Show that $d(x+y,A+B) \leq d(x,A) + d(y,B).$

Question

Problem: Let $(X,d)$ be a metric space. For each nonvoid subsets $A$ of $X$ let $$d(x,A) = \inf_{y \in A} d(x,y), \quad x \in X.$$ Show that:

1. $d(x,A) = d(x,\bar{A})$. In particular, $d(x,A) = 0 \Leftrightarrow x \in \bar{A}$

2. $|d(x,A) - d(y,A)| \leq d(x,y), \quad \forall x,y \in X$

3. In case $X$ is a normed space (on field $\mathbb{K}$) then $\forall \lambda \in \mathbb{K}$, $\forall x,y \in X$ and $\forall A,B \subset X; A,B \neq \emptyset$ we have $$d(\lambda x, \lambda A) = |\lambda| d(x,A) \text{ and } d(x+y,A+B) \leq d(x,A) + d(y,B).$$ Furthermore, if $A$ is convex then the function $f(x) := d(x,A)$ is a convex function.

I have done with $1.$ and $2.$, in $3.$, $d(\lambda x, \lambda A) = \inf d(\lambda x, \lambda y) = |\lambda| \inf d(x,y) = |\lambda| d(x,A)$. How can I prove the second expression.

Answer 1

Let $a \in A, b\in B$. Then $\|x+y-(a+b)\| \leq \|x-a\|+\|y-b\|$ so $d(x+y,A+B) \leq \|x-a\|+\|y-b\|$. Taking infimum over all $a \in A$ and all $b \in B$ we get $d(x+y,A+B) \leq d(x,A)+d(y,B)$.