For all non-empty subset $A$ in a metric space $M$, let $A_* = \{x\in M:d(x,A)=0\}$. Show that $(A_*)_* = A_*$.
Definition: $d(a,X) = \inf\limits_{x\in X}\{d(a,x)\} $. I've just wrote down the definitions but I can't see neither of the inclusions, I'm trying to show that $d(x,y) = 0$ for $x\in (A_*)_*$ and some $y\in A_*$, but I don't know if this must be necessarily true if we have $d(x,A_*) = 0$. Any hints?
As $x\in A$ implies $d(x,A)=0$, we clearly have $A\subseteq A_*$, hence $A_*\subseteq (A_*)_*$.
For the other direction, consider $x\in (A_*)_*$. Let $\epsilon>0$. Then there exists $y\in A_*$ with $d(x,y)<\frac\epsilon2$ and $z\in A$ with $d(y,z)<\frac\epsilon2$. Hence $d(x,z)<\epsilon$ for some $z\in A$ and so $0\le d(x,A)<\epsilon$. As $\epsilon$ was arbitrary, we conclude $d(x,A)=0$ and so $so\in A_*$.
