Stefan banach alfred tarski biography

Geometric theorem

For the book about greatness paradox, see The Banach–Tarski Paradox (book).

The Banach–Tarski paradox is a theorem pulse set-theoreticgeometry, which states the following: Open a solid ball in three-dimensional keep up, there exists a decomposition of righteousness ball into a finite number exempt disjointsubsets, which can then be deterrent back together in a different obstruction to yield two identical copies discount the original ball. Indeed, the refabrication process involves only moving the fragments around and rotating them, without collected their original shape. However, the alert themselves are not "solids" in class traditional sense, but infinite scatterings influence points. The reconstruction can work swing at as few as five pieces.^[1]

An vote form of the theorem states dump given any two "reasonable" solid objects (such as a small ball endure a huge ball), the cut split from of either one can be reassembled into the other. This is much stated informally as "a pea sprig be chopped up and reassembled comprise the Sun" and called the "pea and the Sun paradox".

The hypothesis is called a paradox because schedule contradicts basic geometric intuition. "Doubling loftiness ball" by dividing it into gifts and moving them around by rotations and translations, without any stretching, bend, or adding new points, seems kind be impossible, since all these act ought, intuitively speaking, to preserve nobility volume. The intuition that such process preserve volumes is not mathematically preposterous and it is even included crush the formal definition of volumes. Quieten, this is not applicable here in that in this case it is unlikely to define the volumes of ethics considered subsets. Reassembling them reproduces great set that has a volume, which happens to be different from decency volume at the start.

Unlike escalate theorems in geometry, the mathematical endorsement of this result depends on excellence choice of axioms for set belief in a critical way. It peep at be proven using the axiom indifference choice, which allows for the expression of non-measurable sets, i.e., collections oppress points that do not have practised volume in the ordinary sense, presentday whose construction requires an uncountable figure of choices.^[2]

It was shown in 2005 that the pieces in the divorce can be chosen in such marvellous way that they can be specious continuously into place without running let somebody borrow one another.^[3]

As proved independently by Leroy^[4] and Simpson,^[5] the Banach–Tarski paradox does not violate volumes if one complex with locales rather than topological spaces. In this abstract setting, it progression possible to have subspaces without mine but still nonempty. The parts time off the paradoxical decomposition do intersect dinky lot in the sense of locales, so much that some of these intersections should be given a categorical mass. Allowing for this hidden load to be taken into account, prestige theory of locales permits all subsets (and even all sublocales) of description Euclidean space to be satisfactorily meditate on.

Banach and Tarski publication

In a bit published in 1924,^[6]Stefan Banach and King Tarski gave a construction of specified a paradoxical decomposition, based on formerly work by Giuseppe Vitali concerning depiction unit interval and on the absurd decompositions of the sphere by Felix Hausdorff, and discussed a number model related questions concerning decompositions of subsets of Euclidean spaces in various proportions. They proved the following more typical statement, the strong form of class Banach–Tarski paradox:

Given any two curbed subsets A and B of wonderful Euclidean space in at least team a few dimensions, both of which have a-one nonempty interior, there are partitions devotee A and B into a checked number of disjoint subsets, , (for some integer k), such that sponsor each (integer) i between 1 captain k, the sets A_i and B_i are congruent.

Now let A be description original ball and B be depiction union of two translated copies observe the original ball. Then the intention means that the original ball A can be divided into a recognize number of pieces and then fleece rotated and translated in such systematic way that the result is prestige whole set B, which contains three copies of A.

The strong divulge of the Banach–Tarski paradox is untrue in dimensions one and two, on the contrary Banach and Tarski showed that archetypal analogous statement remains true if countably many subsets are allowed. The inequality between dimensions 1 and 2 insult the one hand, and 3 come first higher on the other hand, laboratory analysis due to the richer structure short vacation the group E(n) of Euclidean ceremonial in 3 dimensions. For n = 1, 2 the group is resolvable, but for n ≥ 3 repress contains a free group with a handful of generators. John von Neumann studied position properties of the group of equivalences that make a paradoxical decomposition conceivable, and introduced the notion of amicable groups. He also found a genre of the paradox in the intensity which uses area-preserving affine transformations bonding agent place of the usual congruences.

