A friend asked me to help him prove that the topological group $\mathrm{Homeo}(0,1)$ (homeomorphism of $(0,1)$ with the compact open topology) is Polish (that is, separable and completely metrizable).

Denote $X=(0,1)$.

We know that for a compact and metrizable space $Y$, the group $\mathrm{Homeo}(Y)$ is Polish. We tried to apply a similar proof to the case of $X$.

The problem starts with separability. $\mathrm{Homeo}(Y)$ inherits its separability from $C(Y,Y)$ (the space of all continuous function). But when we loose compactness, we can't see how one can show that $C(X,X)$ is separable.

So we decided to ask for help:

for $X=(0,1)$

  1. Is $C(X,X)$ separable?

  2. Is $\mathrm{Homeo}(X)$ Polish?

Hints and/or counterexamples will be highly appreciated!

Thank you!


We thought to use the theorem of Birkhoff-Kakutani, according to which in order to prove metrizability it is enough to show that the identity element has a countable nbhd basis. But it only gives us metrizability, not complete metrizability. And even then, suppose that we know that there is some metric that makes this group completely metrizable, what would this imply about this group with the compact open topology?

  • $\begingroup$ A direct proof for separability without resorting to inheritance seems more rational. For metrizibility it's enough to find a countable base around a point in $\operatorname{Homeo}$. $\endgroup$ Commented Mar 30, 2014 at 12:10
  • $\begingroup$ @user138171 Separability: do you have a particular dense (countable) subset in mind? Metrizability: I will edit the post. $\endgroup$
    – Ludolila
    Commented Mar 30, 2014 at 12:21
  • $\begingroup$ No, but it seems if metrizability is proved, proof for denseness will be simpler in a metric space and it seems more probable for Homeo to be separable; I hardly remember some approximation theorems in Analysis which included an uncountable number of approximation functions for (uniformly) continuous functions. $\endgroup$ Commented Mar 30, 2014 at 12:33

3 Answers 3


Since $X$ is a metric space, the compact-open topology on $C(X,X)$ coincides with the topology of compact convergence. With the latter viewpoint, it is in some respects easier to treat.

We choose a normal exhaustion of $X$ by compact sets, say $K_n = \left[3^{-n},1-3^{-n}\right]$ for $n \geqslant 1$, and since $X$ is homeomorphic to $\mathbb{R}$, we see that $C(X,X)$ is homeomorphic to a (closed) subspace of

$$\prod_{n=1}^\infty C(K_n,\mathbb{R}).$$

By Weierstraß' approximation theorem, the space of polynomials is a dense subspace of $C(K_n,\mathbb{R})$, and it is easy to see that the space of polynomials with rational coefficients is then also dense. Hence $C(K_n,\mathbb{R})$ is separable, therefore also the countable product of such spaces, and since it is metrisable, also its subspace (corresponding to) $C(X,X)$. Since $\mathbb{R}$ is complete in the usual metric, we now have the

Proposition: $C(X,X)$ is a Polish space.

So it remains to see whether $\operatorname{Homeo}(X)$ is a $G_\delta$-set in $C(X,X)$, or equivalently some Polish subspace of that, since by Mazurkiewicz' theorem, a subspace of a Polish space is Polish if and only if it is a $G_\delta$.

Lemma: $\overline{\operatorname{Homeo}(X)} = \operatorname{Mon}(X)$, where $\operatorname{Mon}(X)$ is the set of monotonic functions in $C(X,X)$.

Proof: The pointwise limit of a sequence of monotonically nondecreasing functions is monotonically nondecreasing, and similarly for monotonically nonincreasing functions. Hence $\operatorname{Mon}(X)$ is closed. To see that the set of homeomorphism is dense in the monotonic functions, let $f \in \operatorname{Mon}(X)$, without loss of generality let $f$ be nondecreasing. Choose a compact $K \subset X$ - without loss of generality $K$ is an interval - and $\varepsilon > 0$. Since $K$ is compact, there is an $\eta > 0$ such that $f(K) \subset [2\eta,1-2\eta]$. Choose $\eta < \varepsilon$. On the interval $K$, define $g(x) = f(x) + \eta\cdot x$. Then $g(K) \subset [2\eta,1-\eta]$, and $g$ is strictly increasing on $K$. Extend $g$ to a homeomorphism of $X$ by linear interpolation on $(0,\min K)$ and $(\max K,1)$. On $K$, we have $\lvert g(x) - f(x)\rvert = \eta\cdot x < \eta < \varepsilon$, hence $g\in U(f;K,\varepsilon)$, and the set of homeomorphisms is seen to be dense in $\operatorname{Mon}(X)$.

