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F continuous at x[sub]0[/sub], prove g is continuous atx[sub]0[/sub]

  • Thread starter kathrynag
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  • #1
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Homework Statement


Suppose f: E--> R is cont at x0 and x0 is an element of F contained in E. Define g:F--->R by g(x)=f(x) for all x elemts of F. Prove g is continuous at x0. Show by example that the continuity of g at x0 need not imply the continuity of f at x0.

Homework Equations


lx-x0l<delta
lf(x)-f(x0)l<epsilon


The Attempt at a Solution


lx-x0l<delta
lg(x)-g(x0)l<epsilon
lf(x)-f(x0l<epsilon
Ok, then it's continuous because g(x)=f(x)?
 

Answers and Replies

  • #2
morphism
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Your proof isn't very convincing. Write it out in words.
 
  • #3
HallsofIvy
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Morphism's point is that "sketched" the proof but you need to say exactly why those statements prove the theorem.
 
  • #4
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Ok, but how would I do the part with showing by example f doesn't need to be continuous. Wouldn't it ahve to be continuous since g(x)=f(x)?
 
  • #5
morphism
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Do g and f have the same domain...?
 
  • #6
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No, so if f is in E, then g could still be continuous?
 
  • #7
Office_Shredder
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F is a subset of E, so we know f is continuous. Hence, for all e>0, there exists d>0 such that
|x-x0<d and x in E implies |f(x)-f(x0|<e

Hence
|x-x0<d and x in F implies |f(x)-f(x0|<e

But f(x)=g(x) and f(x0)=g(x0)

Hence |x-x0<d and x in F implies |g(x)-g(x0|<e

This is a much clearer way of writing the proof

For a counterexample, I would suggest looking at step functions
 
  • #9
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Ok, i'm confused on finding an example. Like what if I choose the function [x]?
 

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