Cardinal Arithmatic

Week 3/2 Cardinal arithmeti 38

Class 3.2

3.2. Denition. The ardinality of a set A is denoted |A|.Temporary denition: If A is innite, then

|A| = {B : B A}.

Problem 3.45. If n is a natural number then |n| < |n+ 1|.

3.3. Denition. A and B are sets.(1)We say that A has ardinality aleph zero (omega) , or the ar-dinality of A is aleph zero /(omega) and write |A| = 0 (|A| = ) iA is ountable.(2) We say that B has ardinality ontinuum, or the ardinality ofB is ontinuum, and write |B| = c or |B| = 2 i B R.

3.4. Denition. (1) If and are ardinals, then pik a set A of

ardinality and a set B of ardinality suh that A B = , andlet

+ = |A B|. (5)

(2) If and are ardinals, then pik a set A of ardinality and aset B of ardinality , and let

= |AB|.

3.5. Proposition. The denition is meaningful, i.e. there are A andB with the required properties, and the denition does not depend onthe hoie of A and B.

Proof. Addition: By denition, there are A and B with = |A| and|B| = . Now let A = A {0} and B = B {1}. Then A A,B B and A B = .Next we show that that denition does not depend on the hoie

of A and B. Indeed, assume that f : A A and g : B B,A B = . Then h : A B A B, where

h(x) =

{f(x) if x Ag(x) if x B

(Atually h = f g). So |A B| = |A B|.Similar argument works for multipliation.

3.6. Proposition. If , , are ardinals, then

(1) + = + ,


(2) (+ ) + = + (+ ).(3) = ,(4) ( ) = ( ) (5) (+ ) = ( ) + ( )

Proof. Let A,B,C be pairwise disjoint sets with ardinalities , , ,respetively.

(1): + = |A B| = |B A| = + .(2) We have + = |A B| and + = |B C|. Moreover,(+)+ = |(AB)C| beause (AB)C = , and (+(+) =|A(BC)| beause A(BC) = . Sine (AB)C = A(BC),we are done.

(3) f : A B B A provided f(a, b) = b, a.(4) Dene f : A(BC) (AB)C by the formula f(a, b, c) =a, b , c.(5) Homework.

3.7. Proposition. (1) + = = and 2+2 = 2 2 = 2.(2) If is an unountable ardinal, then + = .

Proof. (1) Straightforward: we proved that NN N and RR R.(2) Let |A| = and |C| = , AC = . We proved that A AC.Thus = |A| = |A C| = + .

3.8. Theorem ( Fundamental theorem of ardinal arithmeti). If and are innite ardinals, then

+ = = max(, ).

We prove it weeks later. The Fundamental theorem of ardinal

arithmeti is equivalent to Axiom of Choie.

3.9. Denition. If and are ardinals, then pik a set A of ar-dinality and a set B of ardinality and let

= |AB|.

3.10. Proposition. The denition of ardinal exponentiation is mean-

ingful, i.e. the denition does not depend on the hoie of A and B.

Proof. It was a homework problem.

Observe that 2 = |2N| = |P(N)| = |R|.

3.11. Proposition. 2 > for any ardinality .


Proof. If = |X|, then 2 = |2X | = |P(X)| > |X| = by CantorTheorem 2.4.

3.12. Proposition. If , , are ardinals, then

(1) = ()

(2) ( ) = .(3) + = .

Proof. Let A,B,C be sets with ardinalities , , , respetively.(1) To prove = () we need or show that

ABC (AB)C .

Dene : (AB)C ABC as follows: if f : C AB let

(f) : B C A

suh that suh that

(f)(b, c) = f(c)(b)

for eah b, c B C.Then : (AB)C ABC .

(2) To prove ( ) = we need to show that

(A B)C AC BC .

This is a hw problem.

(3) To prove + = we need to show if B C = then

ABC AB AC .

This is a hw problem.

3.13. Proposition. Assume that and are ardinals.Then

(1) + + ,(2) ,

(3) if > 0 or = 0 then .

Proof. (1) Let A,B,C,D be pairwise disjoint sets with ardinalities, , , , respetively.Then + = |A C| and + = |B D|. Sine , there is

an injetion f : A B. Sine , there is an injetion g : C D.Now let h = fg. Then h : AC BD. Thus |AC| |BD|.(2) and (3) are homeworks.


