Let $C$ be a convex disc in the plane, and $C_1$ and $C_2$ be two translates of $C$. Prove that $C_1$ and $C_2$ are non-crossing, that is, it isn't possible that both $C_1 - C_2$ and $C_2 - C_1$ are non-connected.

I need this result for a different problem, but my understanding of topology and connectedness is pretty weak. What would an effective solution look like?

EDIT: A convex disc is a compact, convex region of the plane with a non-empty interior.

  • $\begingroup$ Can you give an example where $C_1 - C_2$ is not connected? $\endgroup$ – Paul Frost May 6 '19 at 10:44
  • $\begingroup$ I don't really know exactly what connected means or how to think about it intuitively $\endgroup$ – Wesley May 6 '19 at 15:23
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    $\begingroup$ @EpsilonDelta I agree! But the OP should clarify what a convex disk is (I think each disk is convex) and whether $C_i - C_j$ is the difference of sets or the set of all differences $c_i - c_j$. $\endgroup$ – Paul Frost May 8 '19 at 22:49
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    $\begingroup$ I think, the difference here means "the complement". An example of two convex sets where both complements are disconnected is given by two suitably chosen ellipses with orthogonal major axes. (Or just take $C_1$ equal to the x-axis and $C_2$ equal to the y-axis in the Cartesian plane.) $\endgroup$ – Moishe Kohan May 11 '19 at 16:30
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    $\begingroup$ You should also clarify what the difference sign means in your question. I think, it means the complement, but there is also an alternative interpretation. $\endgroup$ – Moishe Kohan May 12 '19 at 21:23

Below a proof in the case when $C$ is strictly convex, i.e. its boundary contains no nondegenerate line segments. One can modify the proof to include general bounded convex subsets. I can do so once I understand the motivation behind the question.

Let $C$ be a strictly convex subset of ${\mathbb R}^2$ and let $v\in {\mathbb R}^2$ be a nonzero vector. Then every line parallel to $v$ intersects the boundary of $C$ in 0, 1 or 2 points. Therefore, one can choose Cartesian coordinates on ${\mathbb R}^2$ so that $v$ is parallel to the y-axis and has coordinates, say, $(0,a)$, $a>0$. With respect to these coordinates, there are two functions $f, g$ defined on a compact interval $I\subset {\mathbb R}$ such that $C$ can be described by inequalities $$ C=\{(x,y): g(x)\le y\le f(x), x\in I\}, $$ where $f$ is concave and $g$ is convex. Similarly, $$ C+v= \{(x,y): g(x)+a\le y\le f(x)+a, x\in I\} $$

The key to the proof is:

Lemma 1. The boundaries of $C$ and $C+v$ intersect in at most two points.

Proof. Note that $C$ is bounded by the graphs of $f$ and $g$, while $C+v$ is bounded by the graphs of $f+a, g+a$. Since $a>0$, graphs of $f$ and $f+a$ are disjoint, the graphs of $g, g+a$ are disjoint and the graphs of $g, f+a$ are also disjoint. Only the graphs of $g+a$ and $f$ can intersect. This, the lemma reduces to the claim that the graph of a strictly convex function $h_1$ can intersect the graph of a strictly concave function $h_2$ in at most two points. This claim follows from the strict convexity of the subgraph of $h_2$ and the epigraph of $h_1$ which implies convexity of the intersection of the subgraph with the epigraph. qed

Now, given two convex sets $C_i=C+v_i, i=1,2$ as in your question, we have that $C_2= C_1 + (v_2-v_1)$. Setting $v=v_2-v_1$, we see that $C_2$ is a translate of $C_1$. Therefore, Lemma 1 applies to the sets $C_1$, $C_2$ and we conclude that either are either equal (if $v=0$) or their boundaries intersect in at most two points.

From now on, we do not need convexity and the problem reduces to:

Lemma 2. Suppose that $D_1, D_2$ are two closed topological disks in $R^2$ bounded by Jordan curves $J_1, J_2$ which intersect in at most two points. Then the complements $$ E_1=D_1\setminus D_2, \quad E_2=D_2\setminus D_1 $$ are both connected.

