There exists a region bounded by a Jordan curve on the plane.

Question

For me, a region is a set bounded by a Jordan curve (homeomorphic image of a circle) on the plane $\mathbb R^2$ together with the boundary (it follows from Jordan curve theorem that the complement of a Jordan curve consists of two connected sets, exactly one of which is bounded). Here $\partial R$ denotes the boundary of a region $R$ (that is a Jordan curve). I need help proving the following.

Given two regions $P,Q$ and a point $x\in \mathrm{int}(P\cap Q)$, there exists a region $R$ such that $\partial R\subseteq \partial P \cup\partial Q$ and $x\in\mathrm{int} R\subseteq \mathrm{int}(P\cap Q)$.

My idea is to define $R$ as the closure of the connected component of the point $x$ in the set $\mathrm{int}(P\cap Q)$, but I don't know how to prove that the boundary of $R$ is a Jordan curve.

Edit (inspired by dsh's answer)

Let $P,Q$ be regions and $x\in \mathrm{int}(P\cap Q)$, Consider the set $Q_1= \partial Q\cap\mathrm{int}\,P$. The set $Q_1$ is open in $\partial Q$ and if $\partial Q\not\subseteq\mathrm{int}P$, then $Q_1$ is a finite or countable union of arcs, each one having the endpoints on $\partial P$.

First assume $L_1,\ldots, L_n$ are all such arcs. Then one can prove by induction that there exist (unique) $n+1$ pairwise disjoint regions $J_1,\ldots,J_{n+1}$ such that $$\partial J_i\subseteq \partial P \cup (L_1\cup\ldots\cup L_n)\text{ and}$$ $$\mathrm{int}\,P=(\mathrm{int}\, J_1\cup\ldots\cup\mathrm{int}\, J_{n+1})\cup (L_1^*\cup\ldots\cup L_n^*),$$ where $L_i^*$ is the arc without its endpoints. Then $x\in \mathrm{int}\,J_i$ for some $i$ and since $x\in\mathrm{int}\, Q$ and $\mathrm{int}\,J_i\cap \partial Q=\emptyset$, we get $\mathrm{int}\,J_i\subseteq\mathrm{int}\,Q$.

Now assume that there are infinitely many such arcs $L_1,L_2,L_3,\ldots$ For each $n\in\mathbb N$ consider the regions $J_1^n,\ldots,J_{n+1}^n$ that exist for arcs $L_1,\ldots, L_n$ as in the finite case, while $J_1^0=P$. For each $n$ the point $x$ belongs to the interior of exactly one of the regions $J_1^n,\ldots,J_{n+1}^n$, call it $C_n$. Then $C_{n+1}\subseteq C_n$ and for each $n$ there are arcs $K$ and $M$ both with endpoints $a,b$ on $\partial C_n$ such that $K^*\subseteq \mathrm{int}\,C_n$ and $K\cup M=\partial C_{n+1}$. Choose a sequence of parametrizations $\gamma_n\colon S^1\rightarrow\partial C_n$ such that $\gamma_{n+1}|I=\gamma_n|I$, where $\gamma_n[I]=M$ and $M$ is as above.

I want to show that for any $s\in S^1$, $\gamma_n(s)$ is eventually stable with respect to $n$ and hence the pointwise limit $\gamma=\lim_{n\to\infty}\gamma_n$ exists and is a parametrization of the required Jordan curve. Any help appreciated.

EDIT

Let me share my other idea based on dsh's answers (which I don't fully understand).

Let $P,Q$ be regions and $x\in \mathrm{int}(P\cap Q)$, I define $R$ to be a connected component of $x$ in $\mathrm{cl}\,\mathrm{int}\,(P\cap Q)$. Then $\partial R\subseteq \partial P\cup\partial Q$.

Next, one can inductively define a sequence $(P_n)$ of regions and a sequence $(\gamma_n)$ of homeomorphisms $\gamma_n\colon S^1\rightarrow \partial P_n$, while $P_0=P$ and $\gamma_0$ is arbitrary, such that we have the following.

1. $P_{n}\subseteq P_{n-1}$ for $n\in\mathbb N$.
2. $R\subseteq \bigcap_{n}P_n$.
3. For each $n\in\mathbb N$ there exists a finite family of arcs $\mathcal I_n$ such that each $I\in\mathcal I_n$ is contained in $\partial R\cap \mathrm{int}\,P$, endpoints of $I$ lie on $P$ and divide $P$ into two arcs $J_I,K_I$, where the Jordan curve $I\cup J_I$ encloses $x$, arcs $K_I$ are disjoint (except possibly for the endpoints) for different $I\in\mathcal I_n$, there exist homeomorphisms $f_I\colon K_I\rightarrow I$, and $$\gamma_n(t)=\begin{cases}f_I(\gamma_0(t)) &, \text{ if }\gamma_0(t)\in K_I\\\gamma_0(t) &,\text{ otherwise}\end{cases},$$ and moreover $\mathcal{I}_{n}\supseteq \mathcal I_{n-1}$ for all $n\in\mathbb N$, while functions $f_I$ are the same for different $n$.
4. If $y\in \partial R$, then $\mathrm{dist}(y,\partial P_n)<\frac1{2^n}$ for $n>0$.

