Boundary of closed, proper subset of $\mathbb{R}^n$ has infinite points when $n \ge 2$ I am attempting to solve Exercise 1.7 from Multidimensional Real Analysis
Let $n \in \mathbb{N}\backslash\{1\}.$ Let $F$ be a closed subset of $\mathbb{R}^n$ with $int(F) \neq \emptyset$. Prove that the boundary of $\partial F$ of $F$ contains infinitely many points, unless $F = \mathbb{R}^n$.
Hint: Assume $x \in int(F)$ (Here $int(F)$ is the interior) and $y \in F^c$. Since both these sets are open in $\mathbb{R}^n$, there exists neighborhoods $U$ of $x$ and $V$ of $y$ with $U \subseteq int(F)$ and $V \subseteq F^c$. Check, for all $z \in V$, that the line segment from $x$ to $z$ contains a point $z'$ with $z' \neq x$ and $z' \in \partial F$.
I managed to prove the hint by taking the set $S = \{t \in [0,1] \mid v(t) \in int(F)\}$, where $v(t) = (1-t)x + ty$. Since $t = 0 \in S,$ (because $x \in int(F)$), $S$ is not empty. Because $S$ is bounded above by 1, there exists a $sup \> S$ = $m$. Using the fact that $int(F)$ and $F^c$ are open, I was able to show that $v(m) \notin int(F)$ and $v(m) \notin F^c$. So $v(m) \in F\backslash int(F) = \partial F$
However, I am stuck here and am unable to show that there are infinite points on the boundary. In particular, I know that there isn't a one-one correspondence between $x,z$ and $v(m)$, i.e: For different $x,z$ as described above there can be the same boundary point lying on the line joing $x$ and $z$.
Any help or a prod in the right direction will be appreciated. Thank you very much.
 A: (1). Take $x=(x_1,...,x_n)\in Int(F).$  Take $r>0$ such that $(y_1,...,y_n)\in  F$ whenever $\sum_{j=1}^n(y_j-x_j)^2<r^2.$
(2). For $t\in [0,2\pi)$ and for $b\in  \mathbb R^+$ let $z(t,b)=(z_1,..., z_n)$ where $z_1=x_1+b\cos t,$ and $z_2=x_2+b\sin t,$ and $z_j=x_j$ if $j>2.$ 
Let $R(t)=\{a>0: \{z(t,b):0<b<a\}\subset F\}.$ Let $M(t)=\sup R(t).$
With $r$ as in (1) we have $0<r\leq M(t)<\infty .$ And we have $\{z(t,b):0<b<M(t)\}\subset F.$
(3).  Since $M(t)>0$ we have $$z(t,M(t))\in Cl(\{z(t,b):0<b<M(t)\})\subset Cl(F)=F.$$ $$\text {Therefore }\quad z(t,M(t))\in F.$$
(4).  For every $s>M(t)$ there exists $s'$ with $M(t)<s'<s$ such that $z(t,s')\not \in F.$
.....  Because, otherwise,  for some $s>M(t)$ the set $\{z(t,b):0<b<s\}=\{z(t,b):0<b<M(t)\} \cup \{z(t,M(t)\}\cup \{z(t,b):M(t)<b<s\}$ would be a subset of $F$, implying, by the def'n of $R(t),$ that $M(t)=\sup R(t)\geq s>M(t)$, which is absurd.
.....Now the   distance from $z(t,M(t))$ to $z(t,s')$ is $s'-M(t)$, which can be arbitrarily small. Therefore $z(t,M(t))$ belongs to the closure of the complement of $F.$ Therefore, as $z(t,M(t))$ also belongs to $F$, we have $$z(t,M(t))\in \partial F.$$ 
(5). Suppose $z(t_1,M(t_1))=z(t_2,M(t_2)).$ Then the ordered pairs $(M(t_1)\cos t_1, M(t_1)\sin t_1)$ and $(M(t_2)\cos t_2, M(t_2)\sin t_2)$ are equal. But since $M(t_1)$ and $M(t_2)$ are positive and $t_1,t_2 \in [0,2\pi),$ this implies $t_1=t_2.$
So the map $\psi (t)=z(t,M(t))$  from $[0,2\pi)$ into $\partial F$ is $1$-to-$1$.  
A: Since $\bar{F}$ is open we have uncountably many points in $\bar{F}$. Fix $x\in int(F)$. Then we have for every point $z\in \bar{F}$ there exists a point of $\partial F$ on the line joining $x,z$. But there are uncountably many lines with different slopes since $n≥2$ and  each such line defines a unique point of $ \partial (F)$ since two straight lines intersect at at most one point. Thus there are uncountably many points.  To be more precise take a ball $B(z,r)\subset \bar{F}$. If $e_1, e_2,... e_n$ be the canonical basis there exists $\delta$ small enough such that $z+\delta e_i$ belongs to $B(z,r)$ for each $i$. Then we have $z-x \notin Span(e_j)$ for some $j$. Take the line joining $z$ and $z+\delta e_j$. And consider the lines joining $x$ and each point of this line. They are all distinct.
