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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 $n>1$?

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

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

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  • $\begingroup$ Okay, I meant a bounded open set. I am editing the question. $\endgroup$
    – MathManiac
    Commented Sep 3, 2015 at 6:06
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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 $(-\pi,\pi]$.

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

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  • $\begingroup$ This, depending on intentions of the author of the problem, may be the bright counterexample or annoyance because of the miswording. $\endgroup$
    – eudes
    Commented Sep 3, 2015 at 14:29
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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 $j<n$ 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 $b_1,\dots,b_{n-1}$. $\Box$

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    $\begingroup$ 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. $\endgroup$ Commented May 5, 2019 at 8:45
  • $\begingroup$ Cool. Only have to be careful to choose $p'\in B$ outside the line $\bar{pq}$. Nice argument! $\endgroup$
    – rts
    Commented May 5, 2019 at 18:54
  • $\begingroup$ Yes. I didn't think of $p'$ possibly being on $\overline {pq}.$ This (obviously) works in any $\Bbb R^n$ with $n>1$. $\endgroup$ Commented May 5, 2019 at 20:19

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