In this blog post, we use connectedness and compactness to characterize the unit interval and the unit circle up to homeomorphism. At the heart of the argument is the idea used in Exercise 24.1 (p. 157) of Munkres’s Topology to distinguish from , looking at when removal of a point disconnects a space. We use the following terminology.
Definition. Let be a connected space. A point is called a cut point when is not connected. A point of that is not a cut point is a non-cut point.
Recall that a topological space is said to be second countable when there is a countable basis for its open sets. For example, it is not too hard to show that compact metric spaces are second countable. We will prove the following theorems.
Theorem 1. A second countable connected compact space with exactly two non-cunt points is homeomorphic to the unit interval.
Theorem 2. Let be a second countable connected compact space such that for any two distinct points and , is disconnected. Then is homeomorphic to the unit circle.
Partial Orders and the Maximum Principle
The unit interval has an order and the order gives the basis for its topology. On the other hand, thinking in terms of how removing a cut point disconnects the interval, we can also recover the order from the topology (up to reversal). So it should be no surprise that the concept of order should be important for the proof of Theorem 1. It will also be useful to have the concept of a partial order.
Definition. Let be a set. A strict partial order on is a relation “” such that:
(i) There is no with .
(ii) If and , then .
We write if or and call the relation “” a partial order. We say that are comparable if either or . A partial order is called a simple order if every two elements of are comparable.
If we have a partial order on , then any subset of inherits a partial order. Even if is not simply ordered, a subset of may be, for example a one element subset will be. We say that is a maximal simply ordered subset if the partial order on is a simple order and there is no simply ordered subset of properly containing . The following is a fundamental principle of set theory.
The Maximum Principle. If is a partially ordered set and is simply ordered, then there is a maximal simply ordered subset with .
The Maximum Principle is equivalent to the famous Axiom of Choice. It is sometimes taken as a convenient replacement for the Axiom of Choice, and this is the approach we take here. In particular, we will not try to prove it. Here is an intuitive argument to make it seem plausible, however. Starting with , build a little at a time as follows. Suppose we have simply ordered. If there is no element of that can be added to while retaining a simple order, then is maximal. If there is such an element, then choose one, say , and let . Keep doing this until we get a maximal simply ordered subset or we use up all the elements of . (If you know about “transfinite induction” another principle equivalent to (but closer to) the Axiom of Choice, you can make this argument precise.)
Do not confuse the notion of “maximal simply ordered subset” with the notion of a “maximal element.” A maximal element of a partially ordered set is an element such that there is no element with . Note that if a simply ordered set has a maximal element, then it is unique.
One final concept we will need is that of “order isomorphism.”
Definition. Let and be partially ordered sets. A function is called an order isomorphism when is a bijection and in if and only if in .
Cut Points and Partial Orders
The key lemma that we will need to prove everything is very intuitive in statement, but difficult to prove. The proof occupies the whole of this section.
Lemma. Let be a connected compact space with more than one point. Then has at least two non-cut points.
The proof of this lemma requires the Maximum Principle described the previous section. We will apply it with the following kinds of partial orders.
Definition. Let be a connected space, and let be a point of . We define a relation on by if and only if either there is some separation of with and or and .
Proposition. In the notation above, is a strict partial order.
Proof. It is clear that does not satisfy , and for any non-cut point , there is no separation of , and so we never have . Suppose and ; we want to show that . This is clear when ; otherwise and must both be cut points. Choose a separation of with ; it suffices to show that is in . Choose a separation of with and . Since and is connected by Exercise 23.12 (p. 152) of Munkres, we must have either contained in or contained in . In fact, it is contained in since both contain ; in particular, and . Since is connected, it is contained in , since both contain . So .
Clearly is a minimal element of for and a point other than is a non-cut point if and only if it is a maximal element for . Before proving the lemma, we need the following observation, which we actually proved as part of the argument above.
Observation. Suppose , and let be any separation of with and . Then any with is in . Furthermore if is a cut point and is a separation of with , then and .
Proof of Lemma. Let . It suffices to show that there is a non-cut point in . Let be a maximal simply ordered subset of . We will show that contains a non-cut point.
Let be the set of cut points in ; we need to show that is nonempty. IF is empty then we are done since must contain at least one element other than . Likewise, if has a maximal element , we are done, since cannot be maximal in (as is a cut-point), and so there must be an element with , which means . So consider the case when is nonempty and has no maximal element. Then for each element , we can choose some element of with , and we can choose a separation of with and . Let . Then since is , is open in , so is open in , and is closed in .
