# The symmetric group of positive integers acting on $\ell^\infty(\mathbb{N})^*$

Let $$\mathfrak{S}_\mathbb{N}$$ be the symmetric group of all positive integers. Let $$\ell^\infty(\mathbb{N})^*$$ be the dual space of $$\ell^\infty(\mathbb{N})$$ equipped with weak*-topology. There is a natural group action of $$\mathfrak{S}_\mathbb{N}$$ on $$\ell^\infty(\mathbb{N})^*$$, that is to define $$\sigma(u)\big(x_i\big):=u\big((x_{\sigma(i)})\big)$$ for any $$\sigma\in\mathfrak{S}_\mathbb{N}$$, $$u\in\ell^\infty(\mathbb{N})^*$$ and $$(x_i)\in \ell^\infty(\mathbb{N})$$.

Now let's equip $$\mathfrak{S}_\mathbb{N}$$ with the permutation topology, i.e. the topology generated by the basis of neighborhood at identity element in form of $$V(F)=\left\{\sigma\in\mathfrak{S}_\mathbb{N}|\sigma(n)=n\text{ for all }n\in F\right\}$$, where $$F$$ is a finite subset of $$\mathbb{N}$$.

My question: is there a chance that the group action defined as above will be continuous under the setting of permutation topology?

My bad idea is to start with $$\ell^1(\mathbb{N})\subset \ell^\infty(\mathbb{N})^*$$ and the permutation $$\sigma$$ acts on $$\ell^1(\mathbb{N})$$ as usual. If a permutation $$\sigma$$ sends a closer position to a very far one (as modelling $$\sigma$$ not in a neighborhood of identity), then $$\sigma(f)-f$$ differs a lot in a sufficiently far position, and this will yield a great difference after applying an element in $$\ell^\infty(\mathbb{N})=\ell^1(\mathbb{N})^*$$. But this does not provide a rigorous proof, besides one knows that $$\ell^\infty(\mathbb{N})^*=\ell^1(\mathbb{N})\oplus c^\perp_0(\mathbb{N})$$.

So I wonder if anyone can give a proof or disproof for it?

Indeed, fix a non-principal ultrafilter $$U$$ supported by the set of even numbers, and define $$m\in\ell^\infty(\mathbf{N})^*$$ by $$m(f)=\lim_{n\to U}f(n)$$.
Now define $$\tau_n$$ as the transposition $$(2n,2n+1)$$ and $$s_n=\prod_{k\ge n}\tau_k$$. Then $$s_n$$ tends to the identity map for the permutation topology.
I claim that $$s_n\cdot m$$ does not tend to $$m$$. Indeed, let $$f$$ be the indicator function of the set of even numbers. It is enough to see that $$(s_n\cdot m)(f)$$ does not tend to $$m(f)$$. Indeed, $$m(f)=1$$, while $$(s_n\cdot m)(f)=m(s_n^{-1}\cdot f)=0$$ since $$s_n\cdot f$$ is eventually supported by odd numbers.