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.

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