Proving $d(x,A)\le d(x,y)+d(y,A) $

Question

This proof should be easy but I get stuck because I don't know how to deal with the infimum in this case.

I want to prove that $$d(x,A)\le d(x,y)+d(y,A) $$ with $$d(x,A)=\text{inf}(\{d(x,a):a\in A\})$$

My attempt: Let $y\in X$. We have $d(x,a)\le d(x,y)+d(y,a) $ for any $a\in A$.

How can I get the inequality for the infimum from this?

Answer 1

For all $a\in A$ we have by triangle inequality

$$d(x,A)\le d(x,a)\le d(x,y)+d(y,a)$$ so $$d(x,A)-d(x,y)\le d(y,a),\;\forall a\in A$$ hence $$d(x,A)-d(x,y)\le d(y,A)$$

Answer 2

Let $\epsilon>0$. Then $d(y,A)+\epsilon$ can not be a lower bound of the set $\{d(y,a):a\in A\}$. Thus, there exists $w\in A$ such that

$$d(y,w)<d(y,A)+\epsilon.$$

Thus, $$ \begin{align} d(x,A)&\leq d(x,w)\\ &\leq d(x,y)+d(y,w)\\ &<d(x,y)+d(y,A)+\epsilon. \end{align} $$ We have shown that $$d(x,A)<d(x,y)+d(y,A)+\epsilon\quad\text{for all }\epsilon>0.$$ Thus, $$d(x,A)\leq d(x,y)+d(y,A).$$