Updated relations and mkdocs.yaml

config/en/mkdocs.yaml

@@ -9,9 +9,21 @@ theme:
     - navigation.tabs
     - toc.follow
   palette: 
-    - scheme: slate
+    - media: "(prefers-color-scheme: light)"
+      scheme: default
+      primary: white
+      accent: teal
+      toggle:
+        icon: material/brightness-7 
+        name: Switch to dark mode
+
+    - media: "(prefers-color-scheme: dark)"
+      scheme: slate
       primary: black
       accent: teal
+      toggle:
+        icon: material/brightness-3
+        name: Switch to light mode
   language: en
 
 extra:

config/nl/mkdocs.yaml

@@ -9,9 +9,21 @@ theme:
     - navigation.tabs
     - toc.follow
   palette: 
-    - scheme: slate
+    - media: "(prefers-color-scheme: light)"
+      scheme: default
+      primary: white
+      accent: teal
+      toggle:
+        icon: material/brightness-7 
+        name: Schakel naar dark mode
+
+    - media: "(prefers-color-scheme: dark)"
+      scheme: slate
       primary: black
       accent: teal
+      toggle:
+        icon: material/brightness-3
+        name: Schakel naar light mode
   language: nl
 
 extra:

docs/en/mathematics/set-theory/relations.md

@@ -1,5 +1,7 @@
 # Relations
 
+## Binary relations
+
 > *Definition*: a binary relation $R$ between the sets $S$ and $T$ is a subset of the Cartesian product $S \times T$. 
 > 
 > * If $(a,b) \in R$ then $a$ is in relation $R$ to $b$, denoted by $aRb$.
@@ -77,4 +79,82 @@ Some relations have special properties
 
     Let $\Pi_R$ be the set of $R$-equivalence classes. Then by reflexivity of $R$ we find that each element $a \in S$ is inside the class $[a]_R$ of $\Pi_R$. If an element $a \in S$ is in the classes $[b]_R$ and $[c]_R$ of $\Pi_R$, then by the previous lemma we find $[b]_R = [a]_R$ and $[c]_R = [a]_R$. Then $[b]_R = [c]_R$, therefore each element $a \in S$ is inside a unique member of $\Pi_R$, which therefore is a partition of $S$.
 
