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康托尔(Georg Cantor,1845 - 1918)关于对角线论证的著名德语论文首次发表在1890年德国数学联盟(Deutsche Mathematiker-Vereinigung)的期刊上。
一,英文译文
In the paper entitled "On a property of a set [Inbegriff] of all real algebraic numbers" (Journ. Math. Bd. 77, S. 258), there appeared, probably for the first time, a proof of the proposition that there is an infinite manifold, which cannot be put into a one-one correlation with the totality [Gesamtheit] of all finite whole numbers 1, 2, 3, …, v, …, or, as I am used to saying, which do not have the power (Mächtigkeit) if the number series 1, 2, 3, …, v, …. From the proposition proved in § 2 there follows another, that e.g. the totality (Gesamtheit) of all real numbers of an arbitrary interval (a ... b) cannot be arranged in the series :
w1 w2, …, wv, …
However, there is a proof of this proposition that is much simpler, and which does not depend on considering the irrational numbers.
Namely, let m and n be two different characters, and consider a set [Inbegriff] M of elements :
E = (x1, x2, … , xv, …)
which depend on infinitely many coordinates x1, x2, … , xv, …, and where each of the coordinates is either m or w. Let M be the totality [Gesamtheit] of all elements E.
To the elements of M belong e.g. the following three:
EI = (m, m, m, m, … ),
EII = (w, w, w, w, … ),
EIII = (m, w, m, w, … ).
I maintain now that such a manifold [Mannigfaltigkeit] M does not have the power of the series 1, 2, 3, …, v, ….
This follows from the following proposition:
"If E1, E2, …, Ev, … is any simply infinite [einfach unendliche] series of elements of the manifold M, then there always exists an element E0 of M, which cannot be connected with any element Ev. »
For proof, let there be
E1 = (a1.1, a1.2, … , a1,v, …)
E2 = (a2.1, a2.2, … , a2,v, …)
Eu = (au.1, au.2, … , au,v, …)
...
where the characters au,v are either m or w. Then there is a series b1, b2, … bv,…, defined so that bv is also equal to m or w but is different from av,v.
Thus, if av,v = m, then bv = w.
Then consider the element
E0 = (b1, b2, b3, …)
of M, then one sees straight away, that the equation
E0 = Eu
cannot be satisfied by any positive integer u, otherwise for that u and for all values of v.
bv = au,v
and so we would in particular have
bu = au,u
which through the definition of bv is impossible. From this proposition it follows immediately that the totality of all elements of M cannot be put into the sequence [Reihenform]: E1, E2, …, Ev, … otherwise we would have the contradiction, that a thing [Ding] E0 would be both an element of M, but also not an element of M.
二,中文译文
在题为“关于所有实数代数的集合的一个属性(On a property of a set of all real algebraic numbers)”(Journ. Math. Bd. 77, S. 258)中,可能第一次出现了命题的证明,即有一个无限的流形,或者如我习惯说的,数列1,2,3,...,v,....,的幂集,它不能与所有整数1,2,3,...,v,...形成一一对应的关系。 从第2节中证明的命题得出,例如,任意区间(a ... b)的所有实数的总体(Gesamtheit)不能被排列成数列:
w1 w2, …, wv, …
然而,这个命题有一个更简单的证明,它不依赖于考虑无理数。
也就是说,让m和n是两个不同的字符,并考虑一个元素的集合M
E = (x1, x2, … , xv, …)
这取决于无限多的坐标x1,x2,…,xv,...,其中每个坐标都是m或w。让M是所有元素E的总体。
在M的元素中,例如属于以下三个:
EI = (m, m, m, m, … ),
EII = (w, w, w, w, … ),
EIII = (m, w, m, w, … ).
我现在指出,这样一个流形M不具有系列1,2,3,...,v的power,..
由以下命这题得出:
"如果E1,E2,...,Ev,...是流形M的任何简单无限元素系列,那么M中总是存在一个元素E0,它不能与任何元素Ev连接。"
为了证明这一点,让我们有
E1 = (a1.1, a1.2, … , a1,v, …)
E2 = (a2.1, a2.2, … , a2,v, …)
Eu = (au.1, au.2, … , au,v, …)
...
然后有一个系列b1, b2, ... bv, ...,定义为bv也等于m或w,但与av,v不同。
因此,如果av,v=m,那么bv=w。
然后考虑M的元素:
E0 = (b1, b2, b3, …)
那么我们可以直接看到,方程:
E0 = Eu
不能被任何正整数u所满足,否则对该u和所有v的值来说都是如此。
bv = au,v
我们将特别有:
bu = au,u
而根据bv的定义,这是不可能的。从这个命题立即可以看出,M的所有元素的整体不能被放入序列: E1, E2, ..., Ev, ...,否则我们就会产生矛盾,即一个事物E0既是M的一个元素,但也不是M的一个元素。
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