Charles S. Chihara | 关于无限过程的可完成性

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ON THE POSSIBILITY OF COMPLETING AN INFINITE PROCESS

关于无限过程的可完成性

Charles S. Chihara

The Philosophical Review, Vol. 74, No. 1. (Jan., 1965), pp. 74-87.

Consider the decimal expansion of pi. Some philosophers have held that it is logically impossible to do so. Unfortunately, the distinction between logical and mere "physical" impossibility is not easy to make out, and on this issue I shall not take a stand. I do not think, however, that calculating all the digits in pi just happens to be impossible for us to do in our present state of technological development, and I doubt that anyone seriously believes we might someday develop a machine which would complete such a calculation. Perhaps we should say that it is inconceivable or unimaginable that an infinite number of calculations should be performed in a finite amount of time. At any rate, it certainly seems that the relevant sense of "impossibility" is a very strong one.

试想圆周率 π 的十进制展开式，一些哲学家认为从逻辑上讲，算出其所有数位是不可能的。遗憾的是，我们很难厘清逻辑上的不可能性与单纯"物理上"的不可能性之间的区别，本文也不会就这一问题表明立场。但我认为，算出圆周率的所有数位，并非只因人类当前的技术发展水平而暂时无法实现；我也怀疑，是否真的有人坚信，人类终有一天能研制出一台完成这一计算的机器。或许我们应当认为，要在有限的时间内完成无穷多次计算，是难以想象、无法构想的。无论如何，这里所说的"不可能性"，其内涵显然具有极强的绝对性。

But now consider the following version of the famous Achilles paradox of Zeno. Achilles runs a race at a rate of one mile per minute, which is twice as fast as the rate of a tortoise having a half-mile head start. Obviously, by the time Achilles arrives at the point at which the tortoise begins the race, the tortoise will have traveled another quarter mile; and by the time Achilles reaches that point, the tortoise will have gone an eighth of a mile; and so on ad infinitum. In order to catch the tortoise, then, Achilles must complete an infinite process, a task which--so it seems--is known to be impossible. Hence, it is impossible for Achilles to catch the tortoise.

但我们不妨来思考芝诺著名的阿喀琉斯悖论的下述版本：阿喀琉斯以每分钟 1 英里的速度参加赛跑，这一速度是乌龟的两倍，而乌龟拥有半英里的领先优势。显然，当阿喀琉斯抵达乌龟的起跑点时，乌龟又向前爬行了 1/4 英里；当阿喀琉斯追到这一位置时，乌龟又爬了 1/8 英里，以此类推，直至无穷。如此一来，阿喀琉斯要追上乌龟，就必须完成一个无限的过程------而这一任务，似乎公认是不可能完成的。因此，阿喀琉斯永远追不上乌龟。

To facilitate analyzing the data, I shall introduce a few distinctions. I shall call a rule for generating a sequence a terminating rule if some term in a sequence produced in accordance with the rule can be correctly called "the last term in the sequence defined by the rule." A nonterminating rule will thus be any rule for generating a sequence that is not a terminating rule. Obviously, in actually writing out a sequence of natural numbers in accordance with a nonterminating rule, some term will have to be the last term in the sequence actually produced, but this term cannot be the last term in the sequence defined by the rule.

为便于分析这一问题，我将引入几个概念区分。若依照某一规则生成的序列中，存在某一项能被准确称作"该规则所定义序列的最后一项"，那么我将这一生成序列的规则称为终止性规则 。相应地，非终止性规则就是指所有不属于终止性规则的序列生成规则。显然，若依照非终止性规则实际书写一个自然数序列，最终写下的某一项必然会成为实际生成序列的最后一项，但这一项绝不是该规则所定义序列的最后一项。

Every nonterminating rule defines an infinite sequence, but not every terminating rule defines a finite sequence: a sequence of order type ω + 1 \omega+1 ω+1 (for example, the sequence 1 , 2 , 3 , . . . ; 1 1, 2, 3, ... ; 1 1,2,3,...;1 ) can be defined by a terminating rule. Every infinite sequence defined by a terminating rule must, however, have a subsequence which can be defined by a nonterminating rule.

每一条非终止性规则都定义一个无穷序列，但并非每一条终止性规则都定义有限序列：序型为 ω + 1 \omega+1 ω+1 的序列（例如序列 1 , 2 , 3 , ... ; 1 1, 2, 3, \dots ; 1 1,2,3,...;1 ）即可由终止性规则定义。不过，由终止性规则定义的任意无穷序列，必然存在一个可由非终止性规则定义的子序列。

Finally, I shall adopt the locution " X X X makes P P P true" in place of the cumbersome " X X X performs an action in virtue of which the proposition P P P is true."

最后，我将使用" X X X 使 P P P 为真"这一表述，替代冗长的" X X X 实施了某一行为，且正是这一行为使得命题 P P P 为真"。

Turning to the Achilles paradox, we should note that we have a nonterminating rule for generating a sequence of numbers of the form 1 − 1 / 2 n 1-1 / 2^{n} 1−1/2n . Now let I n I_{n} In denote the closed interval with the end points 1 − 1 / 2 n − 1 1-1 / 2^{n-1} 1−1/2n−1 and 1 − 1 / 2 n 1-1 / 2^{n} 1−1/2n ; and let T n T_{n} Tn denote the proposition "Achilles travelled through I n I_{n} In ." Obviously, we have a nonterminating rule generating a sequence of propositions of the form T n T_{n} Tn . Now we can take Zeno as arguing that if Achilles is to catch the tortoise, he must make true T 1 T_{1} T1 , and then T 2 T_{2} T2 , and then T 3 . . . T_{3} ... T3... until ( n ) T n (n) T_{n} (n)Tn is made true.

