14 | 9 月 | 2017 | FreeSandal

聽聞德國人認為『萬事有起頭』。因『香腸有兩頭』故愛好吃香腸乎？這與東方俗諺所說『萬事起頭難』，不知該如何比較哩！就像投影幾何『公設化』多種多樣，當真有『公理』的『選擇』嗎？？

在《λ 運算︰計物數《上》》一文中，我們說到『皮亞諾』 Peano 提出了『自然數』之五條公設的系統。用著『未定義』的『基元』數『零 0』，以及『後繼數』successor 的『概念』，打造了『一階算術』系統，現今稱之為『皮亞諾算術系統』。在《布林代數》文章裡，我們對比了『布林代數』、『集合論』與『邏輯學』之間的『密切關係』。整個『布林代數』是可以建立在一個二元運算『孤虛』 ── Sheffer豎線 $|$ ── 之上。也就是說一個『系統』的『公設化』往往不只一種『選擇』，或許是因為雖然兩個看起來『不同』的『概念』，它們彼此之間的『邏輯關係』卻是『等價的』，所以『甲可以推導出乙』，而且『乙能夠演繹出甲』，在此處，我們僅以與『無窮小』概念的『親疏遠近』編排次序，並不論及何者更為『基本』這樣的判斷。

『羅素悖論』在『集合論』的發展史上產生了重大的影響，因此『集合之集合』的構造勢必得『避免矛盾』，『坎特爾』 Cantor 證明了『實數集合』的『元素』是『不可數』的多，這再次引發了如何『列舉』的『難題』，也就是說既然『實數』如果無法『一一指定』，那你又怎麽能夠確定『所說之數』是『存在的』呢？比方說 $\bigcap \limits_{0}^{\infty} \ (0, 2^{-n}) = \phi$ 。所以我們從『言之有物』的觀點，就直接『同意』所謂的『選擇公理』︰一個【集合族】是指由非『空集合』 $\phi$ 所組成的一個『集合』。『存在』著一種【選擇函數】，它是個定義在某個『集合族』 $X$ 上的函數，對於這個函數來講，所有在『集合族』 $X$ 中的『集合元素』 $S$ ，都能夠『選擇』 $f_{choice}(S) \in S$ 。也就是說 $f_{choice} (S)$ 可以『指定』某一個 $S$ 中的『元素』。這裡所說的『同意』之意思就不過是想要在『直觀』中『簡化』討論之事項，比方講像某些『探討』著一條平面上的『封閉曲線』，它到底是不是能夠將『平面』分割成『曲線內』與『曲線外』之兩個部分的此類『議論』。這樣『實數分析』中所謂的『疊套閉區間』 $I_n = [a_n, b_n]$
$\ a_n < a_{n+1}, b_{n+1} < b_n, \ n=1 \cdots n$ 的『概念』，就『超實數』 $r^{*} = r \pm \delta x$ 來講，就是『標準部份函數』 $st(r^{*}) = r$ ，所以說如果那個 $r, \ a_n \leq r \leq b_n, \ |b_n - a_n| \approx 0$ 的話，一定會有 $\bigcap \limits_{1}^{\infty} I_n \neq \phi = r$ ，也就是說這就『確定』了那個『實數值』。假使我們換個觀點來看 $a_n$ 與 $b_n$ 都各自構成一個了『單調』之『上升』以及『單調』之『下降』的『序列』，而且 $|b_n - a_n| \approx 0$ ，那麼難到不該 $a_{\infty} = b_{\infty} = r$ 的嗎？如果說自然界之『物理量』總是來自於『度量』，因此那兩個稱作『柯西序列』 Cauchy sequence 的 $a_n, b_n$ 的『序列』，它們所『代表』的就是『量測』的『極限分析』的啊！假使用『超整數』與『巨量』 $H, K$ 可敘述為 $st (a_H-b_K)= 0$ 。假使設想著對於『可列舉』之物，至少可以說在『無窮遠』處之時﹐『那些‧哪些』的物將會是距離『這麼‧這麼』之『無窮近』的吧？然而對於『不可列舉』之物，我們真的還能夠講述著『某個‧某個』的『什麼』的嗎？因為說不定它還框陷在『難計』的『侑限』裡，你又怎麽可能得到『遠近』之『結論』的呢？？或許這就是『有理數』的『可數性』很適合用來『建構』那種『不可數性』的吧！！

選擇公理 axiom of choice

分球怪論

巴拿赫-塔斯基定理
Banach–Tarski paradox

疊套區間

柯西序列

非柯西序列

threekindnumber

─ 摘自《【Sonic π】電路學之補充《四》無窮小算術‧下下‧上》

若是『任性隨意』選擇，果無法『一通全通』耶！！

4.2.1 Axioms and Basic Definitions for Plane Projective Geometry Printout
Teachers open the door, but you must enter by yourself.
—Chinese Proverb

Undefined Terms. point, line, incident

Axiom 1. Any two distinct points are incident with exactly one line.

