banach tarski banach tarski before
Post on 23-Nov-2021
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Measurement
To keep things simple , let's talk about measuringangle segments in the unit circle
$ = { z e Q : 12-1=1 }
O. .For a given subset E s we'd like
assign a number
PCE ) =
Propertiesoftleasurementsl. .Pls ) =
,p : 2$ →-
2.
If Ei,EzE 8. are-
then PIE,o =
2 ! If { Ejlj? , are-then IPL Ej)
-3.
If E, ,Ez e-Stare-
then PCE,) BLED
Theorem-
:
Proof . If E s $ and ut $ then I and
UE =
are congruent . i . by 133,ME ) = BLUE ) Fut $ .
Now,consider $0 IT :-. { G
" it: te Q } (countable)
$ IT = f equivalence classes in $
where z n w ⇒ 2- = UW for some UET } .
Choose exactly t representative element q from each
equivalence class , and let Io = { 9 } a $ be the collectionof all these representatives .
Claim : $ =
u
u #.
Contradicts ?The measurement function P
, satisfyingID
,L2 ' )
,133 is used daily in calculus !
PCE ) =
So how can it fail to exist ?
The answer lies in an important subtlety :the definition of the Riemann integralonly works over " nice
" sets. The set Io
is not nice !
Much of this quarter will be spentextending the Riemann integral . BUTthere's only so far it can be extended .
tftheSyp : 2$ → co
.
I ]
⑧
This might seem like a bad sign . . .
but it is actually a foundational truthfor Kolmogorov 's probability theory(that we now embark on developing l .
In short : we don't always havecomplete information about the world ,which means there may be some eventswe simply cannot assign probabilitiesto
.
As to the unmeasurable sets . . .
Banaoh-Tarski_Paadx ( 1942) or Rd for d> 3
Given anytwo subsets E
,F E 1123% ith
nonempty interior, there are finite partitions
E = E,u Ez U - - - U En
F = F,
U E U- -
- u Fn
such that Ej is congruent to Fj for lsjsn .
RobinsubngTheen ( 1947 )
If E is a solid ball in 1123,
and F is the disjoint ballsof the same radius
,then
Banach -Tarski works explicitlywith n = 5
.
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