4

One main reason why the law of excluded middle can fail is that some operations are simply undefined in some contexts. This doesn't even mean that they are undefinable in principle, it just means that they are not defined in the current context. If I have a real computer and a real bound on the time I'm willing to wait, then some programs will simply fail to give a definitive answer for some inputs under these constraints. Forcing the answer arbitrarily to some definite value in such cases will often only obscure the true structure of the problem. This doesn't mean that logical inconsistencies will arise if we insist to do so, but we may be trading a finite context for an infinite context just to avoid undefined operations.

There is good evidence that we can avoid partially undefined operations in most practically relevant cases, if we are willing to cope with infinity. But what about the opposite? Can we avoid infinity in most practically relevant cases, if we are willing to cope with partially undefined operations?

| improve this question | |
  • When you say "the law of the excluded middle can fail" are you referring to propositions which are not decidable. For example, the truth of The Continuum Hypothesis of Set Theory is not decidable. Also, can you give an example of an "undefined operation". It is not clear what you mean here. – Nick Sep 14 '14 at 18:17
  • @NickR Yes and no. Yes, because the continuum hypothesis is not decidable in the context of ZFC, but may be decidable the context of other set theories. No, because I was thinking of much simpler and mundane examples. I was thinking about a family of (totally defined) semi-norms on a vector space, and wondered how to nicely represent the topology on the dual-space of linear function continuous with respect to the family of semi-norms. I guess that a family of partially defined norms could be such a representation. I also had partial symmetries and inverse semigroups in mind. – Thomas Klimpel Sep 14 '14 at 18:39
  • Wondering how to nicely represent a topology on any space, whether using normed or seminormed measures, does not seem to raise any issues to do with the Law of the Excluded Middle or partially undefined operations. I guess I am missing something here. – Nick Sep 14 '14 at 18:50
  • @NickR Well, you are right that there is something missing. I added an explanation of the missing link now, but if you read it you will probably understand why I initially decided against including it. – Thomas Klimpel Sep 14 '14 at 19:21
  • 2
    Completion by adding a point at infinity is the most elegant solution. Completion using partially defined operations would seem rather arbitrary and lacks the universality of a point at infinity. What operations are required would depend on the context. Removing the point at infinity removes the closure of the semilattice and so makes it clearly distinguishable. I still miss the link with excluded middle. Love the last two sentences! I'm sure they make perfect sense to you, but they are rather baffling to me. – Nick Sep 14 '14 at 20:00
6

Your question trades on a mistake. Truth and falsity are properties of sentences. The law of the excluded middle therefore is a rule about sentences. An undefined operation isn't a sentence hence it is neither true or false, but this doesn't imply that it has some kind of third truth value. It simply lacks a truth value all together. My toaster also lacks a truth value, in just the same sense.

This is why in logic, we don't regard formula with free variables as sentences. To see why, think of, "I gave the ball to . . . " This is not a sentence---and so neither true or false---until we fill in the dots with something to be the direct object of the verb.

A function that is undefined over some range is like that incomplete sentence. The function isn't taking the values in that range and spitting out "Undefined". The function isn't taking values from that range at all. (We program our calculators and software to spit out "undefined" to warn us that we can't input that value to that function.)

| improve this answer | |
  • There are operations in propositional (and predicate) logic, most notably "and", "or", and "implies". Or take "negation" instead of "implies", if you prefer. And these operations don't necessarily operate only on the two element set {"true", "false"}, but also on a Boolean algebra (or Heyting algebra in case of intuitionistic logic). A formula with free variables is not a sentence, because a sentence is defined to be a formula with no free variables. But a formula can still be interpreted as a proposition in a suitable Boolean algebra. – Thomas Klimpel Sep 14 '14 at 15:53
  • Regarding your clarification: "The function isn't taking the values in that range and spitting out "Undefined". The function isn't taking values from that range at all." This is already a very good first step towards accepting partially defined functions, and is elaborated admirably well in category theory. But it isn't the whole story about partially (un)defined functions. I was wondering about inverse-semigroups while writing the question, especially Robin Cocket's answer to mathoverflow.net/questions/179711/…. – Thomas Klimpel Sep 14 '14 at 16:01
  • I'm out of my league in answering questions about abstract algebra, but it seems to me that the cases you mention there represent a semantic point. If you want to call a formula with free variables a proposition for the purpose of some algebra, that's fine. But then I'd just say that again, for the same reasons I give in my answers that these propositions are still not sentences and hence the principle of excluded middle won't apply to them. I don't know anything about category theory (someday I hope to get time to learn!). Sorry to not be more help! – user5172 Sep 14 '14 at 16:50
  • In classical logic, "P(x) or not P(x)" is unconditionally true, i.e. the principle of excluded middle applies to formulas with free variables. This question is not so much about algebra, but more about why mathematicians will happily explain to you that infinity is much more complicated than just adding an additional symbol representing "infinity", but dismiss partially (un)defined operations as if adding an additional symbol representing "undefined" would be all that is required. – Thomas Klimpel Sep 14 '14 at 17:32
  • @klimpel: Inverse semigroups are defined by the two symmetric axioms: x=yxy and y=xyx. From a categorical adjunction, F left adjoint to G we can deduce F=GFG and G=FGF; hence its possible to see adjunctions, in a sense, as a categorification of the notion of an Inverse Group; and more useful: for example Galois theory is about characterising the equivalence within an adjunction between posets; and which is so important it get its own name - a Galois connection. – Mozibur Ullah Sep 15 '14 at 8:26
2

