In other words Xander would be right, that my castle is illusory and I'm just using big words that don't mean anything so my objective now is to get complete closure on this specific question.
well let's just say @YourLordJoyBoy that I personally believe that the probability declines with producing a great work as one increases in age. but I will speak no more on said matter.
In Jacobson's Basic Algebra I, the exercise 1 in section 1.1 states: "Let $S$ be a set and define a product on $S$ by $ab=b$. Prove that $S$ is a semigroup". Jacobson defines a semigroup as a pair $(M,p)$ where $M$ is a non vacuous set and $p$ is a binary associative operation on $M$. Maybe I am being too much pedantic, but shouldn't the exercise 1.1 have the additional hypothesis that $S$ is non empty?
frieren: offhand, i don't see anything about the semigroup idea that "requires" S/M to be nonempty, e.g. the empty set is a perfectly good associative binary operation on the empty set. but if Jacobson is stuffing nonemptiness into his definition of what a semigroup is, then he should certainly also stuff it into the hypothesis of that exercise
@leslietownes You mean that since the definition of associativity states $\forall a,b,c \in S, (ab)c=a(bc)$, if $S=\emptyset$ the definition of associativity is vacuously true in $\emptyset$ and so, if we define semigroup differently allowing $S$ to be the empty set too, because the associativity holds in $\emptyset$ for any operation we deduce that $\emptyset$ is a semigroup with respect to any operation according to this different definition?
@leslietownes thanks. Again, I will ask another question just to be sure I am understanding correctly: the unique operation you are referring to is some kind of "empty operation"? Because, in the case of the emtpy set, the domain of the operation should be $\emptyset \times \emptyset$ (which equals $\emptyset$) and the codomain should be $\emptyset$ because there aren't elements in the domain.
the structure of what this binary operation "is" as a set might depend on how you encode functions as sets. e.g. if a function from X to Y is an ordered triple (X, Y, subset of XxY having the function property), this operation might be modeled as a nonempty set [whose structure depends on how you do ordered triples as sets]
but i do mean the 'function' you get by considering the empty set as a set of ordered pairs that vacuously satisfies the function property
People say all the time in the context of Fraisse theory for example that amalgamating over the empty structure shows that the amalgamation property implies the joint embedding property for relational languages
if you don't allow the empty semigroup, statements like "the intersection of two subsemigroups of a semigroup is a subsemigroup" have to be changed to "the intersection of two subsemigroups of a semigroup is a subsemigroup if it is non-empty"
@copper.hat thanks for your answer, I appreciate it! :) However, I think I need some time to think about what you wrote. I only briefly looked at your answer and could not understand much. I understood the other answer more, but that answer, as I commented, used the fact that the Lebesgue measure of the oblique rectangle equals its volume, which we don't know yet. For now I can't say if or when I'll be able to look at your answer in more detail, but I know it's there and thanks nevertheless.
