So far we have only defined the GK dimension for finitely generated algebras. We now extend the definition to all algebras.

** Definition. **Let be a -algebra. We define where runs over all finitely generated -subalgebras of

**Fact 1**. Let be a -algebra, a -subalgebra of and let be a two-sided ideal of Then

*Proof*. Let and be any finitely generated subalgebras of and respectively. Let and be frame of and respectively, which both contain 1. Then is also a frame of $A_0$ and so we may assume that Thus for all and so Fixing and taking supremum over all finitely generated subalgebras of gives us Now taking supremum over all finitely generated subalgebras of gives us For the second inequality, let be a finitely generated subalgebra of Let be the natural homomorphism. Then is a finitely generated subalgebra of Let be a frame of Then is a frame of and clearly for all Thus

Taking supremum over all finitely generated subalgebras gives us

**Fact 2**. Let be a -algebra. Then if and only if is locally finite, i.e. every finitely generated subalgebra of is finite dimensional, as a vector space, over

*Proof*. Suppose first that is locally finite and let be a finitely generated subalgebra of Then is finite dimensional over and so is a frame of Clearly and thus

because does not depend on Conversely, suppose that and let be a finitely generated subalgebra of Let be a frame of and suppose for now that for all Then and thus Hence and in general Therefore

which is absurd. So our assumption that for all is false. Hence for some integer and so Thus is finite dimensional and so is locally finite.

**Fact 3**. Let be a -algebra. If then

*Proof*. So there exists a finitely generated -subalgebra of such that Let be a frame of If for some integer then

and so is finite dimensional. But then by Fact 2, which is false. Thus and hence for all Therefore

and so

So if then there is no algebra with We will see later that for every integer there exists an algebra such that