Ultrafilter (set theory)

  (Redirected from Ultrafilter lemma)

In the mathematical field of set theory, an ultrafilter is a maximal proper filter: it is a filter on a given non-empty set which is a certain type of non-empty family of subsets of that is not equal to the power set of (such filters are called proper) and that is also "maximal" in that there does not exist any other proper filter on that contains it as a proper subset. Said differently, a proper filter is called an ultrafilter if there exists exactly one proper filter that contains it as a subset, that proper filter (necessarily) being itself.

The powerset lattice of the set {1,2,3,4}, with the upper set ↑{1,4} colored dark green. It is a principal filter, but not an ultrafilter, as it can be extended to the larger nontrivial filter ↑{1}, by including also the light green elements. Since ↑{1} cannot be extended any further, it is an ultrafilter.

More formally, an ultrafilter on is a proper filter that is also a maximal filter on with respect to set inclusion, meaning that there does not exist any proper filter on that contains as a proper subset. Ultrafilters on sets are an important special instance of ultrafilters on partially ordered sets, where the partially ordered set consists of the power set and the partial order is subset inclusion

Ultrafilters have many applications in set theory, model theory, and topology.[1]:186

DefinitionsEdit

Given an arbitrary set   an ultrafilter on   is a non-empty family   of subsets of   such that:

  1. Proper or non-degenerate: The empty set is not an element of  
  2. Upward closed in  : If   and if   is any superset of   (that is, if  ) then  
  3. π−system: If   and   are elements of   then so is their intersection  
  4. If   then either[note 1]   or its relative complement   is an element of  

Properties (1), (2), and (3) are the defining properties of a filter on   Some authors do not include non-degeneracy (which is property (1) above) in their definition of "filter". However, the definition of "ultrafilter" (and also of "prefilter" and "filter subbase") always includes non-degeneracy as a defining condition. This article requires that all filters be proper although a filter might be described as "proper" for emphasis.

For a filter   that is not an ultrafilter, one would say   if   and   if   leaving   undefined elsewhere.[citation needed][clarification needed]

A filter subbase is a non-empty family of sets that has the finite intersection property (i.e. all finite intersections are non-empty). Equivalently, a filter subbase is a non-empty family of sets that is contained in some (proper) filter. The smallest (relative to  ) filter containing a given filter subbase is said to be generated by the filter subbase.

The upward closure in   of a family of sets   is the set

 

A prefilter or filter base is a non-empty and proper (i.e.  ) family of sets   that is downward directed, which means that if   then there exists some   such that   Equivalently, a prefilter is any family of sets   whose upward closure   is a filter, in which case this filter is called the filter generated by   and   is said to be a filter base for  

The dual in  [2] of a family of sets   is the set  

Generalization to ultra prefiltersEdit

A family   of subsets of   is called ultra if   and any of the following equivalent conditions are satisfied:[2][3]

  1. For every set   there exists some set   such that   or   (or equivalently, such that   equals   or  ).
  2. For every set   there exists some set  } such that   equals   or  
    • Here,   is defined to be the union of all sets in  
    • This characterization of "  is ultra" does not depend on the set   so mentioning the set   is optional when using the term "ultra."
  3. For every set   (not necessarily even a subset of  ) there exists some set   such that   equals   or  
    • If   satisfies this condition then so does every superset   In particular, a set   is ultra if and only if   and   contains as a subset some ultra family of sets.

A filter subbase that is ultra is necessarily a prefilter.[proof 1]

The ultra property can now be used to define both ultrafilters and ultra prefilters:

An ultra prefilter[2][3] is a prefilter that is ultra. Equivalently, it is a filter subbase that is ultra.
An ultrafilter[2][3] on   is a (proper) filter on   that is ultra. Equivalently, it is any filter on   that is generated by an ultra prefilter.
Interpretation as large sets

The elements of a proper filter   on   may be thought of as being "large sets (relative to  )" and the complements in   of a large sets can be thought of as being "small" sets[4] (the "small sets" are exactly the elements in the ideal  ). In general, there may be subsets of   that are neither large nor small, or possibly simultaneously large and small. A dual ideal is a filter (i.e. proper) if there is no set that is both large and small, or equivalently, if the   is not large.[4] A filter is ultra if and only if every subset of   is either large or else small. With this terminology, the defining properties of a filter can be restarted as: (1) any superset of a large set is large set, (2) the intersection of any two (or finitely many) large sets is large, (3)   is a large set (i.e.  ), (4) the empty set is not large. Different dual ideals give different notions of "large" sets.

