Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition of a function by a metric.

  1. Feb 2, 2008 #1
    Let (X,p) be a metric space with p metric on X, define for each subset A of X
    [tex]p(x,A)=inf_{y\in A}p(x,y)[/tex]
    prove that: f_A(x)=p(x,A) f:X->R is continuous.
    basically, if U is open in R, we need to show that f^-1_A(U) is open in X, i.e that
    {x in X|p(x,A) in U} is open in X, now because U is open in R, then there's an open interval or half open interval contained in U that it contains p(x,A), i.e p(x,A) is bounded, but from this to show that f^-1(U) is open I'm kind of in a stuck.

    any hints?

    thanks in advance.
  2. jcsd
  3. Feb 3, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    It's probably easier to use the good old epsilon-delta definition of continuity here. The following observation will be helpful: p(x,A) <= p(x,z) + p(z,A), for any z in X.
  4. Feb 3, 2008 #3
    Morphism is right. You have to do what he suggested twice, in order to get the absolute value on the left side. This is assuming that you've already learned that the epsilon-delta definition of continuity applies to metric spaces in general (and it does, calculus clases specialize in R^n, but it can be proven that it applies to metric spaces in general).
  5. Feb 3, 2008 #4
    ok, I have another question, let [tex]S^n[/tex]={[tex]x\in R^n|||x||=1[/tex]}
    let x_0=(1,0,...,0) (n entries), prove that S^n-{x_0} is homeomorphic to R^(n+1).
    well I thought of dividing by the norm, i.e defining a homeomorphism such as this:
    x=(x1,...,x_n+1) in R^n+1 f(x)=(x2,...,x_n-1,1)/||(x2,...,1)|| when (x1,...,xn)=(1,0,...,0)
    when (x1,...,x_n)!=(1,0,...,0) then f(x)=(1,x2,...,x_n-1)/||(1,x2,...,x_n-1)||
    I know that somewhere I need to divide by the norm in order to keep it that the norm of the vector is 1, but don'y know how, anyone has some hints on this?

    thanks in advance.
  6. Feb 3, 2008 #5


    User Avatar

    hint: stereographic projection.
    btw, the set you defined as S^n is more usually written as S^{n-1}, and you need to show that removing a point makes it homeomorphic to R^{n-1}. It can't be homeomorphic to R^{n+1}, as they have different dimensions.
  7. Feb 8, 2008 #6
    so basically in order to prove it, all you need is to use g(x)=x/||x|| and the fact that S^n-1 is bijective to S^n-1\{x0}, so there's a bijection f between them then gof is the appropiate function.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?