GrandMinnow

"well-ordering the complex plane is something I don't know how to do but the Axiom of Choice implies is possible" [aleph_1] Even the set of real numbers. The axiom of choice entails that there is a well ordering of the set of real numbers. But also it happens that in ZFC there is no definition of a particular such well ordering. "I have still not quite resolved that Oism and axiomatic notions at the foundations of math are incompatible. Oism has its own axiomatics." [aleph_1] The notion of 'axiom' in Objectivism is quite different as the notion of 'axiom' in modern mathematical logic. I don't think Objectivism is entirely hostile to formalizations such as found in ordinary mathematics; otherwise Objectivism would reject computer languages, and I don't think Objectivism does reject computer languages. However, I suspect (and only suspect, since I can't speak for Objectivism) that formalizations such as ZFC (which provides a formalization of analysis) would be (are?) seen by Objectivism as irrational - as either platonism or nominalism. In any case, I am not aware of any Objectivist proposal for a formal (formal in the manner of a recursive syntax) arithmetic, analysis, and other branches of mathematics other than computer languages. The only Objectivist I know of that works in mathematical logic and set theory is Steven Simpson. In my personal conversation with him he admitted that Objectivism has a lot of work to do in order to provide such foundational explanations. Moreover, as to Objectivist posters in this particular forum, aside from the late Steven Speicher, most response to mathematical logic and set theory is hostile.

"Why couldn't you just as well say choose the left sock?" [Eiuol] Just socks. No left or right. "My thinking is that as long as there is something to select, then any arbitrary selection is sufficient." [Eiuol] Then in that regard you agree with the axiom of choice. Roughly speaking, "arbitrary selection is sufficient" is the axiom of choice. Lets's step back first: In formal set theory, we only can prove things that the axioms allow us to prove and everything proven is stated and proven in the formal language. Now, For all S, if the empty set is not a member of S then there is a function C such that the domain of C is S and for all x in S we have C(x) in x is pretty much how, in the formal language, we express the more informal version you mention: "arbitrary selection is sufficient". But the ZF (a formal set theory) axioms do NOT allow us to prove: For all S, if the empty set is not a member of S then there is a function C such that the domain of C is S and for all x in S we have C(x) in x. So we have to take the above statement as an additional axiom, which we call 'the axiom of choice". In other words, ZF does not allow us to prove that "arbitrary selection is sufficient", so whoever thinks arbitrary selection is sufficient will add to ZF the axiom of choice.

What you say does seem to fit your general argument. As far as I can tell, the axiom of infinity does not meet your criterion of reducibility to the extent I understand your notion of reducibility. But even more fundamentally, as far as I can tell, formal axiomatics such as these are quite different from Objectivist notions. It seems to me that there's not even much sense in talking about this formal mathematics in context of Objectivist requirements.

Or, paraphrasing the language of set theory: From the axioms of ZF there is no proof of this statement: For all S, if the empty set is not a member of S then there is a function C such that the domain of C is S and for all x in S we have C(x) in x.

I don't think so. It is the very issue at hand that there is not in general a way to distinguish a next "choice". Aside from a technical explanation, Russell himself gave the problem in anecdotal form (so this is not a technical description but rather a non-technical visualization of the problem): Suppose we have an infinite set of pairs of shoes and we wish to choose one member from each pair. In this case, we could say, "Choose the right shoe" and this is fine. But then suppose we have an infinite set of pairs of socks...oops...there's not a way to go through all the pairs to specify which sock to choose. And it's even worse really. (I'm using "laundry" here just as illustration; of course, set theory does not itself mention such things as laundry). Imagine that we have infinitely many piles of laundry (and some of these piles are themselves infinite). From each pile we want to choose exactly one piece of laundry. It is proven that with the axioms of ZF there is no general description of a "way" to do this.

Personally (aside from whatever Objectivism might say about the matter; and I am not aware of any authoritative Objectivist literature that explicity addresses such things as the axiom of choice), I don't find it required to allow only constructive mathematics. One may regard adopting the set theoretic axioms (including infinity and choice) as a means for study of those abstract mathematical structures in which said axioms happen to hold. To work with such axioms it is not required to take such axioms as assertions that must hold in all mathematical situations, but rather only that the axioms provide a context of study of those mathematical situations in which the axioms do hold.

The well ordering principle is that every non-empty set of natural numbers has a least member. That is not equivalent to the axiom of choice. What you mean is that well ordering theorem: that every non-empty set has a well ordering. But still, your objection to the axiom of choice is that it is non-constructive, yet so is the well ordering theorem. If you object that the axiom of choice provides a choice function but does not show a construction of a choice function, then we might as well observe that the well ordering theorem provides a well ordering but does not show a construction of a well ordering.

