This post isn’t related to anything I’ve been focusing on as of late, but it’s in response to a student’s question. Last week, a student asked me about proving that every path-connected space is connected–at least, I think that’s what he asked. I asked if he had tried to prove it, to which he responded “not yet.” Not wishing to deprive him of self-discovery and being a tad lazy, I told him to try on his own and get back to me if he ran into trouble.
As I have not heard back from him on this matter, I assume that he proved the result on his own. To make up for my laziness, I have decided to write a short exposition of the topological property of connectedness. We begin by reviewing some definitions.
We say a topological space is connected if the only nonempty open and closed subset of is itself. If are two points in , then a path between and is a continuous function such that and . If every two points in are joined by a path, then we say that is path-connected.
There are a number of equivalent definitions of a connected space, and we briefly list some here:
- The only subsets of both open and closed are .
- cannot be written as the union of two disjoint, nonempty open sets.
- cannot be written as the union of two disjoint, nonempty closed sets.
- cannot be written as the union of two separated sets.
- If is continuous, then is constant.
What are some examples of connected spaces? Your intuition might suggest to you that the closed unit interval is connected, and indeed, your intuition would be correct. To see this, recall that an equivalent definition of a connectedness space is that the only subsets, which are both open and closed, are and . Let $E$ be a nonempty subset of which is both open and closed. Define real number and by
I claim that . Note that since is closed, both and belong to . If , then since is open, there exists such that . Taking smaller if necessary, we may assume that . By definition of , there exists a point which does not belong to (in particular, ). We see that since this interval is a subset of . Similarly, since this interval is also a subset of , by choice of . We arrive at a contradiction.
A completely analogous argument shows that , and we conclude that .
We can prove a more general result concerning connected subsets of the real line. This result can be found in Rudin, Principles of Mathematical Analysis Theorem 2.47.
Lemma 1. is connected if and only if
Proof. Suppose there exists (w.l.o.g. ) and . We can then write as the union of two disjoint open subsets and defined by
and so these two sets are nonempty. Hence, is the union of nonempty, disjoint open sets, which implies that is not connected.
Suppose $E$ is not connected. Then there exist two disjoint, open (with respect to the subspace topology) sets and such that . Choose and , where w.l.o.g. . Define a point by
Denote the closures of and in by and , respectively. It is clear that and therefore by definition of the subspace topology. If , then and , since and partition . If , then , so there exists such that . Choosing in this interval, we obtain a point which is in neither nor . Hence, .
“Path-connected implies connected” follows readily from the lemma.
Proposition 2. If is a path-connected topological space, then it is also connected.
Proof. Suppose there exist two nonempty, disjoint open sets and such that
Let be points in and , respectively. By hypothesis that is path-connected, there exists a continuous function such that and . By continuity, the pre-images and are open. But then is the union of two disjoint, nonempty open sets since
which contradicts the connectedness of .
The converse is not true, and there are two well-known examples of connected topological spaces which are not path-connected: the deleted comb space and the topologist’s sine curve. We will only cover the second example. Before doing so, I want to describe some properties of connected spaces.
First, continuity preserves connectedness and path-connectedness.
Proposition . If and are topological spaces, where is connected (path-connected), and is continuous, then is a connected (path-connected) subspace of .
Proof. Let be a nonempty subset of which is both open and closed in equipped with the subspace topology. By definition of the subspace topology, we can write
where is open in and . Since is continuous, and are open in , so that is open in . Similarly, and are closed in , so that is closed in . Since is connected, we must have that , which implies that .
Now suppose that is path-connected. Fix two points and in , and let be a path between and . Then is a continuous function between and .
The following proposition is a generalization of an elementary result in Calculus known as the intermediate value theorem.
Proposition . Suppose is a connected topological space, and is continuous. If , for two points , then there exists a point such that .
Proof. Suppose there exist and some between and (w.l.o.g. ), such that . Since is continuous, the two pre-images
form a partition of into two nonempty, open sets. But this contradicts our hypothesis that is connected.
Let’s go back to the topologist’s sine curve. Consider endowed with the standard topology. We define the topologist’s sine curve by taking the set
and endowing it with the subspace topology. What topological properties does have? You might be tempted to say that is closed in , but this is incorrect; is neither open nor closed. The first claim is obvious, so consider the second. For any , there exists some such that . Since is -periodic, for any ,
Hence, is a limit point of that does not belong to .
If we adjoin the set to , then we obtain the closed topologist’s sine curve. Indeed, let be a point . If , then it’s clear that there exists an open disk about not intersecting , so assume and therefore . Since is continuous on , there exists such that
For , implies that
which is a contradiction. Hence, the open square does not intersect . Since is also bounded, the Heine-Borel theorem implies that is compact.
To see that is connected, let be a nonempty subset, which is both open and closed. Suppose . Since is also open, there exists satisfying . Define
Since is closed, . There exists . Otherwise , which implies that , since it is an open and closed set. Define
Since is closed, . Since is open, there exists such that
But by definition of , for any , there exists such that , which is a contradiction.
To show that is not path-connected, we will prove that does not contain a path from to . Assume the contrary, and let denote such a path. Since is continuous, there exists such that
Let denote the projection of onto the first cooridnate. Being the composition of continuous functions, is a continuous function . Choose sufficiently large so that
By the intermediate value theorem, there exists a point such that
But, , which is a contradiction.