-<br>
+<br>
+
+## Composition of relations
+
+If $R_1$ and $R_2$ are two relations between a set $S$ and $T$, new relations can be formed between $S$ and $T$ by taking the intersection $R_1 \cap R_2$, the union $R_1 \cup R_2$ or the complement $R_1 \backslash R_2$. Furthermore a relation $R^\top$ from $T$ to $S$ can be considered as the relation $\{(t,s) \in T \times S \;|\; (s,t) \in R\}$ and the identity relation from $T$ to $S$ is given by $I = \{(s, t) \in S \times T \;|\; s = t\}$
+
+Another way of making new relations out of existing ones is by taking the composition.
+
+> *Definition*: if $R_1$ is a relation between $S$ and $T$ and $R_2$ is a relation between $T$ and $U$ then the composition $R = R_1;R_2$ is the relation between $S$ and $U$ defined by $sRu$ for $s \in S$ and $u \in U$, if and only if there is a $t \in T$ with $sR_1t$ and $tR_2u$.
+
+<br>
+
+> *Proposition*: suppose $R_1$ is relation from $S$ to $T$, $R_2$ a relation from $T$ to $U$ and $R_3$ a relation from $U$ to $V$. Then $R_1;(R_2;R_3) = (R_1;R_2);R_3$. Composing relations is associative.
+
+??? note "*Proof*:"
+
+    Suppose $s \in S$ and $v \in V$ with $sR_1;(R_2;R_3)v$. Then a $t \in T$ with $sR_1t$ and $t(R_2;R_3)v$ can be found. Then there is also a $u \in U$ with $tR_2u$ and $uR_3v$. For this $u$ there is $sR_1;R_2u$ and $uR_3v$ and hence $s(R_1;R_2);R_3v$.
+
+    Similarly, if $s \in S$ and $v \in V$ with $s(R_1;R_2);R_3v$. Then a $u \in U$ with $s(R_1;R_2)u$ and $uR_3v$ can be found. Then there is also a $t \in T$ with $sR_1t$ and $tR_2u$. For this $t$ there is $tR_2;R_3u$ and $sR_1t$ and hence $sR_1;(R_2;R_3)v$.
+
+<br>
+
+## Transitive closure
+
+> *Lemma*: let $\ell$ be a collection of relations $R$ on a set $S$. If all relations $R$ in $\ell$ are transitive, reflexive or symmetric then the relation $\bigcap_{R \in \ell} R$ is also transitive, reflexive or symmetric respectively.
+
+??? note "*Proof*:"
+
+    Let $\bar R = \bigcap_{R \in \ell} R$. Suppose all members of $\ell$ are transitive. Then for all $a,b,c \in S$ with $a \bar R b$ and $b \bar R c$ there is $aRb$ and $bRc$ for all $R \in \ell$. Thus by transitivity of each $R \in \ell$ there is also $aRc$ for each $R \in \ell$. Thus there is $a \bar R c$. Hence $\bar R$ is also transitive.
+
+    Proof for symmetric relation will follow.
+
+    Proof for reflexive relation will follow.
+
+The above lemma makes it possible to define the reflexive, symmetric or transitive closure of a relation $R$ on a set $S$. It is the smallest reflexive, symmetric or transitive relation containing $R$. 
+
+For example suppose $R = \{(1,2), (2,2), (2,3), (5,4)\}$ is a relation on $S = \{1, 2, 3, 4, 5\}$. 
+
+:   The reflexive closure of $R$ is then the relation
+
+    $$
+    \big\{(1,1), (1,2), (2,2), (2,3), (3,3), (4,4), (5,5), (5,4) \big\},
+    $$
+
+    the symmetric closure of $R$ is then the relation
+
+    $$
+    \big\{ (1,2), (2,1), (2,2), (2,3), (3,2), (4,5), (5,4) \big\},
+    $$
+
+    and the transitive clusure of $R$ is then the relation
+
+    $$
+    \{(1,2), (1,3), (2,2), (2,3), (5,4)\}.
+    $$
+
+It may be observed that the reflexive closure of $R$ equals the relation $I \cup R$ and the symmetric closure equals $R \cup R^\top$. For the transitive closure there is:
+
+> *Proposition*: $\bigcup_{n > 0} R^n$ is the transitive closure of the relation $R$ on a set $S$. 
+
+??? note "*Proof*:"
+
+    Define $\bar R = \bigcup_{n>0} R^n$, to show that $\bar R$ is the least transitive relation containing $R$, $\bar R$ must contain $R$, must be transitive and must be the smallest set with both of those properties. 
+
+    Since $R \subseteq \bar R$, $\bar R$ contains all of the $R^i, i \in \mathbb{N}$, so in particular $\bar R$ contains $R$.
+
+    If $(s_1, s_2), (s_2, s_3) \in \bar R$, then $(s_1, s_2) \in R^j$ and $(s_2, s_3) \in R^k$ for some $j,k$. Since composition is [associative](#composition-of-relations), $R^{j+k} = R^j ; R^k$ and hence $(s_1, s_3) \in R^{j+k} \subseteq \bar R$. 
+
+    We claim that if $T$ is any transitive relation containing $R$, then $\bar R \subseteq T$. By taking $R^n \subseteq \bar R \subseteq T \; \forall n \in \mathbb{N}$ . 
+
+    :   We first check for $n=1$
+
+        $$
+            R^1 = R \subseteq T.
+        $$
+
+    :   Now suppose that for some $k \in \mathbb{N}$ we have $R^k \subseteq T$. Then by assumption $R^{k+1} \subseteq T$. Let $(s_1, s_3) \in R^{k+1} = R^k ; R$, then $(s_1, s_2) \in R$ and $(s_2, s_3) \in R^k$ for some $s_2$. Hence $(s_1, s_2), (s_2, s_3) \in T$ and by transitivity of $T$, $(s_1, s_3) \in T$. 
+
+    Hence if the claim holds for some $k \in \mathbb{N}$ then it also holds for $k+1$. The principle of natural induction implies now that $\forall n \in \mathbb{N}$ we have $R^n \subseteq \bar R \subseteq T$.