There are two ways in which a function in $\operatorname{Mon}(X)$ can fail to be a homeomorphism, it can be not injective, hence constant on some subinterval of $X$, or it can be not surjective. The first failure mode is easy to deal with:

Lemma: $\operatorname{SMon}(X)$, the set of strictly monotonic functions in $C(X,X)$ is a $G_\delta$ in $\operatorname{Mon}(X)$ (and hence also a $G_\delta$ in $C(X,X)$).

Proof: For any $0 < a < b < 1$, the set $Co(a,b) := \{ f\in C(X,X) : f\lvert_{[a,b]} \equiv \operatorname{const}\}$ is closed. Since there are only countably many pairs of rational numbers, and every non-injective monotonic function is constant on some non-degenerate interval with rational endpoints,

$$\operatorname{SMon}(X) = \bigcap_{\substack{0 < a < b < 1 \\ a,b \in\mathbb{Q}}} \left(\operatorname{Mon}(X)\setminus Co(a,b)\right)$$

is a $G_\delta$ in $\operatorname{Mon}(X)$.

Now it remains to see that $\operatorname{Homeo}(X)$ is a $G_\delta$ in $\operatorname{SMon}(X)$. A strictly monotonic $f \in C(X,X)$ is a homeomorphism if and only if it is surjective, that is if and only if $\inf f(X) = 0$ and $\sup f(X) = 1$.

For any $a \in X$, the sets $A(a) = \{ f\in C(X,X) : \sup f(X) \leqslant a\}$ and $B(a) = \{ f \in C(X,X) : \inf f(X) \geqslant a\}$ are closed. Hence

$$\operatorname{Homeo}(X) = \bigcap_{a \in X\cap\mathbb{Q}} \left(\operatorname{SMon}(X) \setminus \left(A(a) \cup B(a)\right)\right)$$

is a $G_\delta$ in $\operatorname{SMon}(X)$.

Wrapping it up, we have proved the

Theorem: $\operatorname{Homeo}(X)$ is a Polish space in the compact-open topology.

(There's probably a more elegant proof, but this is what I came up with.)

  • $\begingroup$ First of all: thank you very much! So many new and interesting ideas! If you don't mind, I have a couple of questions. I am not familiar with the notion "normal exhaustion" and also couldn't find the definition on the internet. From what I understand, you need countably many compact sets to cover $X$. Is it $\sigma$-compactness that you are using? Or is there something else you require? Also, can you maybe explain how you deduce the first proposition after the sentence "since $\mathbb R$ is complete in the usual metric..."? $\endgroup$
    – Ludolila
    Commented Mar 31, 2014 at 17:32
  • $\begingroup$ A normal exhaustion of (a locally compact $\sigma$-compact space) $X$ is a sequence of compact sets with $K_n \subset \overset{\Large\circ}{K}_{n+1}$ and $X = \bigcup K_n$. Picking such a sequence instead of an arbitrary sequence on compacta covering $X$ makes some proofs easier, it doesn't matter here, that was just out of habit. The completeness of $\mathbb{R}$ gives the completeness of $C(K_n,\mathbb{R})$, and with the separability, we have the Polishness. A counatble product of Polish spaces is polish, so $\prod C(K_n,\mathbb{R})$ is polish, and thus also the closed subspace $C(X,X)$. $\endgroup$ Commented Mar 31, 2014 at 18:02
  • $\begingroup$ Technically, the embedded image of $C(X,X)$ in the product. But since it is an embedding, the Polishness of $C(X,X)$ follows. $\endgroup$ Commented Mar 31, 2014 at 18:03

I saw this very old post and thought I could provide a rather elementary answer, based on the following fact:

Fact: $\text{Homeo}(0,1)$ is isomorphic, as a topological group, to $\text{Homeo}[0,1]$.