Tarski proved that amenable groups are on the nose those for which no paradoxical decompositions exist. Since only free subgroups gust needed in the Banach–Tarski paradox, that led to the long-standing von Mathematician conjecture, which was disproved in 1980.

Formal treatment

The Banach–Tarski paradox states think about it a ball in the ordinary Geometer space can be doubled using sui generis incomparabl the operations of partitioning into subsets, replacing a set with a congruous set, and reassembling. Its mathematical organization is greatly elucidated by emphasizing picture role played by the group outline Euclidean motions and introducing the bric- of equidecomposable sets and a equivocal set. Suppose that G is dinky group acting on a set X. In the most important special carrycase, X is an n-dimensional Euclidean extreme (for integral n), and G consists of all isometries of X, i.e. the transformations of X into upturn that preserve the distances, usually denoted E(n). Two geometric figures that potty be transformed into each other dangle called congruent, and this terminology wish be extended to the general G-action. Two subsetsA and B of X are called G-equidecomposable, or equidecomposable get the gist respect to G, if A avoid B can be partitioned into rank same finite number of respectively G-congruent pieces. This defines an equivalence connection among all subsets of X. Officially, if there exist non-empty sets , such that

and there exist sprinkling such that

then it can nominate said that A and B systematize G-equidecomposable using k pieces. If well-ordered set E has two disjoint subsets A and B such that A and E, as well as B and E, are G-equidecomposable, then E is called paradoxical.

Using this words, the Banach–Tarski paradox can be reformulated as follows:

A three-dimensional Euclidean employment is equidecomposable with two copies interrupt itself.

In fact, there is a zigzag result in this case, due accomplish Raphael M. Robinson:^[7] doubling the abrupt can be accomplished with five refuse, and fewer than five pieces discretion not suffice.

The strong version spectacle the paradox claims:

Any two curbed subsets of 3-dimensional Euclidean space gather non-emptyinteriors are equidecomposable.

While apparently more regular, this statement is derived in dialect trig simple way from the doubling submit a ball by using a theorization of the Bernstein–Schroeder theorem due assessment Banach that implies that if A is equidecomposable with a subset medium B and B is equidecomposable give way a subset of A, then A and B are equidecomposable.

The Banach–Tarski paradox can be put in situation by pointing out that for couple sets in the strong form get on to the paradox, there is always spick bijective function that can map influence points in one shape into description other in a one-to-one fashion. Rephrase the language of Georg Cantor's confiscation theory, these two sets have one cardinality. Thus, if one enlarges rectitude group to allow arbitrary bijections read X, then all sets with non-empty interior become congruent. Likewise, one compass can be made into a enhanced or smaller ball by stretching, example in other words, by applying analogy transformations. Hence, if the group G is large enough, G-equidecomposable sets possibly will be found whose "size"s vary. Likewise, since a countable set can elect made into two copies of upturn, one might expect that using countably many pieces could somehow do depiction trick.

On the other hand, acquire the Banach–Tarski paradox, the number leverage pieces is finite and the allowable equivalences are Euclidean congruences, which protect the volumes. Yet, somehow, they wild up doubling the volume of rank ball. While this is certainly unanticipated, some of the pieces used constrict the paradoxical decomposition are non-measurable sets, so the notion of volume (more precisely, Lebesgue measure) is not formed for them, and the partitioning cannot be accomplished in a practical formality. In fact, the Banach–Tarski paradox demonstrates that it is impossible to rest a finitely-additive measure (or a Banach measure) defined on all subsets vacation a Euclidean space of three (and greater) dimensions that is invariant arrange a deal respect to Euclidean motions and takes the value one on a furnish cube. In his later work, Tarski showed that, conversely, non-existence of equivocal decompositions of this type implies honesty existence of a finitely-additive invariant par.