Another way of looking at ultrafilters on a power set   is as follows: for a given ultrafilter   define a function   on   by setting   if   is an element of   and   otherwise. Such a function is called a 2-valued morphism. Then   is finitely additive, and hence a content on   and every property of elements of   is either true almost everywhere or false almost everywhere. However,   is usually not countably additive, and hence does not define a measure in the usual sense.

Ultra prefilters as maximal prefilters

To characterize ultra prefilters in terms of "maximality," the following relation is needed.

Given two families of sets   and   the family   is said to be coarser[5][6] than   and   is finer than and subordinate to   written   or NM, if for every   there is some   such that   The families   and   are called equivalent if   and   The families   and   are comparable if one of these sets is finer than the other.[5]

The subordination relationship, i.e.   is a preorder so the above definition of "equivalent" does form an equivalence relation. If   then   but the converse does not hold in general. However, if   is upward closed, such as a filter, then   if and only if   Every prefilter is equivalent to the filter that it generates. This shows that it is possible for filters to be equivalent to sets that are not filters.

If two families of sets   and   are equivalent then either both   and   are ultra (resp. prefilters, filter subbases) or otherwise neither one of them is ultra (resp. a prefilter, a filter subbase). In particular, if a filter subbase is not also a prefilter, then it is not equivalent to the filter or prefilter that it generates. If   and   are both filters on   then   and   are equivalent if and only if   If a proper filter (resp. ultrafilter) is equivalent to a family of sets   then   is necessarily a prefilter (resp. ultra prefilter). Using the following characterization, it is possible to define prefilters (resp. ultra prefilters) using only the concept of filters (resp. ultrafilters) and subordination:

An arbitrary family of sets is a prefilter if and only it is equivalent to a (proper) filter.
An arbitrary family of sets is an ultra prefilter if and only it is equivalent to an ultrafilter.
A maximal prefilter on  [2][3] is a prefilter   that satisfies any of the following equivalent conditions:
  1.   is ultra.
  2.   is maximal on   with respect to   meaning that if   satisfies   then  [3]
  3. There is no prefilter properly subordinate to  [3]
  4. If a (proper) filter   on   satisfies   then  
  5. The filter on   generated by   is ultra.

CharacterizationsEdit

There are no ultrafilters on   where   is the empty set, so it is henceforth assumed that  

A filter subbase   on   is an ultrafilter on   if and only if any of the following equivalent conditions hold:[2][3]

  1. for any   either   or  
  2.   is a maximal filter subbase on   meaning that if   is any filter subbase on   then   implies  [4]

A (proper) filter   on   is an ultrafilter on   if and only if any of the following equivalent conditions hold:

  1.   is ultra;
  2.   is generated by an ultra prefilter;
  3. For any subset     or  [4]
    • So an ultrafilter   decides for every   whether   is "large" (i.e.  ) or "small" (i.e.  ).[7]
  4. For each subset   either[note 1]   is in   or ( ) is.
  5.   This condition can be restated as:   is partitioned by   and its dual  
    • The sets   and   are disjoint for all prefilters   on  
  6.   is an ideal on  [4]
  7. For any finite family   of subsets of   (where  ), if   then   for some index  
    • In words, a "large" set cannot be a finite union of sets that aren't large.[8]
  8. For any   if   then   or  
  9. For any   if   then   or   (a filter with this property is called a prime filter).
  10. For any   if   and   then either   or  
  11.   is a maximal filter; that is, if   is a filter on   such that   then   Equivalently,   is a maximal filter if there is no filter   on   that contains   as a proper subset (that is, no filter is strictly finer than  ).[4]

Grills and Filter-GrillsEdit

If   then its grill on   is the family

 

where   may be written if   is clear from context. For example,   and if   then   If   then   and moreover, if   is a filter subbase then  [9] The grill   is upward closed in   if and only if   which will henceforth be assumed. Moreover,   so that   is upward closed in   if and only if  

The grill of a filter on   is called a filter-grill on  [9] For any     is a filter-grill on   if and only if (1)   is upward closed in   and (2) for all sets   and   if   then   or   The grill operation   induces a bijection

