COMP 2711H: Lecture 29

Date: 2024-11-06 09:01:05

Reviewed:

Topic / Chapter:

summary

❓Questions

Notes

Size of Sets

Size of sets
- $∣ A ∣ = ∣ B ∣$ is there is bijection $f : A \to B$
- $∣ A ∣ \leq ∣ B ∣$ is there is injection $f : A \to B$
- set $A$ : finite if $\exists n \in N ∣ A ∣ = ∣ {0, 1, \dots, n - 1} ∣$
  - infinite if it's not finite
- suppose: $f : A \to B$ and $X \subseteq A$
- $f [X] = {y \in B ∣\exists x \in X f (x) = y}$
  - 👨‍🏫 an image of $X$
  - result in applying $f$ on all elements of $X$
- set $A$ : countable if $∣ A ∣ = ∣ N ∣$
  - 👨‍🏫 at least for this course!
Theorem
- theorem: countable sets are the smallest infinite set
 - i.e. if $A$ finite: $\exists X \subseteq A ∣ X ∣ = ∣ N ∣$
 - lemma: if $A$ is infinite and $a \in A$ , then
 A{a} is also infinite
 - suppose: $A {a} is finite$
 - which means: $\exists n \in N ∣ A {a} ∣ = ∣ {0, 1, 2, \dots, n - 1} ∣$
 - $f : A {a} \to {0, 1, 2, \dots, n - 1} s.t. it's 1-1 and onto$
 - $g : A \to {0, 1, 2, \dots, (n + 1) 01}$
 
 s.t. it's also 1-1 and onto
 - thus: $A$ is finite (contradiction)
 - pick a smallest element $a_{0} \in A$
 - by lemma above: $A_{1} = A {a_{0}} is infinite$
 - pick $a_{1} \in A$ , $A_{2} = A_{1} {a_{1}} is infinite$
 - ... can be continued
 - above step: result in an infinite sequence
 - as it's a sequence: it's countable
 - then $X = {x ∣\exists i x = a_{i}}$
- theorem: if $A$ is countable and $x \neq \in A$ , then $A \cup {x}$ is countable
 - $f : A \to N$
 - let new mapping:
 - $g : A \to N$
 - $g (x) = 0$
 - $\forall a \in A, g (a) = f (a) + 1$
 - 👨‍🏫 similarly, by adding $k$ , or finite, items to countable $A$ , the result is still countable
 - proof: use one by one
- theorem: if $A, B$ countable: so is $A \cup B$
 - assume $A \cap B = ϕ$
 - or else: let $B = B - A$
 - then $A \to f N \to f^{'} E$
 - i.e. even numbers
 - similarly, $B \to g N \to g^{'} O$
 - i.e. odd numbers
 - 👨‍🏫 then combine!
 - suppose: $f : A \to N$ and $g : B \to N$ are bijection
 - let $h : A \cup B \to N$ be defined as
 - $\forall a \in A h (a) = 2 \cdot f (a)$
 - $\forall b \in B h (b) = 2 \cdot g (b) + 1$
 - 👨‍🏫 $A, B$ 's disjoint condition is used here: cannot assign two values!
- theorem: let $X = {x_{1}, x_{2}, \dots}$ be a countable set
 - whose every element $x_{i}$ is also a countable set
 - then $⋃ X$ is also countable
 - proof
 - $f : X \to N$ : a bijection
 - $\forall x \in X, g_{x} : x \to N$ : a bijection
 - consider $h (y) : ⋃ X \to N \times N$
 - $\exists x \in X \exists y \in x, y \mapsto h (f (x), g_{x} (y))$
 - two indices, and each map: size $∣ N ∣$
 - 👨‍🏫 finally: $h$ is a bijection
 - again, $x_{i}, x_{j}$ must be disjoint for it to be one-to-one
 - yet: every case can be reduced to disjoin case
 - and $N \times N \to h^{'} N$
 - where $h^{'}$ is bijection
 - 👨‍🎓 use similar idea as in $N^{2}$ being countable
- $Q$ is also countable
 - and $Q$ :
 {(a,b)∣a∈Z,b∈N{0},gcd(a,b)=1}
 - i.e. $\frac{a}{b}$
 - $Q \subseteq Z \times N$
 - $Q \to 1 - 1 Z \times N \to 1 - 1, onto N \times N \to 1 - 1, onto N$
 - $∣ Q ∣ \leq ∣ N ∣$ , $∣ Q ∣ \geq ∣ N ∣$
 
