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?



The symmetric group of positive integers acting on $\ell^\infty(\mathbb{N})^*$
No, it's not continuous.


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.


