Proving Continuity of g(x) on a Metric Space T with f(x)=x

[l888l888l888]

Homework Statement

T is a compact metric space with metric d. f:T->T is continuous and for every x in T f(x)=x. Need to show g:T->R is continous, g(x)=d(f(x),x).

Homework Equations

The Attempt at a Solution

f is continuous for all a in T if given any epsilon>0 there is a delta>0 st d(x,a)<delta implies d(f(x),f(a))<epsilon. Need to show there is a delta st d(x,a)< delta implies |g(x)-g(a)|<epsilon.
by a previous problem i did... |g(x)-g(a)|=|d(f(x),x)-d(f(a),a)|<= d(f(x),f(a))+d(x,a)<d(f(x),f(a)) + delta. This is where I got stuck. from the assumption that f is continuous we got from there that d(f(x),f(a))<epsilon. but if i say that then i would have...|g(x)-g(a)|<epsilon + delta and inorder for |g(x)-g(a)|to be <epsilon we would have to choose delta to be 0 which it can't be because it has to be greater than 0. any suggestions on what i should do?

 

[micromass]

Hi l888l888l888!

Homework Statement

T is a compact metric space with metric d. f:T->T is continuous and for every x in T f(x)=x. Need to show g:T->R is continous, g(x)=d(f(x),x).

Homework Equations

The Attempt at a Solution

f is continuous for all a in T if given any epsilon>0 there is a delta>0 st d(x,a)<delta implies d(f(x),f(a))<epsilon. Need to show there is a delta st d(x,a)< delta implies |g(x)-g(a)|<epsilon.

This must be true for every epsilon, so also for \varepsilon/2. So, what happens to the following if you take

|g(x)-g(a)|&lt;\varepsilon/2

by a previous problem i did... |g(x)-g(a)|=|d(f(x),x)-d(f(a),a)|<= d(f(x),f(a))+d(x,a)<d(f(x),f(a)) + delta. This is where I got stuck. from the assumption that f is continuous we got from there that d(f(x),f(a))<epsilon. but if i say that then i would have...|g(x)-g(a)|<epsilon + delta and inorder for |g(x)-g(a)|to be <epsilon we would have to choose delta to be 0 which it can't be because it has to be greater than 0. any suggestions on what i should do?

 

[l888l888l888]

hello micromass! I am not really understanding your question. can you clarify? did you mean d(f(x),f(a))<epsilon/2?

 

[micromass]

I'm so sorry! What was I thinking...

But yes, that is what I meant. You can take d(f(x),f(a))&lt;\varepsilon/2

 

[l888l888l888]

Actually, I think I have it now. correct me if I am wrong...
f is continuous for all a in T if given any epsilon>0 there is a delta>0 st d(x,a)<delta implies d(f(x),f(a))<epsilon/2. Need to show there is a delta st d(x,a)< delta implies |g(x)-g(a)|<epsilon.
|g(x)-g(a)|=|d(f(x),x)-d(f(a),a)|<= d(f(x),f(a))+d(x,a)<epsilon/2 + delta. choose delta to be epsilon/2. so therefore |g(x)-g(a)|=|d(f(x),x)-d(f(a),a)|<= d(f(x),f(a))+d(x,a)<epsilon/2 + epsilon/2=epsilon, as desired...

 

[micromass]

Seems good!