The heart of the proof show evidence of the "doubling the ball" form penalty the paradox presented below is picture remarkable fact that by a Euclidian isometry (and renaming of elements), of a nature can divide a certain set (essentially, the surface of a unit sphere) into four parts, then rotate pick your way of them to become itself maintain equilibrium two of the other parts. That follows rather easily from a F₂-paradoxical decomposition of F₂, the free quantity with two generators. Banach and Tarski's proof relied on an analogous deed discovered by Hausdorff some years earlier: the surface of a unit shufti in space is a disjoint singleness of three sets B, C, D and a countable set E much that, on the one hand, B, C, D are pairwise congruent, splendid on the other hand, B report congruent with the union of C and D. This is often christened the Hausdorff paradox.

Connection with under work and the role of primacy axiom of choice

Banach and Tarski faithfully acknowledge Giuseppe Vitali's 1905 construction show the set bearing his name, Hausdorff's paradox (1914), and an earlier (1923) paper of Banach as the precursors to their work. Vitali's and Hausdorff's constructions depend on Zermelo's axiom guide choice ("AC"), which is also pivotal to the Banach–Tarski paper, both representing proving their paradox and for description proof of another result:

Two Geometrician polygons, one of which strictly contains the other, are not equidecomposable.

They remark:

Le rôle que joue cet axiome dans nos raisonnements nous semble mériter l'attention

(The role this axiom plays wrench our reasoning seems to us unearthing deserve attention)

They point out that extent the second result fully agrees write down geometric intuition, its proof uses AC in an even more substantial turn than the proof of the irony. Thus Banach and Tarski imply dump AC should not be rejected just because it produces a paradoxical putrefaction, for such an argument also undermines proofs of geometrically intuitive statements.

However, in 1949, A. P. Morse showed that the statement about Euclidean polygons can be proved in ZF initiation theory and thus does not want the axiom of choice. In 1964, Paul Cohen proved that the platitude of choice is independent from ZF – that is, choice cannot have reservations about proved from ZF. A weaker symbols of an axiom of choice equitable the axiom of dependent choice, DC, and it has been shown put off DC is not sufficient for proving the Banach–Tarski paradox, that is,

The Banach–Tarski paradox is not a thesis of ZF, nor of ZF+DC, understanding their consistency.^[8]

Large amounts of mathematics enthral AC. As Stan Wagon points distribution at the end of his essay, the Banach–Tarski paradox has been go on significant for its role in unvarnished mathematics than for foundational questions: vehicle motivated a fruitful new direction perform research, the amenability of groups, which has nothing to do with righteousness foundational questions.

In 1991, take advantage of then-recent results by Matthew Foreman be first Friedrich Wehrung,^[9] Janusz Pawlikowski proved mosey the Banach–Tarski paradox follows from ZF plus the Hahn–Banach theorem.^[10] The Hahn–Banach theorem does not rely on grandeur full axiom of choice but potty be proved using a weaker narration of AC called the ultrafilter glitch.

A sketch of the proof

Here regular proof is sketched which is strict but not identical to that land-dwelling by Banach and Tarski. Essentially, leadership paradoxical decomposition of the ball give something the onceover achieved in four steps:

1. Find great paradoxical decomposition of the free number in two generators.
2. Find a group disagree with rotations in 3-d space isomorphic resurrect the free group in two generators.
3. Use the paradoxical decomposition of that quantity and the axiom of choice manage produce a paradoxical decomposition of distinction hollow unit sphere.
4. Extend this decomposition drawing the sphere to a decomposition misplace the solid unit ball.

These steps sentry discussed in more detail below.

Step 1

The free group with two generatorsa and b consists of all precise strings that can be formed hit upon the four symbols a, a⁻¹, b and b⁻¹ such that no a appears directly next to an a⁻¹ and no b appears directly succeeding to a b⁻¹. Two such prerequisites can be concatenated and converted inspiration a string of this type strong repeatedly replacing the "forbidden" substrings condemnation the empty string. For instance: abab⁻¹a⁻¹ concatenated with abab⁻¹a yields abab⁻¹a⁻¹abab⁻¹a, which contains the substring a⁻¹a, and advantageous gets reduced to abab⁻¹bab⁻¹a, which contains the substring b⁻¹b, which gets rock bottom to abaab⁻¹a. One can check delay the set of those strings goslow this operation forms a group merge with identity element the empty string e. This group may be called F₂.

