# Points in the boundary of a compact set $K\subset\mathbb{R}^2$ reachable by a path in $K^c$

Let $K\subset\mathbb{R}^2$ be compact. Let the path boundary of $K$ denote the set of points in $z\in K$ such that for some point $w\in K^c$, there is a continuous path $\gamma:[0,1]\to\mathbb{R}^2$ such that

1. $\gamma(0)=w$.
2. $\gamma((0,1))\subset K^c$.
3. $\gamma(1)=z$.

Of course the path boundary of $K$ is contained in the boundary of $K$, and it is not hard to find a set whose path boundary is a strict subset of its boundary. Take for example the block $[-1,1]\times[-1,1]$, and remove the sets $\left\{(x,y):y>0,\dfrac{1}{n^2}<x<\dfrac{2}{n^2}\right\}$, for $n\geq2$. The resulting set $K$ is compact (countably many open sets have been removed), and the set $\{(0,y):0\leq y<1\}$ is contained in the boundary of $K$ but not in the path boundary of $K$.

My question is: Is there a nice characterization of the sets for which the path boundary is equal to the boundary? Even a characterization for the case in which both $K$ and $K^c$ are connected would be welcome.

As a cautionary tale, I will say that the answer cannot depend just on smoothness of $\partial K$, since if $g:[0,1]\to[0,\infty)$ is any continuous function, then the path boundary of the set $K=\{(x,y):0\leq x\leq1,0\leq y\leq g(x)\}$ is equal to the boundary of $K$, and of course this $K$ can have quite jagged boundary.

PS: If someone has a much better name than "path boundary", feel free to change the question.

• Maybe being simply connected is strong enough? I can't think of a counterexample off the top of my head Commented Feb 23, 2016 at 2:46
• @TylerHG, The example I gave in the problem is simply connected, but the path boundary is a strict subset of the boundary. Commented Feb 23, 2016 at 2:49
• local compactness may have something to do with this Commented Feb 28, 2016 at 2:35

This is more of a long comment than a full answer.

Local connectedness of $\partial K$ is a sufficient condition for the topological boundary of $K$ to coincide with the path boundary of $K$.

Proposition: Suppose $K \subset \Bbb R^2$ is compact and connected. If $\partial K$ is locally connected, then every point on $\partial K$ is on the path boundary of $K$. In particular, the two concepts coincide.

Proof: Assume that $K^c$ is connected (this will result in no loss of generality). Let $\hat{\Bbb C}=\Bbb C \cup \{ \infty\}$ denote the Riemann sphere, and $\Bbb U:=\{z: |z|<1\}$ the unit disk. Define $U:=\hat{\Bbb C} \backslash K$, which is a simply connected subset of $\hat{\Bbb C}$. By Theorem 2.1 of this book, there exists a conformal map $f$ from $\Bbb U \to U$ which extends continuously to a (surjective) map from $\partial \Bbb U \to \partial U=\partial K$. Now take any point $w \in \partial K$. Let $z\in f^{-1}(\{w\})$. Consider the path $\gamma$ defined by $\gamma(t):=f(zt)$. Then $\gamma[0,1) \subset f(\Bbb U) = U$ and $\gamma(1)=f(z)=w$. We can assume without loss of generality that $\gamma[0,1)\not\ni \infty$, and hence $w$ is in the path boundary of $K$. $\Box$

Remark 1: The converse of this proposition is not true, i.e, local connectedness is not a necessary condition for the topological boundary of $K$ to coincide with the path boundary of $K$. For example, if we define $K$ to be the closure of the set of all points $(x,y)$ for which $0 \leq x \leq 1$ and $-2 \leq y \leq \sin(1/x)$, then the boundary and path boundary coincide even though $\partial K$ is not locally connected. However, you'll notice that the points on $\partial K$ of the form $(0,y)$ with $0 \leq y \leq 1$ are only accessible from the "left" and not from the "right." Hence local connectedness still plays a role in terms of accessibility of the points.

Remark 2: The reason why the assumption that $K^c$ was connected resulted in no loss of generality was because all of the connected components of $K^c$ play a symmetric role (as subsets of $\hat{\Bbb C}$), hence it suffices to assume that an arbitrary one contains $\infty$, which is precisely what we did.

• Thanks very much for the interesting answer. It is not quite the full answer, but I decided to award the bounty anyway; why let 100 points go to waste? Your example was also instructive. Commented Mar 3, 2016 at 16:14
• Okay thanks. I suspect that some form of a converse to the above proposition holds, it might just require an appropriate tightening on the condition of being a path boundary point. Unfortunately I haven't been able to think of one. Commented Mar 3, 2016 at 17:08

It looks like your path boundary is more commonly known as the set of accessible points. As long as the boundary is a Jordan curve, it is accessible (from either the inside or outside). This follows from the Jordan-Schoenflies theorem, as explained here or here.

For the converse, here is Theorem 1 of the first paper I referenced (it is attributed to Schoenflies): If F is a compact set in $\mathbb{R}^2$ with precisely two regions such that every point of F is accessible from each of those regions, then F is a simple closed curve.

• Thank you for the references. I am more curious about the case where the boundary is not a Jordan curve as well. Do you know if this is treated in the literature? Commented Feb 29, 2016 at 18:01
• Are you able to access the first paper I referenced? Here is Theorem 1 (which is attributed to Schoenflies): If F is a compact set in $\mathbb{R}^2$ with precisely two regions such that every point of F is accessible from each of those regions, then F is a simple closed curve. Commented Feb 29, 2016 at 18:16
• @MichaelHarrison That sounds like something that could be integrated into the answer (the statement, not the proof, unless it's particularly short). Commented Feb 29, 2016 at 21:42
• True, I've added it. Commented Feb 29, 2016 at 21:57
• I am not able to access the first paper, but I have looked through the second paper, and found the first result you mentioned. I will note that the accessible points are in general a strict subset of the "path boundary". Indeed, the set of accessible points may be empty, while the path boundary of a compact set is never empty. This is because each accessible point in $K$ must be reachable by a path from every point in $K^c$. Commented Mar 2, 2016 at 13:49