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.
On
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 $C_2 - C_1$.
On
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.
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