TITLE: General Triangle Inequality, distance from a point to a set
QUESTION [5 upvotes]: I am trying with no luck to prove:

Let $(X,d)$ be a metric space and $A$ a non-empty subset of $X$. For $x,y\in X$,
prove that
$$d(x,A) < d(x,y) + d(y,A)$$

REPLY [9 votes]: If you have $\leq$ instead of $<$, then you have:
$d(x,A) = \inf_{z\in A} d(x,z)$. Now, say $z_0\in A$ and $y\in X$. Then $d(x,z_0)\leq d(x,y) + d(y, z_0)$. Taking infimum over all $z\in A$ of the left hand side, we obtain:
$$
d(x, A) = \inf_{z\in A}d(x,z) \leq d(x,z_0) \leq d(x,y) + d(y, z_0).
$$
Observe that $d(x, A)$ is now independent of $z_0$. Hence taking the infimum over all $z$ in $A$ of the right hand side, we get:
$$
d(x,A) \leq d(x,y) + \inf_{z\in A}d(y,z) = d(x,y) + d(y, A).
$$