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The only thing I can think of is that the circle $\mathbb{T}$ in its discrete topology is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ in its discrete topology, and the Pontryagin dual of this will be a continuum power of the Pontryagin dual $\mathbb{Q}_\mathrm{disc}^\prime$ (as you were just describing), times $(\mathbb{Q}/\mathbb{Z})_\mathrm{disc}^\prime$. The latter can be identified with the product of all additive groups of $p$-adic integers $\mathbb{Z}_p$, with $p$ ranging over all primes. I somehow doubt one can do any better than that.

Added: To see the claim about $(\mathbb{Q}/\mathbb{Z})_\mathrm{disc}^\prime$, write

$$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}[\tfrac1{p}]/\mathbb{Z},$$ a direct sum over Pruefer groups (using partial fraction decompositions for rational numbers, not rational functions! it works the same way though). So the answer will be a product over primes $p$ of Pontryagin duals of Pruefer groups. Then, each Pruefer group will be a colimit (or direct limit, in older terminology) of cyclic groups,

$$\mathbb{Z}/(p) \hookrightarrow \mathbb{Z}/(p^2) \hookrightarrow \mathbb{Z}/(p^3) \hookrightarrow \ldots,$$ and each finite cyclic group is self-dual, so the Pontryagin dual will be a corresponding inverse limit

$$\mathbb{Z}/(p) \leftarrow \mathbb{Z}/(p^2) \leftarrow \mathbb{Z}/(p^3) \leftarrow \ldots$$ which gives you the compact abelian group of the $p$-adic integers.

The only thing I can think of is that the circle $\mathbb{T}$ in its discrete topology is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ in its discrete topology, and the Pontryagin dual of this will be a continuum power of the Pontryagin dual $\mathbb{Q}_\mathrm{disc}^\prime$ (as you were just describing), times $(\mathbb{Q}/\mathbb{Z})_\mathrm{disc}^\prime$. The latter can be identified with the product of all additive groups of $p$-adic integers $\mathbb{Z}_p$, with $p$ ranging over all primes. I somehow doubt one can do any better than that.

Added: To see the claim about $((\mathbb{Q}/\mathbb{Z})_\mathrm{disc})^\prime$, write

$$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}[\tfrac1{p}]/\mathbb{Z},$$ a direct sum over Pruefer groups (using partial fraction decompositions for rational numbers, not rational functions! it works the same way though). So the answer will be a product over primes $p$ of Pontryagin duals of Pruefer groups. Then, each Pruefer group will be a colimit (or direct limit, in older terminology) of cyclic groups,

$$\mathbb{Z}/(p) \hookrightarrow \mathbb{Z}/(p^2) \hookrightarrow \mathbb{Z}/(p^3) \hookrightarrow \ldots,$$ and each finite cyclic group is self-dual, so the Pontryagin dual will be a corresponding inverse limit

$$\mathbb{Z}/(p) \leftarrow \mathbb{Z}/(p^2) \leftarrow \mathbb{Z}/(p^3) \leftarrow \ldots$$ which gives you the compact abelian group of the $p$-adic integers.

The only thing I can think of is that the circle $\mathbb{T}$ in its discrete topology is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ in its discrete topology, and the Pontryagin dual of this will be a continuum power of the Pontryagin dual $\mathbb{Q}_\mathrm{disc}^\prime$ (as you were just describing), times $\mathbb{Q}/\mathbb{Z}_\mathrm{disc}^\prime$. The latter can be identified with the product of all additive groups of $p$-adic integers $\mathbb{Z}_p$, with $p$ ranging over all primes. I somehow doubt one can do any better than that.

Added: To see the claim about $\mathbb{Q}/\mathbb{Z}_\mathrm{disc}^\prime$, write

$$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}[\frac1{p}]/\mathbb{Z},$$ a direct sum over Pruefer groups (using partial fraction decompositions for rational numbers, not rational functions! it works the same way though). So the answer will be a product over primes $p$ of Pontryagin duals of Pruefer groups. Then, each Pruefer group will be a colimit (or direct limit, in older terminology) of cyclic groups,

$$\mathbb{Z}/(p) \hookrightarrow \mathbb{Z}/(p^2) \hookrightarrow \mathbb{Z}/(p^3) \hookrightarrow \ldots,$$ and each finite cyclic group is self-dual, so the Pontryagin dual will be a corresponding inverse limit

$$\mathbb{Z}/(p) \leftarrow \mathbb{Z}/(p^2) \leftarrow \mathbb{Z}/(p^3) \leftarrow \ldots$$ which gives you the compact abelian group of the $p$-adic integers.

