Consider $\ell^\infty$ as a subspace of the Polish space $\mathbb{R}^\omega$. It is easy to check that $\ell^\infty$ is not Polish in the subspace topology, as it is countable union of the compact nowhere dense sets $\{f\in\mathbb{R}^\omega:\forall m(|f(m)|\leq n)\}$, and hence is not Baire.

Recall a topological group $(G,\tau_0)$ is *Polishable* if there is a Polish group topology $\tau_1$ on $G$ having the same Borel sets as $\tau_0$.

Is $\ell^\infty$, with the subspace topology above, Polishable? (I believe the answer is no.)

Note that it is easy to see, using Pettis' Theorem, that any such Polish topology would be a refinement of the subspace topology.

An elementary proof would be prefferred to one which shows that the Borel equivalence relation $E_1$ embeds into to the coset equivalence relation $\mathbb{R}^\omega/\ell^\infty$, and then cites Theorem 4.2 in Kechris-Louveau, *The Classification of Hypersmooth Borel Equivalence Relations*.