回到阿喀琉斯悖论，我们会发现，存在一条非终止性规则，可生成形如 1 − 1 / 2 n 1-1/2^n 1−1/2n 的数的序列。现令 I n I_n In 表示以 1 − 1 / 2 n − 1 1-1/2^{n-1} 1−1/2n−1 和 1 − 1 / 2 n 1-1/2^n 1−1/2n 为端点的闭区间；令 T n T_n Tn 表示命题"阿喀琉斯经过了区间 I n I_n In "。显然，我们也有一条非终止性规则，可生成形如 T n T_n Tn 的命题序列。如此，我们便可将芝诺的论证理解为：阿喀琉斯要追上乌龟，就必须先使 T 1 T_1 T1 为真，再使 T 2 T_2 T2 为真，接着使 T 3 T_3 T3 为真......直至使 ( n ) T n (n)T_n (n)Tn 为真。

But how is it possible to make true all of an infinite sequence of propositions? How is it possible to finish such a task when one cannot come to the end of the sequence of tasks to be performed? How is it possible to traverse each of the infinitely many intervals of the form I n I_{n} In when the supply of intervals is unlimited and inexhaustible? Such considerations led Herman Weyl to claim "it is incompatible with the character of the infinite as the 'incompletable' that Achilles should have been able to traverse them all."

但如何才能使一个无穷命题序列中的所有命题都为真呢？当待执行的任务序列永无终点时，又该如何完成这一任务？当形如 I n I_n In 的区间无穷无尽、源源不断时，又怎能遍历每一个这样的区间？基于这些思考，赫尔曼·外尔断言："无限的本质是'不可完成的'，而阿喀琉斯能遍历所有这些区间，这与无限的这一本质相悖。" ∗ ^* ∗

Can one ignore this difficulty and simply insist that it must be possible to complete an infinite process of this sort? Consider the following example. As Achilles reaches the end of I 1 I_{1} I1 , suppose that an object emits a flash which lasts until Achilles is halfway through I 2 I_{2} I2 , then another flash as Achilles reaches the end of I 2 I_{2} I2 , which lasts until Achilles is halfway through I 3 I_{3} I3 , and so on ad infinitum. If it is possible that Achilles might complete the infinite process, is it not also possible that an object might give off an infinite number of flashes in a finite amount of time?

我们能否无视这一难题，径直坚称这类无限过程必然是可以完成的？不妨来看下述例子：当阿喀琉斯抵达 I 1 I_1 I1 的终点时，假设一个物体发出一道闪光，这道闪光持续到阿喀琉斯走到 I 2 I_2 I2 的中点；当阿喀琉斯抵达 I 2 I_2 I2 的终点时，该物体又发出一道闪光，持续到阿喀琉斯走到 I 3 I_3 I3 的中点，以此类推，直至无穷。如果阿喀琉斯有可能完成这一无限过程，那么一个物体为何就不可能在有限的时间内发出无穷多次闪光呢？

Weyl argues that if one admits the possibility that Achilles could complete such a task "there is no reason why a machine should not be capable of completing an infinite sequence of distinct acts of decision within a finite amount of time; say, by supplying the first result after 1 / 2 1/2 1/2 minute, the second after another 1 / 4 1/4 1/4 minute, the third 1 / 8 1/8 1/8 minute later than the second, etc."

外尔认为，若承认阿喀琉斯有可能完成这样的任务，那么"就没有理由否认，一台机器也能在有限的时间内完成无穷多次独立的判定行为；比如，在 1/2 分钟后给出第一个计算结果，再过 1/4 分钟给出第二个，第三个比第二个晚 1/8 分钟，依此类推"。

Thus we seem to be faced with a dilemma: either we reject the possibility of completing an infinite process (in which case we are faced with the conclusion that Achilles cannot catch the tortoise) or we must allow the possibility of performing an infinite number of calculations in a finite amount of time.

由此，我们似乎陷入了一个两难困境：要么否定完成无限过程的可能性（如此一来，我们就不得不接受阿喀琉斯追不上乌龟的结论），要么就必须承认，在有限的时间内完成无穷多次计算是可能的。

It has been argued by many, including such eminent philosophers as Whitehead, Russell, and Peirce, that the paradox dissolves once one realizes that the infinite series 1 / 2 + 1 / 4 + 1 / 8 + . . . 1 / 2+1 / 4+1 / 8+... 1/2+1/4+1/8+... has the value 1, since this mathematical fact shows that the distance to be traversed by Achilles and the amount of time required to traverse this distance must be finite, even though the distance can be divided into an infinite number of subdistances. Needless to say, this "mathematical solution" to the paradox does not enable one to evade the above dilemma.

包括怀特海、罗素、皮尔士等著名哲学家在内的许多人都认为，一旦认识到无穷级数 1 / 2 + 1 / 4 + 1 / 8 + ... 1/2+1/4+1/8+\dots 1/2+1/4+1/8+... 的和为 1，这一悖论便迎刃而解。因为这一数学事实表明，尽管阿喀琉斯需要跨越的距离可被分割为无穷多个子距离，但这段总距离以及跨越它所需的时间都是有限的。毋庸置疑，这一对悖论的"数学解法"，并不能让我们避开上述两难困境。

One might also suspect that what underlies the Achilles paradox is a simple quantifier confusion: the proposition (t)(Achilles cannot catch the tortoise at t ) is true, given that the bound variable t ranges over only those instants at which Achilles arrives at a point previously occupied by the tortoise at the preceding stage in the process specified in Zeno's description of the race; but clearly the conclusion of the Zenonian argument requires that ( t ) (t) (t) (Achilles cannot catch the tortoise at t) be true for t ranging over all instants of time. Again, this "solution" in no way addresses itself to the above dilemma.

