Is a set together with an operation always a relational structure?

Question

I'm reading Algebraic Methods in Philosophical Logic, an introductory book on Universal Algebra by by J. Michael Dunn, Gary Hardegree.
This book start its presentation by introducing the notions of "relational structure" and "operational structure" (An algebra).
An operation on a set is considered a particular case of a relation on that set,and so, an "operational structure" is a particular instance of a "relational structure".
However, an "operational structure" must have its carrier set closed under the operations of the structure. This might have confused me.
In my understanding an operation on a set is a kind of relation which combines,modifies, operates on some elements of the set.
This "definition" has nothing to do with the notion of closure, which we can consider an extra proprety of an operation on a set.
So given that the notion of "operational structure" requires a "closed operation". What structure is a set with an opertion, but not closed under that operation?
Can we say that is just a relational structure? (if we look at it through a "relational lens", in the sense that we consider that operation a relation)
Is there a similiar notion of "closure" for relations?
I'll try to illustrate my doubts with an example. Let's take the set A={0,1,2,3} together with the operation of standatd integer addition. We can easily see that this is not an algebra, since A is not closed under integer addition.
Let's now consider our operation as a relation.
We still have our set A={0,1,2,3} but this time we a define a 3-place relation on A such that a1, a2, and, a3, are related if a3=a1+a2.
Is this considered a relational structure?

Answer 1

I don't have a hardcopy, but my guess would be:

• A relational structure is a set $A$ endowed with an n-ary relation subset of $A^n$ (short for $A \times \ldots \times A$).
• An operational structure additionally enforces the relation viewed as a binary relation subset of $A^{n-1} \times A$ to be left-total and right-unique (i.e. representing a function $A^{n-1} \to A$).

With these definitions it is impossible to have a non-closed operation. In fact your first example $(\{0, 1, 2, 3\}, +_{\mathbb{N}})$ does not match either definition since the second component is not a relation on that set.

However, usually in writings on math, you will see something like:

(A, +), with $A$ being a set and $+$ being a binary function, is an algebra. [...] We consider the subalgebra induced by $B \subseteq A$. Indeed that is a subalgebra since its operation $+$ is closed on $B$.

Formally, this means that $+$ viewed as a ternary relation subset of $(A \times A) \times A$ restricted on $B \times B$ is a relation subset of $(B \times B) \times B$. Equivalently in logical symbols:

• $+ \subseteq (A \times A) \times A$
• $\forall b_1, b_2 \in B.\ \not\exists a\notin B. (b_1, b_2, a) \in +$

Indeed note formally we can only speak of the subalgebra once we proved this closedness property. But then again the subalgebra's operation is trivially closed, so formally it doesn't make sense to speak of a closed or non-closed operational algebra with the definitions I guessed above.