3.14. Denition. Let I be a set and let i be a ardinal for i I.(1) Let {Ai : i I} be a set of pairwise disjoint sets suh that|Ai| = i for i I. Let

iI

i ={Ai : i I

}.(2) Let {Ai : i I} be a set of sets suh that |Ai| = i for i I. LetiI

i ={f : f is a funtion, dom(f) = I, and f(i) Ai for i I}

.3.15. Proposition. The denitions above are meaningful.

Proof. For any set {Bi : i I} of sets there is a set {Ai : i I} ofpairwise disjoint sets suh that |Ai| = |Bi|: let Ai = Bi {i}.

3.16. Proposition. (1) Let I be a set, be a ardinal, and let i = for i I. Then

iI

i = |I| andiI

i = |I|.

Proof. Let |A| = , and put Ai = A {i} for i I. TheniI

i = |iI

Ai| = |A I| = |A| |I| = |I|,

beause

iI Ai = A I. Moreover,

iI

i =

{f : f is a funtion, dom(f) = I, and f(i) A for i I} =

|IA| = |I|.

3.17. Proposition. (1) Let I be a set and let i i be ardinalsfor i I. Then

iI

i iI

i andiI

i iI

i.

(2) The assumption i < i for i I does not implies either

iI i


Proof. (2) Let ei = 2i and oi = 2i + 1 for i . Then ei < oi,but

i ei =

i oi = and

i>0 ei =

i>0 oi = 2

. Indeed, we

prove

3.17.1. Claim. If X \ {0} is innite, then

iX i = andiX i = 2

.

Indeed,

= 1 =iX

1 iX

i iX

= |X| = ,

and

2 =iX

2 iX

i iX

= |X| = (2) = 2 = 2.

3.18. Proposition. 2

iIi =

iI 2

i.

Proof. Let {Ai : i I} be a set of pairwise disjoint sets suh that

|Ai| = i for i I. Let A ={

Ai : i I}. Then

iI i = |A|,

and so 2

iIi = |2A|.

Moreover, 2i = |Ai2| for i I, soiI

2i = |{f : f is a funtion, dom(f) = I, and f(i) Ai2 for i I}.

Dene a bijetion

: 2A {f : f is a funtion, dom(f) = I, and f(i) 2Ai for i I}

as follows: if h 2A, then for all i I let (h)(i) = h Ai, where

h X = {x, h(x) : x X}.

3.19. Proposition. If {An : n N} P(N N), |An| < 2, then

nNAn 6= P(N N).

Proof. Let Bn = {n} N [N N

]N. Let Bn = {ABn : A An}.

Then |Bn| |An| < 2 = |P(Bn)|, so there is Cn Bn suh that

Cn / Bn. Then let C ={Cn : n N}. Then C /

{An : n

N}

Sine |P(NN)| = |P(N)| = |R|, we obtain the following orollary:


3.20. Corollary. If R =nNAn, then |An| = |R| for some n N.

Proof. Let f : R P(N N), and put Ai = fAi for i < . Then

P(N N) =iAi, so |Ai| = 2

for some i by Theorem 3.19.

So |Ai| = 2as well.

3.21. Theorem (Knig's Theorem). Let I be a set and let i < ibe ardinals for i I. Then

iI

i 0 or = 0 then .

Problem 3.57. Prove that

(i) 2 = (2),(ii) 2 = ,(iii) (2

) > .

Week 3/HW Cardinal arithmeti 46

Extra problems

Problem 3.58. There are exatly 2 many open subsets of the reals.

Problem 3.59. There is a ardinal > suh that >

Problem 3.60. There is an open subset U of (0, 1)2 suh that forevery open subset V (0, 1) there is x (0, 1) suh that

V = {y : x, y U}.

Problem 3.61. Let G = (V,E) be a graph. A partition (A,B) of Vis alled unfriendly i every vertex has at least as many neighbor

in the other lass as in its own.

(a) Prove that every nite graph has an unfriendly partition.

(b) Assume that every vertex in a graph G = (N, E) has innitedegree. Prove that G has an unfriendly partition.

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