A Set in the plane is never isometric to a proper subset

[Mr Davis 97]

Homework Statement

Is there a bounded set $H$ in the plane which is isometric to a proper subset $S \subset H$?

Homework Equations

The Attempt at a Solution

I'm thinking that the answer is no. Here are my ideas, that by no means constitute a proof. Every isometry of the plane is either a reflection, a rotation, a translation, or a glide reflection. Now, let $\phi$ be an isometry. Then $\phi(H) = S$. Now, anyone of these isometry types will result in an S such that there is at least one element of $S$ that is not in $H$, meaning that $S$ can never be a proper subset of $H$.

This is my thinking, but it is very sketchy. Am I on the right track? How could I get a rigorous proof out of this?

 

[fresh_42]

What happens to the unit disc, if you map a point with distance $r$ to the origin to the point with distance $r/2$?

 

[Mr Davis 97]

What happens to the unit disc, if you map a point with distance $r$ to the origin to the point with distance $r/2$?

Could you clarify what you mean by the "if you map a point with distance $r$ to the origin to the point with distance $r/2$?"

 

[fresh_42]

As said: Let $(r,\varphi)$ be the coordinates of a point on the disc, then map it to $(r/2,\varphi)$.

 

[Mr Davis 97]

As said: Let $(r,\varphi)$ be the coordinates of a point on the disc, then map it to $(r/2,\varphi)$.

How is that an isometry? For example, let $f$ be the proposed map that halves the x-coordinate. Then $d ((3/4,0),(1/4,0)) = 1/2$. However, $d (f(3/4,0),f(1/4,0)) = d((3/8,0),(1/8,0)) = 1/4$

 

[fresh_42]

Sorry, I read isomorphic and didn't quite know what you meant by it, my bad.

 

[Mr Davis 97]

Sorry, I read isomorphic and didn't quite know what you meant by it, my bad.

Do you have any input on the problem given that it's isometric and not isomorphic?

 

[fresh_42]

Not really. I was thinking about extremes as a discrete metric, or not connected sets. With an arbitrary metric and an arbitrary set - bounded only says we can draw it on our finite pieces of paper - I have not really an idea whether there cannot be some pathological constructions. Also isometry doesn't even require the same metric, so maybe we can find a counterexample with two non equivalent metrics.

Ok, I think we can assume the same metric, as subset means the same space, not only an embedding which I thought of. If you have this decomposition of isometries, then your idea looks good, as those constitutes do not change shape, but is this true for any metric?

 

[WWGD]

Ok, I think we can assume the same metric, as subset means the same space, not only an embedding which I thought of. If you have this decomposition of isometries, then your idea looks good, as those constitutes do not change shape, but is this true for any metric?

Since we are talking about the plane, don't we assume it is the Euclidean 2-metric by default?

 

[mathwonk]

your proof assumes the isometry from the set S to the set H extends to an isometry of the whole plane, which seems unwarranted to me. your conjecture however seems plausible to me, but I am not sure.

 

[WWGD]

A thing is that isometries do not have to be homeomorphisms.

 

[mfb]

Now, anyone of these isometry types will result in an S such that there is at least one element of $S$ that is not in $H$, meaning that $S$ can never be a proper subset of $H$.

To me that just sounds like a different way to say what you are supposed to prove.