# Isometry Proof

1. Mar 1, 2010

### JG89

1. The problem statement, all variables and given/known data

Prove that [0,1] is not isometric to [0,2].

2. Relevant equations

3. The attempt at a solution

Suppose that [0,1] is infact isometric to [0,2]. Then there exists a homeomorphism $$f: [0,1] \rightarrow [0,2]$$ such that $$d(x,y) = D(f(x), f(y))$$. $$d$$ is the metric on [0,1] and $$D$$ is the metric on [0,2]. Let's assume for a second that both metrics are the one we would expect on these closed intervals, the usual metric that is on R.

Consider the inverse of f, $$f^{-1} : [0,2] \rightarrow [0,1]$$.

Take x = 2 and y = 0. Then $$2 = |2-0| = |f^{-1}(2) - f^{-1} (0)|$$. Note that $$\exists a, b \in [0,1] : f(a) = 2, f(b) = 0$$. So we have $$2 = |2-0| = |f^{-1}(2) - f^{-1}(0)| = |f^{-1}(f(a)) - f^{-1}(f(b))| = |a-b|$$.

But a and b are both values in [0,1], and so it's impossible for their difference to be equal to 2. This is a contradiction and so [0,1] is not isometric to [0,2].

The only thing is, in the question, they didn't specify what metrics were to be used, so I'm guessing it's general, and my proof only covers the usual metric on R.

Any ideas?

2. Mar 1, 2010

### VeeEight

Construct an 'isometry', f, from [0,2] to [0,1]. Then |2-0|=2.
Where can f(2) go? Where can f(0) go?
What is |f(2)-f(0)|?

3. Mar 1, 2010

### JG89

Aren't you assuming the metric we're using is the usual one on R, in which case my proof is okay?

4. Mar 1, 2010

### VeeEight

Yes, and our proofs are almost the same. What I meant was that you can simplify your argument. It is not necessary that the function be a homeomorphism. And you don't need to talk about inverse functions. Just realize (and I believe you do), that the only places for 2 and 0 to go would contract the distance preserving.

5. Mar 1, 2010

### JG89

Thanks VeeEight.

6. Mar 1, 2010

Cheers :)