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Let $U$ be a bounded simply connected domain in the plane. Let $K$ be the boundary (or frontier) of $U$. For every $\varepsilon>0$ is there a simple closed curve $S\subset U$ such that the Hausdorff distance between $S$ and $K$ is $<\varepsilon$?

I'm aware that there is an $S$ such that every point of $S$ is within $\varepsilon$ of some point of $K$. That is the Zoretti Theorem. But what I'm asking here is slightly different.

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  • $\begingroup$ With the notations of the given answer, the sequence of sets $h(\overline B_r)$ gives an exhaustion of $U$ by compact sets, and it is known that the Hausdorff distance behaves well under increasing sequence of nonempty compact sets, namely $d_H(h(\overline B_r),\overline U)\to0$ as $r\to1$. In particular $d_H(\Gamma_r,\partial U)\to0$. $\endgroup$
    – user111
    Commented Dec 7, 2023 at 10:02
  • $\begingroup$ But the last implication is not immediate, although it's based on standard facts. Since it is the point of the question it seems safer to give a complete proof. $\endgroup$
    – Pietro Majer
    Commented Dec 7, 2023 at 14:36
  • $\begingroup$ Sure, I agree ! $\endgroup$
    – user111
    Commented Dec 7, 2023 at 17:31
  • $\begingroup$ (for instance $X_k:=\{x: 1/k\le \|x\|\le 1-1/k\}$ is increasing and converges to $\overline B_1$ but $\partial X_k$ does not converge to $\partial B_1$) $\endgroup$
    – Pietro Majer
    Commented Dec 9, 2023 at 11:52

1 Answer 1

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Let $B_r(y)\subset\mathbb C$ denote the open disk of radius $r$ and center $y\in\mathbb C$, and $B_r:=B_r(0)$. Let $h:B_1\to U$ a homeomorphism (e.g. a Riemann mapping) . Then $\Gamma_r:=h(\partial B_r)$ is a simple closed curve in $U$ for $0<r<1.$ Moreover

  • For any $\epsilon>0$ there exists $0<\rho<1$ such that for all $\rho<r<1$ there holds $\Gamma_r \subset \partial U + B_\epsilon$. Indeed $h^{-1}\big(U\setminus(\partial U + B_\epsilon)\big)$ is a compact subset of $B_1$, so it is included in some $B_\rho$ for $\rho<1$, whence $U\setminus(\partial U + B_\epsilon)\subset h(B_\rho)$, that is $\Gamma_r\subset h(B_1\setminus B_\rho)\subset \partial U + B_\epsilon$ for all $\rho<r<1$.
  • For any $\epsilon>0$ there exist $\sigma>0$ such that for all $\sigma<r<1$ there holds $\partial U\subset \Gamma_r + B_\epsilon$. This follows easily by compactness, arguing by contradiction: otherwise, there would exist $\epsilon_0>0$ such that for a sequence $r_k <1$, $r_k\to 1$ and $y_k\in\partial U$ one has $\emptyset\neq \partial U\setminus (\Gamma_{r_k} + B_{\epsilon_0})\ni y_k$, that is $ B_{\epsilon_0}(y_k)\cap \Gamma_{r_k}=\emptyset.$ Up to extracting a sequence , $y_k\to y_\infty\in\partial U$ and $ B_{\epsilon_0/2}(y_\infty)\cap \Gamma_{r_k}=\emptyset$ eventually, a contradiction because since $y_\infty\in\partial U$, any nbd of $y_\infty$ contains points of $\Gamma_r$ for all $r<1$ close enough to $1$.

Therefore the Hausdorff distance is $d_{\mathcal H}(\partial U, \Gamma_r)=o(1)$ as $r\to 1$.

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  • $\begingroup$ Why is the last sentence of the second bullet true? That is, why are all points of $\partial U$ approached by $\Gamma_r$ as $r$ tends to one? I see this holds when $h$ extends continuously to $\partial B_1$. But there are examples where that does not happen... $\endgroup$
    – Sam Nead
    Commented Dec 7, 2023 at 8:21
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    $\begingroup$ In the notations above, let $A$ be the connected component of $y_\infty$ in $U\cap B_{\epsilon_0/2}(y_\infty)$. Then $h^{-1}(A)$ is a connected open subset of $B_1$ whose closure meets $\partial B_1$. It follows that the set of $r<1$ such that $\partial B_r\cap V\neq \emptyset$ is a non-empty interval $(a,1)$, which gives the last sentence applying $h$. $\endgroup$
    – Pietro Majer
    Commented Dec 7, 2023 at 9:51
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    $\begingroup$ Since $y_\infty$ is not a point of $U$ it does not lie in the intersection $U \cap B_{\epsilon_0/2}(y_\infty)$. So $A$ is not well-defined. Perhaps the following works: "let $A$ be any connected component of $U \cap B_{\epsilon_0/2}(y_\infty)$." Also, is $V = h^{-1}(A)$, on the last line? $\endgroup$
    – Sam Nead
    Commented Dec 7, 2023 at 11:15
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    $\begingroup$ Yes, exactly, that’s what I was also thinking. I think it is OK because any connected component $A$ has a closure that meets $\partial U$. Yes, $V=h^{-1}(A)$. Thank you! $\endgroup$
    – Pietro Majer
    Commented Dec 7, 2023 at 11:27

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