Any example of this kind of algebraic structure?

Question

Can someone tell me if there is some natural algebraic structure given by an infinite (co)product that has no neutral element ( where a neutral element is one that witnesses $\exists u \forall a [u * a = a = a * u]$), but such that every "finite reduction" (somewhat intuitive, but I could try and come up with something formal if pressed) actually has a (not necessarily unique, not necessarily even 2-sided if you want) neutral element?

I am thinking something like this:

https://en.wikipedia.org/wiki/Rng_(algebra)#Example:_Quinary_sequences

but I cannot see why the infinite product does not have neutral element the (infinite) sequence of all $1$'s, and, furthermore, it seems a bit specific to $5$ arbitrarily, so I would like something more general. Also I don't know what it means when it talks about the supposed "identity/idempotent" element (it seems to switch between sentences) in the relevant "finite reductions" (again, my own terminology) is actually the neutral element (with respect to either multiplication or addition in their rng).

Answer 1

You're right that they chose that ring a bit arbitrarily. It has nothing to do with the property you're interested in.

If you take any infinite collection of nonzero rings (with identity of course) $\{R_i\mid i\in I\}$ then $R=\oplus_{i\in I} R_i$ is a rng without identity, but you can say that it has local identities. That is for every $x\in R$, there is an $e\in R$ such that $ex=xe$.

An identity for $R$ would have to be nonzero on every coordinate, and of course no element in that set has that property. So there's no identity.

To answer your first question after having said this, you might be thing of something like a "semigroup with local identities."

Answer 2

Can someone tell me if there is some natural algebraic structure given by an infinite (co)product that has no neutral element $\ldots$, but such that every "finite reduction" $\ldots$ actually has a $\ldots$ neutral element?

The first-order sentence $(\exists u)(\forall a)(u*a=a=a*u)$ is preserved by products. This means that if a family of structures satisfies this sentence, then their Cartesian product will also satisfy it. This shows that the answer to the question is No if one uses Cartesian products, at least if each individual factor in the product is considered to be a "finite reduction" of the product.

On the other hand, let $L = \langle \{0\}; \vee,\wedge\rangle$ be the $1$-element lattice in the category of distributive lattices. Define $x*y:=x\wedge y$. The coproduct $C$ of $\omega$-many copies of $L$ is the $\omega$-generated free distributive lattice. A neutral element for $*$ would have to be a largest element for $L$. But the infinitely generated free distributive lattice has no largest element, so there is no neutral element for $*$ in $C$. However, every finite distributive lattice has a largest element (= the join of all elements), hence every finite distributive lattice has a neutral element for $*$. Does this mean every "finite reduction" of $C$ has a neutral element for $*$? It depends on the meaning of the quoted phrase, but every finitely generated sublattice of $C$ is finite, and every coproduct of finitely many copies of $L$ is finite, so these types of finite reductions will have a neutral element for $*$.