Boundary of a bounded open set in $\mathbb{R}^2$ Does the boundary of a bounded open set in $\mathbb{R}^2$ necessarily have infinite points? How do we prove that, or is there a counterexample?
It seems true to me, but I haven't been able to find a way to justify it.
Can we generalize this to $\mathbb{R}^n$ for $n>1$?
 A: The boundary of the open set $\mathbb R^2 \backslash \{ (0,0) \}$ has only one point.
A: As @DanielWainfleet already pointed out, the answer for your question is yes.
Lets see how far we can go on this.
For starters, we ask: when does happen that $X\subset\mathbb{R}^2$ has a finite boundary? 
Two obvious cases are when $X$ or its complement are finite;
and it turns out these are the only possibilities.
The argument I came up does 
depend on two facts:


*

*Let $X,C$ be subsets of any topological space $T$. If $C$ is connected and contains points of $X$ and $T\setminus X$, then $C$
intersects the boundary of $X$.

*$\mathbb{R}^2$ minus any enumerable subset is path connected.
The first is well-known. The second it is intuitively clear, and it may be fun to try yourself. I leave it for you.
And here we go:
Let $X\subset\mathbb{R}^2$ so that both $X$ and $\mathbb{R}^2\setminus X$ are infinite. Then the boundary of $X$ is also infinite.
To see this, choose $x_1, x_2, \dotsc \in X$ and $y_1, y_2,\dotsc \in \mathbb{R}^2\setminus X$ 
pairwise distinct points. Let $C$ be a path joining $x_1$ to $y_1$ and avoiding
$x_i, y_i$ for $i>1$. Then $C$ meets the boundary of $X$, say at a point $b_1$.
Now, choose another path, this time joining $x_2$ to $y_2$ but avoiding $b_1$ and
$x_i, y_i$ for $i>2$; this is possible because $b_1 \not\in\{x_2,y_2\}$.
This new path will meet the boundary of $X$ at a point $b_2$ distinct from $b_1$.
Proceed inductively: for each $n$, take a path joining $x_n$ to $y_n$ but avoiding $b_j$ for $j<n$ and $x_i, y_i$ for $i>n$; as before, such a path exists and
meets the boundary of $X$ at a point $b_n$ distinct from $b_1,\dots,b_{n-1}$. $\Box$
A: Let $ S$ se a non-empty bounded open set in $R^2$. (Digression : the boundary of $\phi$  is  $\phi$.) Choose some   $P=(x,y) \in S$ . For each  $\theta \in (-\pi, \pi]$ there exists  $d>0$ such that $$\{ ( x+r \cos \theta , y+r \sin \theta ) : r \in [0,d) \} \subset S.$$ Let $ D(\theta)$ be the least upper bound of such $ r$ .  It  exists because $S$ is bounded. Then $$P_{\theta}= (x+D(\theta) \cos \theta, y+D(\theta) \sin \theta) \not \in S$$ but $P_\theta $ is in the closure of S. (Is this obvious?) Therefore, since $S$ is open, $ P_{\theta}$ belongs to the boundary of S. Obviously $\theta (1) \ne \theta (2)$  implies $P_{\theta (1)} \ne P_{\theta (2)}$, so there are " at least as many" boundary points of $S$ as points in $(-\pi,\pi]$.
A: The boundary of $\emptyset$ is $\emptyset$. 