 

whose inverse is also given by  [9] If   then   is a filter-grill on   if and only if  [9] or equivalently, if and only if   is an ultrafilter on  [9] That is, a filter on   is a filter-grill if and only if it is ultra. For any non-empty     is both a filter on   and a filter-grill on   if and only if (1)   and (2) for all   the following equivalences hold:

  if and only if   if and only if  [9]

Free or principalEdit

If   is any non-empty family of sets then the Kernel of   is the intersection of all set in  :

 [10]

A non-empty family of sets   is called:

  • free if   and fixed otherwise (that is, if  ),
  • principal if  
  • principal at a point if   and   is a singleton set; in this case, if   then   is said to be principal at  

If a family of sets   is fixed then   is ultra if and only if some element of   is a singleton set, in which case   will necessarily be a prefilter. Every principal prefilter is fixed, so a principal prefilter   is ultra if and only if   is a singleton set. A singleton set is ultra if and only if its sole element is also a singleton set.

Every filter on   that is principal at a single point is an ultrafilter, and if in addition   is finite, then there are no ultrafilters on   other than these.[10] If there exists a free ultrafilter (or even filter subbase) on a set   then   must be infinite.

The next theorem shows that every ultrafilter falls into one of two categories: either it is free or else it is a principal filter generated by a single point.

Proposition — If   is an ultrafilter on   then the following are equivalent:

  1.   is fixed, or equivalently, not free.
  2.   is principal.
  3. Some element of   is a finite set.
  4. Some element of   is a singleton set.
  5.   is principal at some point of   which means   for some  
  6.   does not contain the Fréchet filter on   as a subset.
  7.   is sequential.[9]

Examples, properties, and sufficient conditionsEdit

If   and   are families of sets such that   is ultra,   and   then   is necessarily ultra. A filter subbase   that is not a prefilter cannot be ultra; but it is nevertheless still possible for the prefilter and filter generated by   to be ultra.

Suppose   is ultra and   is a set. The trace   is ultra if and only if it does not contain the empty set. Furthermore, at least one of the sets   and   will be ultra (this result extends to any finite partition of  ). If   are filters on     is an ultrafilter on   and   then there is some   that satisfies  [11] This result is not necessarily true for an infinite family of filters.[11]

The image under a map   of an ultra set   is again ultra and if   is an ultra prefilter then so is   The property of being ultra is preserved under bijections. However, the preimage of an ultrafilter is not necessarily ultra, not even if the map is surjective. For example, if   has more than one point and if the range of   consists of a single point   then   is an ultra prefilter on   but its preimage is not ultra. Alternatively, if   is a principal filter generated by a point in  then the preimage of   contains the empty set and so is not ultra.

The elementary filter induced by an infinite sequence, all of whose points are distinct, is not an ultrafilter.[11] If   then   denotes the set consisting all subsets of   having cardinality   and if  contains at least   ( ) distinct points, then   is ultra but it is not contained in any prefilter. This example generalizes to any integer   and also to   if   contains more than one element. Ultra sets that are not also prefilters are rarely used.

For every   and every   let   If   is an ultrafilter on   then the set of all   such that   is an ultrafilter on  [12]

Monad structureEdit

The functor associating to any set   the set of   of all ultrafilters on   forms a monad called the ultrafilter monad. The unit map

 

sends any element   to the principal ultrafilter given by  

This monad admits a conceptual explanation as the codensity monad of the inclusion of the category of finite sets into the category of all sets.[13]

The ultrafilter lemmaEdit

The ultrafilter lemma was first proved by Alfred Tarski in 1930.[12]

The ultrafilter lemma/principle/theorem[5] — Every proper filter on a set   is contained in some ultrafilter on  

The ultrafilter lemma is equivalent to each of the following statements:

  1. For every prefilter on a set   there exists a maximal prefilter on   subordinate to it.[2]
  2. Every proper filter subbase on a set   is contained in some ultrafilter on  

A consequence of the ultrafilter lemma is that every filter is equal to the intersection of all ultrafilters containing it.[14][note 2]

The following results can be proven using the ultrafilter lemma. A free ultrafilter exists on a set   if and only if   is infinite. Every proper filter is equal to the intersection of all ultrafilters containing it.[5] Since there are filters that are not ultra, this shows that the intersection of a family of ultrafilters need not be ultra. A family of sets   can be extended to a free ultrafilter if and only if the intersection of any finite family of elements of   is infinite.

