Because operations can be defined on sets? I really don't see why my remarks are so bewildering. If you're only talking about some set P, then surely the only "properties" of P that exist without introducing an operation such as + are set theoretic properties such as [math]P\subset\mathbb{Z}[/math], no? But no matter, because as you point out...
By the definition of a binary operation that I cited, P is not an algebraic set, because as I said, "Right, because + is not a binary operation on P, "to which you responded...
Who said that? Not me.
Yes and then again, one mightn't say it. The definition I cited is certainly sensitive to the set S.
That is not how I defined a binary operation. I defined it as a map [math]b:S\times S \rightarrow S[/math]. My definition necessarily entails closure of S under b, while yours doesn't.