Metric Spaces: The dist function

Question

Given that $A$ is defined as non-empty subset of $(X,d)$

The distance function is defined as such:

$dist(x,A)=$ inf $_{y\in A} \lbrace d(x,y) \rbrace $

Given the above we are asked to prove the following:

$ x \in A \Rightarrow dist(x, A ) =0$.

This seems so obvious given the following property of metric spaces :

$ d(x ,y)= 0 \iff x=y $ $\forall x ,y \in X $

However, I'm not quite sure what constitutes a rigorous proof in these cases.

Answer 1

Both of them are very obvious

First part: $$\inf_{y\in\{x\}}\{d(x,y)\}=\inf_{y=x}\{d(x,y)\}=d(x,x)=0$$

Second part:

Fix $x\in A$. Because $d(x,x)\in\{d(x,y):\ y\in A\}$ so $$\{d(x,x)\}\subset\{d(x,y):\ y\in A\}$$ and consequently $$0\leq d(x,A)=\inf \{d(x,y):\ y\in A\}\leq \inf \{d(x,x)\}=d(x,x)=0$$ Hence $d(x,A)=0$