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If $A subseteq X$, $B = X setminus A, $ and $a in X$

• (i) $; a$ is a contact point of $A iff$ (iii)


• (ii) $;a in overlineAiff$(iv)


• (iii) $a$ is not an interior point of $Bimplies $ (vi)


• (iv) $,a notin B^circ iff$ (iii)


• (v) $;$there exists a sequence $(a_n) subseteq A$ with $a_n to a$ $implies$ (i)


• (vi) $,d(a, A) = 0$, where $d(a, Z) = inf,d(a, z) : z in Z implies$ (v)

(i)$iff$(iii) is from definitions: $a$ is a contact point of $Aiff$ any ball centered on $a$ intersects
$Aiff$ any ball centred on $a$ is not a subset of $B = X setminus A iff$ no ball centred on $a$ is
a subset of $Biff$ $a notin B^circ $

Indeed $a in overlineA$ iff (by definition?) $a$ is a contact point of $A$ (often rather called an adherence point of $A$) iff every open ball around $a$ intersects $A$ (definition of contact point) iff there exists no open ball around $a$ that is a subset of $B = Xsetminus A$ (just by logic and definition of inclusion and complement) iff $a notin B^circ$ (definition of interior).

These facts hold in all topological spaces (if you replace open ball around $a$ by "open set containing $a$" the same simple proof holds).

The others are more metric specific: From (i) ($a$ is a contact point) we directly prove (v) by picking $a_n in B(a, frac1n) cap A$ by the contact point definition and then showing that $a_n to a$ by the definition of convergence. (v) to (vi) is rather easy too: if $varepsilon>0$ and $a_n to a$ we can pick some $a_n$ from the sequence (so $a_n in A$) with $d(a,a_n) < varepsilon$ and then $0 le d(a,A) le d(a, a_n) < varepsilon$. As $varepsilon >0$ is arbitrary, $d(a,A)=0$ follows. (vi) to (i) is similar: if $B(a,r), r>0$ is an open ball around $a$, $r$ cannot be a lower bound for $d(a,x): x in A$ as this would imply $0< rle d(a,A)$ while $d(a,A)=0$, so for some $x in A$, $d(a,x) < r$ and so $A cap B(a,r) neq emptyset$ and (i) holds.

This second loop $(i) to (v) to (vi) to (i)$ connects all equivalent formulations together.