# General Topology – The Interval and the Circle

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 ${\mathbb{R}^n}$ from ${\mathbb{R}}$, looking at when removal of a point disconnects a space. We use the following terminology.

Definition. Let ${X}$ be a connected space. A point ${x \in X}$ is called a cut point when ${X\setminus\{x\}}$ is not connected. A point of ${X}$ 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 ${T_1}$ space with exactly two non-cunt points is homeomorphic to the unit interval.

Theorem 2. Let ${X}$ be a second countable connected compact ${T_1}$ space such that for any two distinct points ${x}$ and ${y}$, ${X \setminus \{x, y\}}$ is disconnected. Then ${X}$ 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 ${A}$ be a set. A strict partial order on ${A}$ is a relation “${\prec}$” such that:

(i) There is no ${a \in A}$ with ${a \prec a}$.
(ii) If ${a \prec b}$ and ${b \prec c}$, then ${a \prec c}$.

We write ${a \preceq b}$ if ${a \prec b}$ or ${a = b}$ and call the relation “${\preceq}$” a partial order. We say that ${a, b \in A}$ are comparable if either ${a \preceq b}$ or ${b \succeq a}$. A partial order is called a simple order if every two elements of ${A}$ are comparable.

If we have a partial order on ${A}$, then any subset of ${A}$ inherits a partial order. Even if ${A}$ is not simply ordered, a subset of ${A}$ may be, for example a one element subset will be. We say that ${B \subset A}$ is a maximal simply ordered subset if the partial order on ${B}$ is a simple order and there is no simply ordered subset of ${A}$ properly containing ${B}$. The following is a fundamental principle of set theory.

The Maximum Principle. If ${A}$ is a partially ordered set and ${B \subset A}$ is simply ordered, then there is a maximal simply ordered subset ${C \subset A}$ with ${B \subset C}$.

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 ${B}$, build ${C}$ a little at a time as follows. Suppose we have ${C' \supseteq B}$ simply ordered. If there is no element of ${A \setminus C'}$ that can be added to ${C'}$ while retaining a simple order, then ${B}$ is maximal. If there is such an element, then choose one, say ${C}$, and let ${C'' = C' \cup \{c\}}$. Keep doing this until we get a maximal simply ordered subset or we use up all the elements of ${A}$. (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 ${a}$ such that there is no element ${b}$ with ${a \prec b}$. 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 ${A}$ and ${B}$ be partially ordered sets. A function ${f: A \rightarrow B}$ is called an order isomorphism when ${f}$ is a bijection and ${a \preceq b}$ in ${A}$ if and only if ${f(a) \preceq f(b)}$ in ${B}$.

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 ${X}$ be a connected compact ${T_1}$ space with more than one point. Then ${X}$ 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 ${X}$ be a connected ${T_1}$ space, and let ${x}$ be a point of ${X}$. We define a relation ${\prec_x}$ on ${X}$ by ${a \prec_x b}$ if and only if either there is some separation ${U, V}$ of ${X \setminus \{a\}}$ with ${x \in U}$ and ${b \in V}$ or ${x = a}$ and ${b \neq x}$.

Proposition. In the notation above, ${\prec_x}$ is a strict partial order.

Proof. It is clear that ${x}$ does not satisfy ${x \prec x}$, and for any non-cut point ${y}$, there is no separation of ${X \setminus \{y\}}$, and so we never have ${a \prec_x a}$. Suppose ${a \prec_x b}$ and ${b \prec_x c}$; we want to show that ${a \prec_x c}$. This is clear when ${a = x}$; otherwise ${a}$ and ${b}$ must both be cut points. Choose a separation ${U_a, V_a}$ of ${X \setminus \{a\}}$ with ${x \in U_a, b \in V_a}$; it suffices to show that ${c}$ is in ${V_a}$. Choose a separation ${U_b, V_b}$ of ${X\setminus\{b\}}$ with ${x \in U_b}$ and ${c \in V_b}$. Since ${b \notin U_a \cup \{a\}}$ and ${U_a \cup \{a\}}$ is connected by Exercise 23.12 (p. 152) of Munkres, we must have either ${U_a \cup \{a\}}$ contained in ${U_b}$ or contained in ${V_b}$. In fact, it is contained in ${U_b}$ since both contain ${x}$; in particular, ${a \in U_b}$ and ${a \notin V_b \cup \{b\}}$. Since ${V \cup \{b\}}$ is connected, it is contained in ${V_a}$, since both contain ${b}$. So ${c \in V_b \subset V_a}$.

