101 lines
5.0 KiB
TeX
101 lines
5.0 KiB
TeX
\input{decls.tex}
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\title{Sets}
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\begin{document}
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\maketitle
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Set comprehensions can be written $\{ x | x \in \N \}$ or $\{ x : x \in \N \}$ - '$:$' or '$|$' \\
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Sets are defined entirely by the values of $x$ for which $x \in A$
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\begin{description}
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\item[Axiom of Extensionality / Set Equality] $A = B \iff \forall x. (x \in A \iff x \in B)$
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\item[$A \subseteq B$] \quad $\forall x \in A. x \in B$ \\
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Is transitive, reflexive, antisymmetric
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\item[$A \subset B$] \quad $(\forall x \in A.~x \in B) \land (\exists x \in B.~x \not\in A)$ \\
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Is transitive, antisymmetric
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\item[$\varnothing$] $\{\}$
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\item[$\cup$] Union
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\item[$\cap$] Intersection
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\item[$A \setminus B$] \quad $\{ x \in A : x \not\in B \}$
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\item[$A^\complement$]\quad $U \setminus A$
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\item[$[a, b)$] \quad $\{ x \in \R : a \leq x < b \}$
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\end{description}
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\begin{align*}
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C \setminus (A \cup B) \equiv & ~ (C \setminus A) \cap (C \setminus B) \\
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C \setminus (A \cap B) \equiv & ~ (C \setminus A) \cup (C \setminus B) \\
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(A^\complement)^\complement \equiv & ~ A \\
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A^\complement \cup B^\complement \equiv & ~ (A \cap B)^\complement \\
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A^\complement \cap B^\complement \equiv & ~ (A \cup B)^\complement
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\end{align*}
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\section*{Families of Sets}
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A family of sets indexed by a set $I$ (the indexing set): $A_i ~~\forall~i\in I (\equiv (A_i)_{i\in I})$ \\
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$A_i$ is a set for every element $i \in I$ \\
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A family of sets indexed by $\N$ is called a sequence of sets. Also written $(B_i)^{\inf}_{i=0}$ or $(B_i)_{i \geq 0}$
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\begin{align*}
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\bigcup_{i \in I}~A_i \equiv \{x | \exists i \in I. x \in A_i \} \\
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\bigcap_{i \in I}~A_i \equiv \{x | \forall i \in I. x \in A_i \} & \text{ Exists iff } \exists~i\in I
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\end{align*}
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\begin{align*}
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& \forall i \in I. A_i \subseteq \cup_{j \in I}A_j \\
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& \forall i \in I. A_i \subseteq B \implies \cup_{j \in I}A_j \subseteq B \\
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& \forall i \in I.\cap_{j \in I}A_j \subseteq A_i \\
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& \forall i \in I. B \subseteq A_i \implies B \subseteq \cap_{j \in I}A_j \\ \\
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& B \cup \cap_{i \in I}A_i = \cup_{i \in I}(B \cap A_i) \\
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& B \cap \cup_{i \in I}A_i = \cap_{i \in I}(B \cup A_i) \\
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& B \setminus \cup_{i \in I}A_i = \cap_{i\in I}(B\setminus A_i) \\
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& B \setminus \cap_{i \in I}A_i = \cup_{i\in I}(B \setminus A_i)
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\end{align*}
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\section*{Cartesian Products}
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Ordered pairs can be represented as $(x, y) \equiv \{x, \{x, y\}\}$ \\
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$X \times Y = \{ (x, y) |~ \forall x, y.~x \in X \land y \in Y\}$ for sets $X$, $Y$ \\
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$X^n$ is $X\times X^{n-1}$ for set $X$ and natural $n$ \\
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$\card{X\times Y} = \card{X} \times \card{Y}$ (for finite $X$, $Y$)
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\section*{Functions}
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For sets $X$, $Y$:
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\begin{description}
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\item[A function $F: X \to Y$] $\subseteq X\times Y$ \text { where } \\
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$\forall x \in X.~\exists \text{ a unique } F(x) \in Y.~ (x, F(x)) \in F$ \\
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There exist $\card{Y}^{\card{X}}$ functions $F: X \to Y$
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\item[$\operatorname{dom}(F)$] The domain of $F$, i.e. $X$
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\item[$\operatorname{incl}^X_A : A \to X$] $= a \quad \forall A,X. \text{ where } A \subseteq X$
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\item[Constant function] $\exists y_0 \in Y.~\forall x \in X.~ f(x) = y_0$
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\item[Characteristic Function of a set $A \subseteq X$: $\chi_A: X \to \{0, 1\}$]
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\[
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\chi_A: X \to \{0, 1\} = \left\{\begin{array}{lr}
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0 & \text{ if } x \not\in A \\
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1 & \text{ if } x \in A
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\end{array} \right.
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\]
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\item[Restriction of a function $f: X \to Y$] $\restr{f}{A}$ is $f$ specialized contravariantly to $A \subseteq X$
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\item[$f(A)$: Image of $A$ under $f$] $f$ mapped over $A$ \quad for function $f: X \to Y$, $A \subseteq X$
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\item[$\operatorname{ran}(f)$ / image of $f$ / range of $f$] $\{ f(x) | x \in X \}$, $f(X)$, i.e. all possible values of $f(x)$
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\item[Preimage of $B$ under $f$] $\{ x \in X ~|~ f(x) \in B \}$ \\
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written $f^{-1}(B)$, but is \emph{not} the inverse of f
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\end{description}
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For $f: X \to Y$, $A \subseteq A' \subseteq X$, $B \subseteq B' \subseteq Y$:
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\begin{align*}
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f(A) \subseteq~& f(A') \\
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f^{-1}(B) \subseteq ~& f^{-1}(B') \\
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f^{-1}(f(A)) \supseteq ~& A \\
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f(f^{-1}(B)) \subseteq ~& B
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\end{align*}
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For set families $(A_i \subseteq X)_{i \in I}, (B_j \subseteq Y)_{j \in J}$:
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\begin{align*}
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f(\cup_{i \in I} A_i) = ~& \cup_{i \in I}f(A_i) \\
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f(\cap_{i \in I} A_i) \subseteq ~& \cap_{i \in I}f(A_i) \\
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f^{-1}(\cup_{j \in J} B_j) = ~& \cup_{j \in J} f^{-1}(B_j) \\
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f^{-1}(\cap_{j \in J} B_j) = ~& \cap_{j \in J} f^{-1}(B_j)
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\end{align*}
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\section*{Function Composition}
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For functions $f: X \to Y$, $g: Y \to Z$:
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\begin{align*}
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& (g \circ f): X \to Z \\
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& (g\circ f)(x) = g(f(x)) & \forall x \in X \\ \\
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\end{align*}
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\section*{Surjection and Injection}
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For $f: X \to Y$, $f$ is
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\begin{description}
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\item[surjective] iff $f(X) = Y$, i.e. $\forall y \in Y.~ \exists x \in X.~ f(x) = y$ \\
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Range is codomain, 'onto'
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\item[injective] iff $\forall x, x' \in X. ~ f(x) = f(x') \implies x = x'$
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\item[bijective] iff $f$ is injective and $f$ is surjective
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\end{description}
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\end{document}
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