有人还会猜测，阿喀琉斯悖论的根源不过是一种量词混淆：若约束变元 t t t 仅取值于芝诺所描述的赛跑过程中，阿喀琉斯抵达乌龟前一阶段所处位置的那些瞬间，那么命题"对所有 t t t ，阿喀琉斯在时刻 t t t 都追不上乌龟"为真；但显然，芝诺论证的结论要求，当 t t t 取值于所有时间瞬间时，该命题依然为真。而这一"解法"，同样无法解决上述两难困境。

If one aspect of the paradox Zeno uncovered over two thousand years ago is the problem of how Achilles could complete the infinite process described by Zeno, it would certainly seem that no generally accepted solution has yet appeared. I do not wish to maintain, however, that the puzzle with which I am concerned in this paper is the core of the Achilles paradox. Even a cursory survey of the vast literature on the topic will show that there is a wide variety of interpretations concerning the nature of the difficulties raised by Zeno. I should merely like to make clear that, in this paper, the Achilles paradox will be treated as essentially a puzzle about completing an infinite process.

如果芝诺在两千多年前提出的这一悖论，其中一个层面的问题是阿喀琉斯如何才能完成芝诺所描述的无限过程，那么显然，这一问题至今仍未出现被普遍接受的解法。但我并不想主张，本文所探讨的这一谜题就是阿喀琉斯悖论的核心。即便粗略梳理一下相关的海量文献，也能发现学界对于芝诺所提出难题的本质，存在多种多样的解读。我只想明确一点：在本文中，阿喀琉斯悖论将被视作一个本质上关于"完成无限过程"的谜题。

Weyl evidently saw no way of cogently evading his dilemma except by simply denying that the segment to be traversed by Achilles "really consists of infinitely many subsegments ..." Max Black arrived at a similar position in his early work, The Nature of Mathematics, claiming: "the points we can name are finite only, and what we name will be a finite number of points, however great."

显然，外尔认为，要令人信服地避开这一两难困境，唯一的办法就是直接否认阿喀琉斯所要跨越的线段"真的由无穷多个子线段构成......"。马克斯·布莱克在其早期著作《数学的本质》中也得出了类似的观点，他声称："我们能命名的点终究是有限的，无论其数量有多大，我们所命名的点的个数都是有限的。"

Now Goodstein is surely wrong. In order to catch the tortoise, Achilles would have to traverse the interval I 2 I_{2} I2 regardless of whether anyone isolated, marked off, or "named" the end points of the interval.

古德斯坦的观点显然是错误的。无论是否有人去分离、标记或"命名" I 2 I_2 I2 的端点，阿喀琉斯要追上乌龟，就必须跨越区间 I 2 I_2 I2 。

But perhaps Weyl, Black, and Goodstein are moving in the right direction. Has Zeno uncovered an incoherence in our usual ways of thinking about motion, compelling us to reject the principle that if a thing travels any distance, however small, it must first travel half that distance?

但或许外尔、布莱克和古德斯坦的研究方向是正确的。芝诺是否揭示出了我们对运动的常规思考方式中存在的矛盾，迫使我们摒弃"一个物体无论跨越多短的距离，都必须先跨越这段距离的一半"这一原则？

Presumably, Weyl, Black, and Goodstein were forced to their respective positions, at least partly, because they could find no other cogent way of avoiding the paradox. But none of the above positions, so far as I understand them, is easy to accept; and rather than significantly change our usual ways of analyzing motion and space, some philosophers, as for example Bertrand Russell, have accepted the other horn of Weyl's dilemma, preferring to believe that it is possible (though, of course, not "medically" [sic] possible) to perform an infinite number of calculations in a finite amount of time.

推测来看，外尔、布莱克和古德斯坦之所以会得出各自的观点，至少在一定程度上是因为他们找不到其他能令人信服地避开这一悖论的方法。但在我看来，上述观点没有一个是容易让人接受的。因此，一些哲学家------比如伯特兰·罗素------不愿大幅改变我们分析运动和空间的常规方式，而是选择接受外尔两难困境的另一个方面，宁愿相信在有限的时间内完成无穷多次计算是可能的（当然，这种可能性并非"实际物理意义上"的可能）。

To many, however, this alternative seems even more repugnant. Certainly philosophers have tried to disprove the conceivability of a machine which actually computes and prints, according to Weyl's prescription, all the digits in the decimal expansion of pi. J. M. Hinton and C. B. Martin go so far as to claim to have proved that such a machine is logically impossible. Suppose a super-machine computes the nth digit in the decimal expansion of pi at each instant Achilles finishes traversing I n I_{n} In . Then consider the question "When will Achilles' journey have lasted for twice the time it took him to get to the halfway position ?" Martin and Hinton assert that one obviously correct answer is: when the duration of Achilles' journey after he passes the halfway position equals the length of time it took him to reach the halfway position. "But this will never be so," they claim.

