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

## 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)?

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

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

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

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

Office_Shredder
Staff Emeritus
Gold Member
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

Thanks!

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