Math 135 Spring 2007 :: Write-up on connectedness and the Cantor set
I’d like to expand on the cryptic statement I made during class, that the Cantor set C is “totally
disconnected”. It gave me the opportunity to learn a few things myself :)
The mathematical definition is the following. A topological space X is said to be disconnected
if it is the union of two disjoint nonempty open sets. Otherwise, X is said to be connected. If we
are testing for disconnectedness of a subspace C of a topological space X, then “open” should be
understood in the sense of the subspace topology, or relative topology, where sets are being defined
as open if they are the intersection of C with an open set of X. To see why this characterization
applies to the Cantor set, choose two different points x and y in it. Since they differ by at least
one digit in base 3, separate the Cantor set into two halves corresponding to any one of these
discrepancy digits. These two sets can each be seen as the intersection of a finite union of open
intervals with C, hence are open in the relative topology of the Cantor set. This implies that the
Cantor set is disconnected.
A space in which all maximally connected components are one-point sets is called totally disconnected. Since the points x and y are arbitrary in the above reasoning, connected components
consisting of more than one point are ruled out, proving that the Cantor set is totally disconnected.
There are many equivalent notions of connectedness. One is the following, brought up by one
of you in class. X is connected if and only if there is no onto continuous function from X to a
discrete space which contains more than one point. In other words, X is disconnected if and only
if there exists at least one such function. A space is discrete if it contains isolated points—points
surrounded by neighborhoods containing no other points. For instance, split the Cantor set into
two halves A ⊂ [0, 1/3] and B ⊂ [2/3, 1], and define f (x) = 0 for 0 ≤ x ≤ 1/3, f (x) = 3x − 1 if
1/3 ≤ x ≤ 2/3, and f (x) = 1 if 2/3 ≤ x ≤ 1. With this choice, f maps C to {0, 1}, which proves
disconnectedness of C again. Since this type of construction extends to all scales, we can show
that two arbitrary points cannot belong to the same connected component, thereby proving total
disconnectedness again.
Another definition of connectedness is the following. A topological space X is said to be pathconnected if for any two points x and y in X there exists a continuous function f from the unit
interval [0, 1] to X with f (0) = x and f (1) = y. This function is called a path from x to y. Note that
subsets of the real line R are connected if and only if they are path-connected, but this equivalence
is not true in higher dimensions. Since the Cantor set C is a subset of [0, 1], and we saw that it
is disconnected, then it is automatically not path connected. There is, however, a more visual of
proving this fact, other than applying the equivalence of notions. Pick any two points x and y in
C, and assume by contradiction that there exists a function f as above. By the intermediate value
theorem, any point x ≤ z ≤ y must belong to the image of [0, 1] under f . By connectedness, the
segment [x, y] must be in C. But it is easy to see that there exist points between x and y that
are not in C: assume w.l.o.g. that x < y and let the first discrepancy digit occur at place n. The
n-th digit of x is 0 and the n-th digit of y is 2. Then define an intermediate z by changing the
n-th digit, and also to be sure also the n + 1-st digit of x into ones. We conclude that C is not
path-connected, or in other words that it does not contain any open interval.
The counter-example to the equivalence of connectedness and path-connectedness in two dimensions is the topologists sine curve, that you can look up on wikipedia or mathworld. It is a set
that is connected but not path-connected.
Let me know if this write-up left questions unanswered, or if you saw mistakes.