# Clarifying Notation (Sets & Functions)

The basic problem is that I want to be extremely clear about the sets that mathematical manipulations and operations are taking place in, I am hoping for someone who really understands this to read what I've written closely and point out what is getting me all mixed up, though of course reading &/or responding isn't mandatory (lol) - but it is a long post even though it's dealing with just one idea.

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.

-
In model theory it is customary that all operations act on the same structure. This is of course a problem when studying $F$-vector spaces since scalar multiplication is not an operation of this kind. We solve it by, instead of thinking of it as a single operation, we think of it as many: We have $f_r:V\to V$ for each $r\in F$ with $f_r(v)=rv$ for each $v$. –  Andres Caicedo Jan 25 '11 at 22:22
Actually, to be honest, this smells of excessive pedantry, which is confusing at times. Sometimes, it's best not to worry about how things are represented as sets and take objects as having certain attributes attached to them but not necessarily part of them. For example, a function has a domain and a codomain, but these are not essential features of the function; rather what is essential is its graph. Similarly, a field has a number of operations attached, and these operations are very important, but it is the elements we are considering when we have a homomorphism of fields. –  Zhen Lin Jan 25 '11 at 22:27
Thanks for reading. Excessive pedantry? Yeah you'd be right :) But let me clarify that we could be pedantic in a lot areas of math if we wanted to & could easily pull it out because we understand it, I don't understand the pedantry here so I don't fully understand the idea, I want to clarify all of the above so that I can ignore it but be safe in the knowledge you know. Unfortunately I don't know anything about model theory but will come back to this thread if I ever get to look at it and try to understand why your post makes sense :p –  user6248 Jan 25 '11 at 23:13
Note that you are using the same symbol for addition in you vectorspace and your field! That's totally confusing! Sorry about the irony :-) edit well actually, it appears you're not, I hadn't noticed the ' at first. Well, what can I say, you're worse than my irony :-) [/edit] –  Myself Jan 26 '11 at 2:43
To clarify that I don't mean "worse" in a pejorative way: what's important in math, in my opinion, is the transfer of ideas from mind to mind. Obviously, you need some notation or expression, but as long as other math-minded human beings understand you, you're all right. So for most purposes: just name the vectorspace V or V(F) and everyone will know that what you really mean is ((V,+),(F,+',·'),·). –  Myself Jan 26 '11 at 2:50

I first restate just the parts pertaining to your first question: if $X$ and $Y$ are sets, and $F \subseteq X \times Y$ is a relation with the properties one wants of a function (namely: for all $x \in X$ there is $y \in Y$ with $(x,y) \in F$, and for all $x \in X$ and $y_1, y_2 \in Y$ we have that $(x,y_1) \in F$ and $(x, y_2) \in F$ implies $y_1 = y_2$), some people do indeed view the triple $f = (X,Y,F)$ as being the the function, and define a function to be such a triple.

You then ask: "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?" The answer to the first part of this question--- "isn't a function itself a relation"--- meaning, isn't the function itself the set $F$--- is no (and hence the second half of the question is dispensed with entirely). In the definition you have stated, a function is not a relation, but an ordered triple whose entries are a set $X$, then another set $Y$, and then finally a relation between $X$ and $Y$ (ie, a subset of $X \times Y$) with the properties one wants of a function.

To get a sense of the purpose of this: think of the notion of equality of functions that it produces: to say that $(X,Y,F)$ and $(X',Y',F')$ are equal is to say more than just $F = F'$, but also that $X=X'$ and $Y = Y'$. The purpose of this definition of a function is to make explicit the roles of the sets $X$ and $Y$ in the specification of a function. Note that if you are given a function $(X,Y,F)$, you can infer what $X$ is from $F$ alone ($F$ is a set of ordered pairs, and by the defining properties of a function, $X$ is nothing more than the set of first entries of elements of $F$) but you cannot infer what $Y$ is from $F$ alone.

An example might help: let $X = \{0,1\}$ and let $Y = X$ and let $Z$ denote the subset $\{0\}$ of $X$.