$\square$

Clearly ${x}$ is a minimal element of ${X}$ for ${\prec_x}$ and a point other than ${x}$ is a non-cut point if and only if it is a maximal element for ${\prec_c}$. Before proving the lemma, we need the following observation, which we actually proved as part of the argument above.

Observation. Suppose ${x \prec_x a \prec_x b}$, and let ${U_a, V_a}$ be any separation of ${X \setminus \{a\}}$ with ${x \in U_a}$ and ${b \in V_a}$. Then any ${c}$ with ${b \preceq_x c}$ is in ${V_a}$. Furthermore if ${c}$ is a cut point and ${U_c, V_c}$ is a separation of ${X \setminus \{c\}}$ with ${x \in U_c}$, then ${a \in U_c}$ and ${V_c \subset V_a}$.

Proof of Lemma. Let ${x \in X}$. It suffices to show that there is a non-cut point in ${X \setminus \{x\}}$. Let ${S}$ be a maximal simply ordered subset of ${X}$. We will show that ${S}$ contains a non-cut point.

Let ${S'}$ be the set of cut points in ${S}$; we need to show that ${S \setminus \left(\{x\} \cup S'\right)}$ is nonempty. IF ${S'}$ is empty then we are done since ${S}$ must contain at least one element other than ${x}$. Likewise, if ${S'}$ has a maximal element ${y}$, we are done, since ${y}$ cannot be maximal in ${S}$ (as ${y}$ is a cut-point), and so there must be an element ${z \in S}$ with ${y \prec_x z}$, which means ${z \in S\setminus \left(S' \cup \{x\}\right)}$. So consider the case when ${S'}$ is nonempty and has no maximal element. Then for each element ${a \in S'}$, we can choose some element ${b_a}$ of ${S'}$ with ${a \prec_x b_a}$, and we can choose a separation ${U_a, V_a}$ of ${X \setminus \{a\}}$ with ${x \in U_a}$ and ${b_a \in V_a}$. Let ${C_a = V_a \cup \{a\}}$. Then since ${X}$ is ${T_1}$, ${X \setminus \{a\}}$ is open in ${X}$, so ${U}$ is open in ${X}$, and ${C_a = X \setminus U_a}$ is closed in ${X}$.

We next show that the collection of closed sets ${\{C_a:a \in S'\}}$ satisfies the finite intersection hypothesis. Given any finite subcollection ${C_{a_1}, \dots, C_{a_n}}$, renumbering if necessary, we can assume without loss of generality that ${b_{a_1} \prec_x b_{a_2} \prec_x \dots \prec_x b_{a_n}}$. Choose ${c \in S'}$ with ${b_n \prec_x c}$. Then by the observation above, we have that ${V_c \subset V_{a_i}}$ for all ${i}$, and so ${V_c \subset C_{a_1} \cap \dots \cap C_{a_n}}$. In particular ${C_{a_1} \cap \dots \cap C_{a_n}}$ is nonempty.

Let ${C = \bigcap_{a \in S'} C_a}$. Then ${C}$ is nonempty; choose ${y \in C}$. Note that ${y \neq x}$ since by construction we have ${a \preceq y}$ for any ${a \in S'}$. We also have ${y \notin S}$ since for every ${a \in S}$, ${a \prec_x b_a \preceq_x y}$. Then ${\{x\} \cup S'\cup \{y\}}$ is a simply ordered set properly containing ${\{x\} \cup S'}$, and so by maximality, ${S\setminus\left(\{x\} \cup S'\right)}$ 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 ${X}$ denotes a second countable connected compact ${T_1}$ space with exactly two non-cut points that we will denote as ${p}$ and ${q}$. We will now write ${\prec}$ for ${\prec_p}$. We begin with the following observation.

