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This question has been called "pedantic" and I agree, it is, but I doubt many people are able to answer it.

The basic problem is that I want to be extremely clear about the sets that mathematical manipulations and operations are taking place in, I find that an author referring to sets and functions just isn't enough when there exists a deeper structure that really clarifies concepts, as you'll see below when I mention linear algebra. Basically I want help exposing the ghost-like machinery in a clear and rigorous way.

The set-theoretic definition of a function is f = (X,Y,F) where F is a subset of ordered pairs of the Cartesian product of X & Y, (i.e. F ⊆ (X x Y) a relation). This is Bourbaki's way of defining a function and he (they) call F the graph.

But isn't a function itself a relation and therefore musn't we write (X,Y,f) as the set in which the function acts? To expand this out: (X,Y,f) = (X,Y,(X,Y,F)). I've come across notation that specifies (X,Y,f) as ((X,Y),f). Here, page 35 of this .pdf file So ((X,Y),f) = ((X,Y),((X,Y),F)) would seem to make sense.

Bourbaki calls f a set & F it's graph but the notation in the .pdf file says that f would be defined in the way I've explained above, i.e. that F is a subset of XxY. The thing is that since a function f is itself a relation shouldn't it be a relation in a set, i.e. ((X,Y),f)?

Assuming that the above is the way to think about these things, how would I think of both F & f? In f = (X,Y,F), F ⊆ (X x Y) so (x,y) ∈ F or xFy, where obviously (x∈X) & (y∈Y).

How about f? I think f ⊆ (X x Y) so (x,y) ∈ f or xfy.

I don't understand how this makes sense because for the set f = (X,Y,F) Bourbaki writes f : X → Y so for (X,Y,f) I'd have to set g = (X,Y,f) and write g : X → Y. This is a weird conclusion but it seems to suggest itself.

The problem of being extremely clear about what sets you are using is particularly interesting when doing linear algebra.

The use of set-theoretic notation in linear algebra both clarifies things for me and brings up similar questions, for a vector space V I could write ((V,+),(F,+',°),•) with the clarification that:

in (V,+) we have + : V × V → V,

in (F,+',°) we have (+' : F × F → F) & ( ° : F × F → F).

In • we have (• : F × V → V) or perhaps [• : (V,+) × (F,+',°) → (V,+)]?

This notation clearly illustrates why the two operations, vector addition and scalar multiplication are used on a vector space and the axioms for each clearly jump out, i.e. (V,+) is abelian, (F,+',°) is a field and • isn't the clearest to me but I think it's similar to the way that + & ° are related in a field, i.e. "multiplication distributes over addition".

Relating all of this to the concerns I had above in a clear manner, in the set (F,+',°) it would make sense that +' is a set of the form (F,+'') where +'' is a subset of the cartesian product of F x F. Similarly with °, and in the set (V,+) you'd have something similar, also in • you'd have a crazy set ((V,+), (F,+',°), •') or including even more brackets (((V,+), (F,+',°)), •') with •' being a subset of the cartesian product of (V,+) & (F,+',°).

There is another problem when you want to give a vector space a norm, would I write ((V,+),(F,+',°),•,⊗) where ⊗ : V x V → F ? Would ⊗ itself suggest the subset ((V,+), (F,+',°), ⊗') in the manner explained above? I don't think so because ⊗' would be the set of ordered pairs (x,a) with x ∈ V and a ∈ F but since V x V → F you've got the map (x,x') ↦ a, it's quite confusing tbh and need help with this.

All this seems crazy but it also makes a lot of sense, I want to be very rigorous about what I'm doing and all of the above seems to suggest itself but it could be a lot of nonsense caused by simple confusion of a particular issue in the post , I'm thinking that (X,Y,F) implying (X,Y,f) is the culprit but again this idea clarifies things. If you read to this point thanks so much, hopefully you recognise the issue.

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closed as too localized by Andres Caicedo, George Lowther, Dmitri Pavlov, Ryan Budney, Franz Lemmermeyer Jan 26 '11 at 7:08

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There are related questions at mathoverflow.net/questions/30381/definition-of-function and also mathoverflow.net/questions/32181/ambiguity-in-ordered-tuples, whose answers seem to resolve the issues in your question. But if not, could you clarify? (By the way, what you describe as "the set-theoretic definition of a function" is not the usual definition of function in set theory.) –  Joel David Hamkins Jan 26 '11 at 0:09
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-1 for (paradoxically) lack of clarity. –  Qfwfq Jan 26 '11 at 0:34
    
Your response to Makarov tells me that my post is confusing things, I am using the Bourbaki ordered triple but calling Bourbaki's F (graph) the set of ordered pairs as specified by the relation (function). If you re-read my post I think you'll see this. Forget Bourbaki, use wikipedia: "One precise definition of a function is that it consists of an ordered triple of sets, which may be written as (X, Y, F). X is the domain of the function, Y is the codomain, and F is a set of ordered pairs" I feel clear with this, the Bourbaki idea of the graph confused me, thank you so much for spotting this! –  sponsoredwalk Jan 26 '11 at 1:26
    
If you just focus on the sets ((V,+),(F,+',°),•) in my post, is the definition [• : (V,+) × (F,+',°) → (V,+)] correct? If not then (• : F × V → V) is correct but wouldn't it be more accurate to write my set as (V,F,(V,+),(F,+',°),•)? Maybe even better would be (V,F,(V,+),(F,+'),(F,°),•)? Perhaps the best one would be (V,F,((V,V),+),((F,F),+'),((F,F),°),•)? –  sponsoredwalk Jan 26 '11 at 1:29
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Sponsoredwalk, I am sorry that your question was closed and excessively downvoted, since I find these sorts of ontological concerns about the detailed foundation of mathematics to be both interesting and important. My view, however, is that the standard set-theoretic development, meaning ZFC rather than Bourbaki, avoids the problems you mention, while enabling one to be precise about what objects really are in the way you seem to desire. I recommend any of the standard set-theory texts, such as Jech's Set Theory, or Enderton's Elements of Set Theory, among others, for a good treatment of this. –  Joel David Hamkins Jan 26 '11 at 11:27
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1 Answer 1

As far as I can see, your problem arises from trying to simultaneously use two conflicting conventions --- one whereby a function is a set of ordered pairs (also called a relation) and one whereby a function is a triple consisting of a domain, a codomain, and a relation. Choose one convention and stick to it. The first of the two links in Joel's comment leads to more information about the two conventions.

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Absolutely right, I just realised that from Joel's comments in his links, I didn't realise the big difference, thanks. –  sponsoredwalk Jan 26 '11 at 1:30
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