# 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 $n>1$?

• In fact, in dimension $n$, the Hausdorff dimension of the boundary of a bounded open set is at least $n-1$: see here. Commented Feb 19 at 22:09

The boundary of the open set $\mathbb R^2 \backslash \{ (0,0) \}$ has only one point.

• Okay, I meant a bounded open set. I am editing the question. Commented Sep 3, 2015 at 6:06

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 $(-\pi,\pi]$.

The boundary of $\emptyset$ is $\emptyset$.

• This, depending on intentions of the author of the problem, may be the bright counterexample or annoyance because of the miswording. Commented Sep 3, 2015 at 14:29

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:

1. 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 $$X$$.

2. $$\mathbb{R}^2$$ 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 $$j and $$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 variant od f this idea: By contradiction, if $X$ and its complement $X^c$ are infinite and $\partial X (=\partial X^c)$ is finite, let $p\in X \setminus \partial X$ and $q\in X^c$ \ $\partial X^c.$ There is an open ball $B$ with $p\in B\subset int(X)$ ( else $p\in X\cap \overline {X^c}\subset \partial X$ ). Let $p\ne p' \in B$ and $U= \{rp+(1-r)p':r\in [0,1]\}.$ Then $U \subset B\subset int (X).$ For each $p''\in U$ choose $f(p'')\in \partial X \cap \{sp''+(1-s)q:s\in (0,1)\}$. Then $f$ is 1-to-1, contrary to $\partial X$ being finite. Commented May 5, 2019 at 8:45
• Cool. Only have to be careful to choose $p'\in B$ outside the line $\bar{pq}$. Nice argument!
– rts
Commented May 5, 2019 at 18:54
• Yes. I didn't think of $p'$ possibly being on $\overline {pq}.$ This (obviously) works in any $\Bbb R^n$ with $n>1$. Commented May 5, 2019 at 20:19