Mathematicians have only become fond of infinity after Cantor; this is a relatively recent innovation; still one can say that the mathematical infinite differs in important ways from the philosophical infinite. When Spinoza for example talks about an infinite substance its quite clear that there is no fundamental relation between this and any conception of cardinality.

It was Aristotle that smuggled in the LEM into philosophy; and it has remained there. It was Brouwer in mathematics and Hegel in Philosophy that reintroduced it; for Brouwer it led to intuitionism; for Hegel it led to the dialectic. Given the apparatus of formal logic, it is usually taken that a contradiction will allow one to prove anything: the principle of explosion - ex falso quodlibet; a question then introduces itself: can one deny this principle by changing the formal structure of the logic? This is indeed possible as the Peruvian Philosopher Quesado showed.

Brouwers Intuitionism also known as Constructivism was put on a formal basis by his student Heyting. There is a correspondance between Boolean algebras and propositional logic; similarly there is one between Heyting Algebra and intuitionistic logic.

Formally, Boolean algebras are exactly bounded distributed lattices that are complemented. Their models (ie semantics) are fields of sets (set theoretically) or Venn diagrams (geometrically).

Correspondingly, Heyting algebras are exactly bounded distributive lattice with an additional implication operation which satisfies: x and a is 'less true' then b iff a is 'less true' than x implies b. Further their models are toposes (set theoretically) or topological spaces (geometrically).

Heyting algebras generalise Boolean algebras since the ones that satisfy the LEM are exactly the Boolean Algebras. In their set theoretic avatar, we can say toposes are set theory when it loses the notion of an element (which may be difficult to comprehend - the canonical book here is Lawveres Sets for Mathematicans); and geometrically when topological spaces lose their points ( this may be easier to understand as typically we consider a space of points together with its space of open sets; one then need only 'forget' the points).

Note: All operations are fully defined.

Philosophically: Contra Leibniz, taking properties of points not to include their location; we can consider the atomic point to be a bare point; one without any distinguishing features. This has been the norm since Descarte; forgetting points amounts to, in one sense, of having atomic points with structure, this is more inline with the Epicurean notion of an atom; and has had a contemporary impact in Physics: String Theory - a string is an atomic point with structure; secondly, it amounts to a relational view of space as there are no 'absolute points' to anchor to; this is a synthesis of Leibnizs view and Aristotle who denied the existence of points in the continuum and relied on cohesion (ie topology in modern language).

| improve this answer | |
  • A nicely written answer, even if it doesn't mention the words partial and undefined. Perhaps this is even intentional, in order to illustrate that mathematics is indeed dismissive towards partially (un)defined operations. – Thomas Klimpel Sep 15 '14 at 7:51
  • Thanks. It wasn't intentional. I do mention defined rather than undefined. Implicitly category theory, thus topos theory deals directly with 'partially (un)defined' operations; its simply because of the way its expressed this isn't immediately apparent - composition through arrows; if one does indeed take a category to be a kind of monoid on its set of objects - composition as a binary operation, then indeed the operation of composition is only partially defined. Perhaps this should be mentioned in he answer as it directly comments on your question. – Mozibur Ullah Sep 15 '14 at 8:10
  • However, category theorists generally use the first paradigm, because in that presentation all operations are in a sense defined; and if we take categories foundationally then notably a monoid is then just a one-object category. – Mozibur Ullah Sep 15 '14 at 8:13
  • From this perspective, I think you are right; that mathematicians do not like partially defined operations; and when they find natural examples of one - they legislate them away! I also note that you do mention Heyting algebras... – Mozibur Ullah Sep 15 '14 at 8:16
  • Yes, category theory is the noble exception that is honest and straight about partially defined functions. That's why I wondered whether you intentionally omitted it, and instead only cited an advanced topic like topos theory, which is probably "way over the head" for most mathematicians and philosophers. – Thomas Klimpel Sep 15 '14 at 9:27
0

