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{{redirect|Zymurgy|the American beer newsletter|Brewers Association#American Homebrewers Association}}
In [[general topology]], a branch of mathematics, a collection ''A'' of subsets of a set ''X'' is said to have the '''finite intersection property''' if the [[intersection (set theory)|intersection]] over any finite subcollection of ''A'' is nonempty.
{{unreferenced|date=November 2011}}
'''Zymology''' is the study of ''zymurgy'' (from the Greek: ζύμωσις+ἔργον, the workings of fermentation), the area of applied science related to [[Ethanol fermentation|fermentation]]. It deals with the biochemical processes involved in fermentation, with [[yeast]] and [[bacteria]] selection and [[physiology]], and with the practical issues of fermentation, including [[brewing]], [[wine making]], [[yogurt|fermenting milk]] and other [[List of fermented foods|fermented foods]].  Zymology is also known as '''zymurgy'''.


== Fermentation ==
A '''centered system of sets''' is a collection of sets with the finite intersection property.
{{see also|Yeast in winemaking}}
Fermentation can be simply defined, in this context, as the conversion of [[sugar]] molecules into [[ethanol]] and [[carbon dioxide]] by [[yeast]].
: <math>
C_6 H_{12} O_6 \Rightarrow 2 CO_2 + 2C_2 H_5 OH
</math>


==History==
==Definition==
[[France|French]] chemist [[Louis Pasteur]] was the first ''zymologist'', when in 1857 he connected yeast to fermentation. Pasteur originally defined fermentation as ''respiration without air''.
Let ''X'' be a set with <math>A=\{A_i\}_{i\in I}</math> a family of subsets of ''X''. Then the collection ''A'' has the finite intersection property (fip), if any finite subcollection ''J'' ⊆ ''I'' has non-empty intersection <math>\bigcap_{i\in J} A_i.</math>


Pasteur performed careful research and concluded, ''"I am of the opinion that alcoholic fermentation never occurs without simultaneous organization, development and multiplication of cells . . .. If asked, in what consists the chemical act whereby the sugar is decomposed . . . I am completely ignorant of it."''.
==Discussion==
Clearly the empty set cannot belong to any collection with the f.i.p.  The condition is trivially satisfied if the intersection over the entire collection is nonempty (in particular, if the collection itself is empty), and it is also trivially satisfied if the collection is nested, meaning that the collection is [[total order|totally ordered]] by inclusion (equivalently, for any finite subcollection, a particular element of the subcollection is contained in all the other elements of the subcollection), e.g. the [[nested sequence of intervals]] (0, 1/''n''). These are not the only possibilities however. For example, if ''X'' = (0, 1) and for each positive integer ''i'', ''X<sub>i</sub>'' is the set of elements of ''X'' having a decimal expansion with digit 0 in the ''i'''th decimal place, then any finite intersection is nonempty (just take 0 in those finitely many places and 1 in the rest), but the intersection of all ''X<sub>i</sub>'' for ''i'' ≥ 1 is empty, since no element of (0, 1) has all zero digits.


The [[Germany|German]] [[Eduard Buchner]], winner of the 1907 [[Nobel Prize]] in chemistry, later determined that fermentation was actually caused by a yeast secretion that he termed ''[[zymase]]''.
The finite intersection property is useful in formulating an alternative definition of [[compact space|compactness]]: a space is compact if and only if every collection of closed sets satisfying the finite intersection property has nonempty intersection itself.<ref>{{planetmathref|id=4181|title=A space is compact iff any family of closed sets having fip has non-empty intersection}}</ref> This formulation of compactness is used in some proofs of [[Tychonoff's theorem]] and the [[uncountable set|uncountability]] of the [[real number]]s (see next section)


The research efforts undertaken by the [[Denmark|Danish]] [[Carlsberg Laboratory|Carlsberg]] scientists greatly accelerated the increase in knowledge about yeast and brewing. The Carlsberg scientists are generally acknowledged{{by whom|date=May 2012}} as jump-starting the entire field of [[molecular biology]].
==Applications==
'''Theorem.''' Let ''X'' be a [[Compact space|compact]] [[Hausdorff space]] that  satisfies the property that no one-point set is open. If ''X'' has more than one point, then ''X'' is uncountable.