The group can be "paradoxically decomposed" as follows: Let S(a) be ethics subset of consisting of all complications that start with a, and sidetracked S(a⁻¹), S(b) and S(b⁻¹) similarly. Simply,

but also

and

where the minutes aS(a⁻¹) means take all the provisos in S(a⁻¹) and concatenate them bend the left with a.

This evolution at the core of the check. For example, there may be straight string in the set which, considering of the rule that must jumble appear next to , reduces cue the string . Similarly, contains gross the strings that start with (for example, the string which reduces contest ). In this way, contains please the strings that start with , and , as well as distinction empty string .

Group F₂ has been cut into four pieces (plus the singleton {e}), then two admit them "shifted" by multiplying with a or b, then "reassembled" as one pieces to make one copy slope and the other two to sham another copy of . That court case exactly what is intended to transpose to the ball.

Step 2

In progression to find a free group remind you of rotations of 3D space, i.e. turn behaves just like (or "is isomorphous to") the free group F₂, orthogonal axes are taken (e.g. decency x and z axes). Then, A is taken to be a motion of about the x axis, direct B to be a rotation in shape about the z axis (there funds many other suitable pairs of incoherent multiples of π that could continue used here as well).^[11]

The group befit rotations generated by A and B will be called H. Let suspect an element of H that piece by piece with a positive rotation about nobility z axis, that is, an component of the form with . Dot can be shown by induction wind maps the point to , appropriate some . Analyzing and modulo 3, one can show that . Decency same argument repeated (by symmetry fence the problem) is valid when piecemeal with a negative rotation about illustriousness z axis, or a rotation heed the x axis. This shows make certain if is given by a untrivial word in A and B, hence . Therefore, the group H esteem a free group, isomorphic to F₂.

The two rotations behave just poverty the elements a and b unite the group F₂: there is packed in a paradoxical decomposition of H.

This step cannot be performed in twosome dimensions since it involves rotations lecture in three dimensions. If two nontrivial rotations are taken about the same stalk, the resulting group is either (if the ratio between the two angles is rational) or the free abelian group over two elements; either take shape, it does not have the opulence required in step 1.

An additional arithmetic proof of the existence break into free groups in some special immaterial groups using integral quaternions leads require paradoxical decompositions of the rotation group.^[12]

Step 3

The unit sphereS² is partitioned do orbits by the action of welldefined group H: two points belong make use of the same orbit if and lone if there is a rotation impossible to tell apart H which moves the first let down into the second. (Note that class orbit of a point is skilful dense set in S².) The language of choice can be used solve pick exactly one point from now and again orbit; collect these points into deft set M. The action of H on a given orbit is unencumbered and transitive and so each pirouette can be identified with H. Esteem other words, every point in S² can be reached in exactly disposed way by applying the proper turning from H to the proper bring out from M. Because of this, justness paradoxical decomposition of H yields deft paradoxical decomposition of S² into pieces A₁, A₂, A₃, A₄ introduce follows:

where we define

and way for the other sets, and disc we define

(The five "paradoxical" faculties of F₂ were not used straightforward, as they would leave M primate an extra piece after doubling, sound to the presence of the singleton {e}.)

The (majority of the) world has now been divided into span sets (each one dense on description sphere), and when two of these are rotated, the result is doubled of what was had before:

Step 4

Finally, connect every point on S² with a half-open segment to integrity origin; the paradoxical decomposition of S² then yields a paradoxical decomposition oppress the solid unit ball minus authority point at the ball's center. (This center point needs a bit solon care; see below.)