Perhaps a better example of a partially undefined operation would be 0^0 (zero to the power of zero). It is possible to leave it undefined (like your high school math teachers taught you) without much in the way of consequences. (See Oh, the ambiguity! at my math blog.) So, we can have infinity and partially undefined operations coexisting quite happily.

| improve this answer | |
  • That's a very good example, in a certain sense. Without context, 0^0 is simply undefined. If the context are polynomials, then x^0=1 for x=0. If the context is a limit in analysis, then the value of 0^0 can be anything, dependent on the context. And even so 0^0 is undefined, we still have (0^0)*0 = 0. The only problem with this example is that it doesn't show what can be won by embracing partially (un)defined operations. Mathematicians already know why they love infinity, so the task would be to give them good reasons to love partially (un)defined operations. – Thomas Klimpel Sep 16 '14 at 18:39
  • At my blog, I make the case that even in the natural numbers, 0^0 should be left undefined. Yes, there is a very vocal faction that argues for 0^0=1, but it seems you have to go outside of purely number theoretic considerations to justify this. As far as I can tell, it cannot be justified starting only from Peano's axioms and the notion of exponentiation as merely repeated multiplication. – Dan Christensen Sep 16 '14 at 18:47
  • 1
    [you man consider me member of that "vocal faction"] The argument you present in the blog is a strawman argument: you present one limited method of extending definitions to borderline cases, and conclude that it cannot give 0^0 a unique value. Indeed for natural number arithmetic only one can live with leaving 0^0 undefined. But other considerations can be used and they invariably lead to making 0^0=1 the best convention (never anything else). For instance for polynomials in X it is very practical to have their general form to be finite sums of c_iX^i for certain i in N... – Marc van Leeuwen Apr 11 '15 at 11:09
  • 1
    ... By the same principles you use, you can see that arithmetic rules leave no choice but to equate X^0 with the unit of multiplication, usually written 1. The same is true for P(X)^0 for any nonzero polynomial P(X). But for polynomials an important principle is evaluation: systematically replacing X by some constant value is compatible with all arithmetic operations (this is behind all polynomial arithmetic). But for that to hold for X^0 or P(X)^0, one is compelled to define c^0=1 for all constants c, with no exception for c=0. It is in this sense that the binomial formula requires 0^0=1. – Marc van Leeuwen Apr 11 '15 at 11:09
  • 1
    By the way your blog says that 0^0 could be defined to have one of infinitely many values. But the arithmetic rule a^b*a^c=a^(b+c) would give (0^0)*(0^0)=0^0, and this already rules out all candidates values except 0 and 1. – Marc van Leeuwen Apr 11 '15 at 11:16
0

I don't know how well received/welcome this will be, but I am a mathematician and would like to share my opinion.

When it comes to practicing mathematics, there is no "should," there is only what is. We worry about whether something exists or does not, and if it does, then it does. If a function is partially undefined, then it is partially undefined. Whether or not this makes us feel good has no bearing on whether or not it is partially undefined. If you decide to define it in places where it is undefined, this does not make the original function any more defined; you have constructed a new function.

In truth we have moved far beyond worrying about the existence of pathological objects and have come to accept that mathematics sometimes produces terrible things. This question may have provoked a lot of interest from mathematicians in the time of Brouwer, but modern mathematicians are well-initiated and can remain unmoved in the face of paradox or the misbehavior of the things we have created. When we write, we make a definition and stick to it. Someone reading the paper may not like the definition, and maybe they'll send a nasty letter about it. More likely, when they cite the paper they will simply use different definitions and translate them into terms they prefer.

| improve this answer | |
  • I would normally upvote such an answer. However, as this is your first post here, it would probably set wrong expectations. If you look at the question marks in the question, can you really say that your answer addresses any of these explicit questions? Your answer doesn't even mention the words "context" or "infinite". On the other hand, you emphasize the words "should" and "existence", but I don't really know why, i.e. I see nobody else here even mentioning these words. – Thomas Klimpel Jan 23 '15 at 15:51
  • @ThomasKlimpel well I appreciate at least your not downvoting it. – Matt Samuel Jan 23 '15 at 15:52
0

Maybe I have to write an answer myself. But this is not really an answer, it is more the background which caused me to ask this question. But it answers some of the objections brought up in comments, even if it could and should be probably elaborated much more thoroughly. (Sorry for misusing the "community answer" feature in that way.)

