This is post # 7 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.
In the previous post The Cantor set, I, we constructed the “middle third” Cantor set and we showed that it has cardinality continuum. In this post we discuss some basic topological properties of the Cantor set.
The Cantor set we defined in the previous post is the “middle third” Cantor set, the traditional version found in many texts. In this “middle third” recipe, the open interval in the middle third of is taken out. Two closed intervals remain. In each of the succeeding stages, the open interval in the middle third of any remaining closed interval is removed. After the inductive process is completed, the remaining points form the Cantor set. Note that the open intervals removed in this process have a total length of since . Yet the points remained have the same cardinality as the unit interval .
There are other striking properties about the Cantor set. For example, the Cantor set contains no open intervals . This makes the Cantor set a very “thin” set (a nowhere dense set). On the other hand, the Cantor set has no isolated points since every point of the Cantor set is a limit point of itself.
Cantor Set Preliminary
Let be the Cantor set as defined in the previous post. Recall that where
and so on.
Each , where is a string of zeros and ones of finite length, is a closed interval and is obtained by removing the middle third of the previous closed interval. For example, and are obtained by removing the middle third of the closed interval . Refer to the previous post The Cantor set, I for more information about the construction.
Each point is the single point in the intersection of a nested decreasing sequence of , i,e, where and agree on the first many coordinates if . The can be obtained by tracing the construction steps. For example, or . Say . Then or . Say, . Now or . Say, . Thus, we have:
As a result, each is associated with , which is a sequence of countably infinite many zeros and ones such that the first coordinates of agree with .
The mapping in the above paragraph is a bijection between the Cantor set and , the set of all countably infinite sequences of zeros and ones. If , we let be the string of zeros and ones that agrees with the first coordinates of . Thus in this notation, for , we have .
To facilitate the subsequent discussion, we have the following lemma.
- For each , the intervals form a local base at , i.e. for each open interval with , there is some integer such that .
- Let be the set of endpoints of the closed intervals obtained in the construction process. We have .
- The set is dense in .
Condition 1 follows from the fact that the lengths of the closed intervals go to zero.
We now prove condition 2. Let be a closed interval obtained in the construction process and let be the left endpoint and be the right endpoint. Both and belong to since:
The fact that is dense in follows from condition 1. Every open interval containing a point of has to contain one closed interval and thus the endpoints of .
We discuss the following proeprties:
- The Cantor set is a compact subset of .
- The Cantor set is a perfect set.
- , the complement of , is dense in .
- The Cantor set contains no open interval.
- The Cantor set is nowhere dense in .
- The Cantor set is totally disconnected.
Discussion of 1
In the real line or in any other topological space, the intersection of closded sets is always a closed set. The Cantor set is the intersection of where each is a closed set (it being a finite union of closed intervals). Thus the Cantor set is a closed subset of . Being a closed and bounded subset of the real line, is compact.
Discussion of 2
Recall that a subset of the real line is a perfect set if it is closed and every point of is a limit point of . Let . Let be an open interval of the real line such that . We want to show that there are infinitely many points of inside . This following from condition 1 of the lemma. The interval contains which would contain infinitely many points in the Cantor set. Note that the construction process of the Cantor set would continue within the interval .
Discussion of 3
This also follows from condition 1 of the lemma. We need to demonstrate that every point in is either a point of or a limit point of . Let . If , then we are done. So suppose . By condition 1 of the lemma, every open interval containing contains some closed interval . Since the middle third of is removed in the construction process, we see that .
Discussion of 4
This is a corollary of 3. Suppose we have . By 3, contains points of the removed open intervals, which is a contraction.
Discussion of 5
Let . The set is nowhere dense in if is dense in no open interval of , or equivalently, for each open interval , there is some subinterval such that . To say that is nowhere dense in , we only need to concern with the open intervals that have points in .
This is a corollary of condition 4. For each open interval in , it cannot be a subset of . So there must some point such that . But the Cantor set is a closed subset of . Then there must be some such that and .
Discussion of 6
The Cantor set is totally disconnected. A subset of the real line is a disconnected set if there are disjoint open sets and , both containing points of , such that . A set is connected if it is not disconnected.
The Cantor set is disconnected. Note that .
A set is totally disconnected if the only connected subsets of are the singleton sets. We show that any subset of the Cantor set with at least two points is not connected. Let with where . Since is nowhere dense, there exists such that has no points of . Let . Then we have:
Links to previous posts on the topology of the real line:
The Euclidean topology of the real line (1)
The Euclidean topology of the real line (2)
The Euclidean topology of the real line (3) – Completeness
The Euclidean topology of the real line (4) – Compactness
The Cantor bus tour
The Cantor set, I