# $\delta(x,A)\leq \delta(x,B)+\sup\limits_{b\in B} \delta(b,A)$

Definition: A function $$\delta: X\times 2^X\to [0,\infty]$$ with $X$ called approach space if it satisfies the following properties for all $x\in X$ and $A,B\subset X$

i) $\delta(x,\{x\})=0$

ii) $\delta(x,\emptyset)=\infty$

iii) $\delta(x,A\cup B)=\min\{\delta(x,A),\delta(x,B) \}$

iv) $\delta(x,A)\leq \delta(x,A^\varepsilon)+\varepsilon, \quad A^\varepsilon=\{a\in A: \delta(a,A)\leq \varepsilon \}$.

I'm trying prove for every $x\in X$ and $A,B\subset X$ $$\delta(x,A)\leq \delta(x,B)+\sup_{b\in B} \delta(b,A).$$

My proof: Suppose for some $x\in X$ the inequality isn't hold, namely
$$\delta(x,A)> \delta(x,B)+\sup_{b\in B} \delta(b,A).$$

If we take $x:=b$, then \begin{align*} \delta(b,A)& > \delta(b,B)+\sup_{b\in B} \delta(b,A)\\ & > \delta(b,B)+\delta(b,A).\\ \end{align*} This leads to a contradiction ($\delta(b,B)<0$).

Are my proof steps correct? Thanks in advance.

I'm afraid this is not correct. There's a collision of notation - if the supremum is over all $b \in B$, you cannot use the letter $b$ outside of the supremum.
Maybe you meant the following: let $b_0 \in B$ is such that $\delta(b_0,A) = \sup_{b\in B} \delta(b,A)$. Then we obtain a contradiction by taking $x := b_0$. This is also wrong in two ways: such $b_0$ might not exist and more importantly, our proof works only for this particular $x = b_0$, not for arbitrary $x \in X$.
Hint. Let $\varepsilon := \sup_{b\in B} \delta(b,A)$. Note that $B \subseteq A^\varepsilon$, then use (iii) to show $\delta(x,A^\varepsilon) \leq \delta(x,B)$ and finally apply (iv).