26-2-25 - Lecture

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Univocal Predicate There exists at most one x such the p{x} $\forall u,\forall v,p\{u\}\cap p\{v\}\Rightarrow u=v$

Functional Predicate There exists one and only one x such that p{x} ∃x: p{x} and p is univocal

$\exists !$ - there exists only one $\exists !x\in R_{+}:x^2=2$

Set Theory

a ∈ A a is in the set A

Def A, B sets A $\subseteq$ B if $\forall x,x\in A\Rightarrow x\in B$

Def A, B sets A=B if $\forall x,x\in A\iff x\in B$ “extensionality”

Empty Set

$\varnothing$ Take a set A. [empty set] in A if $\forall x,x\in \varnothing \Rightarrow x\in A$

Infinite Sets

$A=\{x|p\{x\}\}$ p{x} = ”$x\in R$ and $x>=3$”

P{x} $x\notin x$ - not valid

A = $\{x|x\notin x\}$ - NOT A SET Either $A\in A\Rightarrow A\notin A$ or $A\notin A\Rightarrow A\in A$ “nasty predicate”

Order

{1,2} = {2,1} Order does not matter in sets “Sets have no order” “Sets have no repetition”

$(p\vee q)\iff (q\vee p)$

A = {x | x=1 or x=2}

{1,1} = {1} - proven using extensionality

Cartesian Product

A,B sets $A\cup B$ = {$x|x\in A$ or $x\in B$} $A\cap B$ = {$x|x\in A$ and $x\in B$} $A\setminus B$ = {$x|x\in A$ and $x\notin B$} $A\times B$ = {$(a,b)|a\in A$ and $b\in B$|}

${\complement }_x(a)=X\setminus A$

Ordered Pair

(a,b) != (b,a) (a,b) = (c,d) ⇒ a=c and b=d

Def (a,b) = {{a}, {a,b}} - needs to give the objects in the set and the order

A = {1,2} B = [2,3] $A\times B$ = {$(x,y)|x\in${1,2} and y $\in$ [2,3]}

$Q=\{\frac{p}{q}|p,q\in Z$ and $q\ne 0$}

[a,b] = {$X\in R|a<=x<=b$} ]a,b[ = (a,b) INTERVAL - NOT ORDERED PAIR

(3,+$\infty$) = {$x\in R|x>3$} (-$\infty$, b] = {$x\in R|x<=b$}

${\mathbb{R}}^{\ast }=\mathbb{R}\setminus \{0\}$ ${\mathbb{R}}_{+}^{\ast }$= {$x\in \mathbb{R}|x>0$}

Functions

Correspondence

A, B, $\Gamma$ sets A domain B codomain $\Gamma$ graph

(A,B,$\Gamma$) is a correspondence if $\Gamma \subseteq A\times B$ A={1,2,3} B={$\alpha$, $\beta$} $A\times B$ = {$(1,\alpha ),(1,\beta ),(2,\alpha ),(2,\beta ),(3,\alpha ),(3,\beta )$} $\Gamma =\{(1,\alpha ),(1,\beta ),(3,\alpha )\}$ Therefore not a correspondence?

Conditions

1. $\forall x\in A,\exists y\in B:(x,y)\in \Gamma$
2. $\forall x\in A,\forall y_1,y_2\in B,(x,y_1)\in \Gamma \text{ and }(x,y_2)\in \Gamma \Rightarrow y_1=y_2$

Domain P: A→B x $\mapsto$ f(x) Functions require a domain and codomain Codomain: e.g. R→R - real numbers to real numbers The range is always contained within the codomain

$\mapsto$ Maps to

Natural domain is the largest possible domain.

A,B sets f: A→B, g:A-B f=g if $\forall x\in A,f(x)=g(x)$

Sin: R→R x $\mapsto$ sin(x)

A,B sets $C\subseteq A$ f:A→B The restriction of f to C is $f{|}_C:c->B,x\mapsto f(x)$

The image of “f” is the set of the range F(A) = {$f(x)|x\in A$} = {$y\in B|\exists x\in A:y=f(x)$}

Continued