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Math Help - quotient map

  1. #1
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    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!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rain07 View Post
    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 (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.
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