Let $X$ be a normed vector space, $x\in X$ and $Z\subseteq X$. Then we define the point-to-set distance function as: $ \|x\|_Z = \inf_{z\in Z} \| x-z\| $ I use the notation $\|\cdot\|_Z$ for convenience without implying that the operation is a norm. Assuming that $Z$ is convex we may not conclude that: $ \|x+y\|_Z \leq \|x\|_Z + \|y\|_Z $ What sort of properties should $Z$ possess so that the triangle inequality holds for all $x\in X$? For example, $Z$ being a singleton brings about the desired inequality but I'm looking for a less trivial example.
Additionally, let $A\in M_n(\mathbb{R})$ be a matrix and let $Z\neq X$. I noticed that we may define:
$ \|A\|_Z:= \sup_{x\in\mathbb{R}^n,x\notin Z}\frac{\|Ax\|_Z}{\|x\|_Z} $
while $\|A\|_X:=0$.
Then, it should be true that: $ \|Ax\|_Z \leq \|A\|_Z\cdot\|x\|_Z $
So, this is one norm-like property for $\|\cdot\|_Z$.
What other properties of $\|\cdot\|_Z$ can we deduce based on assumed properties of $Z$?
Update: As Davide Giraudo pointed out, the triangle inequality holds true in case $Z$ is a linear subspace of $X$. It seems that it holds for convex cones too (for nonconvex cones it's easy to find a counter example on the plane). Let us assume that $Z$ is a convex cone. Then:
$ \|x+y\|_Z=\inf_{z\in Z}\|x+y-z\|\leq \|x+y-(z_1+z_2)\| $
where $z_1,z_2\in Z$ and we proceed as in David's answer. Here, we used the fact that whenever $Z$ is a convex cone, then $z_1,z_2\in Z \Rightarrow z_1+z_2\in Z$ so any vector in $Z$ can be decomposed into a sum of vectors of $Z$ and vice versa.
Question: Would be nice to identify other classes of sets for which the induced point-to-set distance satisfies the triangle inequality.