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I want to split an Euclidean disk (it should make no difference whether it's open or closed) into two non-intersecting sets of points which are identical in a sense that one set can be transformed into another (and vice versa) by using only shift and rotation.

Can that be done?

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don't understand the question what is wrong by just cutting the disk by any diameter, then mirroring over this line swaps the sets, the same for rotating a half circle over the midpoint. (Guess i aam overlooking something fundamental here) –  Willemien Feb 22 at 11:58
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because when you cut the disk along the diameter, the diameter itself has to go to one of the parts(since the two parts are non-intersecting) so the two parts will be non equivalent –  viplov_jain Feb 22 at 12:15

2 Answers 2

up vote 4 down vote accepted

No.

Let $D$ be the (open, closed or whatever) unit disk and let $T\colon \mathbb R^2\to\mathbb R^2$ be a rotation or trnslation such that $D=A\cup T[A]$ for some set $A$ with $A\cap T[A]=\emptyset$. If $T[\partial D]=\partial D$, the point $0$ must be a fixed point of $T$; but then $0\in A$ implies $0=T(0)\in B$ and vice versa, contradiction. Hence $T[\partial D]\ne \partial D$, i.e. these two circles intersect in two distinct points $a,b$. Let $C_a$, $C_b$ be the circles through $a,b$ around the center of rotation (whch is on the line $ab$) - or in case of a translation the lines orthogonal to $ab$ through $a, b$, respectively. These circles/lines intersect (not touch) $\partial D$, hence chop the interior of $D$ into three parts with nonempty interior. For the two "outer" parts we have that their images under $T$ are disjoint from $D$, hence thy must be $\subseteq B$. But they are also disjoint from $T[D]$ hence must be $\subseteq A$, contradiction.

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This is not possible for closed disks. Say it is possible and you can divide the disk into two parts which are non-intersecting(say $A$ and $B$) and one can be transformed into another (and vice versa) by using only shift and rotation . Use the transformations $$g_1:A \rightarrow B $$ $$g_2:B \rightarrow A $$ Extend them to $\Bbb R^2$ call the extensions $g_1'$ and $ g_2'$.This can be done because they are given to be composition shifts and rotations. Now $g_1'og_2'$ and $g_2'og_1'$ are both $identity$. Hence $g_1og_2$ and $g_2og_1$ are both $identity$. The extension of $g_1$ to the disk will have no fixed point. This is a contradiction to brouwer's fixed point thoerem.

For any other disk use a similar proof on the closure of the disk.

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If $D=A\cup B$ with $B=f[A]$, this does not easily give us $f\colon D\to D$. –  Hagen von Eitzen Feb 22 at 12:28
    
f is a bijective function –  viplov_jain Feb 22 at 12:30
    
$f$ needs to be defined in the first place. –  Hagen von Eitzen Feb 22 at 12:30
    
Yes you are right its not that easy to extend.I edited the solution.It's completely different though –  viplov_jain Feb 22 at 12:56
    
$g_1$ will not necessarily map $D\to D$. NB, we can find continuous maps $f\colon \mathbb R^2\to \mathbb R^2$ such that the open disk $D=A\cup f[A]$ and $A\cap f[A]=\emptyset$ –  Hagen von Eitzen Feb 22 at 13:06

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