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.


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 $\gamma$.

  • $\begingroup$ Wlog $P$ is a disc i.e., $P = B(x,r)$. Now take a connected component $C$ of $\text{int}(P \cap Q)$. Now take a point $b \in \partial P \cap \partial Q \cap \partial C$ and from $b$ travel along $\text{int}(P) \cap \partial Q$. At some point the traversal will end at $a \in \partial P \cap \partial Q$. Now traverse along the circle bounding $P$ from $a$ to $b$. This will be your boundary jordan curve. I think this can be made rigorous. Cheers ! $\endgroup$
    – Balaji sb
    Sep 12, 2023 at 16:38

1 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.


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

  1. 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.


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


  1. 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 $\gamma_n.$


  1. (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.
  • 1
    $\begingroup$ Can you provide more detail in regard to point 6? I can't see the proof for the claim. Also I am not sure what you mean by 'finite number of intervals from $\partial R\cap (Q_1 \cup P_1)$'. $\endgroup$
    – Kulisty
    Sep 13, 2023 at 20:35
  • $\begingroup$ I have updated answer and removed confusing reference to 'finite number' (it is actually explained in p.7). $\endgroup$
    – dsh
    Sep 14, 2023 at 22:32
  • $\begingroup$ @Kulisty Note that in your updated post with infinite $L_i,$ you need essentially use compactness of $\partial Q$ somehow. Consider function, which is equal to 0 on Kantor set and hump of height 1 on removed intervals. This is not continuous function (as it is not uniformly continuous on compact) and if you connect points (0,0) and (1,0) of the graph of this function then you obtain limit of closed simple curves, which is not simple curve. $\endgroup$
    – dsh
    Sep 15, 2023 at 13:34
  • 1
    $\begingroup$ Point 7 is still unclear to me. Corresponding endpoints of what? Please explain in more detail how the curve $\gamma_n$ is constructed in the finite step. $\endgroup$
    – Kulisty
    Sep 20, 2023 at 8:02
  • $\begingroup$ Let me know if there are unclear points remain. $\endgroup$
    – dsh
    Sep 20, 2023 at 13:18

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