 👨‍🏫 we haven't proved that $a \leq b \land a \geq b ⟹ a = b$ though
 - can't we just say $N^{2}$ is countable
 
 and $Q \subseteq N^{2}$ ?
- theorem: let $X = N \cup N^{2} \cup N^{3} \cup \dots$
 - $X = {(x_{1}, x_{2}, \dots) ∣}$
 - every natural number: w/ unique factorization
 - $P = {p_{1}, p_{2}, \dots}$
 - and $p_{1} < p_{2} < \dots$
 - define $f : X \to N$ s.t. it's one-to-one and onto
 - first try: $f (x_{1}, x_{2}, \dots, x_{k}) = p_{1}^{x_{1}} p_{2}^{x_{2}} \dots p_{k}^{x_{k}}$
 - problem: $1 = f (0) = f (0, 0) = \dots$
 - or $f (a) = f (a, 0) = f (a, 0, 0) = \dots$
 - thus: not really one-to-one
 - second try: if $x \in X$ and $x$ ending w/ $i_{x}$ 0s
 - define $\overset{x}{ˉ}$ as $x$ w/ its last $i_{x}$ elements removed (no extra 0)
 - $g : X \to N \times N$
 - $g (x) = (i_{x}, f (x))$
 - now: it's $N \times N$ , which is countable
 - 👨‍🏫 corner case on all-0 and one-0 exists
- theorem: let $F = {f ∣ f : N \to {0, 1}}$
 - $F$ : infinite, and not countable
 - 👨‍🏫 some infinities: are larger than other infinities
 - 👨‍🎓~~animal farm~~
 - let $g : N \to F$ : be bijection
 - each function: can be written as binary mapping
 - $g (0) = 0011010101 \dots$
 - $g (1) = 0110101001 \dots$
 - $g (2) = 1110 \dots$
 - $g (n) = \dots$
 - one can make: $t (i) = 1 - g (i) (i)$
 - or reversing at $i$ -th index
 - 👨‍🏫 diagonalization
 - $\forall i, g (i) (i) \neq = t (i)$ , thus $t \neq \in F$
 - thus: $g$ is not onto: contradiction
 - conclusion: $g$ is not bijection
- theorem of Cantor
 - for every set $A$ : $∣ P (A) ∣ \neq = ∣ A ∣$
 - 👨‍🏫 even for $ϕ$ : $∣ P (A) ∣ = 1, ∣ A ∣ = 0$
 - proof: suppose $g : A \to P (A)$ is a bijection
 - $T = {a \in A ∣ a \neq \in g (a)} \in P (A)$
 - consider: $a \in A$
 - then: either $a \in T$ , or $a \neq \in T$
 - case 1: $a \neq \in g (a) ⟹ a \in T$
 
 $g (a) \neq = T$
 - case 2: $a \in g (a) ⟹ a \neq \in T$
 
 $g (a) \neq = T$
 - thus: $g$ is not onto: not covering $T$
 - 👨‍🏫 doing the same thing, without relying on countability
 - 👨‍🏫 as we can apply it multiple times
 - there are infinite sizes of infinite
- theorem of Tarski
 - let $X$ : be a set and $h : P (X) \to P (X)$
 - s.t. $A \subseteq B ⟹ h (A) \subseteq h (B)$
 - there exists $c \subseteq X$ s.t. $h (C) = C$
 - proof
 - set $A \in P (x)$ : expansive if $A \subseteq h (A)$
 - let $\overset{ˉ}{A}$ be a set of expansive sets
 - then every $A \in \overset{ˉ}{A}$ expansive
 - thus $⋃ \overset{ˉ}{A}$ : expansive
 - let $x \in ⋃ \overset{ˉ}{A}$ , thus $x \in h (A) A \subseteq ⋃ \overset{ˉ}{A}} ⟹ x \in h (A) A \subseteq ⋃ \overset{ˉ}{A} ⎭ ⎬ ⎫ ⟹ x \in h (A)$
 - $⋃ \overset{ˉ}{A} \subseteq h (⋃ \overset{ˉ}{A})$
 - $C = ⋃ {A \in P (x) ∣ A \subseteq h (A)}$
 - union of all expansive sets
 - show: $h (C) = C$
 - $C$ is expansive, thus $C \subseteq h (C)$
 - claim: $h (C)$ is expansive
 
 $h (c) \subseteq h (h (C))$
 
 however: $h (C) \subseteq C$
 - and thus $C = h (C)$
- theorem (Schroder-Bernstein): if $∣ X ∣ \leq ∣ Y ∣$ and $∣ Y ∣ \leq ∣ X ∣$
 - then $∣ X ∣ = ∣ Y ∣$
 - $f : X \to 1 - 1 Y$
 - $g : Y \to 1 - 1 X$
 - proof:
 - cut $X$ into
 $Z, X Z$
 
 $Z$ : domain of $f$
 - similarly: cut $Y$ into
 $W, Y W$
 
 $Y$ : domain of $g$
 - goal: find $X \subseteq X$ and $W \subseteq Y$ s.t.
 - $f [Z] = Y - W$
 - $g [W] = X - Z$
 - $Z = X - g [y - f [Z]]$
 - then let $h : P (X) \to P (X)$
 - $h (A) = A - g [Y - f [A]]$
 - 👨‍🏫 show that this function is increasing, and use Tarski's fixed point theorem
 
 $\exists Z, Z = h (Z)$

COMP 2711H: Honors Discrete Mathematical Tools for Computer Science

COMP 2711H: Lecture 29

Date: 2024-11-06 09:01:05

Reviewed:

Topic / Chapter:

Notes

Size of sets

Theorem