Show that a function defined on a subset of a metric space is well-defined and continuous. 
Suppose $(X, d)$ is a metric space, and $A \in X$ is a bounded set.
Prove that the function $$f(x) := \text{sup}\{d(x, a); a \in A\}, \:\:x \in X$$
is well-defined and continuous.


$A$ is bounded $\Rightarrow$ there exists $x_0 \in X,\: M > 0$
such that $d(x, x_0) \leq M \:\:\forall\: x \in A$.
For any fixed $x$
$$d(x, a) \leq d(x, x_0) + d(x_0, a) \leq d(x, x_0) + M, \:\:\forall a \in A$$
Thus
$$f(x) := \text{sup}\{d(x, a); a \in A\} \leq d(x, x_0) + M < \infty$$
and $f(x)$ is well-defined.

I am not sure about how to prove that $f(x)$ is continuous.
Something with $\epsilon-\delta$.
Can you point me in the right direction for the next step to take?
Thanks.

Answer concerning continuity.
For each $a\in A$ we have $d(x,a)\leq d(x,y)+d(y,a)\leq d(x,y)+f(y)$.
This implies $f(x)\leq d(x,y)+f(y)$.
Likewise $f(y)\leq d(x,y)+f(x)$.
Conclusion: $|f(x)-f(y)|\leq d(x,y)$.
So $d(x,y)<\epsilon\implies|f(x)-f(y)|<\epsilon$.
It appears that actually you can take $\delta=\epsilon$.
