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

Suppose there is an isometry $f:[0,1]\to[0,2]$. Since f is continuous and surjective, the only values for $f(0)$ and $f(1)$ are $f(0)=0$ and $f(1)=2$, or $f(0)=2$ and $f(1)=0$. In either case, $|f(1)-f(0)|=2$. This contradicts $f$ being distance-preserving.

Is this right?

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    $\begingroup$ Do both spaces have the same metric? $\endgroup$
    – gebruiker
    Jul 4, 2014 at 9:36
  • $\begingroup$ It is not true that necessarily $\{f(0),f(1)\}=\{0,2\}$. $\endgroup$ Jul 4, 2014 at 9:41
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    $\begingroup$ @gebruiker Yes, both spaces have the usual metric on $\mathbb{R}$ $\endgroup$
    – user153582
    Jul 4, 2014 at 9:52
  • $\begingroup$ I suspected as much. I implicitely tried to say: "You should mention this, because it formally makes a greats difference." :) $\endgroup$
    – gebruiker
    Jul 4, 2014 at 10:00

2 Answers 2

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This

Since f is continuous and surjective, the only values for f(0) and f(1) are f(0)=0 and f(1)=2, or f(0)=2 and f(1)=0.

is not a rigorous argument. You should argue it more precisely (and I doubt it is easy to do...). But actually you don't need to state such a strong result.

Suppose $[0,1]$ and $[0,2]$ are endowed with the absolute value metric $|\cdot |$, and suppose $f:[0,1] \to [0,2]$ is a surjective isometry. Then there are $x,y  \in [0,1]$ such that $f(x) = 0$ and $f(y) = 2$, and since $f$ is an isometry we have $$|x-y| = |f(x)-f(y)| = |0-2|=2.$$ This is absurd since $$\sup_{v,w \in [0,1]} |w-v| = |0-1| = 1<2. $$

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The diameter of a metric space is the supremum, taken over all pairs of points in the space, of the distance between the points of the pair. For a compact metric space the diameter is finite and is attained for some pair of points in the space. Isometric metric spaces have the same diameter. The spaces $[0, 1]$ and $[0, 2]$ with the Eucliden metric have the distinct diameters $1$ and $2$ (obviously), so are not isometric.

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