Let $f: X \to Z$ denote the only possible function (namely the one whose rule is given by $f(x) = 0$ for both possible values of $x$). Let $g: X \to Y$ denote the function given by $g(0) = 0$ and $g(1) = 0$. If you think about it for a moment you see that the rules of $f$ and $g$ are both encoded by the same set of ordered pairs, namely $F = \{(0,0), (1,0)\}$. The only difference between the two is that in one case we are regarding $Z$ as the codomain, and in the other we are regarding the larger set $Y$ as the codomain. With the above set-theoretic definition of "function", $f$ and $g$ are different functions, because one is the set $(X,X,F)$ and the other is the set $(X,Z,F)$, and these two ordered triples differ in their second coordinate, so they are different ordered triples, so the functions (by this definition) are different.

If you think of the set $F$ as "being" the function, then you cannot distinguish between $f$ and $g$ (and more generally, the concept of "codomain" becomes hazy). Of course, for many applications of the function concept there is no conceivable reason to distinguish between $f$ and $g$--- but most mathematicians, I think, would at least want the option of seeing $f$ and $g$ as different, and if you want to formalize functions within set theory, then, the definition above allows you to do so. (Some people do take as a definition that a function from $X$ to $Y$ is a subset of $X \times Y$ with the defining properties of a function; it is a perfectly good definition but it does not allow one to distinguish between $f$ and $g$ above, and the concept of codomain is not encoded in this definition, although if you are not formally minded, you might not notice.)

I want to reiterate: people generally do not think of functions as being ordered triples, consisting of the domain set, the codomain set, and then the relation giving the rule--- it's very common to think of the function as just being the rule. But if you want to formalize the notion, you quickly find that equating a function with the relation that determines its rule leaves a useful distinction (namely the concept of codomain, or the set a function is "to") out. So the purpose of the Bourbaki definition is to explicitly build it in: a function is not just a relation with the defining properties one wants of a function, but it is an explicit choice of domain and codomain together with this data. (As I remarked earlier, the domain could actually be inferred from the relation, so this is somewhat "redundant"--- but the object of this definition is merely to formalize a way of viewing functions within set theory, not to do so in a "minimal" way.)

Hopefully the rest of your questions are slightly cleared up by this.

-
I didn't realise that a triple, as I defined f = (X,Y,F) is Bourbaki notation and it is DIFFERENT to the notation I used when I described (X,Y,f). When I spoke about (X,Y,f) I was using the idea of "f" as a set of ordered pairs, but with f = (X,Y,F) I was using Bourbaki's notation of a triple but CONFUSING things by giving Bourbaki notation the idea of ordered pairs. Basically it all comes from confusing the uses of notation. I want to forget about Bourbaki's idea of triples and just use the idea of sets with (X,Y,F) where F is not the graph but is the set of ordered pairs defined... –  user6248 Jan 26 '11 at 2:58
... by a relation, be it a function, a binary relation, an ordering relation etc... It just seems far clearer to me. Honestly I just didn't recognise there was such a difference. If you don't mind having a look at 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),°),•)? –  user6248 Jan 26 '11 at 3:00
Wow, I've re-read your post and you say that in my definition of a function (in this thread) a function is NOT a relation. I have read that a function is simply a special type of relation. I will just go and read Suppes Set Theory and another set theory book properly & get the definitions correct, I am clearly mixing things from different sources together, what with mixing Bourbaki and other things etc... Thanks a lot for the help –  user6248 Jan 26 '11 at 3:16
@sponsoredwalk: There is a slight disconnect, yes. But essentially, you can think of a function as either an ordered triple $f=(X,Y,R)$, where $X$ and $Y$ are sets, $R$ is a subset of $X\times Y$ that satisfies certain conditions, etc. Or simply as a relation $f\subseteq X\times Y$ that satisfies certain conditions. The former is the usually the most useful notion for several reasons (eg, it allows you to talk about surjectivity), so we only think about the function as a relation when we want to ignore those properties; in that case, we "forget" about the $X$ and $Y$ and just keep $R$. –  Arturo Magidin Jan 26 '11 at 4:51
@sponsoredwalk: And as anon is saying: different people may give slightly different definitions of functions. Yes, some people say (or even define) "function" as a "special kind of relation"; but Bourbaki is most emphatically not doing that. You shouldn't bring in the definition that someone else is giving (especially if that someone else is giving it informally) and stick it into the middle of the formal definition that Bourbaki is giving. That's like trying to stick a recipe from Better Homes and Gardens into the middle of a recipe from American's Test Kitchen. You'll just get a mess. –  Arturo Magidin Jan 26 '11 at 4:58