What topology makes $\Bbb Z$ homeomorphic to $\mathbb{R}/R$, where $R$ is the relation defined by the integer part function?

Consider the map $I:\mathbb{R}\to \mathbb{Z}, x \mapsto I(x)$ where $I(x)$ is the integer part of $x$. On $\mathbb{R}$ we define an equivalence relation $xRy \Leftrightarrow I(x)=I(y)$. Then $I$ defines a bijection $\tilde{I}:\mathbb{R}/R \to \mathbb{Z}$. What is the topology on $\mathbb{Z}$ that makes $\tilde{I}$ an homeomorphism?

The map $I$ is not continuous since $\{1,2,3\}$ are open in $\mathbb{Z}$ endowed with subspace topology, but $I^{-1}\{1,2,3\}=[1,4)$ which is not open. Therefore, the standard topology on $\mathbb{Z}$ won't make $\tilde{I}$ an homeomorphism.

My intuition tells that it's the indiscrete topology. Is this correct?

• The indiscrete topology will make the map continuous, but will it be a homeomorphism? You need to look at the open sets of $\Bbb R/R$ and see what they look like. – Arthur Feb 15 '17 at 14:26
• @Arthur open sets in $\mathbb{R}/R$ are empty set, $\mathbb{R}/R$ and $\pi (-\infty,n]$ where $n\in \mathbb{Z}$. Then how to make $\tilde{I}$ continuous? – Kenneth.K Feb 15 '17 at 14:39

Hint Let $\pi : \Bbb R / R$ denote the quotient map, and suppose $U \subseteq \Bbb R / R$ is open. By definition, this is the case iff $\pi^{-1}(U)$ is open.
The key idea here is that each fiber of $\pi$, say, $[n, n + 1)$, respectively contains the point $n$ which is close to points in other fibers, respectively, $[n - 1, n)$, and so this is reflected in the topology $\pi$ determines on $\Bbb R / R$, which we want to identify with $\Bbb Z$ via a homeomorphism.
More precisely: Suppose $U$ is nonempty, say it contains $\pi(n)$ for some $n \in \Bbb R$; by definition of $R$ we may replace $n$ with its floor and hence assume $n$ is an integer. Now (assuming $\Bbb R$ is endowed with the standard topology) $\pi^{-1}(U)$ contains an interval $(n - \epsilon, n + \epsilon)$ for some $\epsilon > 0$ (and we may as well take $\epsilon < 1$). Thus, our open set $U$ must also contain $\pi(n - \epsilon) = \pi(n - 1)$. What does induction now tell us?
• open sets in $\mathbb{R}/R$ are empty set, $\mathbb{R}/R$ and $\pi(-\infty,n]$ for $n\in \mathbb{Z}$. – Kenneth.K Feb 15 '17 at 14:32
• Then I see with the standard topology on $\mathbb{Z}$, $\tilde{I}^{-1}$ is continuous. But $\tilde{I}$ is not. – Kenneth.K Feb 15 '17 at 14:37
• That sounds right to me! Notice that it remains to show that $\pi((-\infty, n])$ is actually an open set. (One could do this by writing $\pi((-\infty, n])$ as $\pi(V)$ for a suitable $\pi$-saturated open set $V$.) – Travis Willse Feb 15 '17 at 14:38
• To show $\tilde{I}$ is an homeomorphism, we need to show that $\tilde{I}$ and its inverse are continuous. Then to show $\tilde{I}$ is continuous, we need to show for every open $U\subset \mathbb{Z}$, $\tilde{I}^{-1}(U)$ is open. – Kenneth.K Feb 15 '17 at 14:42
• I don't know how to choose a right topology for $\mathbb{Z}$ that makes above happen. – Kenneth.K Feb 15 '17 at 14:43