# Re: Can relational alegbra perform bulk operations?

Date: Wed, 30 Sep 2009 21:17:21 -0700 (PDT)

Message-ID: <babfc59c-9999-42e9-a986-b07d8bae6fa5_at_z4g2000prh.googlegroups.com>

On Oct 1, 11:25 am, Marshall <marshall.spi..._at_gmail.com> wrote:

*> On Sep 30, 7:37 pm, David BL <davi..._at_iinet.net.au> wrote:*

*>*

*>*

*>*

*> > I would have thought that set theory itself cannot be regarded as an*

*> > algebraic structure - because it is not possible to form the set of*

*> > all sets (by Russell).*

*>*

*> No, that's not why. I don't actually see any reason to exclude*

*> sets from the study of algebraic structures. Perhaps the*

*> one reason would be that sets are generally at or near the*

*> foundational bottom of the mathematical hierarchy, and*

*> we risk circularity if we treat them as algebraic structures.*

*> Normally the "is-an-element-of" predicate is the only operator*

*> and it is left undefined.*

*>*

*> Russel's paradox is actually not difficult to deal with and only*

*> comes in to play if the system admits unrestricted set*

*> comprehension. That is, if you allow the construction of*

*> sets via a description of their members. A low-cost alternative*

*> to unrestricted comprehension is separation: start with a set,*

*> and construct a new set that contains those members of the*

*> original set that satisfy some predicate. (Instead of constructing*

*> a set that contains all sets that satisfy some criteria.)*

Yes, an axiomatic system like ZFC is believed to be free from contradiction because it doesn't allow unrestricted comprehension.

However assuming an algebraic structure by definition involves a *set* of abstract elements, then I can't see how ZFC can itself be regarded as an algebraic structure.

*> > Operators like 'union', 'intersection' and*

*> > 'element-of' don't have a domain and therefore are not functions.*

*>*

*> But they do have a domain: sets. In set theories such as*

*> ZFC, there aren't any members of the domain of discourse*

*> that aren't sets.*

I'll try and show my reasoning as explicitly as possible...

Claim: Under ZFC there is no set which is defined as the

set of all sets.

Proof: Suppose there exists U such that for all x, x is an

element of U. By the axiom of separation one can take the subset R of U R = { x in U : not (x in x) } It then follows that if R is an element of R then it can't be an element of R, and if R is not an element of R then it must be an element of R. Either way this is a contradiction which proves there is no such U.

Claim: The intersection operator is not a binary function

Proof: Every function has a domain which is by definition a

set. If the intersection operator is a binary function then its domain is UxU where U is the set of all sets, but U is not a set under ZFC. It follows that the intersection operator is not a binary function.

*> > Wouldn't the RM suffer from the same limitation (well at least an*

*> > untyped version of the RM)?*

*>*

*> I don't think so. It is entirely possible to have a relational theory*

*> that only admits the existence of relations.*

*>*

*> I don't generally like the approach of having a set theory that*

*> only admits sets, but it's actually the more popular approach.*

I suspect there is no such thing as the set of all relations. Received on Wed Sep 30 2009 - 23:17:21 CDT