Consider the equivalence relation defined in $\mathbb{R}$ defined by $x \sim y$ if $x-y\in \mathbb{Z}$.
Describe the quotient space that results from the partition of $\mathbb{R}$ into the equivalence classes result from the equivalence relation.
First assume that $\sim$ is a relation of equivalence, following tha definition of class of equivalence
$[x]=\lbrace y\in \mathbb{R}:x \sim y\rbrace = \lbrace x-n:n\in \mathbb{Z} \rbrace$ this for each $x\in \mathbb{R}$
For find the quotient space i need to know the open sets in the quotient space, fix $[x]\in \mathbb{R/ \sim}$ then $p^{-1}([x])$ where $p:\mathbb{R} \longrightarrow \mathbb{R/ \sim}$ is just
$p^{-1}([x])=\lbrace y\in \mathbb{R}:p(y)\in [x]\rbrace= \lbrace x-n:n\in \mathbb{Z} \rbrace$ therefore since $<\mathbb{Z}>=\mathbb{Z}$ we can change of representant of the class and claim that the quotient is homeomorphic to the $[0,1)$ but i can prove it.
Someone that can give me a proof or contraexample, any help wasvery useful.
Best Answer
The quotient space can be naturally identified with $[0,1)$, true, but not as a group and not as a topological space. The quotient space is $\mathbb R/\mathbb Z$, which is isomorphic (as a group) and homeomorphic to the circle $S^1$. You can map from $[0,1)$ via $t\mapsto e^{2\pi i t}$, and this actually gives you the quotient map from $\mathbb R$ that induces the topology.
If you're not looking for any kind of structure-preserving map, then identifying it with $[0,1)$ is easy and fine. But you mention isomorphism, for which $[0,1)$ would not be appropriate algebraically or topologically.