Lecture 003

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Intersection(\cap): common elements between two sets

Union(\cup): A\cup B = \{x\in A \text{ or } x\in B\}

Difference(- or $\backslash $): A\backslash B = \{x\in U | x\in A \text{ and } x\notin B\}=\{x\in A | x\notin B\}

Complement(A^\complement or \overline{A}, or A^\mathsf{c}): \overline{A}=\{x\in U | x\notin A\}=U\backslash A TODO: :question: can I wrote as ^\complement?

Index set: i\in I in which i denotes the index of set of all possible variation of A.

Indexed intersections(\bigcap): \bigcap_{i\in I}A_i = \{x\in U | \text{ for all } i\in I, x\in A_i\}

Indexed unions(\bigcup): \bigcup_{i\in I}A_i = \{x\in U | \text{ for some } i\in I, x\in A_i\}

Cartesian product(A\times B): A\times B = \{(a, b) | a\in A \text{ and } b\in B\}

Because (A \times B) \times C = \{((a, b), c) | a\in A, b\in B, c\in C\} A\times (B\times C) = \{(a, (b, c)) | a\in A, b\in B, c\in C\} so the above is the same thing as A\times B\times C = \{(a, b, c) | a\in A, b\in B, c\in C\} because no information is lost (the information is tuples or triples are the direction of the operation.)

Logic and Proofs

s.t. for such that holds means it yields true

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