Observation. For ${a \in X \setminus \{p, q\}}$ and any separation ${U, V}$ of ${X \setminus \{a\}}$ with ${p \in U}$ must have ${p \in V}$.

Proof. Let ${C = V \cup \{a\}}$. Then ${C}$ is closed in ${X}$ and so is a connected compact ${T_1}$ space. If ${z \neq a}$ is a point of ${C}$ that is a cut-point of ${X}$, then ${z}$ must also be a cut point of ${C}$. Given any separation ${U_z, V_z}$ of ${X \setminus \{z\}}$, we must have ${V_z \subset V}$ by the observation in the last section. We also have ${U_z \cap C }$ nonempty since it contains ${a}$. Then ${U_z \cap C, V_z}$ is a separation of ${C \setminus \{z\}}$. On the other hand, we know from the lemma of the last section that ${C}$ has at least two non-cut points. So we must have that ${q}$ is in ${C}$.

We have the following consequence.

Proposition. ${X}$ is simply ordered. For ${a \in X\setminus\{p, q\}}$ and any separation ${U, V}$ of ${X \setminus \{a\}}$ with ${p \in U}$, every element ${b}$ in ${U}$ satisfies ${b \prec a}$.

Proof. Since from the observation we have that ${p \prec a \prec q}$ for any ${a \in X\setminus\{p, q\}}$, the first statement is an immediate consequence of the second. Now let ${a, U, V}$ be as in the statement, and let ${b \in U}$. If ${b = p}$, then ${b \prec a}$; otherwise ${b}$ is a cut point, and we can choose a separation ${U_b, V_b}$ of ${X \setminus\{b\}}$. Since ${b \notin V}$ and ${V \cup\{a\}}$ is connected, we must have ${V \cup \{a\} \subset V_b}$ since these both contain ${q}$. This shows that ${b \prec a}$.

The previous proposition implies that for the separation ${U, V}$ of ${X \setminus \{a\}}$, ${U}$ is exactly the set of elements that are less than ${a}$ and ${V}$ is exactly the set of elements that are greater than ${a}$. For ${a, b \in X}$, write ${(a, b)}$ for the set of elements ${c \in X}$ with ${a \prec c \prec b}$, write ${[p, b)}$ for the set of elements ${c \in X}$ with ${p \preceq c \prec b}$, and write ${(a, q]}$ for the set of elements ${c \in X}$ with ${a \prec c \preceq q}$. Then we have that the sets ${(a, b)}$, ${[p, b)}$, and ${(a, q]}$ are all open in ${X}$. In fact, these sets form a basis.

Proposition. The sets ${(a, b)}$, ${[p, b)}$, and ${(a, q]}$ for ${p \prec a \prec b \prec q}$ form a basis of the topology of ${X}$.

Proof. Let ${y \in X}$ and let ${U}$ be an open set around ${y}$. First consider the case when ${y \neq p, q}$. It suffices to show that ${U}$ contains ${(x, y]}$ and ${[y, z)}$ for some ${x \prec y \prec z}$. For ${a \prec b}$, let ${[a, b]}$ denote the set of ${c \in X}$ with ${a \preceq c \preceq b}$. This is a closed set since its complement ${[p, q) \cup (b, q]}$ is open. Consider the collection of nonempty closed sets

$\displaystyle \mathscr{C} = \{[a, y]\setminus U: a \prec y\text{ and }[a, y]\setminus U\text{ is nonempty}\}.$

If ${\mathscr{C}}$ is empty, then ${U}$ contains ${[p, y]}$ and we can take ${x = p}$. Otherwise ${\mathscr{C}}$ is simply ordered by inclusion and so satisfies the finite intersection hypothesis. Let ${C}$ be the intersection of the elements of ${\mathscr{C}}$, and let ${x \in C}$. We have that ${x \in X \setminus U}$, but for any ${a \succ x}$ we can not have ${a \in X \setminus U}$ since then ${[a, y] \setminus U}$ would be in ${\mathscr{C}}$ but would not contain ${x}$. Thus, ${(x, y]}$ is contained in ${U}$. An analogous argument produces a ${z \succ y}$ with ${[y, z)}$ contained in ${U}$. The argument for ${y = p, q}$ is entirely similar.

