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 $ f: E \ rightarrow E $ metric space with metric $ \ rho (\ cdot, \ cdot) $ called isometry if for any $ x, y \ in E $ fair equality $ \ rho (x, y) = \ rho (f (x), f (y)) $. We prove here the following statement:

Theorem. If a$ f: E \ rightarrow E $ mapping of a compact metric space into itself, such that

$ \ rho (x, y) \ leq \ rho (f (x), f (y)) (1) $

for any $ x, y \ in E $then mapping $ f $ - isometry.

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

Through $ | A | $ we denote the number of elements of a finite set $ A $.

For $ x \ in E $ and $ \ varepsilon> 0 $ lots of $ Q_ {x, \ varepsilon} = \ {y: y \ in E, \ rho (x, y) <\ varepsilon \} $ let's call $ \ varepsilon $- point neighborhood $ x $ (or an open ball with a center at $ x $ and radius $ \ varepsilon $).

Final set $ A \ subset E $ let's call $ \ varepsilon $-network in $ E $ (or simply $ \ varepsilon $network) if for any point $ x \ in E $ there is a point $ y \ in A $ such that $ \ rho (x, y) <\ varepsilon $. Lots of $ B \ subset E $ let's call $ \ varepsilon $- sparse if $ \ rho (x, y) \ geq \ varepsilon $ for any $ x, y \ in B $such that $ x \ neq y $.

For any finite set $ A = \ left \ {a_1, \ ldots, a_m \ right \} \ subset E $ denote by $ l (A) $ amount of $ \ sum _ {i \ leq j} \ rho \ left (a_i, a_j \ right) $. Magnitude $ l (A) $ call the length of the set $ A $.

1. Let the sequence $ \ left \ {a_n \ right \} $, $ \ left \ {b_n \ right \} $ elements of the set $ E $converge respectively
to the points$ a, b \ in E $. Then $ \ rho \ left (a_n, b_n \ right) \ rightarrow \ rho (a, b) $ at $ n \ rightarrow \ infty $.

Proof . Consider obvious inequalities.

$ \ rho \ left (a_n, b_n \ right) \ leq \ rho (a, b) + \ rho \ left (a_n, a \ right) + \ rho \ left (b_n, b \ right) (2) $

$ \ rho \ left (a_n, b_n \ right) + \ rho \ left (a_n, a \ right) + \ rho \ left (b_n, b \ right) \ geq \ rho (a, b) (3) $

Because $ a_n \ rightarrow a $, $ b_n \ rightarrow b $ at $ n \ rightarrow \ infty $then for $ \ varepsilon> 0 $ there is such a natural $ N $that for all $ n> n $ will be

$ \ rho \ left (a_n, a \ right) <\ frac {\ varepsilon} {2}, \ rho \ left (b_n, b \ right) <\ frac {\ varepsilon} {2} (4) $

Of $ (2), (3), (4) $ follows that $ \ left | \ rho (a, b) - \ rho \ left (a_n, b_n \ right) \ right | <\ varepsilon $ for all $ n> n $.

2. For each $ \ varepsilon> 0 $ at $ E $ there is an ultimate $ \ varepsilon $-network.

Proof . Open Ball Family$ \ left \ {Q_ {x, \ varepsilon} \ right \} $where $ x $ runs through $ E $is a coating $ E $. T. to.$ E $ compact, choose the final family of balls $ \ left \ {Q_ {x_1, \ varepsilon}, \ ldots, Q_ {x_m, \ varepsilon} \ right \} $also covering $ E $. Clearly a lot$ A = \ left \ {x_1, \ ldots, x_m \ right \} $ - the ultimate $ \ varepsilon $-network.

3. Space$ E $limited. Namely, there is such a number $ d> 0 $, what $ \ rho (x, y) <d $ for any $ x, y \ in E $.

The proof immediately follows from 2. Indeed, we set$ g = \ underset {i \ neq j} {\ max} \ left (x_i, x_j \ right) $where $ x_i $, $ x_j $ - elements $ \ varepsilon $-networks $ A $. It's clear that$ \ rho (x, y) \ leq g + 2 \ varepsilon $.

