Added several of sections in set theory.

config/en/mkdocs.yaml

@@ -71,6 +71,10 @@ nav:
       - 'Relations': mathematics/set-theory/relations.md
       - 'Maps': mathematics/set-theory/maps.md
       - 'Permutations': mathematics/set-theory/permutations.md
+      - 'Orders': mathematics/set-theory/orders.md
+      - 'Recursion and induction': mathematics/set-theory/recusrion-induction.md
+      - 'Cardinalities': mathematics/set-theory/cardinalities.md
+      - 'Additional axioms': mathematics/set-theory/additional-axioms.md
     - 'Calculus': 
       - 'Limits': mathematics/calculus/limits.md
       - 'Continuity': mathematics/calculus/continuity.md

docs/en/mathematics/set-theory/additional-axioms.md

@@ -0,0 +1 @@
+# Additional axioms

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

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+# Cardinalities

docs/en/mathematics/set-theory/recursion-induction.md

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+# Recursion and induction
+
+## Recursion
+
+A recursively defined function $f$ needs two ingredients:
+
+* a *base*, where the function value $f(n)$ is defined, for some value of $n$.
+* a *recursion*, in which the computation of the function in $n$ is explained with the help of the previous values smaller than $n$.
+
+For example, the sum
+
+$$
+    \begin{align*}&\sum_{i=1}^1 i = 1,\\ &\sum_{i=1}^{n+1} i = (n + 1) + \sum_{i=1}^{n} i.\end{align*}
+$$
+
+Or the product
+
+$$
+    \begin{align*}&\prod_{i=0}^0 i = 1,\\ &\prod_{i=0}^{n+1} i = (n+1) \cdot \prod_{i=0}^{n} i.\end{align*}
+$$
+
+## Induction
+
+> *Principle* **- Natural induction**: suppose $P(n)$ is a predicate for $n \in \mathbb{Z}$, let $b \in \mathbb{Z}$. If the following holds
+>
+> * $P(b)$ is true,
+> * for all $k \in \mathbb{Z}$, $k \geq b$ we have that $P(k)$ implies $P(k+1)$. 
+>
+> Then $P(n)$ is true for all $n \geq b$.
+
+For example, we claim that $\forall n \in \mathbb{N}$ we have
+
+$$
+    \sum_{i=1}^n i = \frac{n}{2} (n+1).
+$$
+
+We first check the claim for $n=1$:
+
+$$
+    \sum_{i=1}^1 i = \frac{1}{2} (1+1) = 1.
+$$
+
+Now suppose that for some $k \in \mathbb{N}$ 
+
+$$
+    \sum_{i=1}^k i = \frac{k}{2} (k+1).
+$$
+
+Then by assumption
+
+$$
+\begin{align*}
+    \sum_{i=1}^{k+1} i &= \sum_{i=1}^k i + (k+1), \\
+                       &= \frac{k}{2}(k+1) + (k+1), \\
+                       &= \frac{k+1}{2}(k+2).
+\end{align*}
+$$
+
+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
+
+$$
+    \sum_{i=1}^n i = \frac{n}{2}(n+1).
+$$