然而，对许多人而言，这一选择更令人难以接受。当然，哲学家们一直试图证明，依照外尔的设想，制造出一台能实际计算并打印出圆周率十进制展开式所有数位的机器，是无法想象的。J·M·欣顿和 C·B·马丁甚至声称，他们已经证明了这样的机器在逻辑上是不可能存在的。假设存在一台超级机器，每当阿喀琉斯走完一个区间 I n I_n In ，这台机器就计算出圆周率十进制展开式的第 n n n 位数字。那么我们来思考这个问题："阿喀琉斯的奔跑时间何时会达到他跑到中点所用时间的两倍？"马丁和欣顿断言，一个显然正确的答案是：当他经过中点后继续奔跑的时间，等于他跑到中点所用的时间时。但他们声称："这一时刻永远不会到来。"

the sequence and is a conjunction of the form D 1 ∗ D 2 ∗ D 3 ∗ . . . ∗ D n ∗ R n D_{1}^{*} D_{2}^{*} D_{3}^{*} ...^{*} D_{n}^{*} R_{n} D1∗D2∗D3∗...∗Dn∗Rn such that:(a) R n R_{n} Rn is act-equivalent to D n + 1 ∗ R n + 1 D_{n+1} * R_{n+1} Dn+1∗Rn+1 ;(b) the same sorts of test could be used for determining that R n R_{n} Rn has been made true as those for determining that D n D_{n} Dn has been made true. Then the sequence D 1 D_{1} D1 , D 2 D_{2} D2 , D 3 D_3 D3 ... will be called an analysis-engendered (or simply a − e a-e a−e ) infinite sequence.

该序列是形如 D 1 ∗ D 2 ∗ D 3 ∗ ... ∗ D n ∗ R n D_1^*D_2^*D_3^*\dots^*D_n^*R_n D1∗D2∗D3∗...∗Dn∗Rn 的合取式，且满足以下条件：（a） R n R_n Rn 与 D n + 1 ∗ R n + 1 D_{n+1}*R_{n+1} Dn+1∗Rn+1 在行为上等价；（b）判定 R n R_n Rn 为真所采用的检验方式，与判定 D n D_n Dn 为真的检验方式属于同一类。此时，序列 D 1 , D 2 , D 3 , ... D_1,D_2,D_3,\dots D1,D2,D3,... 将被称为分析衍生的（简称为 a-e）无穷序列。

The reader undoubtedly will be tempted to object that the statement of condition (b) is not very precise. I agree; but I am here relying on the principle that the amount of precision required in making a distinction is a function of the use to which the distinction is put. And there is, I think, sufficient precision in the above definition for my purposes. This can be seen by employing the above definition to deal with the dilemmas proposed by Weyl and Owen.

读者无疑会想要反驳，条件（b）的表述并非十分精确。我对此表示认同，但我在此遵循的原则是：进行概念区分所需的精确程度，取决于这一区分的应用目的。而且我认为，上述定义的精确程度，足以满足本文的研究需求。我们可以运用这一定义来分析外尔和欧文提出的两难困境，以此验证这一点。

The reader can easily verify that there are good reasons for believing that the T n T_{n} Tn sequence is a − e a-e a−e infinite, although there may possibly be some question about the satisfaction of condition (b). Given our present conception of space and motion, however, there does not seem to be any reason for supposing that the mere fact that one interval is to the right of another of equal width would require that the sorts of test used to determine whether the former has been traversed would essentially differ from those dealing with the latter.

读者可轻易验证，我们有充分的理由认为 T n T_n Tn 序列是分析衍生的无穷序列，尽管条件（b）的满足情况或许尚存疑问。但基于我们当前对空间和运动的理解，似乎没有任何理由认为，仅仅因为一个区间位于另一个等宽区间的右侧，判定前者是否被跨越的检验方式，就必须与判定后者的检验方式存在本质区别。

Now I am not committing myself to holding either that the T n T_{n} Tn sequence is definitely a − e a-e a−e infinite or that no difficulties are to be found in our present notions (precritical or otherwise) of space and motion. All I am suggesting is that there are grounds for believing that the T n T_{n} Tn sequence is a − e a-e a−e infinite. And if one keeps in mind the location of the burden of proof, it should be evident that there is no need to maintain a stronger thesis.

我并非要坚定地主张， T n T_n Tn 序列必然是分析衍生的无穷序列，也并非认为我们当前关于空间和运动的观念（无论是否经过批判性反思）不存在任何问题。我只是想指出，我们有依据认为 T n T_n Tn 序列是分析衍生的无穷序列。而且如果考虑到举证责任的归属，就会很明显地发现，我们无需提出更强的论点。

But consider the sequence of tasks described by Owen. We can, of course, devise a method of analysis for this case which engenders a nonterminating sequence of propositions analogous to the T n T_{n} Tn sequence. Let σ n = O 1 ∗ O 2 ∗ O 3 ∗ . . . ∗ O n ∗ L n \sigma_{n}=O_{1}^{*} O_{2}^{*} O_{3}^{*} ...^{*} O_{n}^{*} L_{n} σn=O1∗O2∗O3∗...∗On∗Ln where: O n O_{n} On = Achilles marked the end point of I n I_{n} In ， L n L_{n} Ln = Achilles marked the end points of I j I_{j} Ij for all j > n j>n j>n . The O n O_{n} On sequence is obviously similar to the T n T_{n} Tn sequence but, unlike the latter, the O n O_{n} On sequence is clearly not a − e a-e a−e infinite. A straightforward consideration of σ 1 \sigma_{1} σ1 will show that condition (b) is not satisfied. Clearly the sorts of test we would use to determine whether Achilles had marked the end point of I 1 I_{1} I1 would be quite different from those used to determine whether Achilles had actually completed the task of marking all the remaining end points. Indeed, although there is no special difficulty about verifying the truth of O 1 O_{1} O1 , we have no clear idea of how we might test the truth of L 1 L_{1} L1 .