4. If $ B = \ left \ {a_1, \ ldots, a_n \ right \} $ - the ultimate $ \ frac {\ varepsilon} {2} $- network in $ E $then for any $ \ varepsilon $- sparse set $ K $ will be $ | K | \ leq | B | $i.e. $ | K | \ leq n $.

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

5. To each$ \ varepsilon $-resolved set $ A \ subset E $ set the number $ l (A) $ - its length. We have already proven that a function that puts any$ \ varepsilon $-resolved set $ A $ in line number $ | A | $is limited. Note that the function that each$ \ varepsilon $-resolved set $ A \ subset E $ matches its length $ l (A) $is also limited.

6. Let$ c = \ sup l (A) $where $ \ sup $ taken on all $ \ varepsilon $- sparse sets $ A \ subset E $. Then fair

Lemma 1. There is$ \ varepsilon $-spine set $ C = \ left \ {a_1, \ ldots, a_k \ right \} $such that $ l (C) = c $, $ C $ is an $ \ varepsilon $-network in $ E $, $ f (C) $ is also $ \ varepsilon $-network in $ E $ and for any $ a_i, a_j \ in C $ will be $ \ rho \ left (a_i, a_j \ right) = \ rho \ left (f \ left (a_i \ right), f \ left (a_j \ right) \ right) $.

7. Lemma 2. Mapping$ f $ continuously on $ E $. More precisely: if $ \ rho (x, y) <\ varepsilon $ for any $ x, y \ in E $then $ \ rho (f (x), f (y)) <5 \ varepsilon $.

Proof . Will consider$ \ varepsilon $-network $ C $ from Lemma 1. If $ x $ does not belong to the ball $ Q_ {a_i, \ varepsilon} $then $ x $ not belong $ Q_ {f \ left (a_i \ right), \ varepsilon} $. This means that there is such$ i $, what $ x \ in Q_ {a_i, \ varepsilon} $ and $ f (x) \ in Q_ {f \ left (a_i \ right), \ varepsilon} $. Similarly, there is such$ j $, what $ y \ in Q_ {a_j, \ varepsilon} $ and $ f (y) \ in Q_ {f \ left (a_j \ right), \ varepsilon} $. Rate$ \ rho (f (x), f (y)) $. It's clear that$ \ rho (f (x), f (y)) <\ rho \ left (f \ left (a_i \ right), f \ left (a_j \ right) \ right) + \ varepsilon + \ varepsilon = \ rho \ left (a_i, a_j \ right) +2 \ varepsilon $. And since$ \ rho (x, y) <\ varepsilon $and $ x \ in Q_ {a_i, \ varepsilon} $, $ y \ in Q_ {a_j, \ varepsilon} $then $ \ rho \ left (a_i, a_j \ right) <3 \ varepsilon $. Consequently,$ \ rho (f (x), f (y)) <5 \ varepsilon $.

So, we proved that $ f $ continuously displays $ E $ at $ E $. From Lemma 1 it follows that for each $ \ varepsilon> 0 $ exists $ \ varepsilon $- network in $ E $ such that $ f $preserves the distance between the elements of this network. So for any points $ x, y \ in E $ can find sequences $ x_n \ rightarrow x $, $ y_n \ rightarrow y $ such that $ \ rho \ left (f \ left (x_n \ right), f \ left (y_n \ right) \ right) = \ rho \ left (x_n, y_n \ right) $. But$ \ rho \ left (x_n, y_n \ right) \ rightarrow \ rho (x, y) $ at $ n \ rightarrow \ infty $. From the continuity of the display$ f $ follows that $ f \ left (x_n \ right) \ rightarrow f (x) $, $ f \ left (y_n \ right) \ rightarrow f (y) $ at $ n \ rightarrow \ infty $. Consequently,$ \ rho \ left (f \ left (x_n \ right), f \ left (y_n \ right) \ right) \ rightarrow \ rho (f (x), f (y)) $ at $ n \ rightarrow \ infty $. And since for any$ n $ equality is fulfilled $ \ rho \ left (x_n, y_n \ right) = \ rho \ left (f \ left (x_n \ right), f \ left (y_n \ right) \ right) $then $ \ rho (x, y) = \ rho (f (x), f (y)) $.

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
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