It is clear that the sequence $(\gamma_n)$ is pointwise convergent. I can't see however why the limit function $\gamma$ is continuous and why $\partial R$ is the image of $\gamma$.

Answer 1

Let me try general approach from books i.e. use approximation of boundary with smooth (Whitney approximation theorem) or piece-wise linear maps and then use uniform convergence of maps.

Using Schoenflies theorem one can assume for clarity that one of the region is unit ball ($P = B$) and $S^1$ is boundary of $B.$

1. $\partial R \subseteq \partial Q \cup \partial P$ and is compact.
2. If $\gamma_B:S^1\to \mathbb{R}^2$ and $\gamma_Q:S^1\to \mathbb{R}^2$ are boundaries of regions $B/Q.$ Assume $Q \neq B.$
3. $Q_1 = \gamma_Q(S^1) \cap \mathrm{int}(B)$ is union of countable number of sets, each is homeomorphic to open subinterval of $[0,1]$ with ends on unit circumference (from description of open subset of $S^1$).
4. Each interval $I$ of $Q_1$ splits $B$ into 2 pieces, one of which contains $x$. Endpoints of $I$ splits $S^1$ to 2 pieces $S, S'.$ Assume $S\cup I$ bounds region containing $x.$ $I$ can be reparametrized using $S'$ ("opposite" arc).
5. $B_1$ is defined similarly and has the same topological structure.
6. If $I$ is connected component of $Q_1$ or $B_1$ and intersects with $\partial R$ then it is contained in $\partial R.$
7. Cover $\partial R$ with finite net of balls of radius $1/2^n.$ If center is in $Q_1$ or $B_1$ replace it with corresponding endpoints. Connect finite set of points with pieces of $\gamma_B/\gamma_Q$ with reparametrization from above (all centers of balls from covering should be on the curve). Use some order of endpoints and centers (lying on boundary of $B$).
8. Obtained Jordan curve $\gamma_n$ approximating $\partial R$ with prescribed accuracy: $\max\{dist(\gamma_n(y), \partial R): y\in S^1\} < 1/2^n.$
9. On the next step, $\gamma_{n+1}$ is changed only on pieces that are NOT from $\partial R\cap(Q_1 \cup B_1).$
10. Limit $\gamma$ of Jordan curves $\gamma_n$ has $\gamma(S^1) \subseteq \partial R$ and is closed continuous curve. See below UPD 4.
11. This p. does not depend on p.10. If $\gamma_n(y)\in\partial B$ then $\gamma_n(y)=y.$ Any interval of $\partial R\cap(Q_1 \cup B_1)$ is eventually (wrt $n$) on $\gamma_n$ and stable from that moment on. So $\gamma_n\to\gamma$ pointwise. $\gamma$ is one-to-one: take $a, b\in S^1$ then $\gamma_n(a)$ and $\gamma_n(b)$ are stable from some moment $N,$ use bijectivity of $\gamma_N.$
12. $\gamma$ is bijective continuous map of compact and so is homemorpthism on image, so $\gamma$ is Jordan curve.
13. $\gamma(S^1)$ bounds region of $x$ and is contained in $\partial R.$ Thus $\gamma(S^1)=\partial R$ for otherwise there are either points of exterior of $B$ or $Q$ inside $\gamma$ or there are points of $x$'s component outside $\gamma.$

Existence and continuity of $\gamma$: $\forall y\in S^1:|\gamma_n(y) - \gamma_m(y) | < 1/2^n + 1/2^m$ so $\gamma(x)=\lim \gamma_n(x)$ exists and continuous as each $\gamma_n$ is continuous. See UPD 4.

I believe that all points are valid and can be clarified.

UPD 1

Replaced $P_1/P$ with $B_1/B$ everywhere to avoid confusion.

6. Assume $I\subseteq B_1.$ Note that for each $y\in \partial R \cap I$ there is some closed ball $B_y$ around $y$ such that $B_y\cap \gamma_Q(S^1)=\emptyset$ (as $\gamma_Q(S^1)$ is closed). $I'=I\cap B_y$ is homeomorphic to closed interval (for sufficiently small radius) and splits $B_y$ into 2 pieces (using good shape of $B$), one piece contains points from $R.$ The only candidates for $\partial R$ is some subset of interval $I'$ by p.1. If the other component of $B_y$ contains points in $R$ then $y$ is not in $\partial R.$ So $I'\subset \partial R.$ If $I' \neq I$ then one can connect $R$ and $\mathrm{int}(\mathbb{R}^2\setminus R)$ using continuous curve. Case $Q_1$ is true by symmetry.

Consideration which avoids usage of Schoenflies theorem need special treatment. So assume $I\subseteq Q_1.$ This is the place where one needs to assume that $Q_1$ is not arbitrary countable set of intervals but obtained from compact $\gamma_Q(S^1).$ Assume from the start that $B_y$ does not intersect $\gamma_B(S^1)$ as $\gamma_B(S^1)$ is closed.

