1. ## quotient map

Recall that the integer part (or integral part) of a real number x is the unique integer n ∈ Z such that n ≤ x < n + 1. We denote it by I(x).
On R we deﬁne the relation xRy ⇔ I(x) = I(y).
(a) Let p : R → R/R be the quotient map, let R/R be endowed with the quotient topology, and let U be an open set in R/R. Prove that if n ∈ Z is such that p(n) ∈ U then p(n − 1) ∈ U .
(b) Deduce that the open sets in R/R are ∅, R/R and the image sets p(−∞, n] , where n ∈ Z.
(c) Consider the map I : R → Z, x ↦→ I(x) . Is the map I continuous (when Z is endowed with the induced topology) ?
Prove that I deﬁnes a bijection I : R/R → Z. What is the topology on Z making I a homeomorphism ?

I can see that p(n) ∈ U, let U =(x,y), then n ∈ (x,y), p(n-1) = n-1 ∈(x-1,y-1), p^-1(U) ∈(x-1, y), but I'm not sure if it's correct and how to proceed with the rest of the questions. Any input is appreciated!

2. Originally Posted by rain07
Recall that the integer part (or integral part) of a real number x is the unique integer n ∈ Z such that n ≤ x < n + 1. We denote it by I(x).
On R we deﬁne the relation xRy ⇔ I(x) = I(y).
(a) Let p : R → R/R be the quotient map, let R/R be endowed with the quotient topology, and let U be an open set in R/R. Prove that if n ∈ Z is such that p(n) ∈ U then p(n − 1) ∈ U .
I'm assuming $\mathbb{R}$ has the usual topology. So, we know that $n\in p^{-1}(U)$, right? But, since $p^{-1}(U)$ is open there exists some neighborhood $V$ of $n$ such that $V\subseteq p^{-1}(U)$. But, whatever neighborhood $V$ is it hits points strictly less than $n$ and so points such that $I(x)=n-1$. So, let $x$ be one of those points. Then, $x\in V\subseteq p^{-1}(U)\implies p(x)=p(n-1)\in p\left(p^{-1}(U)\right)=U$ the last part gotten since every quotient map is surjective.
(b) Deduce that the open sets in R/R are ∅, R/R and the image sets p(−∞, n] , where n ∈ Z.
You can do this.

(c) Consider the map I : R → Z, x ↦→ I(x) . Is the map I continuous (when Z is endowed with the induced topology) ?
Prove that I deﬁnes a bijection I : R/R → Z. What is the topology on Z making I a homeomorphism ?
Of course the map $I$ isn't continuous. For, given any $z\in\mathbb{Z}$ we have that $\left(z-\tfrac{1}{2},z+\tfrac{1}{2}\right)\cap\mathbb{Z}=\ {z\}$ is open and so $\mathbb{Z}$ with the subspace topology is discrete. And so, if $I$ were continuous then $I^{-1}(\{0\})$ should be open, but $I^{-1}(\{0\})=[0,1)$

I think you have your map a little screwed up. So you're saying that $I:\mathbb{R}/\sim\to\mathbb{Z}(x)\mapsto I(x)" alt="I:\mathbb{R}/\sim\to\mathbb{Z}(x)\mapsto I(x)" />, right? It doesn't go without saying that the map is well defined. But, if $p(x)=p(y)$ then $x=n+\varepsilon$ for some $0\leqslant \varepsilon<1$ and $y=n+\delta,\text{ }0\leqslant\delta<1$ from where it follows that $I(p(x))=I(n+\varepsilon)=n=I(y+\delta)=I(p(y))$. Now, to see it's an injection suppose that $I(x)=I(y)$ then $x=n+\varepsilon$ and $y=n+\delta$ just as before and so $p(x)=p(y)$. To see it's a surjection you merely must note that $I(p(z))=z,\text{ }\forall z\in\mathbb{Z}$.

I would say you need to give it the quotient topology determined by $\overset{\sim}{I}:\mathbb{R}\to\mathbb{Z}:x\mapsto I(x)$

So, to see now that $I:\mathbb{R}/\sim\to\mathbb{Z}$ is continuous we need merely note that a mapping out of a quotient space is continuous if and only if it's composition with the definining quotient map is continuous. But, $I\circ p:\mathbb{R}\to\mathbb{Z}:x\overset{p}{\longmapsto }p(x)\overset{I}{\mapsto}I(x)$ and of course this is continuous, it is the quotient map used to define the topology on $\mathbb{R}/\sim$!

So, why is it's inverse $I^{-1}:\mathbb{Z}\to\mathbb{R}/\sim:x\mapsto z\mapsto p(z)$ continuous? Well, let's compose it with the quotient map defined in the making of $\mathbb{R}/\sim$. We get....I'll let you finish this.