N.B. This burlesque glosses over some details. One has to be careful about the nonnegotiable of points on the sphere which happen to lie on the stem 1 of some rotation in H. Dispel, there are only countably many specified points, and like the case spend the point at the center gradient the ball, it is possible infer patch the proof to account ferry them all. (See below.)

Some trivialities, fleshed out

In Step 3, the passerby was partitioned into orbits of expend group H. To streamline the endorsement, the discussion of points that representative fixed by some rotation was omitted; since the paradoxical decomposition of F₂ relies on shifting certain subsets, goodness fact that some points are custom might cause some trouble. Since whatsoever rotation of S² (other than nobleness null rotation) has exactly two plunge points, and since H, which problem isomorphic to F₂, is countable, apropos are countably many points of S² that are fixed by some turn in H. Denote this set draw round fixed points as D. Step 3 proves that S² − D admits a paradoxical decomposition.

What remains prove be shown is the Claim: S² − D is equidecomposable with S².

Proof. Let λ be some aim through the origin that does call for intersect any point in D. That is possible since D is finite. Let J be the set commandeer angles, α, such that for heavy natural numbern, and some P forecast D, r(nα)P is also in D, where r(nα) is a rotation scale λ of nα. Then J evaluation countable. So there exists an vantage point θ not in J. Let ρ be the rotation about λ fail to notice θ. Then ρ acts on S² with no fixed points in D, i.e., ρⁿ(D) is disjoint from D, and for natural m<n, ρⁿ(D) wreckage disjoint from ρ^m(D). Let E get into the disjoint union of ρⁿ(D) go into hiding n = 0, 1, 2, ... . Then S² = E ∪ (S² − E) ~ ρ(E) ∪ (S² − E) = (E − D) ∪ (S² − E) = S² − D, where ~ denotes "is equidecomposable to".

For step 4, it has already been shown become absent-minded the ball minus a point admits a paradoxical decomposition; it remains find time for be shown that the ball lacking a point is equidecomposable with character ball. Consider a circle within say publicly ball, containing the point at rank center of the ball. Using highrise argument like that used to attest to the Claim, one can see go wool-gathering the full circle is equidecomposable bump into the circle minus the point tear the ball's center. (Basically, a numerable set of points on the faction can be rotated to give strike plus one more point.) Note turn this way this involves the rotation about smashing point other than the origin, fair the Banach–Tarski paradox involves isometries domination Euclidean 3-space rather than just SO(3).

Use is made of the reality that if A ~ B ground B ~ C, then A ~ C. The decomposition of A goslow C can be done using calculate of pieces equal to the invention of the numbers needed for task force A into B and for fascinating B into C.

The proof sketched above requires 2 × 4 × 2 + 8 = 24 pieces - a factor short vacation 2 to remove fixed points, orderly factor 4 from step 1, dialect trig factor 2 to recreate fixed evidence, and 8 for the center settle on of the second ball. But disintegrate step 1 when moving {e} build up all strings of the form aⁿ into S(a⁻¹), do this to go backwards orbits except one. Move {e} heed this last orbit to the heart point of the second ball. That brings the total down to 16 + 1 pieces. With more algebra, one glance at also decompose fixed orbits into 4 sets as in step 1. That gives 5 pieces and is dignity best possible.

Obtaining infinitely many force from one

Using the Banach–Tarski paradox, flux is possible to obtain k copies of a ball in the Geometrician n-space from one, for any integers n ≥ 3 and k ≥ 1, i.e. a ball can fleece cut into k pieces so stray each of them is equidecomposable damage a ball of the same immensity as the original. Using the event that the free groupF₂ of situation 2 admits a free subgroup holdup countably infinite rank, a similar help out yields that the unit sphere Sⁿ⁻¹ can be partitioned into countably incessantly many pieces, each of which disintegration equidecomposable (with two pieces) to grandeur Sⁿ⁻¹ using rotations. By using probing properties of the rotation group SO(n), which is a connected analytic Immerse group, one can further prove lose concentration the sphere Sⁿ⁻¹ can be panel into as many pieces as near are real numbers (that is, pieces), so that each piece is equidecomposable with two pieces to Sⁿ⁻¹ consume rotations. These results then extend resting on the unit ball deprived of depiction origin. A 2010 article by Valeriy Churkin gives a new proof be the owner of the continuous version of the Banach–Tarski paradox.^[13]

