Hello, Veeky Forums. This will be the first lecture in a series I will be giving here on Brave New Algebra. We will be starting from rudimentary set theory. This thread is meant to be a discussion place, so if anybody has any questions, feel free to ask. Shitposting is accepted but discouraged.
A set is naively a collection of things. We will start by imposing the axioms of Zermelo-Fraenkel set theory, coupled with the axiom of choice (sue me). But first, some definitions.
Two sets are equal, written A=B, if, given any element x, x is in A if and only if x is in B.
The product of two sets A and B is the set AxB:={(a,b) : a in A, b in B}. We will impose axioms to ensure that these two operations are allowed and produce sets.
A set A is a subset of another set B if, given any element x, x in A implies that x is in B. Exercise: prove that A=B if and only if A and B are subsets of one another. Exercise: show that the empty set {} is a subset of every set.
The union of two sets A and B is the set AUB:={x : x in A or x in B}.
The intersection of two sets A and B is the set A∩B:={x: x in A and x in B}.
A function f:A->B is a subset of AxB such that if (a,b) and (a,b') are both in f, then b=b'. Intuitively, a function assigns each element in its domain (A) to a unique element in its codomain (B). A function is injective if (a,b) and (a',b) in f imply that a=a'. A function is surjective if for every element b in B, there is an element (a,b) in f. Notation-wise, if (a,b) is in f, we write f(a)=b. Exercise: can there be a function that is both injective and surjective?