Motivation---Zorn's lemma ←,History---well-ordering[Zermelo's- ] theorem
←< https://en.wikipedia.org/wiki/Well-ordering_theorem >[< https://shinichiwanko2000.livedoor.blog/archives/28531408.html >2025年05月17日 ordinal numbers順序数、序数---well-ordered整然とした(秩序立った(整った? ! ), よく整頓された? ){unlimited potential, infinitude of possibilities無限の可能性がimagine, have the intimate(暗に知らせる、暗示する、ほのめかす, 詳しい、詳細な、奥深い?? )秘められればイイのｶﾅ？？・? ?,, }, < https://shinichiwanko2000.livedoor.blog/archives/28523739.html >2025年05月16日 Furtherさらなる{さらに遠い(く )?、一層遠く、さらにずっと、さらに先に } completeness conditions(Types of completeness properties )---completeness完備性（英語版 ）properties特性? ?{It's just a translation of what the computer is thinking into human language.計算機が考えていることを？・人間の言葉にしただけ、may beナノカモ？・?！・ } ]
Georg Cantor< https://en.wikipedia.org/wiki/Georg_Cantor >ｺﾉﾋﾄﾀﾞｹ?カナ? ? considered the well-ordering theorem to be a "fundamental principle「基(根? )本的な原則 」 of thought見解、思い付き,(考え、考えること、思考（力）、思索、思想、熟考? ? ) ". However, it is considered difficult or even impossible to visualize a well-ordering of R , the set of all real numbers; such a visualization would have to incorporate the axiom of choice. In 1904, Gyula Kőnig claimed to have proven that such a well-ordering cannot exist. A few weeks later, Felix Hausdorff found a mistake in the proof. It turned out, though, that in first-order logic the well-ordering theorem is equivalent to the axiom of choice, in the sense that the Zermelo-Fraenkel axioms with the axiom of choice included are sufficient to prove the well-ordering theorem, and conversely, the Zermelo?Fraenkel axioms without the axiom of choice but with the well-ordering theorem included are sufficient to prove the axiom of choice. (The same applies to Zorn's lemma<< https://en.wikipedia.org/wiki/Zorn%27s_lemma > >. ) In second-order logic, however, the well-ordering theorem is strictly stronger than the axiom of choice: from the well-ordering theorem one may deduce the axiom of choice, but from the axiom of choice one cannot deduce the well-ordering theorem.
There is a well-known joke about the *three statements[1.Hausdorff maximal principle, 2. Axiom of choice, 3.Well-ordering theorem.(1./ is a nonempty subset of R{Because T contains at least one element, and that element contains at least 0, the union I contains at least 0 and is not empty. Every element of T is a subset of R, so the union I only consists of elements in R. }, 2.For every x, y ∈ /, the sum x + y is in /{Suppose x and y are elements of /. Then there exist two ideals J, K ∈ T such that x is an element of J and y is an element of K. Since T is totally ordered, we know that J ⊆ K or K ⊆ J. Without loss of generality, assume the first case. Both x and y are members of the ideal K, therefore their sum x + y is a member of K, which shows that x + y is a member of /. }, 3.For every r ∈ R and every x ∈ /, the product rx is in /.{Suppose x is an element of /. Then there exists an ideal J ∈ T such that x is in J. If r ∈ R, then rx is an element of J and hence an element of /. Thus, / is an ideal in R. Now, we show that / is a proper ideal. An ideal is equal to R if and only if it contains 1. (It is clear that if it is R then it contains 1; on the other hand, if it contains 1 and r is an arbitrary element of R, then r1 = r is an element of the ideal, and so the ideal is equal to R. ) So, if I were equal to R, then it would contain 1, and that means one of the members of T would contain 1 and would thus be equal to R ? but R is explicitly excluded from P. The hypothesis of Zorn's lemma has been checked, and thus there is a maximal element in P, in other words a maximal ideal in R. } ) ], and their relative amenability to intuition:
The axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?
< https://en.wikipedia.org/wiki/Well-ordering_theorem >
To prove the existence of a mathematical object that can be viewed as a maximal element in some partially ordered set(poset[半順序集合｛はん じゅんじょ しゅうごう｝◇【語源 】partially orderedを略｛りゃく｝したpo + set◇【同 】partially ordered set ] for short < https://en.wikipedia.org/wiki/Partially_ordered_set > ) in some way, one can try proving the existence of such an object by assuming there is no maximal element and using transfinite induction and the assumptions of the situation to get a contradiction. Zorn's lemma tidies up the conditions a situation needs to satisfy in order for such an argument項?? to work and enables mathematicians to not have to repeat the transfinite induction argument by hand each time, but just check the conditions of Zorn's lemma.