$\square$

The proof of Theorem 1 is now completed by the following proposition.

Proposition. ${X}$ is order isomorphic to the unit interval.

Proof. Choose a countable basis ${\mathscr{B} = \{B_1, B_2, \dots\}}$ for ${X}$. Since ${X}$ is ${T_1}$, every nonempty open set in ${X}$ is infinite, since otherwise ${X}$ would not be connected. So we can choose elements ${b_1, b_2, \dots}$ such that each ${b_n \in B_n}$, and ${b_n \notin \{p, q, b_1, \dots, b_{n-1}\}}$. We begin by choosing an order isomorphism ${f}$ from ${\{b_1, b_2, \dots\}}$ to the dyadic rational numbers (the rational numbers whose denominator is a power of ${2}$) in ${(0, 1)}$ as follows.

First note that given any two elements ${x \prec y}$ in ${X}$, we can find an element ${b_n}$ with ${x \prec b_n \prec y}$. If not ${(x, y)}$ would be an open subset of ${X}$ not containing a basis element of ${\mathscr{B}}$, and so would have to be empty. Then ${[p, y), (x, q]}$ would be a separation of ${X}$, which contradicts the assumption that ${X}$ is connected.

Let ${f(b_1) = 1/2}$. Let ${n_1, n_2}$ be the first integers such that ${b_{n_1} \prec b_1}$ and ${b_1 \prec b_{n_2}}$, and define ${(b_{n_1}) = 1/4}$ and ${f(b_{n_2}) = 3/4}$. let ${n_3, n_4, n_5, n_6}$ be the first integers with

$\displaystyle b_{n_3} \prec b_{n_1} \prec b_{n_4} \prec b_1 \prec b_{n_5} \prec b_{n_2} \prec b_{n_6},$

and define ${f(b_{n_3}) = 1/8}$, ${f(b_{n_4}) = 3/8}$, ${f(b_{n_5}) = 5/8}$, and ${f(b_{n_6}) = 7/8}$. And so on. A little thought shows that this is an order isomorphism.

Now we extend ${f}$ to a function ${g: X \rightarrow [0, 1]}$ as follows. We define ${g(p) = 0}$ and for ${x \in X \setminus \{p\}}$, we define

$\displaystyle g(x) = \sup\{f(b_n): b_n \preceq x\}.$

Note ${g(b_n) = f(b_n)}$ and ${g(q) = 1}$. Now if ${x \prec y}$, we can find ${b_m}$, with ${x \prec b_m \prec y}$ and we can find ${b_n}$ with ${bm \prec b_n \prec y}$, and so we must have ${g(x) < g(y)}$. So ${g}$ is strictly order preserving (and hence injective). We need to see that ${g}$ is surjective.

We know ${0}$, ${1}$, and all the dyadic rationals in ${(0, 1)}$ are in the image of ${g}$, and so consider ${r \in (0, 1)}$ not a dyadic rational. Let ${S = \{b_n : f(b_n) < r\}}$; since ${f}$ is an order isomorphism, ${f(S)}$ is the set of dyadic rational numbers in ${(0, r)}$ and so ${r = \sup\{f(b_n) : b_n \in S\}}$. Now it suffices to find an ${x \in X}$ such that ${S = \{b_n : b_n \prec x\}}$. Consider the collection of nonempty closed sets

$\displaystyle \mathscr{C} = \{[b_n, y]: b_n \in S\text{ and }b_m \prec y\text{ for all }b_m \in S\}.$

