Is there a easy way of doing these by contradiction in a short concise proof that connects them all.

Question

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

• (i) $\;\ a$ is a contact point of $A \iff$ (iii)
• (ii) $\;a \in \overline{A}\iff$(iv)
• (iii) $a$ is not an interior point of $B\implies $ (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 $A\iff$ any ball centered on $a$ intersects $A\iff$ any ball centred on $a$ is not a subset of $B = X \setminus A \iff$ no ball centred on $a$ is a subset of $B\iff$ $a \notin B^\circ $

Answer 1

Indeed $a \in \overline{A}$ 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 $B = X\setminus A$ (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 $a_n \in B(a, \frac{1}{n}) \cap A$ 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 $0< r\le d(a,A)$ 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.