Terminology: Semigroups, only their "binary operations" aren't closed.

Question

Motivation:

Consider $\mathcal{X}=(X, +)$, where $X=\{-1, 0, 1\}$ and $+$ is standard addition. Then $\mathcal{X}$ is associative (where defined) but not closed.

NB: There is an identity element in $X$ and inverses exist in $X$ all with respect to $+$.

This example is taken from here.

The Question:

This question seems difficult to pose due to certain subtleties so, to make life easier, here's the rough idea first.

What d'you call a "magma" that's associative but not closed?

An attempt at refining the question:

What do you call the mathematical objects $\mathcal{S}=(S, T, \ast)$ for which $S$ is a set and $\ast$ is some function with domain $S\times S$ and codomain some set $T$ with $S\subset T$, such that

• for all $s,t,u\in S$ we have $$s\ast (t\ast u)=(s\ast t)\ast u$$ whenever $t\ast u, s\ast t\in S$ (or $T$ if that's necessary to keep the question in spirit) and

• there exist $x, y\in S$ such that $x\ast y\in T\setminus S$?

(Please disregard this attempt if it complicates the idea of the question needlessly.)

Thoughts:

I'm not sure whether naming these things is necessary. I'm interested in them out of curiosity. Whether the question even makes sense, I don't know.

Are they simply subsets of semigroups?

I made sure to say function and not binary operation above, since the latter implies closure by definition.

Answer 1

You might have put together two different questions due to some subtleties. As I see it, you have two main possibilities. Starting from the one you wrote explicitly:

1. A subset $S$ of a magma $(T,\cdot)$ for which $$ (s\cdot t)\cdot u = s\cdot (t\cdot u) \qquad \forall s,t,u\in S $$

Note that this does not raise any doubts about existance of the products as the equality holds in $T$. I'm not sure this has a name but a fair bet would be something like: " $S$ is an associative subset of the magma $T$ " or maybe " $T$ is locally associative over the subset $S$ ".

2. A set $S$ with a partial operation $\star: S\times S \rightarrow S$ for which, for all $s,t,u\in S$, $$ \text{ if } s\star t\text{ , }t\star u\text{ , }(s\star t)\star u \text{ , }s\star(t\star u)\in S \qquad \text{then } (s\star t)\star u = s\star(t\star u) $$ (Meaning that: if all the products involved exist, then $\star$ is associative)

   This seems to be quite similar to the notion of a partial semigroup.

Using this view you could also see $S$ in definition 1. as a "partial subsemigroup" of the magma $T$ (similarly to what is done for the notion of a subgroup of a semigroup).

Answer 2

It is not really an answer to your question, but I could not enter the picture in a comment. The point that if you "duplicate" the zero into $0_A$ and $0_B$, you could define a category which is very close to your structure:

$\hskip 15pt$

Now you have an associative structure with $1 + (-1) = 0_A$ and $(-1) + 1 = 0_B$.

Answer 3

Such a structure is not always a subset of a semigroup.

Take the following structure:

ab = ba = c

bc = cb = a

ca = ac = b

aa = bb = cc = e

ee = f

This is basically the Klein group, except that I do not define ea, eb, ec, ef and that I willfully set ee different from e.

Now, this is clearly associative, because the only problem e cannot interact with the rest due to lack of definition. But when I want to expand the structure to a hopefully closed associative structure I get

$ae = a (bb) = (ab)b = cb =a $, therefore $f= ee = (aa)e= a(ae) = aa=e$ which is a contradiction.