## Continuous distance (p13 (E))

Let $(X, d)$ be a metric space.  Let $y \in X$.  Let $d_y:X \to \mathbf{R}$
by $d_y(x) = d(y, x)$.
Then $d_y$ is continous.

Proof.
Let $x \in X$.  Let $V$ be a basic neighborhood of $\alpha = d(y,x)$ in
$\mathbf{R}$.  Then $V = (\alpha - \epsilon, \alpha + \epsilon)$, for some $\epsilon > 0$.
Let $z \in X$, $d(z, x) < \epsilon$.  Then $d(y,z) \le d(y,x) + d(x,z) < d(y,x) + \epsilon$.
Also, $d(y,x) \le d(y,z) + d(z,x) < d(y,z) + \epsilon$, so
$d(y,z) > d(y,x) - \epsilon$.

Therefore, $d_y$ is continuous.