Equal shapes when one is part of other

[honzik]

Hi,

I would like to know whether there exist two shapes A and B (in 2D plane) that are equal (ie. equal transformation between A and B exists) but A is proper subset of B. The definition of "being equal" (equal transformation preseves lengths - eg. rotation, traslation etc.) and "proper subset" (the sets are not equal) are I think commonly known.

 

For infinite shapes these shapes A and B exist: eg. half-line, half-plane and so on but what about shapes that are bounded by say circle with the radius of 1? Does some examples exist (maybe some fractals) or can be proven that it is not possible?

 

Thank you in any thoughts.

 

Honzik

[the tree]

I don't know what you mean by equal - do you mean congruent or isomorphic or just of equal measure? Closed and open sets with the same boundaries tend to have the same measure and an isomorphism should exist between them but they are different shapes.

[honzik]

Equal transformation f in the 2-dimensional plane (E2) is transformation that satisfies: d(X,Y) = d(f(X),f(Y)) for every points X, Y from E2 (d is eukleidian distance - classical distance of two points).

 

K is proper subset of L if and only if (iff) K is part of L but L is not part of K (ie. K <> L)

 

Than the question is: Do exist sets A and B in E2 so that equal transformation exists so that f(A) = B (set equation) and A is proper subset of B?

[the tree]

So, an isometry then. I think there may be, or at least I'd struggle to think why not. The preservation of measure between closed an open sets could hold a workable example - although I'm not sure of what a general bijection between them would look like or if it'd preserve structure.

[honzik]

So, an isometry then. I think there may be, or at least I'd struggle to think why not. The preservation of measure between closed an open sets could hold a workable example - although I'm not sure of what a general bijection between them would look like or if it'd preserve structure.

 

General bijection will not work because equal transformation is much stronger - it preserves lengths but bijection in general not. The length of every line segment has to be preserved so equal transformation of (0,1> into <0,1> must be identity or 1-x.

[the tree]

Yeah, I never said it would. Read post again plzthx.