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.

  • $\begingroup$ Re: (iii), do you know how to show that $\mathcal{N}$ itself is not locally compact? $\endgroup$ – Noah Schweber Mar 30 at 19:23
  • $\begingroup$ @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? $\endgroup$ – 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 $G_\delta$:

$$\mathsf{INJ}=\bigcap_{n<m}A_{n,m}^c;\quad A_{n,m}=\{f\in\mathbf{N}^{\mathbf{N}}:f(n)=f(m)\};$$

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

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.)

  • $\begingroup$ 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. $\endgroup$ – Alex M. Mar 30 at 21:09
  • $\begingroup$ 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. $\endgroup$ – Alex M. Mar 30 at 21:10
  • $\begingroup$ @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. $\endgroup$ – YCor Mar 30 at 21:32
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    $\begingroup$ @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 $g\mapsto (g,g^{-1})$. $\endgroup$ – YCor Apr 1 at 15:37
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    $\begingroup$ @LBJFS I don't get your argument, so I added a short paragraph addressing this. $\endgroup$ – YCor Apr 1 at 16:18

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