We next show that the collection of closed sets satisfies the finite intersection hypothesis. Given any finite subcollection , renumbering if necessary, we can assume without loss of generality that . Choose with . Then by the observation above, we have that for all , and so . In particular is nonempty.
Let . Then is nonempty; choose . Note that since by construction we have for any . We also have since for every , . Then is a simply ordered set properly containing , and so by maximality, cannot be empty.
Proof of Theorem 1
We are now in a position to start work on the proof of Theorem 1. For the remainder of this section denotes a second countable connected compact space with exactly two non-cut points that we will denote as and . We will now write for . We begin with the following observation.
Observation. For and any separation of with must have .
Proof. Let . Then is closed in and so is a connected compact space. If is a point of that is a cut-point of , then must also be a cut point of . Given any separation of , we must have by the observation in the last section. We also have nonempty since it contains . Then is a separation of . On the other hand, we know from the lemma of the last section that has at least two non-cut points. So we must have that is in .
We have the following consequence.
Proposition. is simply ordered. For and any separation of with , every element in satisfies .
Proof. Since from the observation we have that for any , the first statement is an immediate consequence of the second. Now let be as in the statement, and let . If , then ; otherwise is a cut point, and we can choose a separation of . Since and is connected, we must have since these both contain . This shows that .
The previous proposition implies that for the separation of , is exactly the set of elements that are less than and is exactly the set of elements that are greater than . For , write for the set of elements with , write for the set of elements with , and write for the set of elements with . Then we have that the sets , , and are all open in . In fact, these sets form a basis.
Proposition. The sets , , and for form a basis of the topology of .
Proof. Let and let be an open set around . First consider the case when . It suffices to show that contains and for some . For , let denote the set of with . This is a closed set since its complement is open. Consider the collection of nonempty closed sets
If is empty, then contains and we can take . Otherwise is simply ordered by inclusion and so satisfies the finite intersection hypothesis. Let be the intersection of the elements of , and let . We have that , but for any we can not have since then would be in but would not contain . Thus, is contained in . An analogous argument produces a with contained in . The argument for is entirely similar.
The proof of Theorem 1 is now completed by the following proposition.
Proposition. is order isomorphic to the unit interval.
Proof. Choose a countable basis for . Since is , every nonempty open set in is infinite, since otherwise would not be connected. So we can choose elements such that each , and . We begin by choosing an order isomorphism from to the dyadic rational numbers (the rational numbers whose denominator is a power of ) in as follows.
First note that given any two elements in , we can find an element with . If not would be an open subset of not containing a basis element of , and so would have to be empty. Then would be a separation of , which contradicts the assumption that is connected.
Let . Let be the first integers such that and , and define and . let be the first integers with
and define , , , and . And so on. A little thought shows that this is an order isomorphism.
Now we extend to a function as follows. We define and for , we define
Note and . Now if , we can find , with and we can find with , and so we must have . So is strictly order preserving (and hence injective). We need to see that is surjective.
We know , , and all the dyadic rationals in are in the image of , and so consider not a dyadic rational. Let ; since is an order isomorphism, is the set of dyadic rational numbers in and so . Now it suffices to find an such that . Consider the collection of nonempty closed sets
If is a finite subcollection of , then
where and . In particular satisfies the finite intersection hypothesis. Thus, the intersection of the elements of is nonempty; let be an element in this intersection. Then for every , we have that since there is some dyadic rational with , and is in and contains only elements greater than . For every , we have that since there is some dyadic rational with , and for any , is in and contains only elements less than . Thus, , and .
Proof of Theorem 2
In this section, denotes a space that satisfies the hypotheses of Theorem 2. First we will show that every point of is a non-cut point.
Proposition. Every point of is a noncut point.
Proof. Suppose had a cut point . Let be a separation of . Then and are compact connected spaces, and so we can find a non-cut point of and a non-cut point of . Note . Now , the union of non-disjoint connected sets. This contradicts that is not connected.
Now choose distinct points in , and let be a separation of . Let and let , compact connected spaces. We have that and are non-cut points of and and is formed by gluing and along . So it suffices to show that and are each homeomorphic to the unit interval.
First we will show that one of or is homeomorphic to the unit interval. If neither of them are homeomorphic to the unit interval, then we can find a non-cut point in and a non-cut point in , and then would be the union of the non-disjoint connected sets and , contradicting the assumption that is not connected.
Now we can assume without loss of generality that is homeomorphic to the unit interval. If were not homeomorphic to the unit interval, then we could find a non-cut point in . Then for any point in , we can write as the union of disjoint nonempty open sets with and . Then is the union of the pairwise non-disjoint connected sets , contradicting the assumption that is not connected.