Show that $d( \; \cdot \; ,A)$ 1-Lipschitz continuous

Question

Let $(X, d)$ be a metric space and $A\subset X$, with $$d(x,A) = \inf_{y \in A} d(x,y)$$

Now my problem is to show the following:$$ \forall_{x,z \in X}\mid d(x,A)-d(z,A)| \le |x-z| \;\; \text{(which is the same as $d(x,z)$)}$$

I've already shown in the same exercise that $d(x,A)=0$ if and only if $x \in \bar A$ (the closure of $A$), maybe this is useful.

I've tried coming from the right and came to $d(x,A)-d(z,A)-d(a,b) \le |x-z|$ with $a$ and $b$ being the elements of $A$ such that $d(x,A)=d(x,a) \text{ and } d(z,A)=d(z,b)$ but i can't get rid of the remaining $d(a,b)$

Thanks in advance.

Answer 1

By definition, for every $y \in A$, we have $$d(x, A) \leq d(x, y) \leq d(x, z) + d(z, y)$$ which implies that (by taking the infimum with respect to $y$ on both sides): $$d(x, A) \leq d(x, z) + d(z, A). \tag{1}$$ Switch the role of $z$ and $x$ above, we have $$d(z, A) \leq d(x, z) + d(x, A). \tag{2}$$ $(1)$ and $(2)$ together imply that $$|d(x, A) - d(z, A)| \leq d(x, z).$$