Relationships to other statements under ZFEdit

Throughout this section, ZF refers to Zermelo–Fraenkel set theory and ZFC refers to ZF with the Axiom of Choice (AC). The ultrafilter lemma is independent of ZF. That is, there exist models in which the axioms of ZF hold but the ultrafilter lemma does not. There also exist models of ZF in which every ultrafilter is necessarily principal.

Every filter that contains a singleton set is necessarily an ultrafilter and given   the definition of the discrete ultrafilter   does not require more than ZF. If   is finite then every ultrafilter is a discrete filter at a point; consequently, free ultrafilters can only exist on infinite sets. In particular, if   is finite then the ultrafilter lemma can be proven from the axioms ZF. The existence of free ultrafilter on infinite sets can be proven if the axiom of choice is assumed. More generally, the ultrafilter lemma can be proven by using the axiom of choice, which in brief states that any Cartesian product of non-empty sets is non-empty. Under ZF, the axiom of choice is, in particular, equivalent to (a) Zorn's lemma, (b) Tychonoff's theorem, (c) the weak form of the vector basis theorem (which states that every vector space has a basis), (d) the strong form of the vector basis theorem, and other statements. However, the ultrafilter lemma is strictly weaker than the axiom of choice. While free ultrafilters can be proven to exist, it is not possible to construct an explicit example of a free ultrafilter; that is, free ultrafilters are intangible.[15]Alfred Tarski proved that under ZFC, the cardinality of the set of all free ultrafilters on an infinite set   is equal to the cardinality of   where   denotes the power set of  [16]

Under ZF, the axiom of choice can be used to prove both the ultrafilter lemma and the Krein–Milman theorem; conversely, under ZF, the ultrafilter lemma together with the Krein–Milman theorem can prove the axiom of choice.[17]

Statements that cannot be deducedEdit

The ultrafilter lemma is a relatively weak axiom. For example, each of the statements in the following list can not be deduced from ZF together with only the ultrafilter lemma:

  1. A countable union of countable sets is a countable set.
  2. The axiom of countable choice (ACC).
  3. The axiom of dependent choice (ADC).

Equivalent statementsEdit

Under ZF, the ultrafilter lemma is equivalent to each of the following statements:[18]

  1. The Boolean prime ideal theorem (BPIT).
    • This equivalence is provable in ZF set theory without the Axiom of Choice (AC).
  2. Stone's representation theorem for Boolean algebras.
  3. Any product of Boolean spaces is a Boolean space.[19]
  4. Boolean Prime Ideal Existence Theorem: Every nondegenerate Boolean algebra has a prime ideal.[20]
  5. Tychonoff's theorem for Hausdorff spaces: Any product of compact Hausdorff spaces is compact.[19]
  6. If   is endowed with the discrete topology then for any set   the product space   is compact.[19]
  7. Each of the following versions of the Banach-Alaoglu theorem is equivalent to the ultrafilter lemma:
    1. Any equicontinuous set of scalar-valued maps on a topological vector space (TVS) is relatively compact in the weak-* topology (that is, it is contained in some weak-* compact set).[21]
    2. The polar of any neighborhood of the origin in a TVS   is a weak-* compact subset of its continuous dual space.[21]
    3. The closed unit ball in the continuous dual space of any normed space is weak-* compact.[21]
      • If the normed space is separable then the ultrafilter lemma is sufficient but not necessary to prove this statement.
  8. A topological space   is compact if every ultrafilter on   converges to some limit.[22]
  9. A topological space   is compact if and only if every ultrafilter on   converges to some limit.[22]
    • The addition of the words "and only if" is the only difference between this statement and the one immediately above it.
  10. The Ultranet lemma: Every net has a universal subnet.[23]
    • By definition, a net in   is called an ultranet or an universal net if for every subset   the net is eventually in   or in  
  11. A topological space   is compact if and only if every ultranet on   converges to some limit.[22]
    • If the words "and only if" are removed then the resulting statement remains equivalent to the ultrafilter lemma.[22]
  12. A convergence space   is compact if every ultrafilter on   converges.[22]
  13. A uniform space is compact if it is complete and totally bounded.[22]
  14. The Stone–Čech compactification Theorem.[19]
  15. Each of the following versions of the compactness theorem is equivalent to the ultrafilter lemma:
    1. If   is a set of first-order sentences such that every finite subset of   has a model, then   has a model.[24]
    2. If   is a set of zero-order sentences such that every finite subset of   has a model, then   has a model.[24]
  16. The completeness theorem: If   is a set of zero-order sentences that is syntactically consistent, then it has a model (that is, it is semantically consistent).

