# $S_\infty$ is a non-locally compact Polish group (Kechris)

Here is example $$7)$$, pp. $$59$$ of Kechris' book "Classical Descriptive Set Theory":

Let $$S_\infty$$ be the group of permutations of $$\mathbb{N}$$. With the relative topology as a subset of $$\mathcal{N}=\Bbb{N}^{\Bbb{N}}$$ (Baire space), it is a topological group and it is a Polish group since $$S_\infty$$ is a $$G_\delta$$ set in $$\mathcal{N}$$ ... Again, $$S_\infty$$ is not locally compact.

I want to prove the above facts. This is my attempting:

(i) $$S_\infty$$ is $$G_\delta$$: let $$INJ=\{(x_n)\mid injective\}$$ and $$SURJ=\{(x_n)\mid surjective\}$$. Then clearly $$S_\infty=INJ\cap SURJ.$$ Let us prove that they are $$G_\delta$$:

$$x\in SURJ\iff\forall n\exists m\colon=x(m)$$ and hence $$SURJ=\bigcap_n\bigcup_m \Bbb{N}^{n-1}\times \{n\}\times \Bbb{N}^{\Bbb{N}-n}.$$ Moreover $$x\in INJ\iff \forall n,m (n\ne m\implies x_n\ne x_m)\iff \forall n,x_n\in \Bbb{N}^{\Bbb{N}}\setminus (\bigcup_{i=1}^{n-1}\pi_i(i))$$ open, where $$\pi_i\colon \Bbb{N}^{\Bbb{N}}\to \Bbb{N}$$ is the $$i$$th-projection. Hence $$INJ=\bigcap_n [\Bbb{N}^\Bbb{n-1}\times (\Bbb{N}\setminus \bigcup_{i=1}^{n-1}\pi_i(i))\times \Bbb{N}^\Bbb{N}-n]$$

(ii) $$S_\infty$$ is a topological group: the compatible metric is $$d(x,y)=\frac{1}{2^{n+1}}$$, where $$n=\mathrm{inf}\{m\mid x(m)\ne y(m)\}$$. Then $$B(x,2^{-n})=\{y\mid x(i)=y(i) \forall i\le n,n\ge m-1\}$$ and the result follows.

(iii) $$S_\infty$$ is not locally compact: I have no successful idea.

• Re: (iii), do you know how to show that $\mathcal{N}$ itself is not locally compact? – Noah Schweber Mar 30 at 19:23
• @NoahSchweber Kechris states that $\mathcal{N}$ is not locally compact at pp. $29$. I tried in the following way: assume by contradiction the existence of $(x_n)$ with compact nhbd $K$. Then every open nhbd $U_1\times\dots\times U_n\times \mathbb{N}^{\mathbb{N}-n}$ contained in $K$ gas compact closure. But its image under the continuous function $\pi_{n+1}$ is $\mathbb{N}$ which is not compact. This is an argument I found some time ago, but probably there is a standard way to do this And however I would like to know other methods. What's your argument? – LBJFS Mar 31 at 8:43

First, $$\mathbf{N}^{\mathbf{N}}$$ is a topological monoid. It has the complete metric $$d(f,g)=\exp(-N(f,g))$$ where $$N(f,g)=\inf\{n:f(n)\neq g(n)\}$$.

The group $$S(\mathbf{N})$$ (often denoted by $$S_\infty$$ but don't recommend this notation since it hides the set on which it acts) has the induced topology, which is a group topology. This topology was initially introduced by L. Onofri (1927), and was apparently rediscovered in the 50's.

A basis of closed neighborhoods of the identity $$\mathrm{id}_\mathbf{N}$$ is $$(V_n)$$, where $$V_n$$ is the (clopen) subgroup of permutations that are identity on $$\mathbf{N}_{\le n}$$.