After reading simultaneously in a number of German introductory quantum mechanics books, I suddenly noticed that not a single one had the courage to say that a Hermitian operator is in general only partially defined on the corresponding Hilbert space. The theory of rigged Hilbert spaces seems simple enough to me, so I started to wonder where this overwhelming fear against partially defined operations comes from. It's a bit ironic too, considering that quantum mechanic itself is so fond of some interpretation which make claims about some sort of fundamental undefinedness. The fear seems to come from mathematics itself, i.e. mathematics is unbelievably dismissive towards partially (un)defined operations. Then I thought a bit about whether the theory of rigged Hilbert spaces could me made even simpler, if the corresponding mathematics embraced partial undefinedness even more fully, and that's the context where families of semi-norms, partially defined norms and partially defined semilattices seemed to paint a simple and beautiful picture.

I had struggled before how to best get the point across that semilattices are different from lattices, and that turning every semilattice into a lattice by "adjoining" the "missing" elements doesn't make live easier. I'm still working on a theory that has a generalized implication operation in addition to meet and join, that allows to "rescue" the original semilattice structure, even if it has been compromised by adding "unnecessary" elements.

If we have an axiom of choice like in ZFC, then every partially ordered structure can be embedded into a complete Boolean lattice. Hence it is pretty clear why we can get away with the LEM, if we really want to. But then we also have to accept all the consequences, i.e. infinite structures with unbelievably big cardinality. But as already the fact that a logic can be seen as a preordered structure where the derivability is the order relation between different propositions or formulas seems to be unbelievably hard to swallow, this logic based approach to these question is likely to be unconvincing.

Both infinity and undefined operations have much more mundane applications where they make life significantly easier. If we adjoin the points at infinity to an affine space, we get a projective space with much nicer properties than the corresponding affine space. But if we adjoin infinity to the possible values of a norm, we don't really do ourself a favor. If we instead allow a norm to be only partially defined, we seem to really get something nice in return, similar to the projective spaces, but of course different. The poorest current "loser" from the fear against partially (un)defined operations are probably inverse semigroups, and semilattices are a special case of inverse semigroups (i.e. idempotent and commutative).

One nice example here would be how an inverse semigroup can represent an equivalence relation with essentially the same amount of information as the corresponding binary relation, while a group requires exponentially more (artificial) elements, if it doesn't want to add structure to the underlying set that wasn't there before. This would also partly explain what I mean by trading a finite context for an infinite one. If I require exponentially more (artificial) elements than the natural numbers, than I have to cope with the continuum. The natural numbers are still a finite context in a certain sense, but the continuum is definitively an infinite context.

The connection between the law of excluded middle and partially defined operations is probably hard to appreciate without some background. Sadly, this background might be slightly too mathematical for a philosophy site, but let me add it nevertheless. A Boolean algebra is a distributive lattice with an involutive negate operation. A lattice is a partially ordered set where any two elements have a greatest lower bound (meet) a least upper bound (join). A semilattice has only one of these two operations guaranteed to be defined everywhere. A semilattice can also be defined as an idempotent commutative inverse semigroup. And an inverse semigroup can be represented as a subsemigroup of the partial one-one transformations of a set.

The connection to the law of excluded middle is that often the structure of a complete semilattice (among propositions, formulas or sentences) arises naturally, but a complete semilattice is nearly indistinguishable from a complete lattice. But you can define an implication operation ("from A follows B" for a Heyting algebra, but in general rather "from A, B, C, ... follows Z") which allows to distinguish "and" and "or" (or "meet" and "join") properly. This is important, because the symmetry between "and" and "or" so typical for classical logic is often just an illusion caused by the fact that complete semilattices are so hard to distinguish from complete lattices. But if the maximal element (infinity) is removed from the complete semilattice, then it becomes much easier to distinguish it from a complete lattice.

| improve this answer | |
  • This makes sense. A quantum mechanics texbook is by no means a mathematics textbook. Physicists have plenty of qualms with basic mathematical precision (they seem to think it's beneath them). A mathematician would specify in advance if something is not defined somewhere. – Matt Samuel Jan 23 '15 at 2:38
  • It may be fear; or lack of courage but it's more likely that physical thinking is different from mathematics. After all the Feynman Path Integral has been used successfully without it being well-defined. – Mozibur Ullah Apr 10 '15 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.