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Math4111Introduction to Real Analysis (Lecture 6)

Lecture 6

Review

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

(a) neither injective nor surjective.

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

(b) surjective but not injective.

A=[0,3π2],B=[1,1]A=[0,\frac{3\pi}{2}],B=[-1,1]

(c) injective but not surjective.

A=[0,π],B=[1,1.1]A=[0,\pi],B=[-1,1.1]

(d) bijective.

A=(0,π),B=(1,1)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:ABf:A\to B" means ”ff is a function from AA to BB

AA:“domain”, and BB: “co-domain”.

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

The “image” of ff is f(A)f(A).

If TBT\subset B, the inverse image (pre-image) of TT under ff is

f1(T)={xA:f(x)T}f^{-1}(T)=\{x\in A: f(x)\in T\}
  • ff is injective or one-to-one if x1,x2A\forall x_1,x_2\in A such that f(x1)=f(x2)f(x_1)=f(x_2), then x1=x2x_1=x_2. (f(x1)=f(x2)    x1=x2x1x2    f(x1)f(x2)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))

  • ff is surjective or onto if yB,xA\forall y\in B, \exists x\in A such that f(x)=yf(x)=y. (f(A)=Bf(A)=B)

  • ff is bijective if it’s both injective and surjective.

Definition 2.3

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

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

Cardinality

Definition 2.4

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

(b) AA is infinite if it’s not finite

(c) AA is countable if ANA\sim \mathbb{N}

(d) AA is uncountable if AA is neither finite nor countable

(e) AA 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 AA is a function f:NAf:\mathbb{N}\to\mathbb{A}

Note: By conversion, instead of f(1),...f(n)f(1),...f(n), we usually write x1,x2,...,x3x_1,x_2,...,x_3 and we say {xn}n=1\{x_n\}_{n=1}^{\infty} is a sequence.

Theorem 2.8

Every infinite subset of countable set AA is countable.

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

Definition 2.9 (arbitrary unions and intersections)

Let AA be a set (called the “index set”). For each αA\alpha\in A, let EαE_{\alpha} be a set.

Union: αAEα={x:αA\bigcup_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A such that xEα}x\in E_{\alpha}\}

Intersection: αAEα={x:αA\bigcap_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A such that xEα}x\in E_{\alpha}\}

Special notation for special cases:

m=1nEm\bigcup^{n}_{m=1}E_m and E1E2...EnE_1\cup E_2\cup ...\cup E_n are by definition α{1,..,n}Eα\bigcup_{\alpha \in \{1,..,n\}}E_{\alpha}

and m=1Em\bigcup^{\infty}_{m=1}E_m and E1E2...EnE_1\cup E_2\cup ...\cup E_n are by definition αNEα\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 {En},n=1,2,3,...\{E_n\},n=1,2,3,... be a sequence of countable sets, and put

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

The SS 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

AA is countable, nNn\in \mathbb{N},

    An={(a1,...,an):a1A,anA}\implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}, is countable.

Proof

Induct on nn,

Base case n=1n=1,

True by assumptions

Induction step: suppose An1A^{n-1} is countable. Note An={(b,a):bAn1,aA}=bAn1{(b,a),aA}A^n=\{(b,a):b\in A^{n-1},a\in A\}=\bigcup_{b\in A^{n-1}\{(b,a),a\in A\}}.

Since bb is fixed, so this is in 1-1 correspondence with AA, so it’s countable by Theorem 2.12.

Theorem 2.14

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

Proof

Let EAE\subset A be a countable subset. We’ll show A\EϕA\backslash E\neq \phi (i.e.tA\exists t\in A such that tEt\notin E)

EE is countable so we can list it’s elements S1,S2,S3,...S_1,S_2,S_3,....

Then we define a new sequence tt which differs from S1S_1‘s first bit and S2S_2‘s second bit,…

This is called Cantor’s diagonal argument.

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