppose $I_s$ and its center $p_s\in P\cap\operatorname{int}(I_s)$ have been chosen, and that $s$ has length $n$. Since $p_s$ is not isolated, there is a point $q\in P$, different from $p_s$, as close to $p_s$ as we wish. Choose it close enough that both $p_s$ and $q$ lie in $\operatorname{int}(I_s)$. Their distance $d=|q-p_s|$ is positive. Choose $\rho>0$ small enough that the closed intervals $$ I_{s0}=[p_s-\rho,p_s+\rho], \qquad I_{s1}=[q-\rho,q+\rho] $$ are contained in $\operatorname{int}(I_s)$, are disjoint, and have length at most $2^{-(n+1)}$. Such a choice is possible: each center has positive distance from the boundary of $I_s$, the distance $d$ between the centers is positive, and $\rho$ can be made smaller than these boundary distances, smaller than $d/2$, and smaller than $2^{-(n+2)}$. Set $p_{s0}=p_s$ and $p_{s1}=q$. Both new centers belong to $P$ and lie in the interiors of their respective intervals. This constructs all levels by induction.

Now let $\alpha=(\alpha_1,\alpha_2,\ldots)$ be any infinite binary sequence, and let $\alpha|n$ be its first $n$ digits. The centers $p_{\alpha|n}$ lie in nested intervals. If $m\geq n\geq1$, both $p_{\alpha|m}$ and $p_{\alpha|n}$ lie in $I_{\alpha|n}$, so $$ |p_{\alpha|m}-p_{\alpha|n}|\leq 2^{-n}. $$ The centers therefore form a Cauchy sequence. By completeness of $\mathbb{R}$, they converge to some $x_\alpha$. Every center belongs to $P$, and $P$ is closed, so the Sequential Criterion for Closedness gives $x_\alpha\in P$. For each $n$, all later centers lie in the closed interval $I_{\alpha|n}$, so their limit $x_\alpha$ also lies there.

If two infinite binary sequences first differ at digit $k$, their corresponding limits lie in the two disjoint intervals $I_{\alpha|k}$ and $I_{\beta|k}$. Thus those limits are distinct. We have constructed an injection from the set of infinite binary sequences into $P$. The set of infinite binary sequences is uncountable: if they could be listed as $\alpha^{(1)},\alpha^{(2)},\ldots$, the sequence whose $n$th digit differs from the $n$th digit of $\alpha^{(n)}$ would not appear in the list. Hence $P$ is uncountable. $\square$

What Perfectness Does—and Does Not—Say

The uncountability theorem explains why the absence of isolated points is more than a local curiosity. A nonempty perfect subset of the real line cannot be finite or countably infinite. At every scale, each chosen point can be replaced by two separated choices, and the repeated branching produces uncountably many distinct points.

The hypotheses should not be weakened carelessly. A closed set may have isolated points, as the finite example shows. A set may also have no isolated points without being closed; the derived-set test makes the missing condition visible. Conversely, being uncountable alone does not make a set perfect: perfectness requires both closedness and the specified local behavior at every member.

When checking a candidate, first establish closedness, using a suitable closed-set criterion. Then take an arbitrary point of the set and an arbitrary radius, and find a different set point inside that ball. This order keeps the two independent requirements clear. Alternatively, compute or characterize the derived set and test whether it equals the original set.

Check Your Understanding

Use the definition, the derived-set characterization, and the nested-interval proof to answer the following questions.

What two properties must a set have to be perfect?
Why does $P=P'$ imply that $P$ has no isolated points?
In the Cantor-set example, why can an endpoint of a retained interval serve as a nearby point of $C$?
Why is the requirement that a perfect set be nonempty necessary in the uncountability theorem?
How do disjoint intervals in the proof ensure that different binary sequences produce different points?
Why does the finite set $\{-1,3\}$ fail the derived-set test for perfectness?

What You'll Learn