The language of set theory does not have such terminology as "I happened to notice". It is proven that we cannot prove in ZF the statement "every non-empty set has a choice function", thus to have the result that every non-empty set has a choice function, we have to add it as an axiom to the axioms of ZF. Resorting to terminology that is not even in the language of set theory is not considered a viable solution since the solution mathematicians demand is that of a formulation in the language of set theory.

That is a very kind thing for you to say.

"It seems that you are testing me." [aleph_1] I'm not. Indeed, you were asking me the series of questions. "We are not talking about first-order theories" [aleph_1] You asked me about consistency and models. I gave you the most exact answer I could regarding first order theories and said that I'd have to know the specifics of any other context. If there is some other context then you're welcome to specify it. "Godel's Incompleteness is operative here" [aleph_1] Yes, but what is your point in that regard? How does the incompleteness theorem bear on anything I've said? "The Intermediate Value Theorem depends on completeness." [aleph_1] Just to be clear, that is 'completeness' in a much difference sense than in either of two other senses mentioned: completeness (as in the model existence theorem) and incompleteness (as in Godel-Rosser). "This is an independent axiom" [aleph_1] It's an axiom for some systems and it's a non-axiomatic theorem in certain other systems. "Your position concerning axioms reminds me of something I read by Novikov, the logicist. His position was that standard logic is valid when the axioms of logic are valid. When axioms such as LEM do not hold then standard logic does not apply." [aleph_1] My point was not about any such provisional status of logical axioms. I was referring to the non-logical axioms that determine theories, which, if consistent, have various models. "You claim that mathematical theorems are things you would not even check for consistency with your philosophical views. I recall Rand writing that your noblest act is the recognition that two plus two make four. I respond that some of what we call math is a recognition of reality and hence is pertinent to epistemology and philosophy more generally. Some math is useful for modeling and hence is provisional, subject to validation beyond mere proof." [aleph_1] I don't claim that criteria beyond mere consistency are not of interest. For example, set theory is of special interest because it provides an axiomatization for the calculus, "the mathematics for the sciences". And of course we may be interested in comparing theories per the criterion of axiomatizing the calculus or many other criteria. But meanwhile, one may have intellectual or even just recreational curiosity in many consistent theories (and even theories in alternative logics) that may or may not do well by various criteria. Then the only point I made is that I don't compare consistent theories for consistency with whatever philosophical views I may have. This does not restrict me from having particular interest in certain theories for various reasons and for comparison upon various criteria. Again, a consistent first order theory has a class of models. I take a model to be an abstraction concerning abstract relations; I don't hold that countenancing a consistent mathematical abstraction requires conformity to any philosophical views I may have. I don't intend to try to convince you to regard mathematical abstractions that way, nor do I hold my own "non-philosophical" view of mathematics as itself a philosophical thesis. Rather, you asked me a question, and to answer it honestly and meaningfully I'm informing you of my own approach. "When axioms such as LEM do not hold then standard logic does not apply." [aleph_1] Probably the most salient reason for working without the law of excluded middle is to ensure that all theorems are constructive in this sense: If I wish to prove merely that there exists an x having property P then classical logic is fine, but if I want to make sure that my approaches to such a proof also produce a particular object as an example of an x having property P, then I'd turn to intuitionistic logic (eschewing the law of excluded middle). "the axiom of infinity is an essential element of the Von Neumann construction of the Natural numbers." [aleph_1] With the von Neumman method, the axiom of infinity is not needed to produce each natural number. What the axiom of infinity is needed for is to have a set that has all the natural numbers as members. "the axiom of Infinity is essential to mathematical induction." [aleph_1] The axiom of infinity is not needed merely to perform mathematical induction. The axiom of infinity is needed to have a set that has all the natural numbers as members. But mathematical induction itself does not depend on the existence of such a set.

Mathematical induction does not require having the set of natural numbers nor the axiom of infinity. The following theorem schema is provable in Z/I (Z set theory without the axiom of infinity). For any formula P: (P[0] & Ax(x is a natural number -> (P[x] -> P[x+1]))) -> Ax(x is a natural number -> P[x]) is a theorem.

I don't begrudge you having your own notion of mathematics (shared by a fair number of other people, Objectivists in general, and even certain mathematicians); my only point was that when you say that infinity is treated in mathematics as you describe, then you're not referring to mathematics as generally practiced (not just by "some" mathematicians but pretty much by the overwhelming majority of them). It's as if I said (just making up a flip example here), "In chemistry, the main table of elements is organized by the continents in which the elements were each first discovered." I am entitled to such a notion of the elements, but it is not the way people who actually study chemistry generally view the matter, in which sense it invites misunderstanding for me to say that this is the way it is in "chemistry".

