ON SETS NOT BELONGING TO ALGEBRAS OF SUBSETS 3

If Mi, M2 e A, then MY U M

2

, M i \ M

2

€ A.

An obvious consequence of this property is that if Mi , M

2

€ *4, then Mi fl M

2

G A. We

do not demand, unless otherwise stated, that singletons and X itself be elements of A. If

A' £ A, then A is commonly called a field2 (see [Si]). By a set that is not a member of an

algebra we mean, of course, a subset of X with that property.

Consider the following question: Can one construct, say, a finite sequence of algebras,

none of which is all of V(X) 3 , such that there is no set which is not a member of any of

these algebras? If A\, A2 are two algebras, none of which is all of V(X), then it is easy

to prove the existence of a set which is a member neither of A\ nor of A2. However, if

we consider three algebras, then the fact that none of them is all of V(X) does not yet

imply the existence of a set which is not a member of any of them. Let X be a set of

three distinct points #i, £

2

, £3. Define three algebras A\, A2, A'3 by specifying all their

members

A 3 ®Axi}Ax2,Xs},X;

A2 9 0,{x

2

},{xi,x

3

},X;

^ 3 3 0,{x

3

},{ari,x

2

},X

Obviously, there is no set that is not a member of any of these three algebras. It is also

obvious that for each of these algebras one can construct only two disjoint sets that do not

belong to it.4

In light of this construction of algebras A[, Ar2, A'3, the following question - which is

indeed the principal subject of this memoir - suggests itself quite naturally:

Question. Let A\,... , Ak, • • • he an at most countable sequence of algebras. What con-

ditions must be imposed on these algebras so that there exists a set which is not a member

of any of them?

We now formulate two results which are obvious corollaries of the two main theorems

of this memoir:

2

In the subsequent sections we will not use the term "field".

3By

V(X) we denote, as usual, the set of all subsets of X.

4

These arguments about two and three algebras are apparently not new.