==Products==
'''Proof.''' We will show that if ''U'' ⊆ ''X'' is nonempty and [[Open set|open]], and if ''x'' is a point of ''X'', then there is a [[Neighbourhood (mathematics)|neighbourhood]] ''V'' ⊂ ''U'' whose [[closure (topology)|closure]] doesn’t contain ''x'' (''x'' may or may not be in ''U''). Choose ''y'' in ''U'' different from ''x'' (if ''x'' is in ''U'', then there must exist such a ''y'' for otherwise ''U'' would be an open one point set; if ''x'' isn’t in ''U'', this is possible since ''U'' is nonempty). Then by the Hausdorff condition, choose disjoint neighbourhoods ''W'' and ''K'' of ''x'' and ''y'' respectively. Then ''K''&nbsp;∩&nbsp;''U'' will be a neighbourhood of ''y'' contained in ''U'' whose closure doesn’t contain ''x'' as desired.<br />


* [[Beer]]
Now suppose ''f'' : '''N''' → ''X'' is a bijection, and let {''x<sub>i</sub>'' : ''i'' ∈ '''N'''} denote the image of ''f''. Let ''X'' be the first open set and choose a neighbourhood ''U''<sub>1</sub> ⊂ ''X'' whose closure doesn’t contain ''x''<sub>1</sub>. Secondly, choose a neighbourhood ''U''<sub>2</sub> ⊂ ''U''<sub>1</sub> whose closure doesn’t contain ''x''<sub>2</sub>. Continue this process whereby choosing a neighbourhood ''U''<sub>''n''+1</sub> ⊂ ''U<sub>n</sub>'' whose closure doesn’t contain ''x''<sub>''n''+1</sub>. Then the collection {''U<sub>i</sub>'' : ''i'' ∈ '''N'''} satisfies the finite intersection property and hence the intersection of their closures is nonempty (by the compactness of ''X''). Therefore there is a point ''x'' in this intersection. No ''x<sub>i</sub>'' can belong to this intersection because ''x<sub>i</sub>'' doesn’t belong to the closure of ''U<sub>i</sub>''. This means that ''x'' is not equal to ''x<sub>i</sub>'' for all ''i'' and ''f'' is not surjective; a contradiction. Therefore, ''X'' is uncountable.
* [[Wine]]
* [[Cider]]
* [[Mead]]
* [[Perry]]


==External links==
All the conditions in the statement of the theorem are necessary:
*[http://www.makewine.com/winemaking/general/yeast/ Fundamentals of winemaking: zymology]


[[Category:Biochemistry]]
1. We cannot eliminate the Hausdorff condition; a countable set with the [[indiscrete topology]] is compact, has more than one point, and satisfies the property that no one point sets are open, but is not uncountable.
[[Category:Brewing]]
[[Category:Oenology]]


{{wine-stub}}
2. We cannot eliminate the compactness condition as the set of all rational numbers shows.
 
3. We cannot eliminate the condition that one point sets cannot be open as a finite space given the [[discrete topology]] shows.
 
'''Corollary.''' Every closed interval [''a'',&nbsp;''b''] with ''a''&nbsp;<&nbsp;''b'' is uncountable. Therefore, '''R''' is uncountable.
 
'''Corollary.''' Every [[Perfect space|perfect]], [[Locally compact space|locally compact]] Hausdorff space is uncountable.
 
'''Proof.''' Let ''X'' be a perfect, compact, Hausdorff space, then the theorem immediately implies that ''X'' is uncountable. If ''X'' is a perfect, locally compact Hausdorff space which is not compact, then the [[one-point compactification]] of ''X'' is a perfect, compact Hausdorff space. Therefore the one point compactification of ''X'' is uncountable. Since removing a point from an uncountable set still leaves an uncountable set, ''X'' is uncountable as well.
 
==Examples==
A [[filter (topology)|filter]] has the finite intersection property by definition.
 
== Theorems ==
Let ''X'' be nonempty, ''F'' ⊆ 2<sup>''X''</sup>, ''F'' having the finite intersection property. Then there exists an ''F''′ [[ultrafilter]] (in 2<sup>''X''</sup>) such that ''F'' ⊆ ''F''′.
 
See details and proof in {{harvtxt|Csirmaz|Hajnal|1994}}.<ref>{{citation|last1=Csirmaz|first1=László|last2=Hajnal|first2=András|author2-link=András Hajnal|title=Matematikai logika|publisher=[[Eötvös Loránd University]]|location=Budapest|year=1994|url=http://www.renyi.hu/~csirmaz/|format=In Hungarian}}.</ref> This result is known as [[ultrafilter lemma]].
 
==Variants==
A family of sets ''A'' has the '''strong finite intersection property''' (sfip), if every finite subfamily of ''A'' has infinite intersection.
 
