I'm currently trying to understand the "hierarchy" of sets / algebraic structures, e.g. things like groups, rings, fields, modules, algebra, vector spaces which I mostly understand, but especially the more technical things like boolean algebras (specific example of an algebra?), boolean ring (specific example of a ring?), algebra over a field (specific example of an algebra?), field of sets (specific example of a field?), algebra of sets (specific example of an algebra?), algebra over a ring (specific example of an algebra?), etc… I would like to figure out how these things are related. I know I have a lot of reading to do which will probably take a lot of time. Thankfully I found a helpful thread that I found I would like to recommend to others: Algebraic structure cheat sheet anyone?
In the meantime, I'm starting with this Wikipedia page, which nicely categorizes these objects based on (1) how many sets and (2) how many binary operations:
https://en.wikipedia.org/wiki/Algebraic_structure
I notice at the beginning of the list is "set", defined as "degenerate algebraic structure having no operations". I have some questions about this:
- Are the well-known "set" operations "union" and "intersection" considered binary operations?
- Does the notion of "set" (at the beginning of this list of algebraic structures, considered as having no binary operations) include a set along with union and intersection?
- If the answer to the previous question was no, do we need to call the triple $(S,\cup,\cap)$ that we usually think of as a set, i.e. $S$ such that if $A \in S$ and $B \in S$, then $A \cup B \in S$? Is this a more complicated structure with a different name, or do we "always" have access to union and intersection when we are talking about sets?
- Similarly as the previous questions, but what about the notions of "complement" and "set difference", i.e. are these "set operations" considered "binary operations"? And are they always included with the notion of a set or do they create an additional structure, e.g. now we're talking about something like $(S, \cup, \cap, c)$ or $(S, \cup, \cap, -)$?
- What about "cartesian product" $\times$ of sets? Is this considered a "binary operation"?
- When I think of a "binary operation", I should think of a function $f:S \times S \to S$? If so, then binary operations defined in terms of the cartesian product. However, isn't the cartesian product itself considered a binary operation? This seems circular.
- Lastly, let's remember that a function itself (a function between two sets) is defined as a special relation between those two sets, and a relation is a subset of the cartesian product of the two sets. So again, function (and hence binary operation) are defined in terms of cartesian product. Yet it seems like cartesian product IS a binary relation… Circular!
Thank you for your help clearing this up!
Best Answer
1: Set operations (union, intersection) define a binary operation on the power set $\mathcal P(S)$ of a set $S$. That is, they give a map $\mathcal P(S) \times\mathcal P(S) \to \mathcal P(S)$. It doesn't make sense to take the intersection of two elements $A \in S$ and $B \in S$ (unless you're thinking of sets in terms of Zermelo-Fraenkel set theory, where "elements" of sets are always themselves sets. In this case, the set $A \cap B$ and $A\cup B$ needn't be elements of $S$, so this still doesn't give a binary operation on $S$).
2: From looking at the Wikipedia page, it seems that they are considering sets with no operations whatsoever. This makes sense, since sets themselves don't come with a natural binary operation.
3+4: The question of whether or not we "always have access to $\cup$ and $\cap$ when we're talking about sets" comes down to the question of what exactly we mean by a "set" to begin with. The question is therefore about set theory, and depends on the formal way that you approach sets-- that is, on the axioms of the set theory to which you subscribe. As far as I know (disclaimer: I am no expert) most widely accepted set theories allow for notions of union, intersection, cartesian product, and complement (that is, these "operations" are axioms or can be constructed from axioms).
5: Given elements $a,b \in S$, it doesn't make sense to take their cartesian product and claim it is also an element of $S$, for the same reason that $\cup$ and $\cap$ don't define binary operations on $S$. In this case, the cartesian product is no even a binary operation on $\mathcal P(S)$! That's because given $A \subset S$, $B\subset S$, $A\times B$ is not a subset of $S$.
6: This is the definition of a binary operation. In algebraic settings, we usually replace $f(x,y)$ with $x*y$, $x\cdot y$, or sometimes simply with juxtaposition (e.g. $xy$) to save space.
7 (and the second part of 6): Circularity in definitions crops up when you try to define $B$ in terms of $A$, and also define $A$ in terms of $B$. The cartesian product is not defined using binary operations or functions, so these definitions are not circular.