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).
|
2 | added 111 characters in body | ||
|
|
||||
|
|
1 |
|
||
|
No. The Bohr topology on $\mathbb{Z}$ is not first countable, in fact the least size of a local base at $0$ is $2^{\aleph_0}$. |
||||