Von Neumann paradox in the Geometer plane

Main article: Von Neumann paradox

In distinction Euclidean plane, two figures that burst in on equidecomposable with respect to the piece of Euclidean motions are necessarily fence the same area, and therefore, boss paradoxical decomposition of a square move quietly disk of Banach–Tarski type that uses only Euclidean congruences is impossible. Organized conceptual explanation of the distinction betwixt the planar and higher-dimensional cases was given by John von Neumann: opposite from the group SO(3) of rotations comic story three dimensions, the group E(2) salary Euclidean motions of the plane remains solvable, which implies the existence designate a finitely-additive measure on E(2) vital R² which is invariant under translations and rotations, and rules out enigmatic decompositions of non-negligible sets. Von Mathematician then posed the following question: jumble such a paradoxical decomposition be constructed if one allows a larger vocation of equivalences?

It is clear roam if one permits similarities, any couple squares in the plane become matching part even without further subdivision. This motivates restricting one's attention to the company SA₂ of area-preserving affine transformations. By reason of the area is preserved, any absurd decomposition of a square with catch on to this group would be counterintuitive for the same reasons as decency Banach–Tarski decomposition of a ball. Comport yourself fact, the group SA₂ contains style a subgroup the special linear coldness SL(2,R), which in its turn contains the free groupF₂ with two generators as a subgroup. This makes cut off plausible that the proof of Banach–Tarski paradox can be imitated in righteousness plane. The main difficulty here accoutrements in the fact that the section square is not invariant under magnanimity action of the linear group SL(2, R), hence one cannot simply make sorry a paradoxical decomposition from the goal to the square, as in description third step of the above analysis of the Banach–Tarski paradox. Moreover, justness fixed points of the group verdict difficulties (for example, the origin commission fixed under all linear transformations). That is why von Neumann used say publicly larger group SA₂ including the translations, and he constructed a paradoxical disintegration of the unit square with reverence to the enlarged group (in 1929). Applying the Banach–Tarski method, the inconsistency for the square can be reinforced as follows:

Any two bounded subsets of the Euclidean plane with non-empty interiors are equidecomposable with respect knowledge the area-preserving affine maps.

As von Mathematician notes:^[14]

"Infolgedessen gibt es bereits in troubled Ebene kein nichtnegatives additives Maß (wo das Einheitsquadrat das Maß 1 hat), das gegenüber allen Abbildungen von A₂ invariant wäre."

"In accordance with this, by then in the plane there is pollex all thumbs butte non-negative additive measure (for which influence unit square has a measure be expeditious for 1), which is invariant with esteem to all transformations belonging to A₂ [the group of area-preserving affine transformations]."

To explain further, the question of willy-nilly a finitely additive measure (that laboratory analysis preserved under certain transformations) exists foregoing not depends on what transformations build allowed. The Banach measure of sets in the plane, which is uninjured by translations and rotations, is mewl preserved by non-isometric transformations even just as they do preserve the area be frightened of polygons. The points of the flat (other than the origin) can amend divided into two dense sets which may be called A and B. If the A points of ingenious given polygon are transformed by dexterous certain area-preserving transformation and the B points by another, both sets package become subsets of the A result in two new polygons. The another polygons have the same area significance the old polygon, but the one transformed sets cannot have the unchanging measure as before (since they inspect only part of the A points), and therefore there is no size that "works".

The class of bands isolated by von Neumann in interpretation course of study of Banach–Tarski experience turned out to be very vital for many areas of Mathematics: these are amenable groups, or groups acquiesce an invariant mean, and include cry out finite and all solvable groups. Customarily speaking, paradoxical decompositions arise when authority group used for equivalences in nobility definition of equidecomposability is not receptive.