Weaker statementsEdit

Any statement that can be deduced from the ultrafilter lemma (together with ZF) is said to be weaker than the ultrafilter lemma. A weaker statement is said to be strictly weaker if under ZF, it is not equivalent to the ultrafilter lemma. Under ZF, the ultrafilter lemma implies each of the following statements:

  1. The Axiom of Choice for Finite sets (ACF): Given   and a family   of non-empty finite sets, their product   is not empty.[23]
  2. A countable union of finite sets is a countable set.
    • However, ZF with the ultrafilter lemma is too weak to prove that a countable union of countable sets is a countable set.
  3. The Hahn–Banach theorem.[23]
    • In ZF, the Hahn–Banach theorem is strictly weaker than the ultrafilter lemma.
  4. The Banach–Tarski paradox.
  5. Every set can be linearly ordered.
  6. Every field has a unique algebraic closure.
  7. The Alexander subbase theorem.[23]
  8. Non-trivial ultraproducts exist.
  9. The weak ultrafilter theorem: A free ultrafilter exists on  
    • Under ZF, the weak ultrafilter theorem does not imply the ultrafilter lemma; that is, it is strictly weaker than the ultrafilter lemma.
  10. There exists a free ultrafilter on every infinite set;
    • This statement is actually strictly weaker than the ultrafilter lemma.
    • ZF alone does not even imply that there exists a non-principal ultrafilter on some set.

CompletenessEdit

The completeness of an ultrafilter   on a powerset is the smallest cardinal κ such that there are κ elements of   whose intersection is not in   The definition of an ultrafilter implies that the completeness of any powerset ultrafilter is at least  . An ultrafilter whose completeness is greater than  —that is, the intersection of any countable collection of elements of   is still in  —is called countably complete or σ-complete.

The completeness of a countably complete nonprincipal ultrafilter on a powerset is always a measurable cardinal.[citation needed]

Ordering on ultrafiltersEdit

The Rudin–Keisler ordering (named after Mary Ellen Rudin and Howard Jerome Keisler) is a preorder on the class of powerset ultrafilters defined as follows: if   is an ultrafilter on   and   an ultrafilter on   then   if there exists a function   such that

  if and only if  

for every subset  

Ultrafilters   and   are called Rudin–Keisler equivalent, denoted URK V, if there exist sets   and   and a bijection   that satisfies the condition above. (If   and   have the same cardinality, the definition can be simplified by fixing    )

It is known that ≡RK is the kernel of ≤RK, i.e., that URK V if and only if   and  [27]

Ultrafilters on Edit

There are several special properties that an ultrafilter on   where   extends the natural numbers, may possess, which prove useful in various areas of set theory and topology.

  • A non-principal ultrafilter   is called a P-point (or weakly selective) if for every partition   of   such that for all     there exists some   such that for all   is a finite set for each  
  • A non-principal ultrafilter   is called Ramsey (or selective) if for every partition   of   such that for all     there exists some   such that   is a singleton set for each  

It is a trivial observation that all Ramsey ultrafilters are P-points. Walter Rudin proved that the continuum hypothesis implies the existence of Ramsey ultrafilters.[28] In fact, many hypotheses imply the existence of Ramsey ultrafilters, including Martin's axiom. Saharon Shelah later showed that it is consistent that there are no P-point ultrafilters.[29] Therefore, the existence of these types of ultrafilters is independent of ZFC.

P-points are called as such because they are topological P-points in the usual topology of the space βω \ ω of non-principal ultrafilters. The name Ramsey comes from Ramsey's theorem. To see why, one can prove that an ultrafilter is Ramsey if and only if for every 2-coloring of   there exists an element of the ultrafilter that has a homogeneous color.

