Proving a function is bounded and continuous in a metric space.

[gottfried]

Homework Statement

Let (X,d) be any metric space. Fix a in X and for each x in X define f_x:X→ℝ by:
f_x(z)=d(z,x)-d(z,a) for all z in X.

Show that f_x(z) is bounded and continuous.

The Attempt at a Solution

I can't figure out how to tell if it is bounded. Any hints? I'm sure you have to use the triangle inequality but I'm not sure how.

So to prove f is continuous I'm going to try and use the fact that if U is open f_x^-1(U) must also be open if f is continuous.

So suppose U is a open subset of ℝ.
then f_x^-1(U) = {z in X: dist(f_x(z),ℝ-U)>0}
Then for all u in f_x^-1(U) it implies dist(f_x(u),ℝ)-U) = r > 0
so then B_x(z,r/2) is contained in f^-1(U)
So f_x^-1(U) is open and f_x is continuous.

 

[mighty2000]

Thank you for your post. I understand your confusion about how to prove that fx(z) is bounded. Here are some hints that may help you:

1. Remember that a function is bounded if its range is contained in a finite interval. In this case, the range of fx(z) is [-d(a,x), d(a,x)]. Can you see why this interval is finite?

2. To prove that fx(z) is bounded, you can use the triangle inequality in the following way: for any z in X, we have d(z,x) ≤ d(z,a) + d(a,x). Can you see how this relates to the range of fx(z)?

I hope these hints help you in your proof. Good luck with your work!