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Bijection is uniformly continuous

  1. Nov 7, 2013 #1
    Let f:N-> Q be a bijection. I want to show that this is uniformly continuous on N. (N is the set of natural numbers, Q the rationals). My first thought was to use induction. Since every point in N is an isolated point, then f is continuous on N.

    Let N1=[1,a_1], where a_1 is a natural number greater than or equal to 1. Then f is u.c. on N1 because N1 is compact. Suppose that f is u.c. on Nn=[1,a_n], where a_n>a_(n-1)>a_(n-2)>...>a_1. Then f is u.c. on [1,a_(n+1)] because this set is compact. Thus, f is u.c. on N by induction.

    Something doesn't feel right though, I don't think this works but am not sure why.
     
  2. jcsd
  3. Nov 7, 2013 #2
    Ah, nevermind. I think I figured it out. I don't need to use induction. Just pick 0<delta<1. Then for all epsilon>0 and for all natural numbers x and y, if |x-y|<delta, then |f(x)-f(y)|=0<epsilon because |x-y|<delta<1 implies x=y.
     
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