Number as sets

• (natural) numbers: can be notated as set
  • $0:=\varphi =\{\}$
  • $s(n):n\cup \{n\}$
  • example
    • $1=\varphi \cup \{0\}=\{\}\cup \{\{\}\}=\{\{\}\}$
    • $2=1\cup \{1\}=\{\{\}\}\cup \{\{\{\}\}\}=\{\{\},\{\{\}\}\}$
    • $3=2\cup \{2\}=\{\{\},\{\{\}\}\}\cup \{\{\{\},\{\{\}\}\}\}=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}$
    • ...
• $n\le m:=n\subseteq m$
• $n\in m:=n\in m$
• def: if $x$ as set $s(x):x\cup \{x\}$
• def: a set $X$: inductive if
  1. $\varphi \in X$
  2. $\forall yy\in X⟹s(y)\in X$
• d

ZFC axioms

7. infinity: there is an inductive set
   • $\exists x\varphi \in X\wedge \forall yy\in X⟹y\cup \{y\}\in X$
8. replacement: if $\phi (n,y)$ is a class function and $X$ a set
   • there is a set $Y$ containing exactly $y^{\prime }s$ s.t. $\exists x\phi (x,y)$
9. foundation: every set $x$ contains an $\epsilon$-minimal element
   • $\forall x\exists y(y\in x\wedge x\cap y=\varphi )$
   • 👨‍🏫 kind of a base case, a set that cannot be opened any more
   • alternatively: $\forall x\exists y(y\in x\wedge \forall zz\in x⟹z\notin y)$
     • creation of Frankel
10. d

Supplementary

• there is a unique set $\mathbb{N}$ s.t.
  • $\mathbb{N}$ is inductive
  • $\mathbb{N}$ is contained in every inductive set
• proof: let $Y$ to be an inductive set (from axiom on infinity)
  • w/ comprehension: $\{y\in Y|\forall X\text{ inductive}(X)⟹y\in X\}$
    • this part: trivial
  • show: $\mathbb{N}$ is inductive
    • $\varphi \in Y$, and $\forall X\text{ inductive}(X)⟹\varphi \in X$
      • $⟹\varphi \in \mathbb{N}$
    • $n\in \mathbb{N},\forall X\text{ inductive}(X)⟹n\in X$
  • $s(n)\in Y,\forall X$
    • as one is subset of each other: two are equal
• let $\phi (x,y)$ be a formula in $\mathbb{L}$ s.t. $\forall x\exists y\phi (x,y)\wedge \exists y^{\prime }\phi (x,y^{\prime })⟹y^{\prime }=y$
  • then $\phi (x,y)$ is called a class function
• theorem: let $x$ be a set, then $x\notin x$
  • suppose $x\in x$
  • let $x=\{n\}$
    • $x\cap X=x$ // i.e. non-empty set
  • also: $x$ cannot be an empty set
    • as it must be a number

Cardinality of the sets

• finite set: set $x$ is finite if there is a function $f:x\to n$
  • s.t. $f$ is one-to-one, onto
• infinite set: set without one-to-one correspondence, if $|x|=n$ is not finite
• let $x,y$ be two sets
  • we can write: $|x|=|y|$ iif there is a 1-1 and onto functions
  • $f:x\to y$
• conditions
  1. $\forall xx\sim x$
  2. $x\sim y\wedge y\sim z⟹x\sim Z$
     • e.g. $f:x\to y,g:y\to z(f\cdot g):x\to z$
  3. $\forall x,yx\sim y⟹y\sim n$
  • $f:x=y;f^{-1}:y\to x$
• theorem: let $\mathbb{E}$ be set of even numbers; $|\mathbb{E}|=|\mathbb{N}|$
  • $f:\mathbb{N}\to \mathbb{E}$
  • $n\mapsto 2n$
  • $\mathbb{E}\cup \mathbb{O}=\mathbb{N}$
  • $\mathbb{E}\cup \mathbb{O}=N$
• theorem: let $\mathbb{P}$: set of primes, $\mathbb{P}\sim \mathbb{N}$
  • as $P=\{p_1,p_2,p_3,\dots \}$
  • $p_1<p_2<p_3<\dots$
  • $f:\mathbb{P}\to \mathbb{N}$
  • $p_i\mapsto i$
• theorem: $|\mathbb{N}\times \mathbb{N}|=|\mathbb{N}|$
  • assign orders in $(x,y)$
    • traverse them by: $(0,0)\to (1,0)\to (0,1)\to (2,0)\to (2,1)\to (2,2)\dots$
    • and we can assign natural number to them
• define: $|X|\le |Y|$ if there exists one-to-one function
  • $f:X\mapsto Y$
  • prove: $|N\times N|\le |\mathbb{N}|$
    • $f:(x,y)\mapsto 2^x{3}^y$
• theorem: $|N^k|=|N|$
  • $N^k=\{(x_1,x_2,\dots ,x_k)|\forall i{x}_i\in \mathbb{N}\}$
  • $A_0=\{(x_1,\dots ,x_k)\in {\mathbb{N}}^k|\sum x_i=0\}=\{(0,0,\dots ,0)\}$
  • $A_1=\{(x_1,\dots ,x_k)\in {\mathbb{N}}^k|\sum x_i=1\}=\{(1,0,\dots ,0),(0,1,\dots ,0),\dots \}$
  • $A_j=\{(x_1,\dots ,x_k)\in {\mathbb{N}}^k|\sum x_i=1\}$
• set: countable if:
  • a set $X$ has same size to natural numbers
  • $|X|=|\mathbb{N}|$
• theorem: let $X$ be a countable set
  • and $\forall y\in X$ also countable
  • then: $\bigcup X$ is also countable
• prove: countable no. of countable elements: countable
• $\mathbb{N}\cup {\mathbb{N}}^2\cup {\mathbb{N}}^3\cup \dots$: countable
  • $={\bigcup }_{i=1}^{\infty }{\mathbb{N}}^i$