To answer your first question, for first order theories, the model existence theorem (aka the completeness theorem) ensures that any consistent theory has a model. So, in that context, my answer to your question is 'no'. For other kinds of theories, I could not answer without specific context. As to the intermediate value theorem, while some proofs may use choice, if I recall correctly, the intermediate value theorem is provable without choice (check me on that; my memory is not perfect and I don't have my notes with me now). As to the greater issue, of course I do recognize that choice is not constructive and leads even to such conundrums as that there is a well ordering of the reals but even in principle no specific well ordering of the reals can be defined. As to axioms, I don't take them as statements that govern ALL mathematical situations, but rather I see a set of axioms as only governing those models in which the axioms are true. For example, the axioms for first order group theory are true for all groups but are false for many other mathematical situations. So too the axioms of ZFC are true for all models of ZFC but are false for many other mathematical situations. In other words, for me, an axiomatization is a "description" of a certain kind of mathematical context; so the axioms do govern in that context but don't govern in other contexts. In that regard, it doesn't make sense to ask me whether mathematical theorems are consistent with my philosophy. Mathematical theorems are not things I would even compare for consistency with my philosophical inclinations. For me, mathematics (putting aside for the moment consideration of applied mathematics, etc.) regards (1) the formal deductions themselves and (2) purely abstract structures. Neither of those are mediated by my philosophical inclinations. As to construction of the natural numbers, again, neither the axiom of infinity nor choice need be involved. In set theory we may define the predicate 'natural number' and construct any particular natural number we wish without ever invoking infinity or choice. Indeed, using the von Neumann method, and even in greater generality, we can do the job with extensionality and three existence principles: that there exists an object, that for any object there is the set whose only member is that object, and that for any two sets there is the union of them. Finitistic and constructive.

Mathematics as practiced by mathematicians (who are not usually Objectivists) treats infinity in a variety of aspects, not merely the particular concept you mention.

"The choice function exists because we want it to, not because reality has constrained us to assert its existence." - aleph_1 In what sense has reality constrained us to assert the existence of the image under a set of a function class (replacement), a set closed under successor (infinity), pairs, unions, or power sets?

"Math induction is a consequence of the Well Ordering Principle (WOP). WOP is equivalent to the axiom of choice." - aleph_1 It is correct that in, for example, Z set theory (perforce, for example, ZF set theory), the well ordering principle is equivalent to the axiom of choice. And usually mathematical induction pertains to sets well ordered by a successor relation. But the well ordering principle is that EVERY set has a well ordering; yet there are many sets that are well ordered by a successor relation without our having to adopt the well ordering principle itself. In particular, the form of mathematical induction most discussed in this thread (weak induction on the natural numbers) does not require adoption of the well ordering principle. And it is correct that all three - weak induction, strong induction, and well ordering - are equivalent: If one of those holds for a set and relation then the other two hold also. But, again, for many sets, their well ordering is provable without having to adopt the well ordering principle (which, again, is that EVERY set has a well ordering). "You need never consider AoC if you reject the Infinity Axiom." - aleph_1 That depends on what you mean by 'reject'. If by 'reject' you mean merely to drop the axiom, then, no, the axiom of choice still has clout, since in this situation it is undetermined whether or not there exist infinite sets. But if by 'reject' you mean dropping the axiom AND adopting its negation, then, yes, the axiom of choice then is superfluous, since in this situation every set is finite and we know that every finite set has a choice function (which, by the way, we prove by mathematical induction on finite sets). Note, though, that this pertains to ZF (where the axiom of infinity is equivalent to the statement that there exists an infinite set (when I say 'infinite' I mean "Tarski infinite" (i.e. not finite) as opposed to "Dedekind infinite" (i.e. 1-1 with a proper subset)) but not Z (where the axiom of infinity implies there exists an infinite set, but the statement that there exists an infinite set does not imply the axiom of infinity); so to make the axiom choice superfluous to Z, we would have to drop the axiom of infinity and adopt an axiom that states that there exists an infinite set. "One would have to take a non-standard approach to constructing the Natural numbers without the Infinity Axiom." - alpeh_1 We don't need the axiom of infinity to construct each natural number. The axiom of infinity though is needed to have a set that has all the natural numbers as members. And the axiom of infinity is not needed to prove the principle of mathematical induction for natural numbers as long as the principle is stated schematically for the property of being a natural number.

Just to be clear, this is what you need to prove: If n is a natural number greater than 2, then there do not not exist natural numbers a, b, and c such each of a, b and c is greater than 0 and a^n+b^n=c^n.

