Sets

• naive comprehension (possibly leading to paradoxes)
  • writing set with property/formula and element (thing)
  • $s=\{x|\phi (x)\}$
  • 👨‍🏫 above notation: only works if condition can be written as $\phi (x)$
• Russell's paradox
  • let $\varphi (x):=x\notin x$
    • $\in$, not $\subset$
  • then let $A=\{x|x\notin x\}$
    • must be a "set", as it's defined by comprehension
  • paradox: $A\in A$?
    • if $A\in A$: it means $A\notin A$ (not in set satisfying $\{x\notin x\}$)
    • if $A\notin A$: it means $A\in A$ (in set satisfying $\{x\notin x\}$)
  • only way out: show that such $A$ doesn't exist

ZFC

• i.e. Zermelo-Fraenkel (Axioms) choice
• creating a new set from an existing set
  • not necessarily all setts hat can be created
• language $\mathbb{L},\mathbb{I}$ must be defined, supporting:
  • arbitrary no. of variables $x,y,z,\dots$ over sets
    • everything: sets
  • $x\in y,x=y$
  • logical & boolean operators
    • only $\wedge ,\neg ,\forall$ is needed
      • others, e.g. $\vee ,⟹,\exists$, can be derived
  • parenthesis

ZFC axioms

1. extensionality: two sets are equal iff they have the same elements
   • $\forall x\forall y,x=y⟺(\forall zz\in x⟺x\in y)$
2. empty set: there exists a set with no elements
   • $\exists x\forall yy\notin x$
     • 👨‍🏫 $\notin$: a syntactic sugar
3. unordered pair: if $x,y$ are sets, there is a set $\{x,y\}$
   • w/ elements are exactly $x,y$
   • $\forall x\forall y\exists z(x\in z\wedge y\in z\wedge \forall ww\in z⟹(w=x\vee w=y))$
4. union: if $x$ a set, $\exists$ set consisting of all elements of all the elements of $x$
   • $\forall x\exists y\forall z(z\in y⟺\exists w(w\in x\wedge z\in w))$
   • e.g. for $x=\{\{1,2,3\},\{a,b\},\{\alpha ,\beta \}\}$
     • $y=\{1,2,3,a,b,\alpha ,\beta \}$
   • 👨‍🏫 uniqueness of ZF3, ZF4 can be shown
   • notation: $y=\bigcup x$
5. comprehension: from a universal set
   • if $\phi (z,w_1,w_2,\dots ,w_k)$ a formula in $\mathbb{L}$
     • w/ free variables $z,w_1,\dots ,w_k$
     • and $x$ a set and $a_1,a_2,\dots ,a_k$ are sets
     • then following is also a set: $\{y\in x:\phi (y,a_1,a_2,\dots ,a_k)\}$
       • subset of $x$ satisfies condition $\phi$
     • 👨‍🏫 limited version of original comprehension
   • $\forall x\forall a_1\forall a_2\dots \forall a_k\exists z[y\in z⟺y\in x\wedge \phi (y,a_1,\dots ,a_k)]$
   • original problem / paradox doesn't exist as
     • there is no "super set" or set-of-all-sets
     • 👨‍🏫 not everything that can be written in $\mathbb{L}$ is a set
     • 👨‍🏫 now, it's Russel's proof instead of paradox
   • $\nexists B\forall xx\in B$
6. powerset: let $x$ be a set. there exists a set $y$ whose elements are subsets of $x$
   • $\forall x\exists y\forall zz\subseteq x⟺z\in y$
   • denote $y=p(x)$

