cognitive conception
completeness property 完整属性
upper and lower bound 上界、下界
least upper bound 最小上界 LUB
subset relations
Natural numbers, Integers, Rational numbers, Real numbers
N ⊂ Z ⊂ Q ⊂ R
Elements of Set Theory(SUPPLEMENT, KEITH DEVLIN: Introduction to Mathematical Thinking)
empty set ∅
different from Φ(phi) the 21st letter of the Greek alphabet
eg. Set A
If $A(x)$ is some property, the set of all those x which satisfy $A(x)$ is denoted by {$x | A(x)$}
Union of A and B
A ∪ B = {x | (x ∈ A) ∨ (x ∈ B)}
the set of all objects which are members of either one of A and B.
The intersection of the sets A, B
A ∩ B = {x | (x ∈ A) ∧ (x ∈ B)}
the set of all members which A and B have in common.
Complement of the set A
the universal set U
$𝐴^′$ = {$x$ ∈ U | $x$ ∉ A}
Venn diagrams
illustrate the below theorem.
the associative laws
(1) $A ∪ (B ∪ C) = (A ∪ B) ∪ C$
(2) $A ∩ (B ∩ C) = (A ∩ B) ∩ C$
结合律
commutative laws
(3) $A ∪ B = B ∪ A$
(4) $A ∩ B = B ∩ A$
交换律
the distributive laws
(5) $A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)$
(6) $A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)$
分配律
the De Morgan laws
(7) $(A∪B)^′$ = $A^′∩ B^′$
(8) $(A∩B)^′$ = $A^′∪ B^′$
the complementation laws
(9) $A ∪ A^′$ = U
(10) $A ∩ A^′$ = ∅
互补律
self-inverse law
(11) $（A^′）^′$ = $A$
自逆律
Beginning of “real analysis”
Lecture 10B - Real Analysis 2
Theorem: the rational line is not complete.
Proof: let A={r∈ Q | r≥0 ∧ $r^2$<2 }
?
As n gets larger
$\frac{n^2}{2n+1}$
$\frac{n^2}{2n+1}\ >\ \frac{p2}{2q^2\ -\ p^2}$
Real number sequences
the number get close to a limit.
$\left{\frac{1}{n}\right}_{\left(n=1\right)}^∞$ = {1, $1/2, 1/3,$ …… }, the number arbitrarily get closer to 0
alternatively sequence
$\left{\left(-1\right)^{\left(n+1\right)}\right}_{\left(n=1\right)}^∞$= {1, -1, 1, -1, …… }
geometric intuition 几何直觉
( Ε, ε )epsilon, the fifth letter of the Greek alphabet.
ASSIGNMENT 10.1 (for Lecture 10A)
1. interval intersection
set there are open intervals (a,b),(c,d), a<c<b<d
so (c,b) is another interval (why not write in this form?)
that (a,c)<(c,b)<(b,d)
this is the same true to closed interval.
it is not the same true for unions, because they don’t have intersection.
⊔, ⊓
let A=(a,b), C=(c,d)
A⊓C = {x| a<x<b } ⊓ {x| c<x<d }
= {x| max(a,c) <x< min(b,d)}
= (max(a,c), min(b,d))
similarly for closed and half-open intervals
it is false for unions, eg. (0,1)⊔(4,5) is not an interval.
- express in intervals or unions of intervals
5. Let A be a set of integers, rationals, or reals. Prove that b is the least upper bound of A iff:
(a) ($∀a$ ∈ A)(a ≤ b); and (b) whenever c < b there is an a ∈ A such that a > c.
Prove:
(a) says b is an upper bound of A.
b is a LUB iff no c<b is an upper bound.(*? что Ето)
iff, for any c<b, c is not an upper bound.
iff, for any c<b, there is an $a∈A$ such that ¬(a≤c).
iff, for any c<b, there is an $a∈A$ such that c<a.
c<a≤b
6.The following variant of the above characterization is often found. Show that b is the lub of A iff:
(a) ($∀a$ ∈ A)(a ≤ b) ; and (b) ($∀e$ > 0)($∃a$ ∈ A)($a>b−e$ )
Prove:
(a) says b is an upper bound of A.
a ≤ b
a > b − e
$e$ > 0
8. Show that any finite set of integers/rationals/reals has a least upper bound.
(a<10), a∈Z , iff LUB = 9
(a<10), a∈Q, LUB = b, iff b∈Q, a<b<10
(a<10), a∈R, LUB = b, iff b∈Q, a<b<10
11. Define the notion of a lower bound of a set of integers/rationals/reals.
a lower bound
(a>0), a∈Z , LB = b, iff b∈Z, 0<b≤a, b=1
(a>0), a∈Q, LB = b, iff b∈Q, 0<b≤a
(a>0), a∈R, LB = b, iff b∈R, 0<b≤a
ASSIGNMENT 10.2 (for Lecture 10C)
- A = {r ∈ Q | r > 0 ∧ $r^2$ > 3}. translate it as :
A = {r ∈ Q | r > 0 ∧ [$r$< -$\sqrt{3}$)V($r$ > $\sqrt{3}$)] }
A = {r ∈ Q | $r$ > $\sqrt{3}$ }
A = {r ∈ Q | ( $\sqrt{3}$, ∞ )}
A has a lower bound in Q, LB= $\sqrt{3}$
but no greatest lower bound and upper bound.
4. Prove that $(\frac{n}{n+1})^2$ → 1 as n → ∞.
give 𝜀>0, we seek an integer N s.t.(such that)
n≥ N ⇒ | $(\frac{n}{n+1})^2$-1 | < 𝜀
i.e, s.t. n≥ N ⇒ | $\frac{n^2-n^2-2n-1}{\left(n+1\right)^2}$ | < 𝜀
i.e, s.t. n≥ N ⇒ $\frac{2n+1}{\left(n+1\right)^2}$ < 𝜀
Pick N so big, that $\frac{\left(n+1\right)^2} {2n+1}$ ≥ $\frac{\left(N+1\right)^2} {2N+1}$ > 𝜀
then, n≥ N ⇒ $\frac{2n+1}{\left(n+1\right)^2}$ ≤ $\frac{2N+1}{\left(N+1\right)^2}$ < 𝜀
so, n≥ N ⇒ | $(\frac{n}{n+1})^2$-1 | < 𝜀