但我们来看看欧文所描述的任务序列。当然，我们可以为这一情况设计一种分析方法，生成一个与 T n T_n Tn 序列类似的非终止性命题序列。令 σ n = O 1 ∗ O 2 ∗ O 3 ∗ ... ∗ O n ∗ L n \sigma_n=O_1^*O_2^*O_3^*\dots^*O_n^*L_n σn=O1∗O2∗O3∗...∗On∗Ln ，其中： O n O_n On 表示"阿喀琉斯标记了区间 I n I_n In 的端点"， L n L_n Ln 表示"阿喀琉斯标记了所有 j > n j>n j>n 的区间 I j I_j Ij 的端点"。 O n O_n On 序列显然与 T n T_n Tn 序列相似，但与后者不同的是， O n O_n On 序列显然并非分析衍生的无穷序列。对 σ 1 \sigma_1 σ1 进行直接分析就会发现，它并不满足条件（b）。显然，判定阿喀琉斯是否标记了 I 1 I_1 I1 的端点所采用的检验方式，与判定他是否实际完成了标记所有剩余端点这一任务的检验方式，存在巨大差异。事实上，尽管验证 O 1 O_1 O1 的真实性并无特殊困难，但我们完全不清楚该如何检验 L 1 L_1 L1 的真实性。

Now if I am correct in claiming that the sequence of tasks specified by Owen is not a − e a-e a−e infinite, then even if there is some sort of conceptual absurdity in the supposition that Achilles finishes Owen's sequence of tasks, it certainly need not follow that there is a corresponding absurdity in the notion that Achilles makes true all the T n T_{n} Tn .

因此，如果我的主张是正确的，即欧文所设定的任务序列并非分析衍生的无穷序列，那么即便假设阿喀琉斯完成了欧文所描述的任务序列在概念上存在荒谬性，也并不意味着阿喀琉斯使所有 T n T_n Tn 为真这一观念，也存在相应的荒谬性。

For a similar reason, even if, as Black suggests at one point, there is a logical incoherence in the claim that Achilles might catch the tortoise by completing an infinite sequence of leaps, this would not show that Achilles cannot catch the tortoise by simply running and making true all the T n T_{n} Tn .

出于类似的原因，即便如布莱克某一观点所言，声称阿喀琉斯能通过完成无穷多次跳跃追上乌龟在逻辑上存在矛盾，这也并不意味着，阿喀琉斯无法通过单纯的奔跑、使所有 T n T_n Tn 为真来追上乌龟。

Returning to Weyl's dilemma, it is now easy to see how one can slip through its horns: again, we need only observe that Weyl's infinite sequence of tasks is not a − e a-e a−e infinite to see the weakness in his claim that Achilles can traverse all the I n I_{n} In only if it is possible to complete an infinite sequence of calculations. Weyl seems to have looked at only those logical features of the sequence of T n T_{n} Tn which tend to strike mathematicians as relevant, and thus mistakenly linked the possibility of making true all the terms in the sequence of T n T_{n} Tn to the possibility of making true all the terms in the similar (but not sufficiently similar) infinite sequence of propositions concerning calculating the digits in the expansion of pi.

回到外尔的两难困境，我们现在能轻易看出该如何摆脱这一困境：我们只需再次指出，外尔所提出的无穷任务序列并非分析衍生的无穷序列，就能发现其主张的漏洞------他认为阿喀琉斯能遍历所有 I n I_n In 的前提，是完成无穷多次计算具有可能性。外尔似乎只关注了 T n T_n Tn 序列中那些在数学家看来具有相关性的逻辑特征，因此错误地将使 T n T_n Tn 序列中所有命题为真的可能性，与使另一类相似（但并非充分相似）的无穷命题序列为真的可能性关联起来，而后者关乎圆周率展开式数位的计算。

Let us now return to the question "Does the Achilles paradox exhibit a logical incoherence in our conception of motion compelling us to reject or revise our basic principles ?" Notice, I am not trying to determine whether our ideas concerning motion are actually correct. Nor am I concerned with considerations, apart from those raised by Zeno, that indicate difficulties or confusions in our conceptualization of motion. These are matters which should be clearly separated and kept distinct from the problem at hand.

现在我们回到这个问题："阿喀琉斯悖论是否揭示出我们的运动观念中存在逻辑矛盾，迫使我们摒弃或修正相关的基本原则？"需要说明的是，我并非要判断我们关于运动的观念是否实际正确，也不关注芝诺之外的其他因素所揭示的、我们在运动概念化过程中存在的困难或困惑。这些问题都应与本文探讨的核心问题明确区分、相互独立。

Let us consider the Achilles paradox again. The puzzle is apt to get started when one thinks of Achilles as passing a number of markers indicating 1 / 2 1/2 1/2 , 1 / 4 1/4 1/4 , 1 / 8 1/8 1/8 , ... as he runs to catch the tortoise. It looks as if Achilles must pass all the markers to finish the journey; but to do that, we feel, is impossible since there is no end to the sequence of markers. At this point, one might try to dissolve the paradox by distinguishing the physical object which marks a position from the position itself. Obviously, Achilles is not required to pass an infinite number of physical markers.

我们再重新审视阿喀琉斯悖论。当人们想象阿喀琉斯在追赶乌龟的过程中，要经过一系列标有 1/2、1/4、1/8......的标记时，这一谜题的困惑性便油然而生。看起来，阿喀琉斯要完成这段路程，就必须经过所有标记；但我们又觉得这是不可能的，因为这些标记构成的序列永无终点。此时，有人可能会试图通过区分标记位置的物理客体与位置本身，来消解这一悖论。显然，阿喀琉斯并非必须经过无穷多个物理标记。

Needless to say, the paradox remains, for we still have Achilles faced with the task of passing through infinitely many points in space. But perhaps this task will not appear quite so formidable when it is recalled that a point or position in space is not a tiny dot through which objects pass. Of course, we do say such things as "Achilles is now at the half-way point," but to say this is not to claim that there are two things halfway to the finish, namely Achilles and a point. One is claiming only that Achilles is halfway to the finish.

