Lecture 003

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Intersection($\cap$): common elements between two sets

• $A\cap B = \{x \in U | x\in A \text{ and } x\in B\}$, let U be Universal set

• $A\cap B = \{x \in A | x\in B\}$

• $A\cap B = \{x \in B | x\in A\}$

• if $A\cap B = \emptyset$ then A and B is called disjoint

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 $A$.

• $\{A_i\}_{i\in I} = \{A_i | i\in I\}$ is an family of sets indexed by $I$

• a set of set is called family

• you can use this to define $A_i = \{i, -i\}$

Indexed intersections($\bigcap$): $<span class="arithmatex"><span class="MathJax_Preview">\bigcap_{i\in I}A_i = \{x\in U | \text{ for all } i\in I, x\in A_i\}</span><script type="math/tex">\bigcap_{i\in I}A_i = \{x\in U | \text{ for all } i\in I, x\in A_i\}$

Indexed unions($\bigcup$): $<span class="arithmatex"><span class="MathJax_Preview">\bigcup_{i\in I}A_i = \{x\in U | \text{ for some } i\in I, x\in A_i\}</span><script type="math/tex">\bigcup_{i\in I}A_i = \{x\in U | \text{ for some } i\in I, x\in A_i\}$

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

• our Cartesian plane is $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R} = \{(x,y) | x, y \in \mathbb{R}\}$

• $A \times \emptyset = \emptyset \times A = \emptyset$ for any A

• $x,y\in \mathbb{Z} \lnot (x, y)\in \mathbb{Z}^2$

Because $<span class="arithmatex"><span class="MathJax_Preview">(A \times B) \times C = \{((a, b), c) | a\in A, b\in B, c\in C\}</span><script type="math/tex">(A \times B) \times C = \{((a, b), c) | a\in A, b\in B, c\in C\}$ $<span class="arithmatex"><span class="MathJax_Preview">A\times (B\times C) = \{(a, (b, c)) | a\in A, b\in B, c\in C\}</span><script type="math/tex">A\times (B\times C) = \{(a, (b, c)) | a\in A, b\in B, c\in C\}$ so the above is the same thing as $<span class="arithmatex"><span class="MathJax_Preview">A\times B\times C = \{(a, b, c) | a\in A, b\in B, c\in C\}</span><script type="math/tex">A\times B\times C = \{(a, b, c) | a\in A, b\in B, c\in C\}$ 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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