Supplementary

• theorem: there is a unique set w/ no elements (w/ ZF1, ZF2)
  • by ZF2, there exists a set $x$ w/ no elements
  • let $y$ be a set w/ no elements
  • for any $z$, $z\notin x$ and $z\notin y$
    • thus, following holds (no premise holds)
    • $\forall zz\in x⟺x\in y$
  • by ZF1, $x=y$
  • empty set: will be notated as $\varphi$
    • e.g. $\forall x\forall y,x\cap y=\varphi$
      • technically not in our language as $\varphi$ is included
    • yet, we can use $\varphi$ as a variable and write equivalent:
      • $\exists \varphi \forall xz\notin \varphi \wedge \forall x\forall yx\cap y=\varphi$
• ZF3: set creation
  • from $\varphi$, $\{\varphi \}$ can be obtained
  • from $\varphi ,\{\varphi \}$, $\{\varphi ,\{\varphi \}\}$ can be obtained
  • and so on...
• ordered pairs: w/ $x,y$: define $(x,y)$
  • $(x,y):=\{\{x\},\{x,y\}\}$
  • $(x,y)\ne (y,x)$
• theorem: let $x,y,a,b$ be sets, $(x,y)=(a,b)⟺x=a\wedge y=b$
  • 👨‍🏫 not so trivial proof
• proof
  • case 1: $x=y$
    • $(x,y)=(x,x)=\{\{x\},\{x,x\}\}=\{\{x\},\{x\}\}=\{\{x\}\}$
    • then $\{\{x\}\}=\{\{a\},\{a,b\}\}$ (claimed by left side of iff)
    • $\{a\}\in \text{RHS}⟹\{a\}\in \text{LHS}$
      • thus $\{a\}=\{x\}$, $a=x$
    • $\{a,\{a,b\}\}\in \text{RHS}⟹\{a,\{b\}\}\in \text{LHS}$
      • thus $\{a,\{b\}\}=\{x\}$, $a=x$
    • $x=y=a=b$
  • case 2: $a=b$
    • 👨‍🏫 trust me, it's case 1
  • case 3: $x\ne y\wedge a\ne b$
    • $\{\{x\},\{x,y\}\}=\{\{a\},\{a,b\}\}$
    • $\{x\}\in \text{RHS}\wedge \{x\}\in \text{LHS}$
    • thus: $\{x\}=\{a\}$ i.e. $x=a$
      • $\{x\}=\{a,b\}$ i.e. $x=a=b$
  • for $\{x,y\}\in LHS$
    • and for $\{x,y\}\in RHS$
    • thus: $\{x,y\}=\{a\}$ i.e. $x=y=a$
      • $\{x,y\}=\{a,b\}$ i.e. $(x=a\wedge w\ne x)⟹y\ne a,y=b$
• extending union
  • $a\cup b:=\bigcup \{a,b\}$
• class
  • class: collection of form $X=\{x:\phi (x)\}$
  • 👨‍🏫 needed as all operations are only defined on sets
  • $V=\{x:x=x\}$
  • $A=\{x\in V:x\notin x\}$
• subset
  • $a\subseteq b:=\forall zz\in a⟹z\in b$
• Cartesian product
  • let $X,Y$ be sets
  • $X\times Y=\{(x,y):x\in X\wedge y\in Y\}$
    • 👨‍🏫 paradox-prone comprehension!
  • use $\{z:\exists x\in X\exists y\in Yx=(x,y)\}$
    • RHS: valid formula in $\mathbb{L}$ now
  • $(x,y)=\{\{x\},\{x,y\}\}\in P(P(X\cup Y))$, $x\in X$, $y\in Y$
  • $x,y\in X\cup Y$
  • $\{x\},\{x,y\}\in P(X,U,Y)$
  • $X\times Y=\{z\in P(PX\cup Y):\exists x\in X\exists y\in Yz=(x,y)\}$
• 👨‍🏫 Russel said: accepting one paradox implies all other paradox
  • e.g. 1=2, then I'm the Pope
  • building system without paradox: important
• relation
  • relation from $X$ to $Y$: subset $R\subseteq X\times Y$
  • $(x,y)\in R⟺xRy$
• function
  • relation $R\subseteq X\times Y$: a function $R:X\to Y$
    • if $\forall x,\exists y\in Y(xRy\wedge \forall y^{\prime }\in Y,xR{y}^{\prime }⟹y=y^{\prime })$