One needs to find $B_y$ with only one component in $B_y\cap I.$ $\mathrm{int}(B_y) \cap I$ consists of countable number of connected open intervals, one of them $I'$ contains $y.$ $dist(y, \gamma_Q(S^1)\setminus I') > 0,$ so one can find sufficiently small radius for $B_y$ such that $B_y\cap I \subset I',$ but new connected components may emerge. Taking points on those redundant components, by compactness there is converging sequence of points in some compact and we get self-intersection of $\gamma_Q.$ Contradiction.

UPD 2

8. $\max\{dist(\gamma_n(S^1), y): y\in\partial R\} < 1/2^n$ from the radii of covering.

.

11. Any interval of $\partial R\cap(Q_1 \cup B_1)$ is eventually (wrt $n$) on $\gamma_n$ and stable from that moment on.

This is due to the fact that at p.7 we use finer and finer covering of $\partial R.$ If $y\in I,$ $I\subseteq\partial R\cap Q_1$ connected interval, then there is some ball $B_y$ such that $B_y\cap\partial R \subseteq I$ (as in p.6). It means that finite covering of p.7 with radius smaller than the radius of $B_y$ should have ball with center in $B_y\cap\partial R.$ From that moment $I$ is in $\gamma_n.$

UPD 3

7. (With 9.)
   1. Assume we have Jordan curve $\gamma'$ bounding the region, which contains path-connected open component of $\mathbb{R}^2\setminus (\partial B\cup\partial Q)$ containing $x.$
   2. Assume $\gamma'$ consists of finite number of open intervals (with their enpoints) and each interval lies in either (1) $B_1$ (with identity parametrization), in (2) $Q_1$ with parametrization from p.3, or in (3) $\partial B$ with identity parametrization. All endpoints of intervals are on $\partial Q\cap\partial B.$
   3. Construction starts with $\gamma'(S^1) = \partial B$ if $Q\not\subseteq B,$ otherwise we are done.
   4. Denote union of open intervals of $\gamma'$ from $B_1$ and $Q_1$ by $C.$ $\partial R\setminus C$ is compact because $C = O \cap \partial R$ where $O\subset \mathbb{R}^2$ is open by proof of p.6.
   5. Cover $\partial R\setminus C$ by balls of radius $\epsilon.$ Take finite subcover with centers $\{x'_1, \ldots, x'_m\}.$ If $x'_i\in Q_1\cup B_1$ by p. 6 replace it with endpoints $y', y''\in \partial R$ using pp.3 (for $Q_1$, trivial for $B_1$) and 6. Add endpoints of intervals from $C.$ Remove duplicate points. We obtained $X=\{x_1, \ldots, x_n\}.$
   6. Note that $X\subseteq \partial B \cap\partial Q\cap \gamma'(S^1)$ because endpoints of intervals from $Q_1/B_1$ lies on $\partial B\cap \partial Q$ and using p.7.1.
   7. Enumerate all points using order inherited from $\partial B.$
   8. All intervals from $Q_1$ and $B_1$ (either from $\gamma'$ or from p.7.5) are consecutive in ordered $X.$ Use p.6 and p.7.1.
   9. Connect points of $X$ using intervals from $B_1/Q_1/\partial B$ and note that newly constructed $\gamma''$ satisfies 7.1 and 7.2.
   10. Also $\gamma''$ satisfies p.8 inequalities for $\epsilon.$

All this about proving that limit curve is continuous (using uniform convergence from p.10) on $\partial R\cap\partial B\cap \partial Q,$ which may look like Cantor set.

UPD 4.

Uniform continuity.

• Assume $\gamma_n$ is not uniform Cauchy sequence. Then $\exists\epsilon, \forall N\in\mathbb{N}, \exists x_N\in S^1, m, n\ge N$ such that $|\gamma_m(x_N) - \gamma_n(x_N)| \ge \epsilon.$ It follows (from eventual stability, p.11) that $x_N=\gamma_m(x_N)\in\partial B$ and $\gamma_n(x_N)\in\partial R \cap Q_1$ or the vice versa. Assume $\gamma_m(x_N)\in \partial B.$
• Extract subsequence and assume WLOG that $x_N\to a\in\partial B$ and $\gamma_n(x_N)\to g\in\partial R\cap\partial Q.$
• $g\in \partial R\cap \partial Q$ and $g\not\in Q_1$ for otherwise (that is if $g\in Q_1$) we would get contradiction with p.11
• Let $I_n$ to denote open interval in $Q_1\cap\partial R$ containing $\gamma_n(x_N).$ $g\not\in \overline{I_n}$ because only 2 intervals may have $g$ as an endpoint.
• This means that there is no ball $B_g$ around $g$ such that $B_g\cap\partial Q$ contains only one connected component (as proved for p.6). Contradiction with assumption that $\gamma_n$ is not uniform Cauchy.