Proving a sequence of functions is continuous on X

Question

The problem is here.

Let $(X,d)$ be a metric space and $z\in X$ is fixed. For each $p\in X$, define $f_p\colon X\to R$ such that
$f_p(x) = d(x,p) - d(x,z)$ for all $x\in X$.
Show that
(1) $f_p\in C(X)$.
(2) For $p,q\in X$, we have $\bar{d}(f_p, f_q) = d(p,q)$.

For the first part I'm thinking I just need to prove $f_p$ is continuous on $X$, which simply means that $X$ is compact....right? If $X$ is compact, then there cannot be any $p\in X$, $x_1,x_2\in X$ which gives $|(d(x_1, p) - d(x_1, z)) - (d(x_2, p) - d(x_2, z))| > \epsilon$ for $|x_1> - x_2| <\text{ some }\delta$.

Is this the correct way of thinking about this problem, or am I way off?

Answer 1

I don't think your approach is correct. To prove (1), notice that: $$ |f_p(x) - f_p(y)| = |d(x,p) - d(y,p) + d(y,z) - d(x,z)| \leq |d(x,p) - d(y,p)| + | d(x,z) - d(y,z) |\leq 2\, d(x,y), $$ by using the triangle inequality twice.

For (2), I suppose that $\bar d(f_p, f_q)$ is defined as $$ \bar d(f_p, f_q) = \sup_{x \in X} | f_p(x) - f_q(x) |.$$ To prove the result, notice that: $$|f_p(x) - f_q(x)| = | d(x,p) - d(x,q) | \leq d(p,q),$$ again by the triangle inequality, so $\bar d(f_p, f_q) \leq d(p,q)$. On the other hand, taking $x = p$, $$|f_p(x) - f_q(x)| = | d(p,p) - d(p,q) | = d(p,q),$$ which concludes.