If you are building a mathematical object in stages and find that (i) you have not finished even after infinitely many stages, and (ii) there seems to be nothing to stop you continuing to build, then Zorn’s lemma may well be able to help you.
William Timothy Gowers< https://en.wikipedia.org/wiki/Timothy_Gowers >decent guy, serious personまじめな人だったI heardトカ？？・？・ , "How to use Zorn’s lemma "
< https://en.wikipedia.org/wiki/Zorn%27s_lemma >
{GODかみさまのpassing idea, hit upon the idea of it? 思いつき? ? で〝Either way, I'll fall. 〟ドチラデモ♪ころんでしまうsee ifｶﾅ？(accurate正確な、精密な、間違いのない!? copy important!! )？・？・ }
(relevant, )
https://plaza.rakuten.co.jp/tadashityutyu/diary/201610150000/
2016年10月15日definition---unique factorization domain (UFD)一意分解環(ふうがわりな？・・因子分解[因数分 解?]-領域[整域(環?・・かんになるからそやなもの？！・・・) ] ) {irreducible elements既約元(整域の 0 でも単元でもない元は、それが2つの非単元の積<きょうえき複素数をいみする？？？>でないとき・・・・・：3 はこの環で素元ではない。なぜならば、例えば、 3 ｜ ( 2 +√{- 5} ) ( 2 - √{- 5} ) = 9 であるが、・・ 3 は2つの因数のいずれも割り切らない・・<素数既約多項式とのギャップみたいなもの？・・>・) } (/.//空でない集合は必ず自分自身と交わらない要素を持つ.,/(,からでない集合 ＜ からの集合(<(はっせいとは、・・なにもないとこと？(宇宙？)からいきなり？その点のような？存在で小宇宙？的なもの？！がはっせいしてくるの？・・意？！？.>)・→・・
e tc??< https://cse.google.com/cse?ie=UTF-8&q=Zorn%27s+lemma+site%3Aplaza.rakuten.co.jp%2Ftadashityutyu&cx=002636997843861491696%3Av3cnak5qsac&siteurl=plaza.rakuten.co.jp%2Ftadashityutyu%2F&ref=&ss= >Zorn's lemma site:plaza.rakuten.co.jp/tadashityutyu
https://shinichiwanko2000.livedoor.blog/archives/28538994.html
2025年05月18日 Motivation---Zorn's lemma ←,History---well-ordering[Zermelo's- ] theorem
Georg Cantor< https://en.wikipedia.org/wiki/Georg_Cantor >ｺﾉﾋﾄﾀﾞｹ?カナ? ? considered the well-ordering theorem to be a "fundamental principle「基(根? )本的な原則 」 of thought見解、思い付き,(考え、考えること、思考（力）、思索、思想、熟考? ? ) ". However, it is considered difficult or even impossible to visualize a well-ordering of R , the set of all real numbers; such a visualization would have to incorporate the axiom of choice. In 1904, Gyula Kőnig claimed to have proven that such a well-ordering cannot exist. A few weeks later, Felix Hausdorff found a mistake in the proof. It turned out, though, that in first-order logic the well-ordering theorem is equivalent to the axiom of choice, in the sense that the Zermelo-Fraenkel axioms with the axiom of choice included are sufficient to prove the well-ordering theorem, and conversely, the Zermelo?Fraenkel axioms without the axiom of choice but with the well-ordering theorem included are sufficient to prove the axiom of choice. (The same applies to Zorn's lemma<< https://en.wikipedia.org/wiki/Zorn%27s_lemma > >. ) In second-order logic, however, the well-ordering theorem is strictly stronger than the axiom of choice: from the well-ordering theorem one may deduce the axiom of choice, but from the axiom of choice one cannot deduce the well-ordering theorem.
