Results 1 to 8 of 8

Math Help - Homeomorphism

  1. #1
    Member
    Joined
    Nov 2007
    Posts
    108

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    There is no need to use contradiction for Part (1). Itís totally straightforward.

    Let x\in M. You want to show that f is continuous at x. Given \epsilon>0, simply set \delta=\epsilon and you have that \forall\,y\in M, \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 x_1,x_2\in M. Then

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

    So f is injective as well. Hence it is a homeomorphism.
    Last edited by JaneBennet; March 2nd 2008 at 10:45 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2007
    Posts
    108
    Quote Originally Posted by JaneBennet View Post
    There is no need to use contradiction for Part (1). It’s totally straightforward.

    Let x\in M. You want to show that f is continuous at x. Given \epsilon>0, simply set \delta=\epsilon and you have that \forall\,y\in M, \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 x_1,x_2\in M. Then

    \mathrm{f}(x_1)=\mathrm{f}(x_2)\ \Rightarrow\ \mathrm{d}_N(\mathrm{f}(x_1),\mathrm{f}(x_2))=0
    \color{white}.\hspace{22mm}. \Rightarrow\ \mathrm{d}_M(x_1,x_2)=0
    \color{white}.\hspace{22mm}. \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; March 2nd 2008 at 01:06 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    Youíre right, one has to check that \mathrm{f}^{-1} is continuous as well.

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

    Let \mathrm{f}(x)\in N. You want to show that \mathrm{f}^{-1} is continuous at \mathrm{f}(x). Given \epsilon>0, simply set \delta=\epsilon and you have that \forall\,\mathrm{f}(y)\in N, \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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2007
    Posts
    108
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    Show that there are no isometries between \mathbb{Q} and \mathbb{N}. Make use of the fact that distances in \mathbb{Q} can be arbitrarily small whereas distances in \mathbb{N} can’t be arbitrarily small. (I mean of course distances between distinct elements.)
    Last edited by JaneBennet; March 2nd 2008 at 06:53 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2007
    Posts
    108
    Quote Originally Posted by JaneBennet View Post
    Show that there are no isometries between \mathbb{Q} and \mathbb{N}. Make use of the fact that distances in \mathbb{Q} can be arbitrarily small whereas distances in \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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    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, \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 \mathrm{f}:M\to N be an isometry, y\in M be a fixed point and (x_n)_{n\,=\,1}^\infty be a sequence in M such that x_n\ne y for any n\in\mathbb{N}. Then if \mathrm{d}_M(x_n,y)\to0 as n\to\infty, \mathrm{d}_N(\mathrm{f}(x_n),\mathrm{f}(y))\to0 as n\to\infty.

    Now suppose \mathrm{f}:\mathbb{Q}\to\mathbb{N} is an isometry. Let y=0 and x_n=\frac{1}{n}. We have shown above that any isometry is injective; hence x_n\ne y\ \Rightarrow\ \mathrm{f}(x_n)\ne\mathrm{f}(y) for any n. But \lim_{n\,\to\,\infty}{\mathrm{d}_\mathbb{Q}(x_n,y)  } = 0, whereas \mathrm{d}_\mathbb{N}(\mathrm{f}(x_n),\mathrm{f}(y  )) can never tend to zero if \mathrm{f}(x_n)\ne\mathrm{f}(y) since the set \{\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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Homeomorphism
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: June 11th 2011, 11:00 AM
  2. Homeomorphism
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 23rd 2011, 11:49 AM
  3. homeomorphism between R^(n^2) and M_n(R)
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: January 21st 2011, 05:20 PM
  4. homeomorphism
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: November 30th 2009, 01:29 PM
  5. Homeomorphism
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: November 29th 2009, 07:18 PM

Search Tags


/mathhelpforum @mathhelpforum