Skip to Content
Math4111Introduction to Real Analysis (Lecture 6)

Lecture 6

Review

Let AA and BB be subset of R\mathbb{R}, and consider the function f:Aβ†’Bf: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=[0,3Ο€2],B=[βˆ’1,1]A=[0,\frac{3\pi}{2}],B=[-1,1]

(c) injective but not surjective.

A=[0,Ο€],B=[βˆ’1,1.1]A=[0,\pi],B=[-1,1.1]

(d) bijective.

A=(0,Ο€),B=(βˆ’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:Aβ†’Bf:A\to B" means ”ff is a function from AA to BB”

AA:β€œdomain”, and BB: β€œco-domain”.

If SβŠ‚AS\subset A, the range of SS under ff is f(S)={f(x):s∈S}f(S)=\{f(x):s\in S\}

The β€œimage” of ff is f(A)f(A).

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

fβˆ’1(T)={x∈A:f(x)∈T}f^{-1}(T)=\{x\in A: f(x)\in T\}
  • ff is injective or one-to-one if βˆ€x1,x2∈A\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)β€…β€ŠβŸΉβ€…β€Šx1=x2≑x1β‰ x2β€…β€ŠβŸΉβ€…β€Šf(x1)β‰ 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 βˆ€y∈B,βˆƒx∈A\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 βˆƒ\exists bijection f:Aβ†’Bf: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 A∼BA\sim B

Cardinality

Definition 2.4

(a) AA is finite if Aβ‰ Ο•A\neq \phi or βˆƒn∈N\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 A∼NA\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:N→Af:\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 EβŠ‚AE\subset A to create a new order function by deleting element βˆ‰E\notin E

Definition 2.9 (arbitrary unions and intersections)

Let AA be a set (called the β€œindex set”). For each α∈A\alpha\in A, let EΞ±E_{\alpha} be a set.

Union: β‹ƒΞ±βˆˆAEΞ±={x:βˆƒΞ±βˆˆA\bigcup_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A such that x∈EΞ±}x\in E_{\alpha}\}

Intersection: β‹‚Ξ±βˆˆAEΞ±={x:βˆƒΞ±βˆˆA\bigcap_{\alpha \in A}E_{\alpha}=\{x:\exists \alpha \in A such that x∈EΞ±}x\in E_{\alpha}\}

Special notation for special cases:

⋃m=1nEm\bigcup^{n}_{m=1}E_m and E1βˆͺE2βˆͺ...βˆͺEnE_1\cup E_2\cup ...\cup E_n are by definition β‹ƒΞ±βˆˆ{1,..,n}EΞ±\bigcup_{\alpha \in \{1,..,n\}}E_{\alpha}

and ⋃m=1∞Em\bigcup^{\infty}_{m=1}E_m and E1βˆͺE2βˆͺ...βˆͺEnE_1\cup E_2\cup ...\cup E_n are by definition β‹ƒΞ±βˆˆNEΞ±\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

β€œA 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=1∞EnS=\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, n∈Nn\in \mathbb{N},

β€…β€ŠβŸΉβ€…β€ŠAn={(a1,...,an):a1∈A,an∈A}\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 Anβˆ’1A^{n-1} is countable. Note An={(b,a):b∈Anβˆ’1,a∈A}=⋃b∈Anβˆ’1{(b,a),a∈A}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 EβŠ‚AE\subset A be a countable subset. We’ll show A\Eβ‰ Ο•A\backslash E\neq \phi (i.e.βˆƒt∈A\exists t\in A such that tβˆ‰Et\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_1β€˜s first bit and S2S_2β€˜s second bit,…

This is called Cantor’s diagonal argument.

Last updated on