The Cantor Set As A Perfect Nowhere-Dense Compactum

From the Finite Stages to the Cantor Set

The middle-thirds construction produces a sequence of simpler sets whose intersection is the Cantor set. Its key topological features can be read directly from the intervals at each finite stage: the intervals become short, their endpoints remain, and the set at every stage is a finite union of closed intervals. We will use these facts to prove that the Cantor set is compact, perfect, and nowhere dense.

Work in \(\mathbb{R}\) with its standard topology. Begin with \(C_0=[0,1]\). To form \(C_{n+1}\) from \(C_n\), remove the open middle third of every interval making up \(C_n\). Define the middle-thirds Cantor set by

The construction gives \(C_0=[0,1]\), \(C_1=[0,1/3]\cup[2/3,1]\), and

At every stage \(n\), the set \(C_n\) consists of \(2^n\) closed intervals, each of length \(3^{-n}\). To verify this, the assertion holds at \(n=0\): there is one interval of length \(1\). If it holds at stage \(n\), each interval \([a,b]\) is replaced by \([a,a+(b-a)/3]\) and \([a+2(b-a)/3,b]\). Thus each interval produces two closed intervals of one-third its length. The number of intervals doubles and their length is divided by \(3\), proving the assertion at stage \(n+1\).

Also, \(C_{n+1}\subseteq C_n\), since each step only removes points. The left and right endpoints of every interval at stage \(n\) remain in \(C_{n+1}\): they are endpoints of the two retained subintervals. Repeating this observation shows that every endpoint created at a finite stage belongs to \(C\).

Closed, Bounded, and Compact

Proof. At each finite stage, \(C_n\) is a finite union of closed intervals, so \(C_n\) is closed. The intersection of any collection of closed sets is closed: its complement is the union of their open complements, which is open. Since \(C=\bigcap_{n=0}^{\infty}C_n\), it follows that \(C\) is closed. Also, \(C\subseteq C_0=[0,1]\), so \(C\) is bounded. By the Heine–Borel theorem, the standard result that a subset of \(\mathbb{R}\) is compact if and only if it is closed and bounded, \(C\) is compact. \(\square\)

Compactness here concerns open covers, not merely the appearance of the construction. The inference uses both properties: closedness alone does not imply compactness in \(\mathbb{R}\), nor does boundedness alone. Heine–Borel supplies the conclusion once both hypotheses have been verified.

Every Point Is a Limit Point

A point \(x\) is a limit point of \(C\) if every open neighborhood of \(x\) contains a point of \(C\) different from \(x\). To prove that \(C\) is perfect, we must find such a point for every \(x\in C\), including endpoints.

Proof. Let \(x\in C\), and let \(U\) be any open neighborhood of \(x\) in \(\mathbb{R}\). By the standard topology on \(\mathbb{R}\), there is an \(r>0\) such that \((x-r,x+r)\subseteq U\). Choose \(n\) large enough that \(3^{-n}<r\). Since \(x\in C\subseteq C_n\), it lies in one of the stage-\(n\) intervals, say \([a,b]\). The endpoints \(a\) and \(b\) belong to \(C\), as endpoints survive every later stage. At least one of \(a,b\) differs from \(x\); choose such an endpoint and call it \(y\). Since \(x,y\in[a,b]\), their distance satisfies \(|y-x|\leq b-a=3^{-n}<r\). Thus \(y\in U\cap C\) and \(y\ne x\). Every open neighborhood of \(x\) therefore contains a point of \(C\) other than \(x\). This holds for every \(x\in C\), so every point is a limit point. The preceding proposition shows that \(C\) is closed, and hence it is perfect by definition. \(\square\)

The Cantor Set Has Empty Interior

The interval lengths shrink at every stage, but that fact alone needs to be connected to open neighborhoods to prove that the final set has no interior. The following argument does so without needing to identify which removed gap lies inside a given neighborhood.

Proof. Suppose, for contradiction, that the interior of \(C\) is nonempty. Then it contains a nonempty open interval \(J=(a,b)\), with \(a<b\), and \(J\subseteq C\). Since \(C\subseteq C_n\) for every \(n\), we have \(J\subseteq C_n\) for every \(n\).

Fix \(n\). The intervals making up \(C_n\) are separated by nonempty gaps: each is formed by removing an open middle third, and later stages remove only more points. If \(J\) met two different intervals of \(C_n\), choose \(u,v\in J\) in two such intervals, with \(u<v\). Between those intervals there is a point \(s\) in a gap of \(C_n\). Then \(u<s<v\). Because \(J=(a,b)\) is an interval, this implies \(s\in J\), contradicting \(J\subseteq C_n\). Thus \(J\) must lie in a single interval of \(C_n\). That interval has length \(3^{-n}\), so \(b-a\leq3^{-n}\).

Choose \(n\) large enough that \(3^{-n}<b-a\). This contradicts \(b-a\leq3^{-n}\). Therefore the interior of \(C\) is empty. As \(C\) is closed, \(\overline{C}=C\), so the interior of its closure is empty. This is exactly the definition of nowhere dense. \(\square\)

Common Pitfall: The Removed Third Is Open

When the middle third is removed, its two endpoints stay. A common error is to treat the removed interval as closed and delete its endpoints as well. For example, deleting \([1/3,2/3]\) from \([0,1]\) would give \([0,1/3)\cup(2/3,1]\). That is not the middle-thirds construction: the removed set is the open interval \((1/3,2/3)\), so the correct first stage is \([0,1/3]\cup[2/3,1]\).

The distinction matters for the later arguments. The equality \(1/3\in[0,1/3]\) shows that \(1/3\) survives stage \(1\), and it remains an endpoint at every later stage. Thus \(1/3\in C\). In the proof of perfection, the endpoint of a sufficiently small stage interval is available as a nearby point of \(C\); incorrectly deleting endpoints would invalidate that witness. The construction keeps the endpoints precisely because it removes open middle thirds.

The proofs give a useful summary of the set’s topology. It is compact because it is closed and bounded in \(\mathbb{R}\); it is perfect because every one of its points is approached by other points of the set; and it is nowhere dense because no open interval fits inside it. Perfectness does not mean that the set contains an interval, and nowhere density does not mean that the set is empty. These properties describe different features and are all verified directly from the nested intervals.