Let A be a non-empty set in R^k. Prove the inequality

Question

Let A be a non-empty in $R^k$. For $\vec{x} \in R^k$, define: $$d(\vec{x},A):=inf|\vec{x} - \vec{a}|: \vec{a} \in A$$ Show that: $$|d(\vec{x},A) - d(\vec{y},A)| \leq |\vec{x} - \vec{y}|$$ I am thinking of using the Triangle in equality, but it is normally applied for vector. This time, A is a set, so I am not so sure if I am doing this right.

Answer 1

Without loss of generality assume $d(x,A) \ge d(y,A)$. (Otherwise, just reverse the roles of $x$ and $y$ below.)

You want to show $d(x,A) - d(y, A) \le |x-y|$.

For any $\epsilon>0$ there exists some $a \in A$ (that depends on $\epsilon$) such that $|y-a| \le d(y,A) + \epsilon$, by the definition of infimum.

Then, \begin{align} d(x,A) - d(y, A) &\le d(x,A) - |y - a| + \epsilon \\ &\le |x - a| - |y - a| + \epsilon. \end{align}

Can you finish from here?