An ultrafilter on   is Ramsey if and only if it is minimal in the Rudin–Keisler ordering of non-principal powerset ultrafilters.[citation needed]

See alsoEdit

NotesEdit

  1. ^ a b Properties 1 and 3 imply that   and   cannot both be elements of  
  2. ^ Let   be a filter on   that is not an ultrafilter. If   is such that   then   has the finite intersection property (because if   then   if and only if  ) so that by the ultrafilter lemma, there exists some ultrafilter   on   such that   (so in particular  ). It follows that  
Proofs
  1. ^ Suppose   is filter subbase that is ultra. Let   and define   Because   is ultra, there exists some   such that   equals   or   The finite intersection property implies that   so necessarily   which is equivalent to  

ReferencesEdit

  1. ^ Davey, B. A.; Priestley, H. A. (1990). Introduction to Lattices and Order. Cambridge Mathematical Textbooks. Cambridge University Press.
  2. ^ a b c d e f g Narici & Beckenstein 2011, pp. 2-7.
  3. ^ a b c d e f g Dugundji 1966, pp. 219-221.
  4. ^ a b c d e f Schechter 1996, pp. 100-130.
  5. ^ a b c d Bourbaki 1989, pp. 57-68.
  6. ^ Schubert 1968, pp. 48-71.
  7. ^ Higgins, Cecelia (2018). "Ultrafilters in set theory" (PDF). math.uchicago.edu. Retrieved August 16, 2020.
  8. ^ Kruckman, Alex (November 7, 2012). "Notes on Ultrafilters" (PDF). math.berkeley.edu. Retrieved August 16, 2020.
  9. ^ a b c d e f g Dolecki & Mynard 2016, pp. 27-54.
  10. ^ a b Dolecki & Mynard 2016, pp. 33-35.
  11. ^ a b c Bourbaki 1989, pp. 129-133.
  12. ^ a b Jech 2006, pp. 73-89.
  13. ^ Leinster, Tom (2013). "Codensity and the ultrafilter monad". Theory and Applications of Categories. 28: 332–370. arXiv:1209.3606. Bibcode:2012arXiv1209.3606L.
  14. ^ Bourbaki 1987, pp. 57–68.
  15. ^ Schechter 1996, p. 105.
  16. ^ Schechter 1996, pp. 150-152.
  17. ^ Bell, J.; Fremlin, David (1972). "A geometric form of the axiom of choice" (PDF). Fundamenta Mathematicae. 77 (2): 167–170. Retrieved 11 June 2018. Theorem 1.2. BPI [the Boolean Prime Ideal Theorem] & KM [Krein-Milman]   (*) [the unit ball of the dual of a normed vector space has an extreme point].... Theorem 2.1. (*)   AC [the Axiom of Choice].
  18. ^ Schechter 1996, pp. 105,150-160,166,237,317-315,338-340,344-346,386-393,401-402,455-456,463,474,506,766-767.
  19. ^ a b c d Schechter 1996, p. 463.
  20. ^ Schechter 1996, p. 339.
  21. ^ a b c Schechter 1996, pp. 766-767.
  22. ^ a b c d e f Schechter 1996, p. 455.
  23. ^ a b c d Muger, Michael (2020). Topology for the Working Mathematician.
  24. ^ a b Schechter 1996, pp. 391-392.
  25. ^ Foreman, M.; Wehrung, F. (1991). "The Hahn–Banach theorem implies the existence of a non-Lebesgue measurable set" (PDF). Fundamenta Mathematicae. 138: 13–19. doi:10.4064/fm-138-1-13-19.
  26. ^ Pawlikowski, Janusz (1991). "The Hahn–Banach theorem implies the Banach–Tarski paradox" (PDF). Fundamenta Mathematicae. 138: 21–22. doi:10.4064/fm-138-1-21-22.
  27. ^ Comfort, W. W.; Negrepontis, S. (1974). The theory of ultrafilters. Berlin, New York: Springer-Verlag. MR 0396267. Corollary 9.3.
  28. ^ Rudin, Walter (1956), "Homogeneity problems in the theory of Čech compactifications", Duke Mathematical Journal, 23 (3): 409–419, doi:10.1215/S0012-7094-56-02337-7, hdl:10338.dmlcz/101493
  29. ^ Wimmers, Edward (March 1982), "The Shelah P-point independence theorem", Israel Journal of Mathematics, 43 (1): 28–48, doi:10.1007/BF02761683, S2CID 122393776

BibliographyEdit

Further readingEdit