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No. The Bohr topology on $\mathbb{Z}$ is not first countable, in fact the least size of a local base at $0$ is $2^{\aleph_0}$. It is also known that this topology is not sequential (because there are no non-trivial convergent sequences).
No. The Bohr topology on $\mathbb{Z}$ is not first countable, in fact the least size of a local base at $0$ is $2^{\aleph_0}$.