Lecture 6

Review

Let $A$ and $B$ be subset of $\mathbb{R}$, and consider the function $f:A \to B$ defined by $f(x)=cos(x)$. Find choices for the domain $A$ and co-domain $B$ which make $f$...

(a) neither injective nor surjective.

$A=\mathbb{R},B=\mathbb{R}$

(b) surjective but not injective.

$A=[0,\frac{3\pi}{2}],B=[-1,1]$

(c) injective but not surjective.

$A=[0,\pi],B=[-1,1.1]$

(d) bijective.

$A=(0,\pi),B=(-1,1)$

injective: y don't repeat, surjective: there exists x for each y

New Materials (Chapter 2: Basic topology (4171). Finite, countable, and uncountable sets)

Functions

Definition 2.1/2.2

"$f:A\to B$" means "$f$ is a function from $A$ to $B$"

$A$:"domain", and $B$: "co-domain".

If $S\subset A$, the range of $S$ under $f$ is $f(S)=\{f(x):s\in S\}$

The "image" of $f$ is $f(A)$.

If $T\subset B$, the inverse image (pre-image) of $T$ under $f$ is

• $f$ is injective or one-to-one if $\forall x_1,x_2\in A$ such that $f(x_1)=f(x_2)$, then $x_1=x_2$. ($f(x_1)=f(x_2)\implies x_1=x_2 \equiv x_1\neq x_2\implies f(x_1)\neq f(x_2)$)

• $f$ is surjective or onto if $\forall y\in B, \exists x\in A$ such that $f(x)=y$. ($f(A)=B$)

• $f$ is bijective if it's both injective and surjective.

Definition 2.3

If $\exists$ bijection $f:A\to B$, we say:

• $A$ and $B$ can be put into 1-1 correspondence
• $A$ and $B$ b oth have the same cardinality
• $A$ and $B$ are equivalent $A\sim B$

Cardinality

Definition 2.4

(a) $A$ is finite if $A\neq \phi$ or $\exists n\in \mathbb{N}$ such that $A\sim \{1,2,...,n\}$

(b) $A$ is infinite if it's not finite

(c) $A$ is countable if $A\sim \mathbb{N}$

(d) $A$ is uncountable if $A$ is neither finite nor countable

(e) $A$ is at most countable if it's finite or countable

Note in some other books call (c) countable infinite, and (e) for countable

Definition 2.7

A sequence in $A$ is a function $f:\mathbb{N}\to\mathbb{A}$

Note: By conversion, instead of $f(1),...f(n)$, we usually write $x_1,x_2,...,x_3$ and we say $\{x_n\}_{n=1}^{\infty}$ is a sequence.

Theorem 2.8

Every infinite subset of countable set $A$ is countable.

Ideas of proof: if $A$ is countable, so we can list its element in a sequence. and we iterate $E\subset A$ to create a new order function by deleting element $\notin E$

Definition 2.9 (arbitrary unions and intersections)

Let $A$ be a set (called the "index set"). For each $\alpha\in A$, let $E_{\alpha}$ be a set.

Union: $\bigcup_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A$ such that $x\in E_{\alpha}\}$

Intersection: $\bigcap_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A$ such that $x\in E_{\alpha}\}$

Special notation for special cases:

$\bigcup^{n}_{m=1}E_m$ and $E_1\cup E_2\cup ...\cup E_n$ are by definition $\bigcup_{\alpha \in \{1,..,n\}}E_{\alpha}$

and $\bigcup^{\infty}_{m=1}E_m$ and $E_1\cup E_2\cup ...\cup E_n$ are by definition $\bigcup_{\alpha \in \mathbb{N}}E_{\alpha}$

Note: Despite the $\infty$ symbol this def makes no references to limits, different from infinite sums.

Countability

Theorem 2.12

"A countable union of countable sets is countable".

Let $\{E_n\},n=1,2,3,...$ be a sequence of countable sets, and put

The $S$ is countable.

Proof by infinite grid, and form a new sequence and remove duplicates.

Corollary

An at most countable union of at most countable sets is at most countable.

Theorem 2.13

$A$ is countable, $n\in \mathbb{N}$,

$\implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}$, is countable.

Proof: Induct on $n$,

Base case $n=1$,

True by assumptions

Induction step: suppose $A^{n-1}$ is countable. Note $A^n=\{(b,a):b\in A^{n-1},a\in A\}=\bigcup_{b\in A^{n-1}\{(b,a),a\in A\}}$.

Since $b$ is fixed, so this is in 1-1 correspondence with $A$, so it's countable by Theorem 2.12.

Theorem 2.14

Let $A$ be the set of all sequences for 0s and 1s. Then $A$ is uncountable.

Proof: Let $E\subset A$ be a countable subset. We'll show $A\backslash E\neq \phi$ (i.e.$\exists t\in A$ such that $t\notin E$)

$E$ is countable so we can list it's elements $S_1,S_2,S_3,...$.

Then we define a new sequence $t$ which differs from $S_1$'s first bit and $S_2$'s second bit,...

This is called Cantor's diagonal argument.