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bluepython508
2024-02-23 11:09:21 +00:00
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9
MA1511/decls.tex
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@@ -22,9 +22,16 @@
\newcommand{\polar}[2]{#1\paren{\cos{\paren{#2}} + i\sin{\paren{#2}}}}
\newcommand{\adj}[1]{\operatorname{adj}#1}
\newcommand{\card}[1]{\left|#1\right|}
\newcommand{\littletaller}{\mathchoice{\vphantom{\big|}}{}{}{}}
\newcommand{\restr}[2]{{% we make the whole thing an ordinary symbol
\left.\kern-\nulldelimiterspace % automatically resize the bar with \right
#1 % the function
\littletaller % pretend it's a little taller at normal size
\right|_{#2} % this is the delimiter
}}
\makeatletter
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
\renewcommand*{\env@matrix}[1][*\c@MaxMatrixCols c]{%
\hskip -\arraycolsep
\let\@ifnextchar\new@ifnextchar
\array{#1}}

67
MA1511/sets.tex
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@@ -29,18 +29,18 @@ A family of sets indexed by a set $I$ (the indexing set): $A_i ~~\forall~i\in I
$A_i$ is a set for every element $i \in I$ \\
A family of sets indexed by $\N$ is called a sequence of sets. Also written $(B_i)^{\inf}_{i=0}$ or $(B_i)_{i \geq 0}$
\begin{align*}
\bigcup_{i \in I}~A_i \equiv \{x | \exists i \in I. x \in A_i \} \\
\bigcap_{i \in I}~A_i \equiv \{x | \forall i \in I. x \in A_i \} & \text{ Exists iff } \exists~i\in I
\bigcup_{i \in I}~A_i \equiv \{x | \exists i \in I. x \in A_i \} \\
\bigcap_{i \in I}~A_i \equiv \{x | \forall i \in I. x \in A_i \} & \text{ Exists iff } \exists~i\in I
\end{align*}
\begin{align*}
& \forall i \in I. A_i \subseteq \cup_{j \in I}A_j \\
& \forall i \in I. A_i \subseteq B \implies \cup_{j \in I}A_j \subseteq B \\
& \forall i \in I.\cap_{j \in I}A_j \subseteq A_i \\
& \forall i \in I. B \subseteq A_i \implies B \subseteq \cap_{j \in I}A_j \\ \\
& B \cup \cap_{i \in I}A_i = \cup_{i \in I}(B \cap A_i) \\
& B \cap \cup_{i \in I}A_i = \cap_{i \in I}(B \cup A_i) \\
& B \setminus \cup_{i \in I}A_i = \cap_{i\in I}(B\setminus A_i) \\
& B \setminus \cap_{i \in I}A_i = \cup_{i\in I}(B \setminus A_i)
& \forall i \in I. A_i \subseteq \cup_{j \in I}A_j \\
& \forall i \in I. A_i \subseteq B \implies \cup_{j \in I}A_j \subseteq B \\
& \forall i \in I.\cap_{j \in I}A_j \subseteq A_i \\
& \forall i \in I. B \subseteq A_i \implies B \subseteq \cap_{j \in I}A_j \\ \\
& B \cup \cap_{i \in I}A_i = \cup_{i \in I}(B \cap A_i) \\
& B \cap \cup_{i \in I}A_i = \cap_{i \in I}(B \cup A_i) \\
& B \setminus \cup_{i \in I}A_i = \cap_{i\in I}(B\setminus A_i) \\
& B \setminus \cap_{i \in I}A_i = \cup_{i\in I}(B \setminus A_i)
\end{align*}
\section*{Cartesian Products}
Ordered pairs can be represented as $(x, y) \equiv \{x, \{x, y\}\}$ \\
@@ -51,9 +51,50 @@ $\card{X\times Y} = \card{X} \times \card{Y}$ (for finite $X$, $Y$)
For sets $X$, $Y$:
\begin{description}
\item[A function $F: X \to Y$] $\subseteq X\times Y$ \text { where } \\
$\forall x \in X.~\exists \text{ a unique } y \in Y.~ (x, y) \in F$ \\
$F(x)$ denotes the unique element $y \in Y$ for which $(x, F(x)) \in F$
$\forall x \in X.~\exists \text{ a unique } F(x) \in Y.~ (x, F(x)) \in F$ \\
There exist $\card{Y}^{\card{X}}$ functions $F: X \to Y$
\item[$\operatorname{dom}(F)$] The domain of $F$, i.e. $X$
\item[$\operatorname{incl}^{X}_{A} : A \to X$] $= a \quad \forall A X. \text{ where } A \subseteq X$
\item[$\operatorname{incl}^X_A : A \to X$] $= a \quad \forall A,X. \text{ where } A \subseteq X$
\item[Constant function] $\exists y_0 \in Y.~\forall x \in X.~ f(x) = y_0$
\item[Characteristic Function of a set $A \subseteq X$: $\chi_A: X \to \{0, 1\}$]
\[
\chi_A: X \to \{0, 1\} = \left\{\begin{array}{lr}
0 & \text{ if } x \not\in A \\
1 & \text{ if } x \in A
\end{array} \right.
\]
\item[Restriction of a function $f: X \to Y$] $\restr{f}{A}$ is $f$ specialized contravariantly to $A \subseteq X$
\item[$f(A)$: Image of $A$ under $f$] $f$ mapped over $A$ \quad for function $f: X \to Y$, $A \subseteq X$
\item[$\operatorname{ran}(f)$ / image of $f$ / range of $f$] $\{ f(x) | x \in X \}$, $f(X)$, i.e. all possible values of $f(x)$
\item[Preimage of $B$ under $f$] $\{ x \in X ~|~ f(x) \in B \}$ \\
written $f^{-1}(B)$, but is \emph{not} the inverse of f
\end{description}
For $f: X \to Y$, $A \subseteq A' \subseteq X$, $B \subseteq B' \subseteq Y$:
\begin{align*}
f(A) \subseteq~& f(A') \\
f^{-1}(B) \subseteq ~& f^{-1}(B') \\
f^{-1}(f(A)) \supseteq ~& A \\
f(f^{-1}(B)) \subseteq ~& B
\end{align*}
For set families $(A_i \subseteq X)_{i \in I}, (B_j \subseteq Y)_{j \in J}$:
\begin{align*}
f(\cup_{i \in I} A_i) = ~& \cup_{i \in I}f(A_i) \\
f(\cap_{i \in I} A_i) \subseteq ~& \cap_{i \in I}f(A_i) \\
f^{-1}(\cup_{j \in J} B_j) = ~& \cup_{j \in J} f^{-1}(B_j) \\
f^{-1}(\cap_{j \in J} B_j) = ~& \cap_{j \in J} f^{-1}(B_j)
\end{align*}
\section*{Function Composition}
For functions $f: X \to Y$, $g: Y \to Z$:
\begin{align*}
& (g \circ f): X \to Z \\
& (g\circ f)(x) = g(f(x)) & \forall x \in X \\ \\
\end{align*}
\section*{Surjection and Injection}
For $f: X \to Y$, $f$ is
\begin{description}
\item[surjective] iff $f(X) = Y$, i.e. $\forall y \in Y.~ \exists x \in X.~ f(x) = y$ \\
Range is codomain, 'onto'
\item[injective] iff $\forall x, x' \in X. ~ f(x) = f(x') \implies x = x'$
\item[bijective] iff $f$ is injective and $f$ is surjective
\end{description}
\end{document}