# Antisubalgebras

## Idea

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.

## Definitions

Let us work in the context of universal algebra, so an algebra is a set $X$ equipped with a family of functions $f_i\colon X^{n_i} \to X$ (where each arity? $n_i$ is a cardinal number) that satisfy certain equational identities (which are irrelevant here). As usual, a subalgebra of $X$ is a subset $S$ such that $f_i(p_1,\ldots,p_{n_i}) \in S$ whenever each $p_k \in S$.

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

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

## Examples

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 $X$). 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 $A$ is an antisubalgebra and $c$ is a constant (given by an operation $X^0 \to X$ or a composite of same with other operations), then $p \ne c$ whenever $p \in A$. 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 $p$ such that $p \ne c$ for each constant $c$ (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, $A$ is an antisubgroup if $p \ne 1$ whenever $p \in A$, $p \in A$ or $q \in A$ whenever $p q \in A$, and $p \in A$ whenever $p^{-1} \in A$. An antisubgroup $A$ is normal if $p q \in A$ whenever $q p \in A$. The trivial antisubgroup is the $\ne$-complement of $\{1\}$.

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

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

Created on November 23, 2011 15:04:29 by Toby Bartels (71.31.209.116)