Axiom 2. Any two distinct lines are incident with at least one point.

Axiom 3. There exist at least four points, no three of which are collinear.

Axiom 4. The three diagonal points of a complete quadrangle are never collinear.

Axiom 5. (Desargues’ Theorem) If two triangles are perspective from a point, then they are perspective from a line.

Axiom 6. If a projectivity on a pencil of points leaves three distinct points of the pencil invariant, it leaves every point of the pencil invariant.

……

縱非不能，恐無益也。

何況利用厚生者，實不必鑽牛角尖的吧◎

此處以『交比守恆』

Cross-ratio preservation

Cross-ratios are invariant under Möbius transformations. That is, if a Möbius transformation maps four distinct points $z_{1},z_{2},z_{3},z_{4}$ to four distinct points $w_{1},w_{2},w_{3},w_{4}$ respectively, then

{\frac {(z_{1}-z_{3})(z_{2}-z_{4})}{(z_{2}-z_{3})(z_{1}-z_{4})}}={\frac {(w_{1}-w_{3})(w_{2}-w_{4})}{(w_{2}-w_{3})(w_{1}-w_{4})}}.

If one of the points $z_{1},z_{2},z_{3},z_{4}$ is the point at infinity, then the cross-ratio has to be defined by taking the appropriate limit; e.g. the cross-ratio of $z_{1},z_{2},z_{3},\infty$ is

{\frac {(z_{1}-z_{3})}{(z_{2}-z_{3})}}.

The cross ratio of four different points is real if and only if there is a line or a circle passing through them. This is another way to show that Möbius transformations preserve generalized circles.

談談『公設六』，

非為且無干於『邏輯次序』︰

假設此莫比烏斯變換為 $w = \frac{a z + b}{c z + d}$ ，那麼 $w_k = \frac{a z_k + b}{c z_k + d}$ 。

借 SymPy 計算簡化

pi@raspberrypi:~ w_i - w_j = \frac{(a \cdot d - b \cdot c)(z_i - z_j)}{(c \cdot z_i + d)(c \cdot z_j +d)}\therefore \frac{(w_1 - w_3)(w_2 - w_4)}{(w_2 - w_3)(w_1 - w_4)} = \frac{ \frac{(a \cdot d - b \cdot c)(z_1 - z_3)}{(c \cdot z_1 + d)(c \cdot z_3 +d)} \frac{(a \cdot d - b \cdot c)(z_2 - z_4)}{(c \cdot z_2 + d)(c \cdot z_4 +d)}} { \frac{(a \cdot d - b \cdot c)(z_2 - z_3)}{(c \cdot z_2 + d)(c \cdot z_3 +d)}\frac{(a \cdot d - b \cdot c)(z_1 - z_4)}{(c \cdot z_1 + d)(c \cdot z_4 +d)}}= \frac{(z_1 - z_3)(z_2 - z_4)}{(z_2 - z_3)(z_1 - z_4)}$ 。


 
祇是反思『歷史先後』︰
Pappus's hexagon theorem
In mathematics, Pappus's hexagon theorem (attributed to Pappus of Alexandria) states that given one set of collinear points A, B, C, and another set of collinear points a, b, c, then the intersection points X, Y, Z of line pairs Ab and aB, Ac and aC, Bc and bC are collinear, lying on the Pappus line. These three points are the points of intersection of the "opposite" sides of the hexagon AbCaBc. It holds in a projective plane over any field, but fails for projective planes over any noncommutative division ring.^[1] Projective planes in which the "theorem" is valid are called pappian planes.
The dual of this incidence theorem states that given one set of concurrent lines A, B, C, and another set of concurrent lines a, b, c, then the lines x, y, z defined by pairs of points resulting from pairs of intersections A∩b and a ∩ B, A ∩ c and a ∩ C, B ∩ c and b ∩ C are concurrent. (Concurrent means that the lines pass through one point.)
Pappus's theorem is a special case of Pascal's theorem for a conic—the limiting case when the conic degenerates into 2 straight lines. Pascal's theorem is in turn a special case of the Cayley–Bacharach theorem.
The Pappus configuration is the configuration of 9 lines and 9 points that occurs in Pappus's theorem, with each line meeting 3 of the points and each point meeting 3 lines. In general, the Pappus line does not pass through the point of intersection of ABC and abc.^[2] This configuration is self dual. Since, in particular, the lines Bc, bC, XY have the properties of the lines x, y, z of the dual theorem, and collinearity of X, Y, Z is equivalent to concurrence of Bc, bC, XY, the dual theorem is therefore just the same as the theorem itself. The Levi graph of the Pappus configuration is the Pappus graph, a bipartite distance-regular graph with 18 vertices and 27 edges.

Pappus's hexagon theorem: Points X, Y and Z are collinear on the Pappus line. The hexagon is AbCaBc.
 
設想什麼情況
『有三不及四』呀⊙

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每日彙整: 2017-09-14

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Cross-ratio preservation

Pappus's hexagon theorem

輕。鬆。學。部落客