If ${[b_{n_1}, y_1], \dots, [b_{n_k}, y_k]}$ is a finite subcollection of ${\mathscr{C}}$, then

$\displaystyle [b_{n_1}, y_1] \cap \dots \cap [b_{n_k}, y_k] = [b_{n_i}, y_j]$

where ${b_{n_i} = \max\{b_{n_1}, \dots, b_{n_k}\}}$ and ${y_j = \max\{y_1, \dots, y_k\}}$. In particular ${\mathscr{C}}$ satisfies the finite intersection hypothesis. Thus, the intersection of the elements of ${\mathscr{C}}$ is nonempty; let ${x}$ be an element in this intersection. Then for every ${b_n \in S}$, we have that ${b_n \prec x}$ since there is some dyadic rational ${s}$ with ${f(b_n) < s < r}$, and ${\left[f^{-1}(s), q\right]}$ is in ${\mathscr{C}}$ and contains only elements greater than ${b_n}$. For every ${b_n \notin S}$, we have that ${x \prec b_n}$ since there is some dyadic rational ${s}$ with ${r < s < f(b_n)}$, and for any ${b_m \in S}$, ${\left[b_m, f^{-1}(s)\right]}$ is in ${\mathscr{C}}$ and contains only elements less than ${b_n}$. Thus, ${S = \{b_n: b_n \prec x\}}$, and ${g(x) = r}$.

$\square$

Proof of Theorem 2

In this section, ${X}$ denotes a space that satisfies the hypotheses of Theorem 2. First we will show that every point of ${X}$ is a non-cut point.

Proposition. Every point of ${X}$ is a noncut point.

Proof. Suppose ${X}$ had a cut point ${a}$. Let ${U, V}$ be a separation of ${X\setminus \{a\}}$. Then ${C = U \cup \{a\}}$ and ${D = V \cup \{a\}}$ are compact connected ${T_1}$ spaces, and so we can find a non-cut point ${y \neq a}$ of ${C}$ and a non-cut point ${z \neq a}$ of ${D}$. Note ${y \neq z}$. Now ${X \setminus \{y, z\} = \left(C \setminus \{y\}\right) \cup \left(D \setminus \{z\}\right)}$, the union of non-disjoint connected sets. This contradicts that ${X \setminus \{y, z\}}$ is not connected.

$\square$

Now choose distinct points ${p, q}$ in ${X}$, and let ${U, V}$ be a separation of ${X \setminus \{p, q\}}$. Let ${C = U \cup \{p, q\}}$ and let ${D = V \cup \{p, q\}}$, compact connected ${T_1}$ spaces. We have that ${p}$ and ${q}$ are non-cut points of ${C}$ and ${D}$ and ${X}$ is formed by gluing ${C}$ and ${D}$ along ${\{p, q\}}$. So it suffices to show that ${C}$ and ${D}$ are each homeomorphic to the unit interval.

First we will show that one of ${C}$ or ${D}$ is homeomorphic to the unit interval. If neither of them are homeomorphic to the unit interval, then we can find a non-cut point ${y \neq p, q}$ in ${C}$ and a non-cut point ${z \neq p, q}$ in ${D}$, and then ${X \setminus\{y, z\}}$ would be the union of the non-disjoint connected sets ${C \setminus \{y\}}$ and ${D \setminus \{z\}}$, contradicting the assumption that ${X \setminus \{x, y\}}$ is not connected.

Now we can assume without loss of generality that ${C}$ is homeomorphic to the unit interval. If ${D}$ were not homeomorphic to the unit interval, then we could find a non-cut point ${z}$ in ${D}$. Then for any point ${y \neq p, q}$ in ${C}$, we can write ${C}$ as the union of disjoint nonempty open sets ${U, V}$ with ${p \in U}$ and ${q \in V}$. Then ${X \setminus \{y, z\}}$ is the union of the pairwise non-disjoint connected sets ${U, D \setminus\{z\}, V}$, contradicting the assumption that ${X \setminus \{y, z\}}$ is not connected.