NoteNextra-origin/content/Math4111/Math4111_L6.md

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^{-1}(T)=\{x\in A: f(x)\in T\}

• 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

S=\bigcup^{\infty}_{n=1} E_n

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.

QED