Is $\ell^\infty$ Polishable?

Question

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.

Answer 1

The answer is no.

Su Gao $\textit{Invariant Descriptive Set Theory}$, Lemma 9.3.3 has a direct proof.

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As people usually do, denote the equivalence relation $\ell_\infty : = \mathbb{R}^\omega \backslash \ell_\infty$ (where of course the latter $\ell_\infty$ refers to the group).

$\ell_\infty$ is universal under $\leq_\text{B}$ for $K_\sigma$ equivalence relations. (Rosendal. See Theorem 8.4.2 of Gao.) $E_1$ is $K_\sigma$. Hence $E_1 \leq_\text{B} \ell_\infty$. If the group $\ell_\infty$ is Polishable, then the equivalence relation $\ell_\infty := \mathbb{R}^\omega \backslash \ell_\infty$ is a Borel orbit equivalence relation. It is well-known that $E_1$ is not Borel reducible to any orbit equivalence relation of a Borel group action (Kechris-Louveau; see Theorem 10.6.1 of Gao).