Proof. If $J_1\cap J_2$ is empty, so is $D_1\cap D_2$; if $J_1\cap J_2$ is a single point, so is $D_1\cap D_2$. (Here and below I am using the Jordan curve theorem.) Suppose, therefore, that $J_1\cap J_2=P=\{p_1, p_2\}$, $p_1\ne p_2$. Then the subset $P$ divides each $J_i$ in two (compact) topological arcs: $a_i, b_i$. After relabelling, $E_1=D_1\setminus D_2$ is bounded by $K_1=a_1\cup b_1$ while $E_2=D_2\setminus D_1$ is bounded by $K_2=a_2\cup b_2$. Since $a_i\cap b_i=P$ ($i=1,2$), is a 2-point set, we conclude that both $K_1, K_2$ are Jordan curves. Each Jordan curve separates ${\mathbb R}^2$ in two components (one bounded and one unbounded). Since each $E_i$ is bounded, it follows that its interior $int(E_i)$ is connected. Moreover, $E_i$ equals the closure of $int(E_i)$ minus the arc $b_i$, hence, $int(E_i)$ is dense in $E_i$. If a topological space is the closure of a connected subset, it is connected itself. Therefore, each $E_i$ is connected. qed

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Note: I'll assume you mean discs in the most literal sense: The inside part of a circle along with the boundary.

In fact both of $C_1-C_2$ and $C_2 -C_1$ are connected.

First observe for $D_1,D_2$ the perimeters of $C_1,C_2$ respectively that $D_1 \cap C_2$ is connected. This is clear if $D_1=D_2$ or $C_1 \cap C_2 =\varnothing$. Otherwise $D_1 \cap D_2$ has either one or two points. In the first case $C_1,C_2$ are tangent and clearly $C_1-C_2$ and $C_2 -C_1$ are connected. In the second case $D_1 \cap D_2 = \{a,b\}$ and $D_1 \cap C_2$ is one of the two arcs in $D_1$ from $a$ to $b$.

Now let $c,d \in C_2 - C_1$ be arbitrary. There exists a disc $C'_1$ with the same centre as $C_1$ but strictly larger radius, with $c,d \notin C_1'$. For example let $r$ be the smaller of the two distances from $c,d$ to $D_1$ and increase the radius by $r/2$. By the same logic $D_1' \cap C_2$ is an arc $A$. Shrinking $A$ slightly to $A'$ we can ensure $A' \subset C_2$.

Now let $l_1,l_2$ be lines segments connecting $c,d$ to some $y \in A'$. Its an exercise to show $l_1,l_2 \subset \overline{(\mathbb R^2 - C_1')} \subset \mathbb R^2 - C_1$. Also since $C_2$ is convex and contains $y,c \in C_2$ it also contains the line segment between them, likewise $l_2 \subset C_2$.

It follows $l_1 \cup l_2 \cup A' \subset C_2 - C_1$ is a connected set that contains $\{c,d\}$. Hence the two points have the same connected component in $C_2 - C_1$. Since the points are arbitrary there is only one connected component.

By symmetry the same holds for $C_2 - C_1$.

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    $\begingroup$ "Otherwise $𝐷_1\cap 𝐷_2$ has either one or two points" is false. $\endgroup$ – Moishe Kohan May 11 '19 at 16:08
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    $\begingroup$ Just take a square and slide it a bit along one of its sides. $\endgroup$ – Moishe Kohan May 11 '19 at 16:31
  • $\begingroup$ I'm taking a disc to mean a circle plus the stuff inside the circle. $\endgroup$ – Daron May 11 '19 at 22:25
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    $\begingroup$ I see: I thought of topological disks (for round disks convexity is a strange assumption since it holds automatically). Let's wait for Wesley to clarify his question. $\endgroup$ – Moishe Kohan May 12 '19 at 1:24
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    $\begingroup$ Just want to notice you the OP has edited the question... $\endgroup$ – YuiTo Cheng May 12 '19 at 14:08

Assume that $C$ is a compact convex set containing open $r$-ball at origin for some $r>0$. Then assume that $$C_1=C,\ C_2=C+(t,0),\ t>0$$

Then assume that $C$ is in $\{ -1\leq y\leq 1\}$ and $y=\pm 1$ is a supporting line.

When $p_\pm \in \partial C$ has a supporting line $y=\pm 1$ s.t. $x$-coordinates of $p_\pm$ are lowest, then there is a curve $c\subset \partial C$ between $p_\pm$ s.t. $c\bigcap C_2=\emptyset$

Then $D\bigcap C$ is connected where $D$ is a domain enclosed by $c,\ c+(t,0),\ y=\pm 1$. So prove that $D\bigcup C$ contains $C-C_2$

Proof : Assume that $q\in C-C_2 $ is not in $D\bigcap C$. Hence there is $T$ s.t. $q$ is in the line $ y=T$. Let $q_c \in c\bigcap \{y=T\}$. Hence a line segment $[q_c q]$ is in $C$ so that the translation $[q_c+(t,0)\ q+(t,0)]$ is in $C_2$. Here $[q_c+(t,0)\ q+(t,0)]$ contains $q$, which is a contradiction.

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