Originally Posted by breck81

My statement does not come close to making that statement or assumption.

"I see schmucks walking down the street" does not mean "All people walking down the street are schmucks."

If A is the set containing "All people walking down the street" and B is the set containing "Schmucks," B is a contained subset of A. Being a member of set B means also being a member of set A, but being a member of set A does not necessarily imply being a member of set B.

Learn English, please.