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Thread: Homeomorphism

  1. #1
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    Homeomorphism

    Can someone help me to show that Q is not homeomorphic to N.

    We know that there exists a bijection between Q and N. How can I show that f:Q onto N is not continuous ?

    I also have trouble with showing any isometry is continuous and is a homeomorphism.

    Part 1)
    We know that f: M onto N is an isometry if d_N (fp, fq) = d_M (p,q) for each p,q in M. So, f preserves distance. We now want to show that f is continuous. Assume that f is not continuous, then there exists x in M such that f is not continuous at x. This implies that there exists epsilon greater than 0 st for each delta greater than 0 and for each y in M, we have d_M (x,y) less than delta implies d_N (f(x), f(y)) greater than or equal to 0.
    I am not able to see a contradiction here. Anyone can help pls?

    Part 2)
    We want to show that f is a bijection, but this is given in the definition of isometry. We need to show that f is continuous, this will be given if I can prove part 1). The last proof is to prove that inverse of f is continuous. Anyone can help?
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  2. #2
    Senior Member JaneBennet's Avatar
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    There is no need to use contradiction for Part (1). Itís totally straightforward.

    Let $\displaystyle x\in M$. You want to show that f is continuous at x. Given $\displaystyle \epsilon>0$, simply set $\displaystyle \delta=\epsilon$ and you have that $\displaystyle \forall\,y\in M$, $\displaystyle \mathrm{d}_M(x,y)<\delta\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x),\mathrm{f}(y))=\mathrm{ d}_M(x,y)<\delta=\epsilon$.

    Part (2):
    f is a function from M onto N; in other words, it is surjective.

    Let $\displaystyle x_1,x_2\in M$. Then

    $\displaystyle \mathrm{f}(x_1)=\mathrm{f}(x_2)\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x_1),\mathrm{f}(x_2))=0$
    $\displaystyle \color{white}.\hspace{22mm}.$ $\displaystyle \Rightarrow\ \mathrm{d}_M(x_1,x_2)=0$
    $\displaystyle \color{white}.\hspace{22mm}.$ $\displaystyle \Rightarrow\ x_1=x_2$

    So f is injective as well. Hence it is a homeomorphism.
    Last edited by JaneBennet; Mar 2nd 2008 at 11:45 AM.
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  3. #3
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    Quote Originally Posted by JaneBennet View Post
    There is no need to use contradiction for Part (1). It’s totally straightforward.

    Let $\displaystyle x\in M$. You want to show that f is continuous at x. Given $\displaystyle \epsilon>0$, simply set $\displaystyle \delta=\epsilon$ and you have that $\displaystyle \forall\,y\in M$, $\displaystyle \mathrm{d}_M(x,y)<\delta\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x),\mathrm{f}(y))=\mathrm{ d}_M(x,y)<\delta=\epsilon$.

    Part (2):
    f is a function from M onto N; in other words, it is surjective.

    Let $\displaystyle x_1,x_2\in M$. Then

    $\displaystyle \mathrm{f}(x_1)=\mathrm{f}(x_2)\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x_1),\mathrm{f}(x_2))=0$
    $\displaystyle \color{white}.\hspace{22mm}.$ $\displaystyle \Rightarrow\ \mathrm{d}_M(x_1,x_2)=0$
    $\displaystyle \color{white}.\hspace{22mm}.$ $\displaystyle \Rightarrow\ x_1=x_2$

    So f is injective as well. Hence it is a homeomorphism.
    Thanks a lot for your help, but I don't know why in part 2) showing f is bijective, then it is homemorphism. Since from what I learned, to show f is homeomorphism, I need to check 3 conditions: f is a bijection, f is continuous, and inverse of f is continuous.
    However, I now see how to approach part 2) . Since f is a bijection, the inverse of f is also a bijection, and inverse of f also preserves distance. Hence, inverse of f is isometry. By applying part 1), inverse of f is continuous
    Last edited by namelessguy; Mar 2nd 2008 at 02:06 PM.
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  4. #4
    Senior Member JaneBennet's Avatar
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    Youíre right, one has to check that $\displaystyle \mathrm{f}^{-1}$ is continuous as well.