The only thing I can think of is that the circle $\mathbb{T}$ in its discrete topology is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ in its discrete topology, and the Pontryagin dual of this will be a continuum power of the Pontryagin dual $\mathbb{Q}_\mathrm{disc}^\prime$ (as you were just describing), times $(\mathbb{Q}/\mathbb{Z})_\mathrm{disc}^\prime$. The latter can be identified with the product of all additive groups of $p$-adic integers $\mathbb{Z}_p$, with $p$ ranging over all primes. I somehow doubt one can do any better than that.

Added: To see the claim about $(\mathbb{Q}/\mathbb{Z})_\mathrm{disc}^\prime$, write

$$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}[\tfrac1{p}]/\mathbb{Z},$$ a direct sum over Pruefer groups (using partial fraction decompositions for rational numbers, not rational functions! it works the same way though). So the answer will be a product over primes $p$ of Pontryagin duals of Pruefer groups. Then, each Pruefer group will be a colimit (or direct limit, in older terminology) of cyclic groups,

$$\mathbb{Z}/(p) \hookrightarrow \mathbb{Z}/(p^2) \hookrightarrow \mathbb{Z}/(p^3) \hookrightarrow \ldots,$$ and each finite cyclic group is self-dual, so the Pontryagin dual will be a corresponding inverse limit

$$\mathbb{Z}/(p) \leftarrow \mathbb{Z}/(p^2) \leftarrow \mathbb{Z}/(p^3) \leftarrow \ldots$$ which gives you the compact abelian group of the $p$-adic integers.

The only thing I can think of is that the circle $\mathbb{T}$ in its discrete topology is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ in its discrete topology, and the Pontryagin dual of this will be a continuum power of the Pontryagin dual $\mathbb{Q}_\mathrm{disc}^\prime$ (as you were just describing), times $\mathbb{Q}/\mathbb{Z}_\mathrm{disc}^\prime$. The latter can be identified with the product of all additive groups of $p$-adic integers $\mathbb{Z}_p$, with $p$ ranging over all primes. I somehow doubt one can do any better than that.

Added: To see the claim about $\mathbb{Q}/\mathbb{Z}_\mathrm{disc}^\prime$, write

$$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}[\frac1{p}]/\mathbb{Z},$$ a direct sum over Pruefer groups (using partial fraction decompositions for rational numbers, not rational functions! it works the same way though). So the answer will be a product over primes $p$ of Pontryagin duals of Pruefer groups. Then, each Pruefer group will be a colimit (or direct limit, in older terminology) of cyclic groups,

$$\mathbb{Z}/(p) \hookrightarrow \mathbb{Z}/(p^2) \hookrightarrow \mathbb{Z}/(p^3) \hookrightarrow \ldots,$$ and each cyclic group is self-dual, so the Pontryagin dual will be a corresponding inverse limit

$$\mathbb{Z}/(p) \leftarrow \mathbb{Z}/(p^2) \leftarrow \mathbb{Z}/(p^3) \leftarrow \ldots$$ which gives you the compact abelian group of the $p$-adic integers.

The only thing I can think of is that the circle $\mathbb{T}$ in its discrete topology is isomorphic to $\mathbb{R} \oplus \mathbb{Q}/\mathbb{Z}$ in its discrete topology, and the Pontryagin dual of this will be a continuum power of the Pontryagin dual $\mathbb{Q}_\mathrm{disc}^\prime$ (as you were just describing), times $\mathbb{Q}/\mathbb{Z}_\mathrm{disc}^\prime$. The latter can be identified with the product of all additive groups of $p$-adic integers $\mathbb{Z}_p$, with $p$ ranging over all primes. I somehow doubt one can do any better than that.

Added: To see the claim about $\mathbb{Q}/\mathbb{Z}_\mathrm{disc}^\prime$, write

$$\mathbb{Q}/\mathbb{Z} = \bigoplus_p \mathbb{Z}[\frac1{p}]/\mathbb{Z},$$ a direct sum over Pruefer groups (using partial fraction decompositions for rational numbers, not rational functions! it works the same way though). So the answer will be a product over primes $p$ of Pontryagin duals of Pruefer groups. Then, each Pruefer group will be a colimit (or direct limit, in older terminology) of cyclic groups,

$$\mathbb{Z}/(p) \hookrightarrow \mathbb{Z}/(p^2) \hookrightarrow \mathbb{Z}/(p^3) \hookrightarrow \ldots,$$ and each finite cyclic group is self-dual, so the Pontryagin dual will be a corresponding inverse limit

$$\mathbb{Z}/(p) \leftarrow \mathbb{Z}/(p^2) \leftarrow \mathbb{Z}/(p^3) \leftarrow \ldots$$ which gives you the compact abelian group of the $p$-adic integers.