Stellar wind - Revision history

en>Strike Eagle: Reverted 1 edit by 184.2.141.250 identified as test/vandalism using STiki

2014-12-09T18:29:27Z

Reverted 1 edit by 184.2.141.250 identified as test/vandalism using STiki

← Older revision		Revision as of 20:29, 9 December 2014
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72.200.151.13 at 13:23, 4 March 2014

2014-03-04T13:23:40Z

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en>Vsmith: unneeded math stuff

2014-01-26T15:38:25Z

unneeded math stuff

← Older revision		Revision as of 17:38, 26 January 2014
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	Hi there. My title is ~~Sophia Meagher although~~ it is ~~not~~ the ~~title~~ on ~~my birth certificate~~. ~~Some time ago she selected~~ to ~~reside~~ in ~~Alaska~~ and ~~her mothers and fathers live close by~~. ~~Invoicing~~ is ~~my profession~~. ~~The favorite pastime~~ for ~~him~~ and ~~his children~~ is ~~style~~ and he'~~ll be beginning something else along with it~~.<br><br>~~Here~~ is ~~my blog ::~~ [~~http:~~//~~3bbc~~.~~com~~/~~index~~.~~php?do~~=/~~profile~~-~~366067~~/~~info~~/ ~~spirit messages~~]		{{other uses\|König's theorem (disambiguation)}}

			In [[set theory]], '''König's theorem''' (named after the Hungarian mathematician Gyula Kőnig, who published under the name [[Julius König]]) colloquially states that if the [[Axiom of Choice]] holds, ''I'' is a [[Set (mathematics)\|set]], ''m<sub>i</sub>'' and ''n<sub>i</sub>'' are [[cardinal number]]s for every ''i'' in ''I'', and <math>m_i < n_i \!</math> for every ''i'' in ''I'' then
			:<math>\sum_{i\in I}m_i<\prod_{i\in I}n_i.</math>

			The ''sum'' here is the cardinality of the [[disjoint union]] of the sets ''m<sub>i</sub>'' and the product is the cardinality of the [[cartesian product]].
			However, without the use of the Axiom of Choice, the sum and the product cannot be defined as cardinal numbers, and the meaning of the inequality sign would need to be clarified.

			== Details ==

			The precise statement of the result: if ''I'' is a [[Set (mathematics)\|set]], ''A<sub>i</sub>'' and ''B<sub>i</sub>'' are sets for every ''i'' in ''I'', and <math>A_i<B_i\!</math> for every ''i'' in ''I'' then
			:<math>\sum_{i\in I}A_i<\prod_{i\in I}B_i,</math>
			where '''<''' means ''strictly less than in [[cardinality]],'' i.e. there is an [[injective]] [[function (mathematics)\|function]] from ''A<sub>i</sub>'' to ''B<sub>i</sub>,'' but not one going the other way. The union involved need not be disjoint (a non-disjoint union can't be any bigger than the disjoint version, also assuming the [[axiom of choice]]). In this formulation, '''König's theorem''' is equivalent to the [[Axiom of Choice]].<ref Name="Rubin 1985">{{cite book\|last=Rubin\|first=H.\|coauthors=Rubin, J.E.\|title=Equivalents of the Axiom of Choice, II\|publisher=[[North-Holland Publishing Company\|North Holland]]\|place=New York, NY\|year=1985\|pages=185\|isbn=0-444-87708-8}}</ref>

			(Of course, König's theorem is trivial if the cardinal numbers ''m<sub>i</sub>'' and ''n<sub>i</sub>'' are [[finite set\|finite]] and the index set ''I'' is finite. If ''I'' is [[empty set\|empty]], then the left sum is the empty sum and therefore 0, while the right hand product is the [[empty product]] and therefore 1).

			König's theorem is remarkable because of the strict inequality in the conclusion. There are many easy rules for the arithmetic of infinite sums and products of cardinals in which one can only conclude a weak inequality ≤, for example: if <math>m_i < n_i \!</math> for all ''i'' in ''I'', then one can only conclude
			:<math>\sum_{i\in I} m_i \le \sum_{i\in I} n_i </math>
			since, for example, setting <math>m_i = 1 </math> & <math>n_i = 2</math> where the index set ''I'' is the natural numbers, yields the sum <math>\aleph_0</math> for both sides and we have a strict equality.

			==Corollaries of König's theorem==
			*If <math>\kappa\,</math> is a cardinal then <math>\kappa < 2^{\kappa}.\!</math>
			If we take ''m<sub>i</sub>'' = 1, and ''n<sub>i</sub>'' = 2 for each ''i'' in κ, then the left hand side of the above inequality is just κ, while the right hand side is 2<sup>κ</sup>, the cardinality of functions from κ to {0,1}, that is, the cardinality of the power set of κ. Thus, König's theorem gives us an alternate proof of [[Cantor's theorem]]. (Historically of course Cantor's theorem was proved much earlier.)

			===Axiom of choice===
			One way of stating the axiom of choice is "An arbitrary Cartesian product of non-empty sets is non-empty.". Let ''B<sub>i</sub>'' be a non-empty set for each ''i'' in ''I''. Let ''A<sub>i</sub>'' = {} for each ''i'' in ''I''. Thus by König's theorem, we have:
			*If <math>\forall i\in I(\{\}<B_i)</math>, then <math>\{\}<\prod_{i\in I}B_i.</math>
			That is, the Cartesian product of the given non-empty sets, ''B<sub>i</sub>'', has a larger cardinality than the sum of empty sets. Thus it is non-empty which is just what the axiom of choice states. Since the axiom of choice follows from König's theorem, we will use the axiom of choice freely and implicitly when discussing consequences of the theorem.