    Since f is a bijection, we can write $\displaystyle N=\{\mathrm{f}(x):x\in M\}$.

    Let $\displaystyle \mathrm{f}(x)\in N$. You want to show that $\displaystyle \mathrm{f}^{-1}$ is continuous at $\displaystyle \mathrm{f}(x)$. Given $\displaystyle \epsilon>0$, simply set $\displaystyle \delta=\epsilon$ and you have that $\displaystyle \forall\,\mathrm{f}(y)\in N$, $\displaystyle \mathrm{d}_N(\mathrm{f}(x),\mathrm{f}(y))<\delta\ \Rightarrow\ \mathrm{d}_M(\mathrm{f}^{-1}(\mathrm{f}(x)),\mathrm{f}^{-1}(\mathrm{f}(y)))=\mathrm{d}_M(x,y)=\mathrm{d}_N( \mathrm{f}(x),\mathrm{f}(y))<\delta=\epsilon$.
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  5. #5
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    Do you have any suggestion for me on how to show Q is not homeomorphic to N. Where Q is the set of rationals, and N is the set of naturals.
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  6. #6
    Senior Member JaneBennet's Avatar
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    Show that there are no isometries between $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{N}$. Make use of the fact that distances in $\displaystyle \mathbb{Q}$ can be arbitrarily small whereas distances in $\displaystyle \mathbb{N}$ can’t be arbitrarily small. (I mean of course distances between distinct elements.)
    Last edited by JaneBennet; Mar 2nd 2008 at 07:53 PM.
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  7. #7
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    Quote Originally Posted by JaneBennet View Post
    Show that there are no isometries between $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{N}$. Make use of the fact that distances in $\displaystyle \mathbb{Q}$ can be arbitrarily small whereas distances in $\displaystyle \mathbb{N}$ canít be arbitrarily small. (I mean of course distances between distinct elements.)
    I thought that isometry is a stronger property than homeomorphism. So, if f is an isometry, then f is a homeomorphism. However, if f is not an isometry, then it doesn't guarantee that f is not a homeomorphism. I think this way because I take the contrapositive statement: if f is not homeomorphism, then f is not isometry.
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  8. #8
    Senior Member JaneBennet's Avatar
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    The correct contrapositive statement is: If f is not an isometry, then f is not a homeomorphism.

    In any case, f can be an isometry without being a homeomorphism. For example, $\displaystyle \mathrm{f}:\mathbb{N}\to\mathbb{Q},\ \mathrm{f}(n) = n+\mbox{$\frac{1}{2}$}$. Hence the statement “if f is not a homeomorphism, then f is not an isometry” is false.

    Try this: Let $\displaystyle \mathrm{f}:M\to N$ be an isometry, $\displaystyle y\in M$ be a fixed point and $\displaystyle (x_n)_{n\,=\,1}^\infty$ be a sequence in M such that $\displaystyle x_n\ne y$ for any $\displaystyle n\in\mathbb{N}$. Then if $\displaystyle \mathrm{d}_M(x_n,y)\to0$ as $\displaystyle n\to\infty$, $\displaystyle \mathrm{d}_N(\mathrm{f}(x_n),\mathrm{f}(y))\to0$ as $\displaystyle n\to\infty$.

    Now suppose $\displaystyle \mathrm{f}:\mathbb{Q}\to\mathbb{N}$ is an isometry. Let $\displaystyle y=0$ and $\displaystyle x_n=\frac{1}{n}$. We have shown above that any isometry is injective; hence $\displaystyle x_n\ne y\ \Rightarrow\ \mathrm{f}(x_n)\ne\mathrm{f}(y)$ for any n. But $\displaystyle \lim_{n\,\to\,\infty}{\mathrm{d}_\mathbb{Q}(x_n,y) } = 0$, whereas $\displaystyle \mathrm{d}_\mathbb{N}(\mathrm{f}(x_n),\mathrm{f}(y ))$ can never tend to zero if $\displaystyle \mathrm{f}(x_n)\ne\mathrm{f}(y)$ since the set $\displaystyle \{\mathrm{d}_\mathbb{N}(m,n):m,n\in\mathbb{N},m\ne n\}$ has a lower bound of 1. This is a contradiction – hence f cannot be an isometry.
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