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Continuity in metric spaces

  • Thread starter gotmilk04
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  • #1
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Homework Statement


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).


Homework 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


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
 

Answers and Replies

  • #2
MathematicalPhysicist
Gold Member
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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.
 
  • #3
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We aren't given that f is continuous, which is why I'm stuck.
 
  • #4
22,097
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We aren't given that f is continuous, which is why I'm stuck.
If f is not continuous, then the statement in the OP is false.
 

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