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?