毋庸置疑，这一悖论依然存在，因为阿喀琉斯仍需完成穿越空间中无穷多个点的任务。但倘若我们想起，空间中的一个点或位置，并非一个物体需要从中穿过的微小圆点，这一任务或许就不会显得那么难以完成了。当然，我们会说"阿喀琉斯现在到了中点"，但说出这句话，并非意味着在终点的中点处存在两个事物------阿喀琉斯和一个点，而只是表明阿喀琉斯处于跑到终点的半路上。

Still, do we not have essentially the same problem with the proposition ( n ) T n (n) T_{n} (n)Tn ? Must not Achilles make true each of the T n T_{n} Tn one after another until all of the infinitely many propositions in the sequence are made true? And how can he do that?

但即便如此，命题 ( n ) T n (n)T_n (n)Tn 不依然让我们面临本质上相同的问题吗？阿喀琉斯难道不是必须依次使每一个 T n T_n Tn 为真，直至使这一无穷命题序列中的所有命题都为真吗？而他又该如何做到这一点呢？

But why does it even seem to be impossible to make true all the T n T_{n} Tn ? I suggest that one reason is this: one tends to think of infinite sequences on the model of finite sequences. Suppose now that all the elements of a finite sequence of propositions P 1 , P 2 , P 3 , . . . P k P_{1}, P_{2}, P_{3}, ... P_{k} P1,P2,P3,...Pk must be made true in the given order. Obviously, one finishes making true all the propositions in the sequence only when one makes true P k P_{k} Pk , the last proposition in the sequence. Now if this sort of situation remains fixed in one's mind, one can easily be led to think that making true all the T n T_{n} Tn is an impossible task, since there is no last element of that sequence to be made true.

但为何我们会觉得，使所有 T n T_n Tn 为真这件事本身是不可能的？我认为原因之一在于：人们总是倾向于以有限序列的模式来思考无穷序列。假设我们需要依照给定的顺序，使一个有限命题序列 P 1 , P 2 , P 3 , ... , P k P_1,P_2,P_3,\dots,P_k P1,P2,P3,...,Pk 中的所有命题都为真。显然，只有当我们使序列的最后一个命题 P k P_k Pk 为真时，才算完成了使该序列中所有命题为真的任务。倘若人们始终囿于这一思维模式，就很容易认为使所有 T n T_n Tn 为真是一项不可能完成的任务，因为这一序列根本没有可供使其为真的最后一项。

But a person might also be led to think that the infinite series 1 / 2 + 1 / 4 + 1 / 8 + . . . 1 / 2+1 / 4+1 / 8+... 1/2+1/4+1/8+... could not have the value 1, since one could never finish "adding up" all the relevant terms, there being no last term in the sequence to be "added up." Of course, to claim that the above infinite series has the value 1 is not to claim that if one "added up" the terms in the sequence 1 / 2 , 1 / 4 , 1 / 8 , . . . 1/2, 1/4, 1/8, ... 1/2,1/4,1/8,... one would eventually get 1, but rather that the sequence of partial sums has the limit 1. And to say that a sequence has a particular limit is to characterize the defining rules as generating terms of a certain sort.

同理，有人也会认为无穷级数 1 / 2 + 1 / 4 + 1 / 8 + ... 1/2+1/4+1/8+\dots 1/2+1/4+1/8+... 的和不可能为 1，因为人们永远无法完成对所有项的"累加"------这一序列根本没有可供"累加"的最后一项。当然，声称上述无穷级数的和为 1，并非指如果人们对序列 1 / 2 , 1 / 4 , 1 / 8 , ... 1/2,1/4,1/8,\dots 1/2,1/4,1/8,... 中的项依次"累加"，最终会得到 1；而是指该级数的部分和序列的极限为 1。而称一个序列具有某一特定极限，是对生成该序列的定义规则所具有的、生成特定类型项的特征的描述。

This example suggests that we have been thinking of Achilles' task in the wrong way. By analogy with the notion of the value of the infinite series, we should not think of Achilles as making true each of the T n T_{n} Tn one after another in the manner of someone making true each proposition in a finite sequence.

这一例子表明，我们一直以来对阿喀琉斯所面临任务的思考方式是错误的。以无穷级数的和这一概念为类比，我们不应将阿喀琉斯使所有 T n T_n Tn 为真的过程，看作是像人们使有限序列中的每个命题依次为真那样，逐个使每个 T n T_n Tn 为真。

When we say that the infinite series 1 / 2 + 1 / 4 + 1 / 8 + ... 1/2 + 1/4 + 1/8 + \dots 1/2+1/4+1/8+... has the value 1, we are not describing a process of addition that has been completed; rather, we are stating a mathematical fact about the series, a fact that is grounded in the defining rules of the series and the concept of a limit. Similarly, when we say that Achilles makes true all the T n T_{n} Tn , we are not describing a process in which Achilles performs an infinite number of distinct acts, each of which makes a particular T n T_{n} Tn true, in a sequential manner. Instead, we are stating a fact about Achilles' motion: that in the course of running the one mile (or the finite distance to the point where he catches the tortoise), Achilles's movement has the property that it makes true every proposition in the infinite sequence T 1 , T 2 , T 3 , ... T_{1}, T_{2}, T_{3}, \dots T1,T2,T3,... .

