# Is the usual metric on $\Bbb{N}^\Bbb{N}$ left invariant on $S(\Bbb{N})$?

Let $$\Bbb{N}^\Bbb{N}$$ be the set of all functions $$(x_n\mid n\in\Bbb{N})$$ from $$\Bbb{N}$$ into itself (I identify sequences with their images, as usual). I know this is a metrizable space with compatible metric defined as follows: $$d(x,y)=2^{-n-1}$$ where $$n=\mathrm{inf}\{m\in\Bbb{N}\mid x_m\ne y_m\}$$ (clearly $$d(x,x)=0$$ for every $$x\in\Bbb{N}^\Bbb{N}$$). Denote by $$S(\Bbb{N})$$ the group of permutations on $$\Bbb{N}$$.

In "Classical Descriptive Set Theory" by Kechris, the author states (Birkhoff-Kakutani Theorem (9.1), pp. $$58$$) that

If $$G$$ is metrazable, $$G$$ admits a compatible metric $$d$$ which is left-invariant: $$d(zx,zy)=d(x,y)$$.

Here is my question: is the metric $$d(x,y)=2^{-n-1}$$ defined above left-invariant on $$S(\Bbb{N})$$?

My attempt: the case $$x=y$$ is trivial; if $$x\ne y$$, then I want to prove that $$x_m\ne y_m \iff (zx)_m\ne (zy)_m.$$ This is equivalent to saying that $$x_m=y_m \iff (zx)_m=(zy)_m$$.

As $$(zx)_m=z_{x_m}$$, it sufficies to show that $$x_m=y_m \iff z_{x_m}=z_{y_m}.$$ Then $$(\implies)$$ is clear and the converse follows by the fact that $$x,y,z$$ are bijections. Am I right?

• You are trying to work with point-wise multiplication (unless you are denoting $\cdot$ the addition of natural numbers), but $\mathbb{N}^{\mathbb{N}}$ is not a group by point-wise multiplication, since sequences that have $0$ in their images don't have inverses. For multiplication you also wouldn't have $x_m=y_m\Leftrightarrow (zx)_m=(zy)_m$. The $\Leftarrow$ doesn't hold if $z_m=0$. – user647486 Apr 20 at 14:29
• Ah, sorry! I'm not working with point-wise multiplication. By $\cdot$ I simply mean the result of the composition of $z$ and $x$: the permutation $(zx)$ works as follows: $m\mapsto x_m\mapsto z_{x_m}$. My notation is wrong. I edit – LBJFS Apr 20 at 14:32
• Composition is not a group either, since only bijections will have inverses. With point-wise addition then you have a group, and $d$ is left (and right) invariant. – user647486 Apr 20 at 14:33
• I don't know if this is what you wanted to prove, but if you let the group $S_{\mathbb{N}}$ of permutations of $\mathbb{N}$ act on the group $\mathbb{N}^{\mathbb{N}}$ by left composition, then, as you proved, $d(x,y)=d(z\circ x, z\circ y)$. Actually, it would be enough for $z$ to be only an injection. – user647486 Apr 20 at 14:50