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Thread: 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 $\displaystyle n \leq x < n+1$).

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

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

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

    Consider the map $\displaystyle I: {R} \rightarrow {Z}$ by $\displaystyle x \rightarrow I(x)$. 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 $\displaystyle n \leq x < n+1$).

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

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

    Quote Originally Posted by Amanda1990 View Post
    Consider the map $\displaystyle I: {R} \rightarrow {Z}$ by $\displaystyle x \rightarrow I(x)$. Is the map I continuous (when Z is endowed with the induced topology)?
    Notice that $\displaystyle 1-\tfrac1n\to1$ as $\displaystyle n\to\infty$. But $\displaystyle p(1-\tfrac1n)=0$ 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 $\displaystyle p(x) = [I(x), I(x) +1)$?

    On a similar note, I agree that I is not continuous, but I does define a bijection, J say, from $\displaystyle {R}/$~ $\displaystyle \rightarrow Z$ (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 $\displaystyle p(x) = [I(x), I(x) +1)$?

    On a similar note, I agree that I is not continuous, but I does define a bijection, J say, from $\displaystyle {R}/$~ $\displaystyle \rightarrow Z$ (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 $\displaystyle \{k\in\mathbb{Z}:k\leqslant n\}$. I suppose I should have written it as $\displaystyle (-\infty,n]\cap\mathbb{Z}$.
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