TITLE: About the quotient map $p : \mathbb{R} \to \mathbb{R} / R$ where $R$ collapses all intervals $[n,n+1)$
QUESTION [1 upvotes]: Recall that the integer part (or integral part) of a real number $x$ is the unique integer $n \in \mathbb Z$ such that $n \le x \lt n+1$. We denote it by $I(x)$.
On $\mathbb R$ we define the relation $xRy \iff I(x) = I(y)$.

Prove that $R$ is an equivalence relation.

Let $p:\mathbb R \to \mathbb R /R$ be the quotient map, let $\mathbb R /R$ be endowed with the quotient topology, and let $U$ be an open set in $\mathbb R /R$. Prove that if $n \in \mathbb Z$ is such that $p(n) \in U$ then $p(n-1) \in U$.

Deduce that the open sets in $\mathbb R /R$ are $\emptyset,\mathbb R /R$ and the image sets $p(-\infty,n]$, where $n \in \mathbb Z$.

Consider the map $I:\mathbb R \to \mathbb Z, x \mapsto I(x)$. Is the map $I$ continuous (when $\mathbb Z$ is endowed with the subspace topology)?



I have proved that $R$ is an equivalence relation and am stuck on part (2). I don't really see how it that is true. And being unable to do part (2) is making it hard to do (3) since they seem to build on each other. Any help with (2) would be very much appreciated.
For (2), since $U$ is an open set in $\mathbb R /R$ then $p^{-1}(U)$ is an open set in $\mathbb R$ because of the definition of quotient topology. So if $p(n) \in U$, then $n \in p^{-1}(U)$? which is open but I don't know how to show something about $n-1$?

REPLY [1 votes]: For (2) note the following important facts.

As you noted, by the definition of a quotient map if $U \subseteq \mathbb{R}/R$ is open, then $p^{-1} [ U ] \subseteq \mathbb{R}$ is open. (The converse holds, too.)
Since the mapping $p$ is constant on each segment $[n,n+1)$ it follows that for any set $A \subseteq \mathbb{R} / R$ and $x \in \mathbb{R}$, if $p(x) \in A$, then $[ I(x) , I(x) + 1 ) \subseteq p^{-1} [ A ]$.

Putting these together, if $U \subseteq \mathbb{R} / R$ is open and $n \in \mathbb{Z}$ is such that $p(n) \in U$, then $p^{-1} [ U ]$ is an open neighbourhood of $n$. Thus there is a $0 < \varepsilon < 1$ such that $(n-\varepsilon , n + \varepsilon ) \subseteq p^{-1} [ U ]$. Taking any $x$ with $n - \varepsilon < x < n$ it must be that $p(x) \in U$, and so $[ I(x) , I(x) + 1 ) \subseteq p^{-1} [ U]$. But we can easily compute $I(x) = n - 1$.
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