antisubalgebra

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 $X$ 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 $S$.)

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)