			===König's theorem and cofinality===
			König's theorem has also important consequences for [[cofinality]] of cardinal numbers.

			*If <math>\kappa\ge\aleph_0</math>, then <math>\kappa<\kappa^{cf(\kappa)}. \!</math>
			Choose a strictly increasing cf(κ)-sequence of cardinals approaching κ. Each of them is less than κ, so their sum which is κ is less than the product of cf(κ) copies of κ.

			According to [[Easton's theorem]], the next consequence of König's theorem is the only nontrivial constraint on the continuum function for [[regular cardinal]]s.

			*If <math>\kappa\geq\aleph_0</math> and <math>\lambda\geq 2</math>, then <math>\kappa<cf(\lambda^\kappa).\!</math>
			Let <math>\mu = \lambda^\kappa \!</math>. Suppose that, contrary to this corollary, <math>\kappa \ge cf(\mu)</math>. Then using the previous corollary, <math>\mu<\mu^{cf(\mu)}\le\mu^{\kappa}=(\lambda^\kappa)^\kappa=\lambda^{\kappa\cdot\kappa}=\lambda^\kappa=\mu</math>, a contradiction. Thus the supposition must be false and this corollary must be true.

			==A proof of König's theorem==
			Assuming [[Zermelo–Fraenkel set theory]], including especially the [[axiom of choice]], we can prove the theorem. Remember that we are given <math>\forall i\in I\quad A_i<B_i</math>, and we want to show :<math>\sum_{i\in I}A_i<\prod_{i\in I}B_i.</math>

			First, we show that there is an [[injective function\|injection]] from the sum to the product. Using the axiom of choice, for each ''i'' we choose an injection ''f<sub>i</sub>'' from ''A<sub>i</sub>'' to ''B<sub>i</sub>''. Notice that ''f<sub>i</sub>'' cannot be a surjection because then its inverse would be an injection from ''B<sub>i</sub>'' to ''A<sub>i</sub>''. So, for each ''i'', there must be an element of ''B<sub>i</sub>'' not in the range of ''f<sub>i</sub>''. Using the axiom of choice again, we choose such an ''x<sub>i</sub>'' for each ''i''. Define ''g'' on the sum by ''g''(''i,a'') (''j'') = ''f<sub>i</sub>''(''a'') when ''j'' = ''i'' and ''a'' is an element of ''A<sub>i</sub>'' and ''g''(''i,a'') (''j'') = ''x<sub>j</sub>'' when ''j'' ≠ ''i'' and ''a'' is an element of ''A<sub>i</sub>''. Since ''f<sub>i</sub>''(''a'') ≠ ''x<sub>i</sub>'' for each ''i'', ''g'' is an injection from the sum to the product.

			Second, we show that there is no injection ''h'' from the product to the sum. Suppose, to the contrary, that such an ''h'' existed. In a similar manner to [[Cantor's diagonal argument]], we will construct an element ''e'' of the product, which cannot have a value under ''h''. For each ''i'' in ''I'', construct a partial function ''f<sub>i</sub>'' from ''A<sub>i</sub>'' to ''B<sub>i</sub>'' by ''f<sub>i</sub>''(''a'') = ''d''(''i'') if there is a ''d'' in the product such that ''h''(''d'') = (''i'',''a''). (This '''is''' a partial function because ''h'' is an injection, so the ''d'' is unique.) If ''f<sub>i</sub>'' were a surjection, then, using the axiom of choice, we could construct an injection ''g'' from ''B<sub>i</sub>'' into ''A<sub>i</sub>'' (''g'' would be a right inverse of ''f''<sub>''i''</sub>), contradicting the hypothesis. Hence, for each ''i'' in ''I'', there are elements of ''B<sub>i</sub>'' not in the image of ''f<sub>i</sub>''. So using the axiom of choice again, we choose ''e''(''i'') in ''B<sub>i</sub>'' but not in the image of ''f<sub>i</sub>''. Consider, now, the value of ''h''(''e'') = (''i'',''c'') with ''c'' in ''A<sub>i</sub>''. But then ''f<sub>i</sub>''(''c'') = ''e''(''i''), contradicting the construction of ''e''. Hence no such injection can exist, and the product is strictly larger in cardinality than the sum.

			==Notes==
			<references/>

			==References==
			*{{cite book \| author=M. Holz, K. Steffens and E. Weitz \| title=Introduction to Cardinal Arithmetic \| publisher=Birkhäuser \| year=1999 \| isbn=3-7643-6124-7}}

			==External links==
			*[http://planetmath.org/encyclopedia/KonigsTheorem.html König's theorem] article on PlanetMath, includes a proof

			{{DEFAULTSORT:Konigs theorem}}
			[[Category:Axiom of choice]]
			[[Category:Theorems in the foundations of mathematics]]
			[[Category:Cardinal numbers]]
			[[Category:Articles containing proofs]]

en>CocuBot: r2.7.2) (Robot: Adding no:Stjernevind

2012-04-02T20:16:47Z

r2.7.2) (Robot: Adding no:Stjernevind

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