Set of all distinct cosets of $\mathbb Z$ in $\mathbb R$ is not equipotent with the set of all distinct cosets of $2\mathbb Z$ in $\mathbb R$?

Question

By third Isomorphism theorem , $\mathbb R/2\mathbb Z \Big/\mathbb Z/2\mathbb Z \cong \mathbb R/\mathbb Z$ ; so can we conclude that set of all distinct cosets of $\mathbb Z$ in $\mathbb R$ is not in bijective correspondence with the set of all distinct cosets of $2\mathbb Z$ in $\mathbb R$ ?

Answer 1

Clearly these sets are not the same. However, there is a bijection between them since they are sets with the same cardinality.

Answer 2

$f\colon \mathbb{R}/2\mathbb{Z}\to \mathbb{R}/\mathbb{Z}\colon [x]\mapsto[x/2]$ is a well-defined bijection (in fact it's a group isomorphism).