Beware that $$S(\mathbf{N})$$ is not closed in $$\mathbf{N}^{\mathbf{N}}$$. However, the embedding $$S(\mathbf{N})\to\mathbf{N}^{\mathbf{N}}\times\mathbf{N}^{\mathbf{N}}$$, $$g\mapsto (g,g^{-1})$$ is a homeomorphism onto its closed image. This yields a complete metric on $$S(\mathbf{N})$$ defining the topology. Thus, $$S(\mathbf{N})$$ is a Polish space, hence is Baire.

$$S(\mathbf{N})$$ is not compact, since the set of $$\sigma_n:k\mapsto k+n$$ is an infinite closed discrete subset. Hence $$V_n$$ is not compact of any $$n$$, and hence $$S(\mathbf{N})$$ is not locally compact.

Added: here's a less direct argument, but rather of measure-theoretic flavor, of the failure of local compactness:

Fact: let $$\mathcal{U}$$ be the Boolean algebra of clopen subsets of $$S(\mathbf{N})$$. For every left-invariant finitely additive measure $$\mu:\mathcal{U}\to [0,\infty]$$, we have $$\mu(V_n)\in\{0,\infty\}$$ for all $$n$$ but at most 1 exception.

Since $$(V_n)$$ is a basis of open neighborhoods of 1, this clearly contradicts the existence of a left-invariant Haar measure.

The fact holds because $$V_{n+1}$$ has infinite index in $$V_n$$ and thus $$V_n$$ contains infinitely many pairwise disjoint left translates (=left cosets) of $$V_{n+1}$$. So, for every $$n$$, $$\mu(V_n)<\infty$$ implies $$\mu(V_{n+1})=0$$, which implies the fact.

An exercise: let $$G$$ be a closed subgroup of $$S(\mathbf{N})$$. Show that $$G$$ is locally compact if and only if there exists $$n$$ such that $$G\cap V_n$$ has only finite orbits on $$\mathbf{N}$$.

Edit: about $$\mathsf{INJ}$$ and $$\mathsf{SURJ}$$ being $$G_\delta$$:

$$\mathsf{INJ}=\bigcap_{n

$$\mathsf{SURJ}=\bigcap_nB_n^c;\quad B_n=(\mathbf{N}\smallsetminus\{n\})^{\mathbf{N}};$$

and each of $$A_{n,m}$$, $$B_n$$ is closed. (Note: $$B_n$$ has empty interior, and hence $$\mathsf{SURJ}$$ is $$G_\delta$$-dense in $$\mathbf{N}^{\mathbf{N}}$$. However $$A_{n,m}$$ is clopen, and clearly $$\mathsf{INJ}$$ is not dense in $$\mathbf{N}^{\mathbf{N}}$$; it's even closed.)

• I was wondering if the following works: if $S$ were locally-compact, it would have a left-invariant Haar measure. Also, there would be some compact closed ball $B$ around the identity, therefore of finite measure. The distance chosen seems to be left-invariant. Now I would try to show that $B$ contains a countable disjoint union of translates of some smaller ball (itself of non-zero measure), contradicting the finiteness of the measure of $B$. This is essentially the same idea as the one used to show that there are no translation invariant measures on infinite-dimensional Banach spaces. – Alex M. Mar 30 at 21:09
• In a Banach space, though, I have independent dimensions ("axes") on which to place the smaller balls - which I don't seem to be able to mimick here. – Alex M. Mar 30 at 21:10
• @AlexM. actually the measure argument to prove the failure of local compactness works here and is even easier than the real Banach case, using the basis of neighborhoods of 1 made of subgroups. I added this argument. – YCor Mar 30 at 21:32
• @LBJFS yes, the metric I defined defines the Polish topology on $S(N)$. However, it is not a complete metric on $S(N)$, since it's not closed. A complete metric defining the same topology is given in the post, namely obtained by diagonally embedding $S(N)$ into $N^N$ by $g\mapsto (g,g^{-1})$. – YCor Apr 1 at 15:37
• @LBJFS I don't get your argument, so I added a short paragraph addressing this. – YCor Apr 1 at 16:18