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

[kathrynag]

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)?

 

[morphism]

Your proof isn't very convincing. Write it out in words.

 

[HallsofIvy]

Morphism's point is that "sketched" the proof but you need to say exactly why those statements prove the theorem.

 

[kathrynag]

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)?

 

[morphism]

Do g and f have the same domain...?

 

[kathrynag]

No, so if f is in E, then g could still be continuous?

 

[Office_Shredder]

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

 

[kathrynag]

Thanks!

 

[kathrynag]

Ok, I'm confused on finding an example. Like what if I choose the function [x]?