# What are the domain and range of this function?

1. Jan 4, 2007

### StatusX

Say I define a function f by sending a well ordered set X to the least element in X. For example, if N is the natural numbers with the usual ordering, then f(N)=1. This function seems to be well defined. I guess its domain is the set of well ordered sets, but this sounds strange. Whats even stranger is the range. What could this possibly be? The union of all well ordered sets?

2. Jan 4, 2007

### d_leet

At the very least you have the power set of the natural numbers in the domain, but I don't really know enough to say if there are other sets in that union or not.

3. Jan 4, 2007

### AKG

It's domain is the class {(X,<) | X is a set and < is a well-order on X}. This is a class, not a set. The range is the universe (everything), which is also a class, not a set.

4. Jan 5, 2007

### StatusX

So what exactly is the difference between classes and sets? Apparently you can define functions between classes as with sets. Is there anything you can't do with them?

5. Jan 5, 2007

### matt grime

How can you define a function between classes? The definition of a function is that its domain is a set. So what you've defined is not an honest to goodness function.

It takes some care, I imagine, to actually rigorously define a generalization of function to the case of classes.

Well, you might - I care about categories so we tend to sidestep the issue.

6. Jan 5, 2007

### AKG

Every set is a class, but not vice versa. In set theory, sets exists always but classes, in general, do not. For example, the class of sets that aren't members of themselves does not exist as a set, for it would lead to contradictions. In set theory functions are, in general, classes of ordered pairs (you're probably familiar with this). "Reasonable" functions are not only classes of ordered pairs, but sets of ordered pairs. The function you're describing is merely a class of ordered pairs. If X denotes a class, you can do most things with X that you could do with a set. You can make true statements like $x \in X$ for some set x, you can talk about the union of X (that is, the union of the sets in X), the intersection of X (that is, the intersection of the sets in X), subclasses of X, the powerset of X, etc. but there's no guarantee that any of these classes you get are sets. The only thing you can't do is say $(\exists y)(y = X)$.

7. Jan 5, 2007

### AKG

That's not what it says in my text. My text simply says that f is a function iff everything in f is an ordered pair, and if (x,y) and (x,z) are ordered pairs in f, then y=z.

8. Jan 6, 2007

### matt grime

And this requires you to define ordered pairs of classes. My point was that I would imagine there is some care required to do properly all of these things that the OP would normally define as set constructions: ordered pairs, element comparison and so on.

Last edited: Jan 6, 2007