Why is it necessary $n\geq 2$ in this problem about boundary and isolated point? Let $A\subset\mathbb{R}^n$ be an open set, where $n\geq2$. Prove that given $a\in\mathbb{R}^n-A$, are equivalent:
(i) the set $A\cup\{a\}$ is open;
(ii) $a$ is an isolated point of the boundary of $A$;
(iii) There exists $r>0$ such that $B(a;r)-\{a\}\subset A$.
Why is it necessary $n\geq 2$ in this problem? I'm confusing because I belive that I've proved (i)$\Leftrightarrow$(iii) and (iii) $\Rightarrow$(ii)  without using this hypothesis.
Furthermore, I would like to know how to prove (ii) $\Rightarrow$(i) (or (ii) $\Rightarrow$(iii)).
Thanks.
 A: You're right, you don't need $n \geqslant 2$ for the equivalence $(i) \iff (iii)$ or the implication $(iii) \Rightarrow (ii)$.
What you need $n \geqslant 2$ for is the implication $(ii) \Rightarrow (iii)$. Let's look at the example $A = (0,\,1) \subset \mathbb{R}^1$ to see where it breaks down for $n = 1$. Let's take $a = 0$. $a$ is an isolated boundary point, for example the $\frac12$-neighbourhood of $0$ doesn't contain another boundary point. Now, any $\varepsilon$-neighbourhood of $0$, for $0 < \varepsilon < 1$, consists of three parts


*

*$(0,\, \varepsilon) \subset A$,

*$\{0\} \subset \partial A$,

*$(-\varepsilon,\,0) \subset (\mathbb{R}\setminus A)^\circ$.


The part inside $A$ and the part "outside" $A$ are separated by the part on the boundary of $A$. That is generally so, for each $M \subset X$, where $X$ is a topological space, you have a disjoint union
$$X = \overset{\circ}{M} \,\dot{\cup}\, \partial M\, \dot{\cup}\, (X\setminus M)^\circ.$$
Now, if you have an isolated boundary point $p$ of $M$, a small enough punctured open neighbourhood of $p$ (that is, a $V \setminus \{p\}$, where $V$ is an open neighbourhood of $p$), splits into two disjoint open parts, one in the interior of $M$, the other in the exterior of $M$, since
$$V = (V \cap \overset{\circ}{M})\, \dot{\cup}\, (V \cap \partial M)\,\dot{\cup}\, \bigl(V \cap (X\setminus M)^\circ\bigr)$$
and $V\cap \partial M = \{p\}$ if $V$ is small enough.
For $X = \mathbb{R}^n$, we have that an $\varepsilon$-neighbourhood minus the centre point is connected when $n \geqslant 2$, but not connected when $n = 1$.
So in the case $n \geqslant 2$, either $(B_\varepsilon(a) \setminus \{a\}) \cap A$ or $(B_\varepsilon(a)\setminus \{a\}) \cap (\mathbb{R}^n \setminus \overline{A})$ must be empty, since by definition a connected set cannot be written as the disjoint union of two nonempty open parts. $(B_\varepsilon(a) \setminus \{a\}) \cap A$ cannot be empty, since $a \in \partial A$ and $a \notin A$, hence $(B_\varepsilon(a) \setminus \{a\}) \subset A$.
