show that $x \in A^o$ if and only if $d(A^c,x) > 0$ show that $x \in A^o \iff d(A^c,x) > 0$ where $d(A^c,x) = \inf_{y\in A^c} \lambda (x,y)$ where $\lambda$ is a metric and $(X,\lambda)$ is a metric space and $A^o$ is the set of interior points of A
I have tried to do the forward implication:
$x \in A^o \subset A \subset [A] = \{ x \in X : d(A,x) = 0 \}$ but then this means $d(A,x) = 0$ hence $d(A^c,x) > 0$ I am having troubles doing the reverse implication.
 A: Notice that this question is perfectly equivalent to your other question (to which I already gave an answer).
A: Consider $x\in A^{\circ}.$ Suppose that $d(A^c,x)=0.$ Then, for any $\epsilon>0$ there exist $y\in A^c$ such that $d(x,y)<\epsilon.$ But, since $x\in A^{\circ}$ there exists $r>0$ such that $B(x,r)\subset A^{\circ}.$ So, if we consider $\epsilon=r$ we get a contradiction because
$$d(y,x)<r \implies y \in B(x,r) \subset A^{\circ} \implies y\in A^{\circ},$$ which is not possible because $y\in A^c.$
Now suppose that $d(A^c,x)>0.$ Then, there exist $r>0$ such that $B(x,r)\cap A^c=\emptyset.$ That is, $B(x,r)\in A,$ which means, $x\in A^{\circ}.$
A: Proof. For the converse, we shall prove the contrapositive statement that $x \notin A^0 \implies d(A^c, x) \leq 0$. Suppose $x \notin A^0$. Note that the case where $x \in A^c$ is trivial. Suppose $x \in A$. Then $x$ is not an interior point of $A$. Then for all $\epsilon>0$, we have some $y \in A^c$ such that $y \in B_\lambda(x, \epsilon)$, giving us $d(A^c, x) \leq \lambda(x, y)<\epsilon$. It follows that $d(A^c, x) \leq 0$. $\square$
Anything I have called "trivial" takes a little bit of poking at the definition as to why. If you're stuck, shoot me a comment.
Hope that helps.