当我们说无穷级数 1 / 2 + 1 / 4 + 1 / 8 + ... 1/2+1/4+1/8+\dots 1/2+1/4+1/8+... 的和为 1 时，我们并非在描述一个已经完成的累加过程；相反，我们是在陈述一个关于该级数的数学事实，这一事实建立在该级数的定义规则与极限概念之上。同理，当我们说阿喀琉斯使所有 T n T_n Tn 为真时，我们并非在描述一个阿喀琉斯依次实施无穷多次独立行为的过程------这些行为中每一次都使一个特定的 T n T_n Tn 为真。相反，我们是在陈述一个关于阿喀琉斯运动的事实：在跑完这 1 英里（或追上乌龟所需的那段有限距离）的过程中，阿喀琉斯的运动具有这样一种属性，即它能使无穷命题序列 T 1 , T 2 , T 3 , ... T_1,T_2,T_3,\dots T1,T2,T3,... 中的每一个命题都为真。

This way of conceiving Achilles' task avoids the dilemma posed by Weyl, for it does not require us to accept the possibility of performing an infinite number of distinct calculations (or any other distinct acts) in a finite time. The key difference is that the T n T_{n} Tn sequence is an analysis-engendered infinite sequence, generated by a mathematical analysis of Achilles' single, continuous act of running. The infinite sequence of calculations described by Weyl, by contrast, is not an analysis-engendered sequence; it consists of a potential infinity of distinct, discrete acts that would need to be performed in sequence, with no single continuous act underlying them.

这种对阿喀琉斯所面临任务的理解方式，能让我们避开外尔提出的两难困境，因为它并不要求我们接受"在有限时间内完成无穷多次独立的计算（或其他任何独立行为）"的可能性。二者的核心区别在于， T n T_n Tn 序列是分析衍生的无穷序列，它源于对阿喀琉斯单一、连续的奔跑行为所做的数学分析。而外尔所描述的无穷计算序列则并非分析衍生的序列：它由潜在无穷多的独立、离散的行为构成，这些行为需要依次实施，且不存在作为其基础的单一连续行为。

In the case of Achilles' motion, the infinite sequence of intervals I n I_{n} In and the corresponding propositions T n T_{n} Tn are not features of the physical world that Achilles must "confront" or "overcome" in some way; they are features of our mathematical description of his motion. We divide the finite distance into an infinite number of subintervals through our analysis, but the distance itself remains finite, and Achilles' traversal of it is a single, continuous physical process. The infinity here is a feature of the description, not of the described process.

就阿喀琉斯的运动而言，无穷的区间序列 I n I_n In 以及对应的命题序列 T n T_n Tn ，并非阿喀琉斯需要以某种方式"面对"或"克服"的物理世界的固有特征；它们是我们对其运动进行数学描述时产生的特征。我们通过分析将这段有限的距离分割为无穷多个子区间，但这段距离本身依然是有限的，阿喀琉斯跨越这段距离的行为，也是一个单一、连续的物理过程。此处的无限，是描述的特征，而非被描述过程的特征。

This distinction between the infinity of the description and the finitude of the described process is crucial to resolving the puzzle of completing an infinite process as it arises in the Achilles paradox. Zeno's paradox gains its force by conflating these two things: it treats the infinite divisibility of the mathematical description of the distance as implying an infinite "divisibility" or "decomposability" of the physical process of traversing the distance. Once we separate the mathematical analysis from the physical reality, the apparent impossibility of Achilles' task vanishes.
描述的无限性 与被描述过程的有限性之间的这一区分，对于解决阿喀琉斯悖论中出现的"完成无限过程"这一谜题而言，至关重要。芝诺悖论之所以具有迷惑性，正是因为它将二者混为一谈：它把对距离的数学描述所具有的无限可分性，等同于跨越这段距离的物理过程所具有的无限"可分性"或"可分解性"。一旦我们将数学分析与物理现实区分开来，阿喀琉斯的任务所显现出的不可能性便会消失。

It is also worth noting that this resolution does not require us to revise our basic principles of motion or space, nor does it require us to deny the infinite divisibility of space or time. We can retain our ordinary conception of motion as continuous movement through space over time, and we can accept that space and time are infinitely divisible in a mathematical sense. What we must reject is the mistaken inference from the infinite divisibility of the mathematical description to the existence of an actual infinity of distinct physical acts or stages in the process of motion.

还值得注意的是，这一解决方式既不需要我们修正关于运动或空间的基本原则，也不需要我们否定空间或时间的无限可分性。我们可以保留对运动的常规理解------即运动是物体随时间在空间中进行的连续移动，也可以接受空间和时间在数学意义上具有无限可分性。我们必须摒弃的，是从"数学描述的无限可分性"推导出"运动过程中存在实际无穷多的独立物理行为或阶段"这一错误推论。

In conclusion, the Achilles paradox, when understood as a puzzle about completing an infinite process, does not reveal a logical incoherence in our conception of motion. The apparent paradox arises from a misinterpretation of the nature of the infinite sequence associated with Achilles' pursuit of the tortoise: we mistake an analysis-engendered infinite sequence, which is a feature of our mathematical description, for a sequence of distinct physical tasks that Achilles must perform in sequence. Once we correct this misinterpretation and distinguish between the infinity of the description and the finitude of the physical process, the puzzle of completing an infinite process in the Achilles paradox is resolved. We can consistently hold that Achilles catches the tortoise by performing a single, continuous act of running, which in virtue of its continuity, makes true all the propositions in the infinite T n T_{n} Tn sequence---without being committed to the impossible idea of performing an infinite number of distinct acts in a finite time.

