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

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?

• Show that for every $\varepsilon > 0$ you have $d(x,A) \leqslant d(x,y) + \bigl(d(y,A) + \varepsilon\bigr)$. Jan 18, 2017 at 12:38

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 $$d(x,A)-d(x,y)\le d(y,A)$$

• You seem to have used $d(y,a)\le d(y,A)$...
– soap
Jan 18, 2017 at 14:19
• Otherwise, how can you justify the last step? Just because $d(x,A)-d(x,y)\le d(y,a),\;\forall a\in A$ does that mean that $d(x,A)-d(x,y)\le d(y,A)$? There may not exist an $a\in A$ such that $d(y,A)=d(y,a)$...
– soap
Jan 18, 2017 at 14:25
• @XicoSimThe answer is correct. Look at the definition of the infimum. Jan 18, 2017 at 14:46
• @XicoSim If you have $C \leqslant d(y,a)$ for all $a\in A$, then you also have $C \leqslant \inf\limits_{a\in A} d(y,a)$. Jan 18, 2017 at 14:59
• @XicoSim If $\inf B < D$, then in particular there is a $b\in B$ with $b < D$. So if $C \leqslant b$ for all $b\in B$, then $C \leqslant \inf B$. Jan 18, 2017 at 15:18

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