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Continuity in metric spaces

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that if (x[itex]_{n}[/itex]) is a sequence in a metric space (E,d) which converges to some x[itex]\in[/itex]E, then (f(x[itex]_{n}[/itex])) is a convergent sequence in the reals (for its usual metric).


    2. Relevant equations
    Since (x[itex]_{n}[/itex]) converges to x, for all ε>0, there exists N such that for all n[itex]\geq[/itex]N, d(x[itex]_{n}[/itex],x)<ε.
    So |x-x[itex]_{n}[/itex]|<ε


    3. The attempt at a solution
    I understand that this will prove continuity, but I'm not sure how to get from d(x[itex]_{n}[/itex],x)<ε to what we want: d(f(x[itex]_{n}[/itex]_,f(x))<ε
     
  2. jcsd
  3. Sep 18, 2012 #2

    MathematicalPhysicist

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    If f is continuous then from [itex]d(x,x_n)\rightarrow 0[/itex] you can deduce [itex]d(f(x_n),f(x))\rightarrow 0[/itex] (by definition).

    Anyway for f general this might not be the case, take f(x)=1/x, x_n =1/n. x_n ->0 but f(0) is not defined.
     
  4. Sep 18, 2012 #3
    We aren't given that f is continuous, which is why I'm stuck.
     
  5. Sep 18, 2012 #4

    micromass

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    If f is not continuous, then the statement in the OP is false.
     
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