综上所述，当阿喀琉斯悖论被视作一个关于"完成无限过程"的谜题时，它并未揭示出我们的运动观念中存在逻辑矛盾。这一悖论的表面迷惑性，源于我们对与阿喀琉斯追赶乌龟相关的无穷序列的本质产生了误读：我们将分析衍生的无穷序列 ------这一属于数学描述的特征，误当作了阿喀琉斯需要依次完成的一系列独立物理任务。一旦我们纠正这一误读，并区分描述的无限性与物理过程的有限性，阿喀琉斯悖论中关于"完成无限过程"的谜题便会迎刃而解。我们可以一致地认为，阿喀琉斯通过单一、连续的奔跑行为追上了乌龟；而正是这一行为的连续性，使其能让无穷的 T n T_n Tn 命题序列中的所有命题都为真------同时我们无需认同那种"在有限时间内完成无穷多次独立行为"的不可能观点。

论完成无限过程的可能性摘要

文章围绕芝诺的阿喀琉斯悖论展开，探讨完成无限过程的可能性问题，先引出悖论所带来的两难困境，再通过引入概念区分与分析，最终厘清悖论的误读并给出解答。

文章以圆周率无穷数位的计算难题为引，提出阿喀琉斯悖论带来的关键两难：要么否定完成无限过程的可能性，接受阿喀琉斯追不上乌龟的结论，要么承认能在有限时间内完成无穷多次计算，而传统的数学解法和量词混淆解读均无法破解这一困境。
为分析悖论，作者引入关键概念区分：终止性与非终止性规则、分析衍生的无穷序列，指出阿喀琉斯经过对应区间的命题序列 T n T_{n} Tn 属于分析衍生的无穷序列，而圆周率计算、标记区间端点等任务序列并非此类序列，二者存在本质差异。
作者指出人们对阿喀琉斯完成无限过程的质疑，根源在于以有限序列的思维模式看待无穷序列，误将使无穷命题序列为真的过程，等同于依次完成有限序列中各命题的过程，这一思维误区也导致人们对无穷级数的和产生错误理解。
文章明确分析衍生的无穷序列是对单一连续行为的数学分析结果，阿喀琉斯悖论中无穷的区间与命题序列，只是对其连续奔跑行为的数学描述特征，而非物理世界中需要逐一完成的独立任务，距离本身与跨越距离的物理过程都是有限的。
芝诺悖论的迷惑性源于混淆了数学描述的无限性与物理过程的有限性，将距离的数学无限可分性等同于物理运动过程的无限可分解性。纠正这一误读后，悖论即可化解，无需修正人们关于运动和空间的基本原则，也无需否定时空在数学意义上的无限可分性。
最终得出结论，阿喀琉斯悖论并未揭示出人类运动观念的逻辑矛盾，阿喀琉斯通过单一、连续的奔跑行为即可追上乌龟，这一连续行为本身能使无穷的 T n T_{n} Tn 命题序列全部为真，无需在有限时间内完成无穷多次独立的物理行为，由此避开了外尔提出的两难困境。

赫尔曼·外尔（Hermann Weyl）简介

赫尔曼·外尔（Hermann Weyl，1885 年 11 月 9 日－1955 年 12 月 8 日）是 20 世纪德国数学家、理论物理学家和哲学家，在多个领域留下了开创性贡献。

一、学术生涯与师承

师从大卫·希尔伯特（David Hilbert），属于哥廷根数学学派成员。
曾在苏黎世联邦理工学院、普林斯顿高等研究院等机构任教，与爱因斯坦等人共事。
研究方向覆盖纯数学、理论物理与数学哲学。

二、主要学术贡献

数学领域
- 发展了李群与李代数理论，对现代微分几何与拓扑学产生影响。
- 提出外尔曲率（Weyl curvature），为广义相对论与规范场论提供数学基础。
- 在分析学与数论中，对狄利克雷级数、特征值问题等方向有推进。
物理学领域
- 是规范场论的先驱，最早提出"规范不变性"思想，为后来的杨-米尔斯理论奠定基础。
- 研究广义相对论，尝试统一引力与电磁相互作用，参与早期统一场论探索。
数学哲学领域
- 持直觉主义/构造主义立场，反对实无限（completed infinity）的存在，将无限描述为"不可完成的"（incompletable）。
- 这一立场体现在相关文本中：他认为阿喀琉斯无法完成遍历无限个区间的任务，因为无限序列无法被穷尽。

三、思想与影响

外尔的哲学思想影响了数学基础论的发展，他对实无限的批判，与布劳威尔（L.E.J. Brouwer）的直觉主义一脉相承，对希尔伯特的形式主义纲领形成挑战。
著作《数学与自然科学的哲学》（Philosophy of Mathematics and Natural Science）是科学哲学领域的经典文本，至今仍被广泛研读。
在物理学上，他的规范理论思想演变为现代粒子物理标准模型的框架。

与文本的关联

赫尔曼·外尔断言："无限的本质是'不可完成的'，而阿喀琉斯能遍历所有这些区间，这与无限的这一本质相悖。"

引用的这段文字，是外尔直觉主义数学哲学的表达：

他不将无限视作"已完成的整体"（completed totality），而将其看作"永远在生成中"的潜在无限（potential infinity）。
因此，他认为阿喀琉斯无法"遍历所有无限个区间"，完成无限任务与无限的描述相悖。

Reference

On the Possibility of Completing an Infinite Process - Charles S. Chihara
https://www.jstor.org/stable/2183531