Since $[0,1]$ is compact and metrisable, it follows that $\text{Homeo}[0,1]$ is Polish. Since Polishness is an invariant of topological group isomorphism, the group $\text{Homeo}(0,1)$ is also Polish (these are the only steps in the proof which actually use Polishness).

The proof of the fact above is a longish exercise in point set topology; here's a few of the details/ideas of the proof.

First note that each $F \in \text{Homeo}[0,1]$ has the property that if $x \in (0,1)$ then $F(x) \in (0,1)$; this is a consequence of the fact that a point of $[0,1]$ is contained in $(0,1)$ if and only if that point separates $[0,1]$ into two connected components.

We now obtain a group homomorphism $\text{Homeo}[0,1] \mapsto \text{Homeo}(0,1)$ defined by the restriction map $F \mapsto F \mid (0,1)$; I'll call this the restriction homomorphism. We need to show that this homomorphism is a topological group isomorphism.

Next show the restriction homomorphism is injective: if $F \mid (0,1) = G \mid (0,1)$ then, choosing $x_i \in (0,1)$ converging to $0$ we obtain $F(0)=G(0)$, and similarly $F(1)=G(1)$.

Next show the restriction homomorphism is surjective: every $f \in \text{Homeo}(0,1)$ is the restriction of some $F \in \text{Homeo}(0,1)$. To do this, given a sequence $(x_n)$ in $(0,1)$, convergence to an endpoint can be detected solely by the sequence's behavior in $(0,1)$: this occurs if and only if for every $a \in (0,1)$ there are only finitely many terms of the sequence in one of the two intervals $(0,a)$, $(a,1)$; and the endpoint it converges to is detected by which of those two intervals it has infinitely many terms in.

Next show that the restriction homomorphism $F \mapsto f = F \mid (0,1)$ is continuous with respect to the compact-open topologies. Consider a basis element of $\text{Homeo}(0,1)$ denoted $B_{(0,1)}(C,U)$ where $C \subset (0,1)$ is compact and $U \subset (0,1)$ is open, i.e. $$B_{(0,1)}(C,U) = \{f \in \text{Homeo}(0,1) \mid f(C) \subset U\} $$ The inverse image of this basis element under the restriction homomorphism is literally the corresponding basis element denoted $B_{[0,1]}(C,U)$, which proves continuity.

Finally show that the restriction homomorphism $F \mapsto f = F \mid (0,1)$ is open. For this one considers a basis element $B_{[0,1]}(C,U)$ of $\text{Homeo}[0,1]$, and shows that it's image under the restriction homomorphism is a union of basis elements of $\text{Homeo}(0,1)$. This is probably the most intricate step of the proof. Note that $U-\{0,1\}$ is always an open subset of $(0,1)$, but $C-\{0,1\}$ may or may not be compact. One should break the proof into cases depending on which endpoints $0,1$ are isolated points of the set $C$. For instance, suppose that $0$ is a non-isolated point of $C$, and that $1$ is not a non-isolated point of $C$ (either $1 \not\in C$ or $1 \in C$ and is isolated), and that $F(0)=0$ and $F(1)=1$. Define $C_n = C \cap [1/n,1)$, and then prove that the image of $B_{[0,1]}(C,U)$ under the restriction homomorphism is equal to $\bigcup_{n} B_{(0,1)}(C_n,U-\{0,1\})$. (Proving that $B_{(0,1)}(C_n,U-\{0,1\})$ is contained in the image of $B_{[0,1]}(C,U)$ requires observing that $F[0,1/n] = [0,F(1/n)]$ and $F[1/n,1] = [F(1/n),1]$).

In general, the correct formula for $C_n$ should be one of $C \cap [1/n,1)$, or $C \cap (0,1-1/n)$, or $C \cap [1/n,1-1/n]$, or $C \cap (0,1)$ depending on which of $0$ or $1$ is a non-isolated point of $C$.


I tried to find a short proof, that $\mathrm{Homeo}(\mathbb{R})$, endowed with the topology of compact convergence, is a polish space.

I found this paper, which answers your question as well - by showing, that several topologies of $\mathrm{Homeo}(\mathbb{R})$ coincide: On the Group of Homeomorphisms of the Real Line That Map the Pseudoboundary Onto Itself


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