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This is a quote from page 1, volume III of Principia Mathematica, by Whitehead and Russell: "A 'well-ordered' series is one which is such that every existent class contained in it has a first term, or, what comes to the same thing, one which is such that every class which has successors has a sequent." The second part of that sentence seems to imply, in modern terminology, that a (linearly ordered) set A's being well-ordered is equivalent to: for every non-empty subset B of A such that B has an upper bound in A not in B, there exists in A an element a such that a is the next element in A after B, i.e., that if there exists a c in A such that c is greater than every d in B, then there exists an a in A such that a is greater than every e in B, and there does not exist an f in A such that f is greater than every g in B and a is greater than f. (Sorry if that was excessively tedious) An obvious counter-example to this statement of PM is Z, the positive and negative integers and zero. Z is such that for every non-empty subset B of Z which has an upper bound in Z not in B, there exists in Z a next element after B, but Z is not well-ordered, since Z is a non-empty subset of Z which has no first term. Link to PM v III: https://quod.lib.umich.edu/u/umhistmath/aat3201.0003.001/2?page=root;rgn=works;size=100;view=pdf;rgn1=title;q1=Principia+Mathematica

Is my interpretation of the terminology of PM correct, and if so, is this elementary error in PM well-known, and what do mathematicians and logicians think of it?

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    $\begingroup$ There is no error, the principia specifically mentions series, where as you mention the set Z as a counter example, here lies your mistake, Z is not a sequence, but a set. A set is not an ordered collection while a sequence( and its sum which is the series)is. If you want to prove leibniz wrong, you would need to provide a counter example using his exact definitions of series. $\endgroup$ Commented Aug 11, 2020 at 8:49
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    $\begingroup$ @dvd280: On the other hand, according to Wikipedia “series” is PM's term for what we call a totally ordered set. And $\mathbb Z$ definitely is a totally ordered set. I don't have access to PM itself to check the claim of Wikipedia. $\endgroup$ Commented Aug 11, 2020 at 9:33
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    $\begingroup$ In $\mathbb{Z}$, $\varnothing$ has successors (every integer is an upper bound), but not a smallest one. $\endgroup$ Commented Aug 11, 2020 at 9:54
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    $\begingroup$ @dvd280 Why do you mention Leibniz? $\endgroup$ Commented Aug 11, 2020 at 17:06
  • $\begingroup$ @Kruckman Its been a long day xD $\endgroup$ Commented Aug 12, 2020 at 8:29

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What is said in the introduction is true. In Principia the concept of "sequent" of a class A for a relation P is quite "general" and abstract. To simplify a bit, if P is a series and A a class that isn't empty and it is included in the field of P (the field is just the union of the domain and co-domain of P) than the sequent of A for P (if any) is the "first" term in the series P that comes after all the elements of A. But, if the class A is empty than the sequent of A, if any, is the first term of the series P. You can easily find the "justification" of my claims by looking at Principia chapter *206 (especially *206.14 and *206.12). Your counter-example doesn't work simply because the null class, in the series of integers for magnitude, doesn't have a sequent. As a matter of fact in *250 it is proved what is claimed in the introduction of the chapter (see *250.122).

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  • $\begingroup$ You are correct. In the modern definition of "well-ordered set", in the quoted PM introduction's definition, and in my paraphrase of PM's "what comes to the same thing" equivalent of that definition, the requirement of having a smallest element is made only for non-empty ("existent" in PM, meaning, I think, non-empty) sets. However, my paraphrase was wrong; PM's equivalent "what comes..." doesn't restrict its requirement, that every subset which has a succesor has a sequent, to non-empty subsets. The empty subset of Z has a successor, bu no sequent, so Z isn't a counterexample. I was careless. $\endgroup$ Commented Oct 28, 2021 at 18:59

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