10.9 Isometries[2C9]

Definition 297

[0TK] Given \((M_ 1,d_ 1)\) and \((M_ 2,d_ 2)\) metric spaces, a map \(𝜑:M_ 1→ M_ 2\) is called an isometry if

\begin{equation} ∀ x,y∈ M_ 1, ~ ~ d_ 1(x,y)=d_ 2(𝜑(x),𝜑(y))~ ~ .\label{eq:isometria} \end{equation}
298

We will see in Sec. 12.2 the same definition in the case of normed vector spaces. Obviously an isometry is Lipschitz, and therefore continuous. Isometries enjoy some properties.

E298

[0TM]Topics:isometry. An isometry is always injective.

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[0TP]If the isometry \(𝜑\) is surjective (and therefore is bijective) then the inverse \(𝜑^{-1}\) is also an isometry. [0TQ]If \((M_ 1,d_ 1)\) is complete then its image \(𝜑(M_ 1)\) is a complete set in \(M_ 2\); and therefore it is a closed in \(M_ 2\).

Hidden solution: [UNACCESSIBLE UUID ’0TR’] Consequently, if the isometry \(𝜑\) is bijective and one of the two spaces is complete then the other is also complete.

[UNACCESSIBLE UUID ’0TS’] [0TT]Topics:isometry. Difficulty:*.Let \((X,d)\) be a compact metric space; let \(T:X→ X\) be an isometry, then \(T\) is surjective.

Provide a simple example of a non-compact metric space and \(T:X→ X\) a non-surjective isometry.

Hidden solution: [UNACCESSIBLE UUID ’0TV’] [0TW]Topics:isometry.Prerequisites:1.Difficulty:*.

Let \((X,d)\) and \((Y,𝛿)\) be two metric spaces of which \(X\) compact, \(T:X→ Y\) and \(S:Y→ X\) two isometries. Prove that \(T\) and \(S\) are bijective.

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Hidden solution: [UNACCESSIBLE UUID ’0TY’] [0TZ]Topics:isometry.Difficulty:*. Find an example of two metric spaces \((X,d)\) and \((Y,𝛿)\) that are not isometric but for which there are two isometries \(T:X→ Y\) and \(S:Y→ X\).

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Hidden solution: [UNACCESSIBLE UUID ’0V1’]

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