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

[kathrynag]

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
lx-x0l<delta
lf(x)-f(x0)l<epsilon


3. 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]?