There is a well-known joke about the *three statements[1.Hausdorff maximal principle, 2. Axiom of choice, 3.Well-ordering theorem.(1./ is a nonempty subset of R{Because T contains at least one element, and that element contains at least 0, the union I contains at least 0 and is not empty. Every element of T is a subset of R, so the union I only consists of elements in R. }, 2.For every x, y ∈ /, the sum x + y is in /{Suppose x and y are elements of /. Then there exist two ideals J, K ∈ T such that x is an element of J and y is an element of K. Since T is totally ordered, we know that J ⊆ K or K ⊆ J. Without loss of generality, assume the first case. Both x and y are members of the ideal K, therefore their sum x + y is a member of K, which shows that x + y is a member of /. }, 3.For every r ∈ R and every x ∈ /, the product rx is in /.{Suppose x is an element of /. Then there exists an ideal J ∈ T such that x is in J. If r ∈ R, then rx is an element of J and hence an element of /. Thus, / is an ideal in R. Now, we show that / is a proper ideal. An ideal is equal to R if and only if it contains 1. (It is clear that if it is R then it contains 1; on the other hand, if it contains 1 and r is an arbitrary element of R, then r1 = r is an element of the ideal, and so the ideal is equal to R. ) So, if I were equal to R, then it would contain 1, and that means one of the members of T would contain 1 and would thus be equal to R ? but R is explicitly excluded from P. The hypothesis of Zorn's lemma has been checked, and thus there is a maximal element in P, in other words a maximal ideal in R. } ) ], and their relative amenability to intuition:
The axiom of choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?
< https://en.wikipedia.org/wiki/Well-ordering_theorem >
To prove the existence of a mathematical object that can be viewed as a maximal element in some partially ordered set(poset[半順序集合｛はん じゅんじょ しゅうごう｝◇【語源 】partially orderedを略｛りゃく｝したpo + set◇【同 】partially ordered set ] for short < https://en.wikipedia.org/wiki/Partially_ordered_set > ) in some way, one can try proving the existence of such an object by assuming there is no maximal element and using transfinite induction and the assumptions of the situation to get a contradiction. Zorn's lemma tidies up the conditions a situation needs to satisfy in order for such an argument項?? to work and enables mathematicians to not have to repeat the transfinite induction argument by hand each time, but just check the conditions of Zorn's lemma.
If you are building a mathematical object in stages and find that (i) you have not finished even after infinitely many stages, and (ii) there seems to be nothing to stop you continuing to build, then Zorn’s lemma may well be able to help you.
William Timothy Gowers< https://en.wikipedia.org/wiki/Timothy_Gowers >decent guy, serious personまじめな人だったI heardトカ？？・？・ , "How to use Zorn’s lemma "
< https://en.wikipedia.org/wiki/Zorn%27s_lemma >
{GODかみさまのpassing idea, hit upon the idea of it? 思いつき? ? で〝Either way, I'll fall. 〟ドチラデモ♪ころんでしまうsee ifｶﾅ？(accurate正確な、精密な、間違いのない!? copy important!! )？・？・ }
(relevant, )
https://plaza.rakuten.co.jp/tadashityutyu/diary/201610150000/
2016年10月15日definition---unique factorization domain (UFD)一意分解環(ふうがわりな？・・因子分解[因数分 解?]-領域[整域(環?・・かんになるからそやなもの？！・・・) ] ) {irreducible elements既約元(整域の 0 でも単元でもない元は、それが2つの非単元の積<きょうえき複素数をいみする？？？>でないとき・・・・・：3 はこの環で素元ではない。なぜならば、例えば、 3 ｜ ( 2 +√{- 5} ) ( 2 - √{- 5} ) = 9 であるが、・・ 3 は2つの因数のいずれも割り切らない・・<素数既約多項式とのギャップみたいなもの？・・>・) } (/.//空でない集合は必ず自分自身と交わらない要素を持つ.,/(,からでない集合 ＜ からの集合(<(はっせいとは、・・なにもないとこと？(宇宙？)からいきなり？その点のような？存在で小宇宙？的なもの？！がはっせいしてくるの？・・意？！？.>)・→・・
e tc??< https://cse.google.com/cse?ie=UTF-8&q=Zorn%27s+lemma+site%3Aplaza.rakuten.co.jp%2Ftadashityutyu&cx=002636997843861491696%3Av3cnak5qsac&siteurl=plaza.rakuten.co.jp%2Ftadashityutyu%2F&ref=&ss= >Zorn's lemma site:plaza.rakuten.co.jp/tadashityutyu
https://shinichiwanko2000.livedoor.blog/archives/28538994.html
2025年05月18日 Motivation---Zorn's lemma ←,History---well-ordering[Zermelo's- ] theorem
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