In constructive mathematics, we often do algebra by equipping an algebra with a tight apartness (and requiring the algebraic operations to be strongly extensional). In this context, it is convenient to replace subalgebras with anti-subalgebras, which classically are simply the complements of subalgebras.


Let us work in the context of universal algebra, so an algebra is a set XX equipped with a family of functions f i:X n iXf_i\colon X^{n_i} \to X (where each arity? n in_i is a cardinal number) that satisfy certain equational identities (which are irrelevant here). As usual, a subalgebra of XX is a subset SS such that f i(p 1,,p n i)Sf_i(p_1,\ldots,p_{n_i}) \in S whenever each p kSp_k \in S.

Now we require SS to have a tight apartness \ne, which induces a tight apartness on each X n iX^{n_i} (via existential quantification), and we require the operations f if_i to be strongly extensional. (There is no need to require that any arity n in_i be finite or that there be finitely many f if_i.)

A subset AA of XX is open (relative to \ne) if pAp \in A or pqp \ne q whenever qAq \in A. An antisubalgebra of XX is an open subset AA such that some p kAp_k \in A whenever f i(p 1,,p n i)Af_i(p_1,\ldots,p_{n_i}) \in A. By taking the contrapositive?, we see that the complement of AA is a subalgebra SS; then AA may be recovered as the \ne-complement of SS (the set of those pp such that pqp \ne q whenever qSq \in S). However, we cannot start with an arbitrary subalgebra SS and get an antisubalgebra AA in this way, as we cannot (in general) prove openness. (We can take the antisubalgebra generated by the \ne-complement of SS, as described below, but its complement will generally only be a superset of SS.)


The empty subset of any algebra is an antisubalgebra, the empty antisubalgebra or improper antisubalgebra, whose complement is the improper subalgebra (which is all of XX). An antisubalgebra is proper if it is inhabited; the ability to have a positive definition of when an antisubalgebra is proper is a significant motivation for the concept.

If AA is an antisubalgebra and cc is a constant (given by an operation X 0XX^0 \to X or a composite of same with other operations), then pcp \ne c whenever pAp \in A. If there are only Kuratowski-finitely many constants (which is needed to prove openness), we define the trivial antisubalgebra to be the subset of those elements pp such that pcp \ne c for each constant cc (the \ne-complement of the trivial subalgebra?). In general, we may also take the trivial antisubalgebra to be the union of all antisubalgebras, although this is not predicative.

Instead of subgroups, use antisubgroups. In detail, AA is an antisubgroup if p1p \ne 1 whenever pAp \in A, pAp \in A or qAq \in A whenever pqAp q \in A, and pAp \in A whenever p 1Ap^{-1} \in A. An antisubgroup AA is normal if pqAp q \in A whenever qpAq p \in A. The trivial antisubgroup is the \ne-complement of {1}\{1\}.

Instead of ideals (of commutative rings), use antiideals (and we also have left and right antiideals of general rings). In detail, AA is an antiideal if p0p \ne 0 whenever pAp \in A, pAp \in A or qAq \in A whenever p+qAp + q \in A, and pAp \in A whenever pqAp q \in A. It follows that an antiideal AA is proper iff 1A1 \in A. AA is prime if it is proper and pqAp q \in A whenever pAp \in A and qAq \in A; AA is minimal if it is proper and, for each pAp \in A, for some qq, for each rAr \in A, pq+r1p q + r \ne 1 (which is constructively stronger than being prime and minimal among proper ideals). The trivial? antiideal is the \ne-complement of {0}\{0\}.

Given any subset BB of XX, the antisubalgebra generated by BB is the union of all antisubalgebras contained in BB. (This construction, unlike those above, is not predicative.)

Created on November 23, 2011 15:04:29 by Toby Bartels (