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| {{other uses|König's theorem (disambiguation)}}
| | I will show you and explain getting to the most effective of the SERPS where your company can utilize tons of research traffic. It Really Is marvelous to show up number one, when someone types in the phrase you would like.<br><br>These would be the hottest types of advertisements utilized, due to its power to create product awareness. These advertisements aid in building brand awareness perhaps the ad is clicked or not. This really is as a result of fact that the business is more interested in the outlook being advised about a product, and thus realizing the manufacturer. This results in people getting more acquainted with the item for future reference once the need arises.<br><br>Another well-known technique you are able to market your business is by content creation. Write about topics you understand about and publish these article to article sites that are taking new articles and new experts. There are many of article directory sites ready to recognize and submit articles.<br><br><br><br>Additionally there are other types of on-line careers that may be worth your while. Medical and legal transcription pay perfectly, and the amount of online transcription firms is increasing. I did appropriate transcription for approximately 8 weeks, and produced a lot of money, but writing happens to be my passion, so I went back to writing. I have so many online and off line writing responsibilities, I can scarcely maintain. Once my university classes finish for that summer, I will likely go back to transcription.<br><br>Every Thursday at 2pm exactly the same theme will be covered again over a LIVE video webcast. That's right - Patrick will be since the same subject detailed with live chat broadcast from OKC. That is another opportunity for one to get involved and request your [http://www.linkedin.com/pub/jordan-kurland/a/618/581 Jordan Kurland] questions. That is where the future is headed. Let Patrick take you there.<br><br>And check out what Andy Jenkins is educating about movie at his website www.TheVideoBoss.com. (this isn't an affiliate link.) Although his master ideas are not as simple as aiming the camera at your personal talking-head, these ideas can help camera-shy entrepreneurs. And, as Andy may say, rock on.<br><br>After being dissatisfied and registering, you next alternative, in desperation, will be to try yet another "program." Ofcourse that next alternative likewise, in the end, proves to function as the just like the others that have burned you in the past.<br><br>The web link within the resource field claims "SEO Massachusetts", and if you click it you'll be directed to my website. Hopefully by the time you discovered this informative article, I've enough different links spread around that say the same thing (SEO Massachusetts), so if you type "SEO Boston" into Google I am no 1. |
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| 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
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| :<math>\sum_{i\in I}m_i<\prod_{i\in I}n_i.</math>
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| 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]].
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| 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.
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| == Details ==
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| 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
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| :<math>\sum_{i\in I}A_i<\prod_{i\in I}B_i,</math>
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| 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>
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| (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).
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| 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
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| :<math>\sum_{i\in I} m_i \le \sum_{i\in I} n_i </math>
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| 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.
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| ==Corollaries of König's theorem==
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| *If <math>\kappa\,</math> is a cardinal then <math>\kappa < 2^{\kappa}.\!</math>
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| 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.)
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| ===Axiom of choice===
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| 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:
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| *If <math>\forall i\in I(\{\}<B_i)</math>, then <math>\{\}<\prod_{i\in I}B_i.</math>
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| 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. | |
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| ===König's theorem and cofinality===
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| König's theorem has also important consequences for [[cofinality]] of cardinal numbers.
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| *If <math>\kappa\ge\aleph_0</math>, then <math>\kappa<\kappa^{cf(\kappa)}. \!</math>
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| 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.
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| 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.
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| *If <math>\kappa\geq\aleph_0</math> and <math>\lambda\geq 2</math>, then <math>\kappa<cf(\lambda^\kappa).\!</math>
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| 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.
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| ==A proof of König's theorem==
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| 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>
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| 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.
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| 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.
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| ==Notes==
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| <references/>
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| ==References==
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| *{{cite book | author=M. Holz, K. Steffens and E. Weitz | title=Introduction to Cardinal Arithmetic | publisher=Birkhäuser | year=1999 | isbn=3-7643-6124-7}}
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| ==External links==
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| *[http://planetmath.org/encyclopedia/KonigsTheorem.html König's theorem] article on PlanetMath, includes a proof
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| {{DEFAULTSORT:Konigs theorem}}
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| [[Category:Axiom of choice]]
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| [[Category:Theorems in the foundations of mathematics]]
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| [[Category:Cardinal numbers]]
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| [[Category:Articles containing proofs]]
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