Updated/finished sets.md

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

@@ -22,14 +22,140 @@ Suppose for example that $B = {x,y,z}$, then $\wp(B) = \{\varnothing,\{x\},\{y\}
 
 > *Proposition*: let $B$ be a set with $n$ elements. Then its power set $\wp(B)$ contains $w^n$ elements.
 
-<details>
+??? note "*Proof*:"
-<summary><em>Proof</em>:</summary>
 
-Let $B$ be set with $n$ elements. A subset $A$ of $B$ is completely determined by its elements. For each element $b \in B$ there are two options, it is in $A$ or it is not. So, there are $2^n$ options and thus $2^n$ different subsets $A$ of $B$.
+    Let $B$ be set with $n$ elements. A subset $A$ of $B$ is completely determined by its elements. For each element $b \in B$ there are two options, it is in $A$ or it is not. So, there are $2^n$ options and thus $2^n$ different subsets $A$ of $B$.
 
-</details>
 <br>
 
+> *Proposition*: suppose $A$, $B$ and $C$ are sets. Then the following hold:
+> 
+> 1. if $A \subseteq B$ and $B \subseteq C$ then $A \subseteq C$,
+> 2. if $A \subseteq B$ and $B \subseteq A$ then $A = B$.
+
 ??? note "*Proof*:"
 
-    Let $B$ be set with $n$ elements. A subset $A$ of $B$ is completely determined by its elements. For each element $b \in B$ there are two options, it is in $A$ or it is not. So, there are $2^n$ options and thus $2^n$ different subsets $A$ of $B$.
+    To prove 1, suppose that $A \subseteq B$. Let $a \in A$, then $a \in B$ therefore $a \in C$.
+
+    To prove 2, every element of $A$ is in $B$ and every element of $B$ is in $A$. As the set is uniquely determined by its elements $A = B$.
+
+<br>
+
+> *Definition*: let $P$ be a predicate with reference set $X$, then
+>$$
+>   \big\{x \in X \;\big|\; P(x) \big\}
+>$$
+> denotes the subset of $X$ consisting of all elements $x \in X$ for which statement $P(x)$ is true.
+
+## Operations on sets
+
+> *Definition*: let $A$ and $B$ be sets.
+>
+> * The intersection of $A$ and $B$ $(A \cap B)$ is the set of all elements contained in both $A$ and $B$.
+> * The union of $A$ and $B$ $(A \cup B)$ is the set of elements that are in at least on of $A$ or $B$.
+> * $A$ and $B$ are disjoint if the intersection $(A \cap B)$ is the empty set $\varnothing$.
+
+<br>
+
+> *Definition*: suppose $I$ is a set (an index set) and for each element $i$ there exists a set $A_i$, then
+> 
+> $$
+> \bigcup_{i \in I} A_i := \big\{x \;\big|\; \text{there is an } i \in I \text{ with } x \in A_i \big\},
+> $$
+>
+> and
+>
+> $$
+> \bigcap_{i \in I} A_i := \big\{x \;\big|\; \text{for all } i \in I \text{ there is } x \in A_i \big\}.
+> $$
+>
+
+Implying unions and intersections taken over an index set. For example suppose for each $i \in \mathbb{N}$ the set $A_i$ is defined as $\{x \in \mathbb{R} \;|\; 0 \leq x \leq i \}$, then
+
+$$
+\bigcap_{i \in \mathbb{N}} A_i = \{0\},
+$$
+
+and
+
+$$
+\bigcup_{i \in \mathbb{N}} A_i = \mathbb{R}_{\geq 0}.
+$$
+
+<br>
+
+> *Definition*: if $C$ is a collection of sets, then
+>
+> $$
+> \bigcup_{A \in C} A := \big\{x \;\big|\; \text{there is an } A \in C \text{ with } x \in A \big\},
+> $$
+>
+> and
+>
+> $$
+> \bigcap_{A \in C} A := \big\{x \;\big|\; \text{for all } A \in C \text{ there is } x \in A \big\}.
+> $$
+
+<br>
+
+> *Definition*: let $A$ and $B$ be sets. The difference of $A$ and $B$ $(A \backslash B)$ is the set of all elements from $A$ that are not in $B$. 
+> 
+>:  The symmetric difference of $A$ and $B$ $(A \triangle B)$ is the set consisting of all elements that are in exactly one of $A$ or $B$.
+>
+>:  If one is working inside a fixed set $U$ and only considering subsets of $U$, then the difference $U \backslash A$ is also called the complement of $A$ in $U$, denoted by $A^*$. In this case the set $U$ is called the universe.
+
+## Cartesian products
+
+Suppose $a_1, a_2, \dots, a_k$ are elements from some set, then the ordered k-tuple of $a_1, a_2, \dots, a_k$ is denoted by $(a_1, a_2, \dots, a_k)$
+
+> *Definition*: the cartesian product $A_1 \times \dots \times A_k$ of sets $A_1, \dots , A_k$ is the set of all ordered k-tuples $(a_1, a_2, \dots, a_k)$ where $a_i \in A_i$ for $1 \leq i \leq k$.
+> 
+>:  If $A$ and $B$ are sets then
+>
+> $$
+> A \times B = \big\{ (a,b) \;\big|\; a \in A,\; b \in B \big\}
+> $$
+
+Notice that for all $1 \leq i \leq k$ and $A_i = A$ then $A_1 \times \dots \times A_k$ is also denoted by $A^k$. 
+
+## Partitions
+
+> *Definition*: let $S$ be a nonempty set. A collection $\Pi$ of subsets is called a partition if and only if 
+> 
+> * $\varnothing \notin \Pi$,
+> * $\bigcup_{X \in \Pi} X = S$,
+> * for all $X \neq Y \in \Pi$ there is $X \cap Y = \varnothing$
+
+For example the set $\{1,2, \dots , 10\}$ can be partitioned into the sets $\{1,2,3\}$, $\{4,5\}$ and $\{6,7,8,9,10\}$.
+
+## Quantifiers
+
+> *Definitions*: the universal quantifier "for all" is denoted by $\forall$ and the existential quantifier "there exists" is denoted by $\exists$.
+
+> *Proposition* **- DeMorgan's rule**: the statement
+>
+> $$
+>   \neg (\forall x \in X \;[P(x)])
+> $$
+>
+> is equivalent with the statement
+>
+> $$
+>   \exists x \in X \;[\neg (P(x))].
+> $$
+>
+> The statement 
+>
+> $$
+>   \neg (\exists x \in X \;[P(x)])
+> $$
+>
+> is equivalent with the statement
+>
+> $$
+> \forall x \in X \; [\neg (P(x))].
+> $$
+
+??? note "*Proof*:"
+
+    will be added later.