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Math Help - Quotient spaces and equivalence relations

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    Quotient spaces and equivalence relations

    Let I(x) denote the integer part of a real number x (ie the unique integer n such that n \leq x < n+1).

    I can show that x ~ y iff I(x) = I(y) is an equivalence relation.

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

    Deduce that the open sets in {R}/~are the empty set, {R}/~ and the image sets p(-\infty, n] where n \in Z.

    Consider the map I: {R} \rightarrow {Z} by x \rightarrow I(x). Is the map I continuous (when Z is endowed with the induced topology)?
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    Quote Originally Posted by Amanda1990 View Post
    Let I(x) denote the integer part of a real number x (ie the unique integer n such that n \leq x < n+1).

    I can show that x ~ y iff I(x) = I(y) is an equivalence relation.

    Let p: {R} \rightarrow {R} / ~ be the quotient map, let {R}/~ be endowed with the quotient topology, and let U be an open set in {R}/~. Prove that if n \in Z is such that p(n) \in U then p(n-1) \in U.
    If p(n) \in U then n\in p^{-1}(U) which, by the definition of the quotient topology, is an open set in R and therefore contains n-\delta (for \delta small enough). Hence n-1 = p(n-\delta) \in U.

    Quote Originally Posted by Amanda1990 View Post
    Consider the map I: {R} \rightarrow {Z} by x \rightarrow I(x). Is the map I continuous (when Z is endowed with the induced topology)?
    Notice that 1-\tfrac1n\to1 as n\to\infty. But p(1-\tfrac1n)=0 and p(1) = 1.
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    Thanks a lot - for the first part of the question I had the right idea but just couldn't write it properly.

    For the last point you made, did you mean to use I instead of p? The map p just takes any real number to its equivalence class right? ie isn't p(x) = [I(x), I(x) +1)?

    On a similar note, I agree that I is not continuous, but I does define a bijection, J say, from {R}/~ \rightarrow Z (I think this i fairly clear). What is the topology on Z making J a homomorphism?
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    Quote Originally Posted by Amanda1990 View Post
    Thanks a lot - for the first part of the question I had the right idea but just couldn't write it properly.

    For the last point you made, did you mean to use I instead of p? The map p just takes any real number to its equivalence class right? ie isn't p(x) = [I(x), I(x) +1)?

    On a similar note, I agree that I is not continuous, but I does define a bijection, J say, from {R}/~ \rightarrow Z (I think this i fairly clear). What is the topology on Z making J a homomorphism?
    Yes, I should have said I rather than p.

    The topology on Z making J a homeomorphism is given by the middle part of the question: The open sets are the intervals (∞,n], together with the empty set and the whole space.
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    Well this is it - it looks as though the middle part of the question should give the answer to this but surely a topology on Z must be a family of subsets of Z (by definition). Then intervals (∞,n] are not subsets of Z and so cannot form a topology (?)
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    Quote Originally Posted by Amanda1990 View Post
    Well this is it - it looks as though the middle part of the question should give the answer to this but surely a topology on Z must be a family of subsets of Z (by definition). Then intervals (∞,n] are not subsets of Z and so cannot form a topology (?)
    They certainly have to be subsets of Z, so you should interpret (∞,n] as meaning \{k\in\mathbb{Z}:k\leqslant n\}. I suppose I should have written it as (-\infty,n]\cap\mathbb{Z}.
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