28-2-25 - Lecture

Functions cont.

Injective Functions

$f:A\to B$

$f$ is injective (one-to-one) if $\forall x_1,x_2\in A$ , $x_1\ne x_2\Rightarrow f(x_1)\ne f(x_2)$
$\forall x_1,x_2\in A$ , $x_1=x_2\Rightarrow f(x_1)=f(x_2)$

$p\Rightarrow q$, Contrapositive: $\neg q\Rightarrow \neg p$

$A=\{a\}$
$f:A\to B$ injective (one-to-one)

Always give domain with function

Surjective Functions

$f$ is surjective (onto) if f(A) = B

• I.e. The image (range) = the domain
• The output is entirely covered by $f$
• The codomain is the same as the image
• All values of A and B are used

Bi-jective

• A function that is both injective and surjective
• “one-to-one correspondence”

Inverse Function

• Must be an bi-jective function $f^{-}:B\to A$

Definition
$f:A\to B$ is invertible if $\exists f^{-1}:B\to A$ such that:

1. $\forall x\in A$, $f^{-}(f(x))=x$ - injective
2. $\forall y\in B$, $f(f^{-1}(y))=y$ - surjective

Composition of Functions

$f:A\to B$
$g:C\to D$
$(g\circ f)(x)=g(f(x))$
$P(A)\subseteq C$
$g\circ f:A\to D$
$x\mapsto (g\circ f)(x)=g(f(x))$

Example 1

$f:(R)\to (R)$
$x\mapsto x^2+1$

$g:(R)\to (R)$
$y\to 3y+2$

$(g\circ f)(x)=g(f(x))=g(x^2+1)=3(x^2+1)+2$
$(f\circ g)(x)=f(g(x))=f(3y+2)=(3y+2{)}^2+1$

- The composition is not commutative

Example 2

$f:A\to B$

$f(x)=-\sqrt{x}$
$f:{\mathbb{R}}_{+}\to \mathbb{R}$
$x\mapsto -\sqrt{x}$

$g(y)=y^{\sqrt{3}}$
$g:{\mathbb{R}}_{+}^{\ast }\to \mathbb{R}$
$y\mapsto y^{\sqrt{3}}$

$g\circ f$ does not exist?

$I_A:A\to A$
$x\mapsto x$
$f^{-1}\circ f=I_A$

Idk Heading?

$f(x)=\frac{1}{x-1}$
$f:\mathbb{R}\setminus 1\to \mathbb{R}$
$x\mapsto \frac{1}{x-1}$

$y=\frac{1}{1-x}$
Injectivity
$\forall y\in B$ the sol x must be unique (I.e. at must 1 solution) $y-yx=1$ $yx=y-1$ $x=\frac{y-1}{y}$ if $y\ne 0$ Subjectivity

• At most one intersection $f$ is not subjective for $y=0$ $f^{-1}:{\mathbb{R}}^{\ast }\to \mathbb{R}\setminus \{1\}$ $y\mapsto \frac{y-1}{y}$

Real Numbers

Groups of Axioms

• Field Axioms
• Order axioms
• Supremum (dedecand???) axioms

Operations

$+:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ $\tau :A\times A\to A$

Axioms

Field Axioms

1. $\forall a,b\in \mathbb{R}$, $a+b=b+a$ and $a\ast b=b\ast a$
2. $\forall a,b\in \mathbb{R}$, $(a+b)+c=a+(b+c)$ and $(a\ast b)\ast c=a\ast (b\ast c)$
3. $\forall a,b\in \mathbb{R}$, $(a+b)\ast c=a\ast c+b\ast c$
4. $\forall a\in \mathbb{R}$, $0+a=a+0=a$
5. $\forall a\in \mathbb{R}$, $1\ast a=a\ast 1=a$
6. $\forall a\in \mathbb{R},\exists (-a)\in \mathbb{R}$ : $a+(-a)=(-a)+a=0$
7. $\forall a\in {\mathbb{R}}^{\ast },\exists a^{-1}\in \mathbb{R}$ : $a\ast a^{-1}=a^{-1}\ast a=1$

By definition: $\forall a,b\in \mathbb{R}$ $a-b=a+(-b)$

$\forall a,b\in \mathbb{R},b\ne 0$ : $\frac{a}{b}=a\ast b^{-1}$

Field axioms are not enough to distinguish real numbers.

Ordering Axioms

Rules out complex numbers

1. $\forall a\in \mathbb{R}$, $\neg (a<a)$

2. $\forall a,b\in \mathbb{R}$, $a<b\Rightarrow \neg (b<a)$

3. $\forall a,b,c\in \mathbb{R}$, $a<b$ and $b<c\Rightarrow a<c$

4. $\forall a,b\in \mathbb{R},a\ne b$, either $a<b$ or $b<a$

5. $\forall a,b,c\in \mathbb{R}$, $a<b\Rightarrow a+b<b+c$

6. $\forall a,b\in \mathbb{R}$, $a<b$ and $0<c\Rightarrow ac<bc$

Continued