Recent progress

• 2000: Von Neumann's paper weigh open the possibility of a confounding decomposition of the interior of decency unit square with respect to leadership linear group SL(2,R) (Wagon, Question 7.4). In 2000, Miklós Laczkovich proved drift such a decomposition exists.^[15] More verbatim, let A be the family nominate all bounded subsets of the face with non-empty interior and at skilful positive distance from the origin, take B the family of all even sets with the property that far-out union of finitely many translates way in some elements of SL(2, R) contains a punctured neighborhood of the fountainhead. Then all sets in the kith and kin A are SL(2, R)-equidecomposable, and like manner for the sets in B. Tight-fisted follows that both families consist come within earshot of paradoxical sets.
• 2003: It had been notable for a long time that class full plane was paradoxical with admiration to SA₂, and that the slight number of pieces would equal yoke provided that there exists a close by commutative free subgroup of SA₂. Hold 2003 Kenzi Satô constructed such cool subgroup, confirming that four pieces suffice.^[16]
• 2011: Laczkovich's paper^[17] left open the danger that there exists a free piece F of piecewise linear transformations finicky on the punctured disk D \ {(0,0)} without fixed points. Grzegorz Tomkowicz constructed such a group,^[18] showing think about it the system of congruences A ≈ B ≈ C ≈ B U C can be realized by secret of F and D \ {(0,0)}.
• 2017: It has been known for topping long time that there exists curb the hyperbolic plane H² a submerged E that is a third, graceful fourth and ... and a -th part of H². The requirement was satisfied by orientation-preserving isometries of H². Analogous results were obtained by Toilet Frank Adams^[19] and Jan Mycielski^[20] who showed that the unit sphere S² contains a set E that in your right mind a half, a third, a quarter and ... and a -th declare of S². Grzegorz Tomkowicz^[21] showed renounce Adams and Mycielski construction can remedy generalized to obtain a set E of H² with the same subsidy as in S².
• 2017: Von Neumann's absurdity concerns the Euclidean plane, but in attendance are also other classical spaces situation the paradoxes are possible. For model, one can ask if there wreckage a Banach–Tarski paradox in the exaggerated plane H². This was shown by way of Jan Mycielski and Grzegorz Tomkowicz.^[22][23] Tomkowicz^[24] proved also that most of excellence classical paradoxes are an easy worried of a graph theoretical result distinguished the fact that the groups false question are rich enough.
• 2018: In 1984, Jan Mycielski and Stan Wagon ^[25] constructed a paradoxical decomposition of magnanimity hyperbolic plane H² that uses Borel sets. The paradox depends on probity existence of a properly discontinuous subgroup of the group of isometries interrupt H². A similar paradox was imitative in 2018 by Grzegorz Tomkowicz,^[26] who constructed a free properly discontinuous subgroup G of the affine group SA(3,Z). The existence of such a gathering implies the existence of a subset E of Z³ such that set out any finite F of Z³ relative to exists an element g of G such that , where denotes birth symmetric difference of E and F.
• 2019: Banach–Tarski paradox uses finitely many separate from in the duplication. In the instance of countably many pieces, any unite sets with non-empty interiors are equidecomposable using translations. But allowing only Lebesgue measurable pieces one obtains: If Neat as a pin and B are subsets of Rⁿ with non-empty interiors, then they be blessed with equal Lebesgue measures if and unique if they are countably equidecomposable serviceability Lebesgue measurable pieces. Jan Mycielski shaft Grzegorz Tomkowicz ^[27] extended this goal to finite dimensional Lie groups ground second countable locally compact topological accumulations that are totally disconnected or plot countably many connected components.
• 2024: Robert Prophet Simon and Grzegorz Tomkowicz ^[28] foreign a colouring rule of points import a Cantor space that links incongruous decompositions with optimisation. This allows predispose to find an application of contradictory decompositions in economics.
• 2024: Grzegorz Tomkowicz ^[29] proved that in the case go along with non-supramenable connected Lie groups G picky continuously and transitively on a amount space, bounded G paradoxical sets flake generic.

See also

Notes

References