== References ==
<references/>
* {{planetmathref|id=4178|title=Finite intersection property}}
 
{{DEFAULTSORT:Finite Intersection Property}}
[[Category:General topology]]
[[Category:Set families]]

Revision as of 15:58, 11 August 2014

In general topology, a branch of mathematics, a collection A of subsets of a set X is said to have the finite intersection property if the intersection over any finite subcollection of A is nonempty.

A centered system of sets is a collection of sets with the finite intersection property.

Definition

Let X be a set with a family of subsets of X. Then the collection A has the finite intersection property (fip), if any finite subcollection J ⊆ I has non-empty intersection

Discussion

Clearly the empty set cannot belong to any collection with the f.i.p. The condition is trivially satisfied if the intersection over the entire collection is nonempty (in particular, if the collection itself is empty), and it is also trivially satisfied if the collection is nested, meaning that the collection is totally ordered by inclusion (equivalently, for any finite subcollection, a particular element of the subcollection is contained in all the other elements of the subcollection), e.g. the nested sequence of intervals (0, 1/n). These are not the only possibilities however. For example, if X = (0, 1) and for each positive integer i, Xi is the set of elements of X having a decimal expansion with digit 0 in the i'th decimal place, then any finite intersection is nonempty (just take 0 in those finitely many places and 1 in the rest), but the intersection of all Xi for i ≥ 1 is empty, since no element of (0, 1) has all zero digits.

The finite intersection property is useful in formulating an alternative definition of compactness: a space is compact if and only if every collection of closed sets satisfying the finite intersection property has nonempty intersection itself.[1] This formulation of compactness is used in some proofs of Tychonoff's theorem and the uncountability of the real numbers (see next section)

Applications

Theorem. Let X be a compact Hausdorff space that satisfies the property that no one-point set is open. If X has more than one point, then X is uncountable.

Proof. We will show that if U ⊆ X is nonempty and open, and if x is a point of X, then there is a neighbourhood V ⊂ U whose closure doesn’t contain x (x may or may not be in U). Choose y in U different from x (if x is in U, then there must exist such a y for otherwise U would be an open one point set; if x isn’t in U, this is possible since U is nonempty). Then by the Hausdorff condition, choose disjoint neighbourhoods W and K of x and y respectively. Then K ∩ U will be a neighbourhood of y contained in U whose closure doesn’t contain x as desired.

Now suppose f : N → X is a bijection, and let {xi : i ∈ N} denote the image of f. Let X be the first open set and choose a neighbourhood U1 ⊂ X whose closure doesn’t contain x1. Secondly, choose a neighbourhood U2 ⊂ U1 whose closure doesn’t contain x2. Continue this process whereby choosing a neighbourhood Un+1 ⊂ Un whose closure doesn’t contain xn+1. Then the collection {Ui : i ∈ N} satisfies the finite intersection property and hence the intersection of their closures is nonempty (by the compactness of X). Therefore there is a point x in this intersection. No xi can belong to this intersection because xi doesn’t belong to the closure of Ui. This means that x is not equal to xi for all i and f is not surjective; a contradiction. Therefore, X is uncountable.

All the conditions in the statement of the theorem are necessary:

1. We cannot eliminate the Hausdorff condition; a countable set with the indiscrete topology is compact, has more than one point, and satisfies the property that no one point sets are open, but is not uncountable.

2. We cannot eliminate the compactness condition as the set of all rational numbers shows.

3. We cannot eliminate the condition that one point sets cannot be open as a finite space given the discrete topology shows.

Corollary. Every closed interval [a, b] with a < b is uncountable. Therefore, R is uncountable.

Corollary. Every perfect, locally compact Hausdorff space is uncountable.

Proof. Let X be a perfect, compact, Hausdorff space, then the theorem immediately implies that X is uncountable. If X is a perfect, locally compact Hausdorff space which is not compact, then the one-point compactification of X is a perfect, compact Hausdorff space. Therefore the one point compactification of X is uncountable. Since removing a point from an uncountable set still leaves an uncountable set, X is uncountable as well.

Examples

A filter has the finite intersection property by definition.

Theorems

Let X be nonempty, F ⊆ 2X, F having the finite intersection property. Then there exists an F′ ultrafilter (in 2X) such that F ⊆ F′.

See details and proof in Template:Harvtxt.[2] This result is known as ultrafilter lemma.

Variants

A family of sets A has the strong finite intersection property (sfip), if every finite subfamily of A has infinite intersection.

References

  1. ↑ Template:Planetmathref
  2. ↑ Many property agents need to declare for the PIC grant in Singapore. However, not all of them know find out how to do the correct process for getting this PIC scheme from the IRAS. There are a number of steps that you need to do before your software can be approved.

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