Suppose $X$ is a simply connected closed subset of $\mathbb R^2$. Let $a,b$ belong to $X$. Is it true that there is at most one curve inside $X$ from $a$ to $b$ such that the length of the curve is minimal?

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  • @XipanXiao And then? (The question assumes a closed region, btw. And it is not asked for existence but for uniqueness in case of existence). – Thomas Mar 25 '14 at 17:55
  • Sorry didn't notice it is "closed". – Xipan Xiao Mar 25 '14 at 19:00
  • Can someone show me why closed is essential here? – M.B. Mar 26 '14 at 18:48
  • @M.B. Might be it is not essential. The closedness might have something to do with the existence of curve of minimal length, not the uniqueness. –  Mar 27 '14 at 03:07
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    @M.B.: It's required for existence. Let $X$ be the plane with the non-negative $x$-axis removed and consider the points $(0,1)$ and $(0,-1)$. You can get arbitrarily close to length 2 by deforming the line segment joining these to the left of the origin. – Anthony Carapetis Mar 27 '14 at 04:23
  • Sure, but the problem states "at most one". – M.B. Mar 27 '14 at 05:09
  • The results of googling [ergodic curves](http://www.google.com/search?q=%22ergodic+curves%22) might be useful. – Dave L. Renfro Mar 27 '14 at 16:12
  • @M.B. If the curve is not a straight line in the neighbourhood of a given point, it will touch the boundary in points where it is not. If the region is not bounded this is not possible (existence is assumed to hold). – Thomas Mar 27 '14 at 17:03
  • @M.B. (I meant if the region is not closed) – Thomas Mar 27 '14 at 17:15
  • @Thomas: OK, but I cannot see why that it implies that the result is not true for open sets. – M.B. Mar 27 '14 at 17:40
  • @MB It is, but it is rather easy to prove then -- you just have to prove the statement I just made, then you know each point is an interior point of $X$ and then you can easily see it must be a straight line. – Thomas Mar 27 '14 at 17:42
  • @Thomas: OK, thanks. I agree. – M.B. Mar 27 '14 at 18:52

1 Answers1


The answer is yes. The idea is outlined in a previous answer. The rigorous proof is worked out now.

Assume the contrary that there are two points $a, b\in X$ and two curves $\gamma_i :[0,1] \to X$ joining $a$ and $b$. Let $\gamma_i :[0,1] \to X$.

Claim one: We can WLOG assume that the two curves do not intersect.

Proof: Consider the set $$ \gamma_1^{-1} \big( \gamma_2[0,1]\big) \subset [0,1]\ .$$ This is a closed set in $[0,1]$. We assume that this set is not the whole $[0,1]$ (or the image of $\gamma_1$ lies completely in $\gamma_2$). Thus the complement is open and contains an open interval $(t_1, t_2)$. We restrict $\gamma_1$ to $[t_1, t_2]$, (and also restrict the domain of $\gamma_2$ to the portion that connects $\gamma_1(t_1)$ and $\gamma_1(t_2)$).

Call both restrictions $\alpha_1$ and $\alpha_2$ and renaming $a = \gamma_1(t_1)$ and $b = \gamma_1(t_2)$. Note that this two curves are still length minimizing (If not, then there is a shorter paths joining these two points, which means that the original curves $\gamma_i$ are also not length minimizing).

As a result, we come up with two points $a, b\in X$ connected by two non-intersecting length minimizing curves. Thus claim one is proved.

Now the circle $C$ formed by first going along $\gamma_1$ and then $\gamma_2$ is a simple closed curve. Thus the Jordan curve theorem states that $\gamma_1 \cup \gamma_2$ bounds a bounded open set $\Omega$ such that $\partial \Omega = \gamma_1 \cup \gamma_2$.

Claim two: $\Omega \subset X$.

Proof: Assume the contrary that there is a point $x\in \Omega$ such that $x\notin X$. By translation we assume $x=0$ Consider the inclusion map

$$\iota : X \to \mathbb R^2 \setminus \{0\}$$

Now as $0 \in \Omega$ and $\Omega$ is bounded by $C$. So any ray starting at $0$ will hit some points in $C$. This implies that $\iota_*[C]$ represents a nontrivial element in $\pi_1(\mathbb R^2 \setminus \{0\})$. But this is impossible as $X$ is simply connected, $\iota_*[C]=0$. This contradiction implies that $x\in X$. Thus $\Omega \subset X$.

Now what remains is elementary geometry.

First, by some rotation assume that $a, b$ are in the $y$-axis. Let $L$ be a line parallel to the $y$-axis which touches $\bar \Omega$ at $c$ and not at $a, b$ (Such a $c$ can be found: Let $L$ comes from the left from $-\infty$, if it touches at $a, b$, then instead let $L$ comes from the right to find $c$) WLOG assume $c$ is in $\gamma_1$.

Let $B$ be a small closed ball around $c$ which does not contain $\gamma_2$. Moves the line $L$ towards $\Omega$ a little bit to form the line $L'$. Consider a conneced component $$J \subset L' \cap \bar\Omega \cap B$$ $J$ is a line segment in $\Omega$ such that the boundary points are in $\gamma_1$ (as $B$ does not contains $\gamma_2$). Thus $\gamma_1$ cannot be length minimizing, as the line segment $J$ is length minimizing. This contradictions implies that there cannot be more than one length minimizing curves joining two points.

Just to note that even for a path connected closed subset in $\mathbb R^2$ there might not be any curve with finite length joining two points.

  • This looks fine to me. While there is one fine point I'd do a bit differently, the approach does give the desired result. Nice. – Thomas Mar 28 '14 at 14:28