Same here. Just to be clear, with, for example, Enderton (which is a typical treatment): Integers are equivalence classes of natural numbers, so each integer is itself an infinite set. An ordered pair of natural numbers, such as <0 1> is finite, but the integer called "negative one" is an equivalence class that is an infinite set; <0 1> is one of the members. Rationals are equivalence classes of integers, so each rational is itself an infinite set. And ordered pair of integers (a fraction), such as <integer_1 integer_2> is finite, but the rational number we could call "the ratio of the integer 1 and the integer 2" is an equivalence class that is an infinite set; <integer_1 integer_2> is one of the members. I feel that there may be some perplexing philosophical problems with set theory, but the notion of perceptual reduction doesn't happen to be very high among my own personal concerns.

In the context of this set theory we're talking about, there is nothing that is called 'infinity'. Rather there is the property 'is infinite'. If you overlook this distinction, then the set theoretic approach will always remain confusing to you. Some sets have the property of being finite (we say these sets are finite). But there is no set that we call "Finity". And some sets don't have the property of being finite (we say these sets are infinite). And there is no set that we call "Infinity".

It might be that Objectivism does not accept even the FRAMEWORK of a formalized (or even unformalized) language and mathematical logic in which a formal axiom of infinity is considered. Also, as you know (so this comment is directed generally, not personally), by "actual infinity" in this context we don't mean "tangible" like a tree and not necessarily even platonically (though many mathematicians and philosophers of mathematics do ascribe to certain forms of realism (aka 'platonism')). Rather, by saying that the axiom of infinity entails the existence of a set that is actually infinite, we may mean as little as that there is (in whatever ontological sense one regards 'is', ranging from platonic through fictionalism through pure formalism, etc.) a set that includes as members all and only the natural numbers, and that we don't obligate ourselves to understand that notion in terms of mere potentiality. Recall, for example, the formalist Abraham Robinson who declares that mention of 'infinity' is meaningless yet he works unashamedly with ZFC (with its axiom of infinity). There may be approaches in which infinity is accepted only as potentiality but where we also formally construct rational numbers, but it's not obvious how to do that. Recall that in the standard approach, even a single rational number is an equivalence class of natural numbers (this equivalence class being an infinite set itself).

Yes, you have the basic idea. The definition of 'countable' is "S is countable if and only if there is a 1-1 correspondence between S and the set of natural numbers or there is a natural number such that there is a 1-1 correspondence between S and said natural number." A set S if countably infinite if and only if there is a 1-1 correspondence between S and the set of natural numbers. So a set is countable if and only if the set is either finite or countably infinite. So it turns out that any countable set S can be ordered (if the set is countably infinite, then use any 1-1 correspondence with the set of natural numbers; and if the set is finite use any 1-1 correspondence with a natural number) so that (1) S has a first member, and (2) if S is infinite, then for any member of S there is exactly one next member, and if S finite then then for any member other than the last member, there is exactly one next member, and (3) for any member x of S, by a finite number of steps, we "arrive" at x by starting with the first member of S and going stepwise to each next member until we "arrive" at x. In contrast, that is not possible with an uncountable set. It does. The set of rational numbers is countable. Even though the standard ordering of the set of rational numbers is a dense ordering, there is a 1-1 correspondence (by "zig zagging" through numerators and denominators) between the set of rational numbers and the set of natural numbers, thus there are other orderings of the set of rational numbers that are not dense orderings but rather are isomorphic to the standard ordering of the set of natural numbers. We need to be careful in the terminology. In this set theory we're talking about, there is no such object that is called 'infinity'. Rather, 'is infinite' is an ADJECTIVE. Being infinite or not infinite is a PROPERTY of sets. Some sets are finite and some sets are infinite, but there is no object itself that we call "infinity" (well, in another context there is, but it's not with regard to cardinality, so let's set that other context aside for now). The definitions are: S is finite if and only if there is a natural number that is in 1-1 correspondence with S. S is infinite if and only if S is not finite. Now, it turns out that there are sets that are infinite but not in 1-1 correspondence with each other, while one set (call it S) is in 1-1 correspondence with a proper subset of the other (call it T). When this obtains we say that T is of greater infinity (has greater infinite cardinality) than S. Again, it is not the case that there is an object called 'infinity', rather it is that there are sets having the property of being infinite and some of them have greater infinite cardinality than others.

Unless you can find an Objectivist treatment of mathematics that goes as far as such matters as complete ordered fields and such, I don't see the point of even asking Objectivists such questions.

Objecitvism accepts infinity only as potential. So I can't even imagine in what sense an Objectivist would take the continuum hypothesis as even a meaningful statement.

In an earlier post, I meant to add the parenthetical as below: "[...] as far as I understand Objectivism, it doesn't reject the notion of sets (or if it does reject the notion of sets, I'd like to read more about that)."

Means infinite but not equinumerous with the set of natural numbers.