Boschernitsana theorem

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The article provides a simple proof that the mapping of a compact metric space into itself, which does not reduce the distance, is an isometry.

Display  metric space with metric  called isometry if for any  fair equality . We prove here the following statement:

Theorem. If a mapping of a compact metric space into itself, such that



for any then mapping  - isometry.

Recall some simple statements about metric compacts and introduce some conventions and definitions necessary for further discussion.

Through  we denote the number of elements of a finite set .

For  and  lots of  let's call - point neighborhood  (or an open ball with a center at  and radius ).

Final set  let's call -network in  (or simply network) if for any point  there is a point  such that . Lots of  let's call - sparse if  for any such that .

For any finite set  denote by  amount of . Magnitude  call the length of the set .

1. Let the sequence ,  elements of the set converge respectively
to the points. Then  at .

Proof . Consider obvious inequalities.





Because ,  at then for  there is such a natural that for all  will be



Of  follows that  for all .

2. For each  at  there is an ultimate -network.

Proof . Open Ball Familywhere  runs through is a coating . T. to. compact, choose the final family of balls also covering . Clearly a lot - the ultimate -network.

3. Spacelimited. Namely, there is such a number , what  for any .

The proof immediately follows from 2. Indeed, we setwhere ,  - elements -networks . It's clear that.

4. If  - the ultimate - network in then for any - sparse set  will be i.e. .

Proof . Pool balls $inline \ underset \left\{i = 1\right\} \left\{\ overset \left\{n\right\} \left\{\ unicode \left\{222a\right\}\right\}\right\} Q_ \left\{a_i, \ frac \left\{\ varepsilon\right\} \left\{2\right\}\right\} inline$ covers . If athen two different elements from  will be in one of the balls that contradicts the fact that  - - sparse set.

5. To each-resolved set  set the number  - its length. We have already proven that a function that puts any-resolved set  in line number is limited. Note that the function that each-resolved set  matches its length is also limited.

6. Letwhere  taken on all - sparse sets . Then fair

Lemma 1. There is-spine set such that ,  is an -network in ,  is also -network in  and for any  will be .

7. Lemma 2. Mapping continuously on . More precisely: if  for any then .

Proof . Will consider-network  from Lemma 1. If  does not belong to the ball then  not belong . This means that there is such, what  and . Similarly, there is such, what  and . Rate. It's clear that. And sinceand , then . Consequently,.

So, we proved that  continuously displays  at . From Lemma 1 it follows that for each  exists - network in  such that preserves the distance between the elements of this network. So for any points  can find sequences ,  such that . But at . From the continuity of the display follows that ,  at . Consequently, at . And since for any equality is fulfilled then .

Comment

This proof of the Boschernitsan theorem is based on conversations with my student comrade, now an American mathematician Leonid Luxemburg, during one of his visits to Moscow and is my presentation of the idea he proposed.

Slobodnik Semen Grigorievich ,
content developer for the application “Tutor: Mathematics” (see the article on Habré ), Candidate of Physical and Mathematical Sciences, math teacher of school 179, Moscow