From 6e4def3f9ffdea9afd63a85b674c679adc63d6d4 Mon Sep 17 00:00:00 2001 From: Luc Date: Wed, 27 Dec 2023 17:10:27 +0100 Subject: [PATCH] Added part of maps in set theory and improved general syntax. --- README.md | 2 +- config/en/mkdocs.yaml | 1 + docs/en/index.md | 4 ++- docs/en/mathematics/set-theory/maps.md | 48 ++++++++++++++++++++++++++ docs/en/programming/start.md | 3 ++ 5 files changed, 56 insertions(+), 2 deletions(-) create mode 100644 docs/en/mathematics/set-theory/maps.md diff --git a/README.md b/README.md index 8867a3a..2313ca2 100644 --- a/README.md +++ b/README.md @@ -1,3 +1,3 @@ # My notes -My notes digitalized. \ No newline at end of file +This is the repository containing my digitalized notes for [wiki.bijl.us](https://wiki.bijl.us). \ No newline at end of file diff --git a/config/en/mkdocs.yaml b/config/en/mkdocs.yaml index 1cf7201..7824462 100755 --- a/config/en/mkdocs.yaml +++ b/config/en/mkdocs.yaml @@ -66,6 +66,7 @@ nav: - 'Set theory': - 'Sets': mathematics/set-theory/sets.md - 'Relations': mathematics/set-theory/relations.md + - 'Maps': mathematics/set-theory/maps.md - 'Calculus': - 'Limits': mathematics/calculus/limits.md - 'Continuity': mathematics/calculus/continuity.md diff --git a/docs/en/index.md b/docs/en/index.md index dd6974f..0812555 100644 --- a/docs/en/index.md +++ b/docs/en/index.md @@ -1 +1,3 @@ -# Welcome \ No newline at end of file +# Welcome + +Welcome to this web page where I have digitalized my notes from several fields. \ No newline at end of file diff --git a/docs/en/mathematics/set-theory/maps.md b/docs/en/mathematics/set-theory/maps.md new file mode 100644 index 0000000..bd49eca --- /dev/null +++ b/docs/en/mathematics/set-theory/maps.md @@ -0,0 +1,48 @@ +# Maps + +## Definition + +> *Definition*: a relation $f$ from a set $A$ to a set $B$ is called a map or function from $A$ to $B$ if for each $a \in A$ there is one and only one $b \in B$ with $afb$. +> +> * To indicate that $f$ is a map from $A$ to $B$ we may write $f:A \to B$. +> * If $a \in A$ and $b \in B$ is the unique element with $afb$ then we may write $b=f(a)$. +> * The set of all maps from $A$ to $B$ is denoted by $B^A$. +> * A partial map $f$ from a $A$ to $B$ with the property that for each $a \in A$ there is at most one $b \in B$ with $afb$. + +For example, let $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = \sqrt{x}$ for all $x \in \mathbb{R}$ is a partial map, since not all of $\mathbb{R}$ is mapped. + +
+ +> *Proposition*: let $f: A \to B$ and $g: B \to C$ be maps, then the composition $g$ after $f$: $g \circ f = f;g$ is a map from $A$ to $C$. +> +> ??? note "*Proof*:" +> +> Let $a \in A$ then $g(f(a))$ is an element in $C$ in relation $f;g$ with $a$. If $c \in C$ is an element in $C$ that is in relation $f;g$ with $a$, then there is a $b \in B$ with $afb$ and $bgc$. But then, as $f$ is a map, $b=f(a)$ and as $g$ is a map $c=g(b)$. Hence $c=g(b)=g(f(a))$ is the unique element in $C$ which is in relation $g \circ f$ with $a$. + +
+ +> *Definition*: Let $f: A \to B$ be a map. +> +> * The set $A$ is called the *domain* of $f$ and the set $B$ the *codomain*. +> * If $a \in A$ then the element $b=f(a)$ is called the image of $a$ under $f$. +> * The subset of $B$ consisting of the images of the elements of $A$ under $f$ is called the image or range of $f$ and is denoted by $\text{Im}(f)$. +> * If $a \in A$ amd $b=f(a)$ then the element $a$ is called a pre-image of $b$. The set of all pre-images of $b$ is denoted by $f^{-1}(b)$. + +Notice that $b$ can have more than one pre-image. Indeed if $f: \mathbb{R} \to \mathbb{R}$ is given by $f(x) = x^2$ for all $x \in \mathbb{R}$, then both $-2$ and $2$ are pre-images of $4$. + +If $A'$ is a subset of $A$ then the image of $A'$ under $f$ is the set $f(A') = \{f(a) \;|\; a \in A'\}$, so $\text{Im}(f) = f(A)$. + +If $B'$ is a subet of $B$ then the pre-image of $B'$, denoted by $f^{-1}(B') is the set of elements $a$ from $A$ that are mapped to an element $b$ of $B'$. + +
+ +> *Theorem*: let $f: A \to B$ be a map. +> +> * If $A' \subseteq A$, then $f^{-1}(f(A')) \supseteq A'$. +> * If $B' \subseteq B$, then $f(f^{-1}(B')) \subseteq B'$. +> +> ??? note "*Proof*:" +> +> Let $a' \in A'$, then $f(a') \in f(A')$ and hence $a' \in f^{-1}(f(A'))$. Thus $A' \subseteq f^{-1}(f(A'))$. +> +> Let $a \in f^{-1}(B')$, then $f(a) \in B'$. Thus $f(f^{-1}(B')) \subseteq B'$. \ No newline at end of file diff --git a/docs/en/programming/start.md b/docs/en/programming/start.md index e69de29..6b95b4f 100755 --- a/docs/en/programming/start.md +++ b/docs/en/programming/start.md @@ -0,0 +1,3 @@ +# Programming + +Welcome to the programming page. \ No newline at end of file