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?

Thank you to @user647486 for helpful comments.

  • $\begingroup$ 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 $z_m=0$. $\endgroup$ – user647486 Apr 20 at 14:29
  • $\begingroup$ 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 $\endgroup$ – LBJFS Apr 20 at 14:32
  • 1
    $\begingroup$ 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. $\endgroup$ – user647486 Apr 20 at 14:33
  • 1
    $\begingroup$ Well, you can work with the group of permutations, yes. You just need to change the setup at the beginning of the question to work only with bijections and not all functions. The rest looks fine. $\endgroup$ – user647486 Apr 20 at 14:39
  • 2
    $\begingroup$ 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. $\endgroup$ – user647486 Apr 20 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.