(adsbygoogle = window.adsbygoogle || []).push({}); 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 [tex] f: [0,1] \rightarrow [0,2] [/tex] such that [tex] d(x,y) = D(f(x), f(y)) [/tex]. [tex] d [/tex] is the metric on [0,1] and [tex] D [/tex] 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, [tex] f^{-1} : [0,2] \rightarrow [0,1] [/tex].

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

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?

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# Isometry Proof

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