Cardinality of Sets (UCLA)

7/21/2019 Cardinality of Sets (UCLA)

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Bootcamp

Christoph Thiele

Summer 2012

Letcure 8: Cardinality of sets

Recall that a relation between from a set Xto a setYis a subset of the set of orderedpairs x, y . Such a relation is called a function if

1. For allx Xthere exists y Y such thatx, y f

2. If for x, x Xand y Y with x, y f and x, y fwe have x= x.

Informally, for every x Xthere exist (first proeprty) a unique (second property)y Y such thatx, y f

A function is called

1. surjective, if for all y Y there exists x Xsuch that x, y f

2. injective, if for every y, y

Y and x X with x, y f and x, y

f wehave y=y .

Note the symmetry between the definition of a function and these two properties.In particular, if and only if a function f is injective and surjective, then the inverserelation g fromY to Xdefined byy, xif and only ifx, y fis a function as well.In this case, this inverse function is injective and surjective as well.

A function is called bijective, if it is injective and surjective.Iff is a function, we write y= f(x) forx, y f.

Definition 1 Iff is a function fromX to Y andg is a function fromY to Z then

g f is defined to be the function fromX to Z defined by(g f)(x) =g(f(x))

Exercise 1 Iff andg are surjective theng f is surjective. Iff andg are injectivetheng f is surjective.

LetA0denote the empty set and assumeAnis already defined then defineAn+1=A{n}. Informally, at elats forn >0, An= 0, 1, 2, . . . , n1 and there aren elementsin An.

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Lemma 1 For everyn and everym An+1 there is a bijective map fn,a fromAn toAn+1\ {m}.

Proof: We define f(k) = k ifk < m and we define f(k) = k+ 1 ifk m. The

inverse map g is defined as g(k) =k ifk < mand g(k) =k 1 ifk > m.

Lemma 2 Letn, mN.

1. There is an injective function fromAn to An+m form0

2. There is a surjective function fromAn+m to An form 0

3. There is no injective function fromAn+m to An form >0

4. There is no surjective function fromAn to An+m form >0

Note that there is a surjective map from Am to A0 if and only ifm= 0. Hence thecase n= 0 is a true exception to the scheme, but since this exception only concernsthe empty set we shall not be too troubled by that.

Proof:1) We fix n and do induction on m. For m= 0 the identity map will do. Assume

we have an injective mapf fromAntoAn+m, then the compositionfn+m,0fis injectivefrom An to An+m+1

2) We fix n and do induction on m. For m= 0 the identity map will do. Assumewe have a surjective map f from An+mtoAn, then the composition f

1n+m,0 f is a

surjective map from An+m+1 to An.

3) We fix m >0 and do induction on n. Ifn = 0, there is no map from Am to A0,since we cannot map 0 Am to any element. Assume there is no injective map fromAn+mtoAn for some n. Assume to get a contradiction that there is an injective mapfrom Am+n+1toAn+1. Ifn+ 1 is not in the range of this map, we may by restrictionobtain an injective map from An+m to An, a contradiction. Assume n+ 1 is in therange of this mapf, then there is a unique k An+m+1such that f(k) =n +1. Thenfn+m,k f is an injective map from An+m to An, a contradiction.

4) We fix m > 0 and do induction on n. If n = 0, there is no surjective mapfrom A0 to Am, since we 0 Am does not have a preimage. Assume there is nosurjective map from AntoAn+m for some n. Assume to get a contradiction that thereis a surjective map from An+1toAn+m+1. Let a = f(n + 1). Then there is a surjectivemap from An to An+m defined by the restriction off to An postcomposed with themap f1n+m,a.

Exercise 2 1. An injective map fromAn to An is surjective.

2. A surjective map fromAn to An is injective.

Definition 2 A setS is called finite, if there is somen N and a bijection fromSto An. If such bjection exists, S is said to have the same cardinality asAn

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Exercise 3 Given a finite set, there is exactly onen N such thatS has the samecardinality asAn.

One may also say simply that S has cardinality n.The set N is not finite. Namely, if it was, then there was an injective map from

N An for some n. This map could be restricted to An+1, giving an injective mapfrom An+1to An, which is impossible.

We have the following observations, destroying hope to have staright forwardgeneralizations of the theorems for finite sets.

Lemma 3 1. There is an injective map fromN toN that is not surjective.

2. There is a surjective map fromN toN which is not injective.

Proof: The map is injective (one Peano axiom) but not surjective (0 is not inthe range, as another Peano axiom states.) Define a map f as follows: f(0) = 0 and

whenever n is of the form (m) (that is whenever n = 0), then m is unique and wedefine f(n) =m. then f is surjective but not injective since 0 has two preimages: 0and 1.

This behaviour is actually typical for every infinite set S. This will follow fromthe following lemma, by defining maps on Sthat are the identity outside the rangeof the map N S given by the lemma and adjusting the desired behaviour insidethe set N.

Lemma 4 Given an infinite setS, there is an injective map N

Proof: The set S is not empty, since it is not finite, hence we find an element

x Sand may define a map A1 S by f(0 = x. Assume we have already defiendan injective mapAn S. This map is not surjective since Sis not finite, hence thereisx otside the range offand we may extend f toAn+1by settingf(n + 1) =x. Thisdefines an injective map from An+1 S. By iterating this extension, we define f(n)for all nN, and this map is injective.

Definition 3 Two sets are said to be equivalent, if there exists a bijection betweenthe two sets.

Note that this is indeed an equivalence relation. It is an equivalence relation onthe class of all sets.

The following will show that not all infinite sets are equivalent.Lemma 5 For every setS, there is no bijection fromSto the set of all subsets ofS

Proof: This is called the Cantor diagonal element. Assume to get a contradictionthat there is such a bijection f. Let Tbe the set of all elements s S in T whichsatisfy s f(s). By surjectivity, there is a t Ssuch that f(t) = T. If t T, thent f(t), a contradiction. IftT, then tf(t) and hence t T, a contradiction.

We define a relation of sets as follows: S T if there exists an injective mapf :S T. This is an order relation on the classof all sets.

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Lemma 6 (Schroder Bernstein) This relation is a partial order in that it satisfies

1. Reflexivity: S S for all setsS.

2. Antisymmetry: S T andTS impliesS=T.

3. TransitivityS T andTU impliesSU.

Proof: Reflexity follows by appealing to the identity map, Transitivity follows bycomposition of two injective maps. The only non-trivial statement is the antisymme-try.

Suppose we have injective maps f : X Y and g : Y X. Define a newinjective map h: X Y X Y byh(z) =f(z) ifx Xand h(z) =g(z) ifz Z.

Define Z0 = XY. Define Z1 = {z Z0 : z Z : f(z) = z} Then this

z is unique and we may define (z) = z. Assume Zn Z is defined, then defineZn+1= {(z) :z Zn Z1}.

Now define the order ofz Z to be the supremum of all nwith z Zn.Now define the map k: Z Z as follows:

1. k(z) =h(z) ifzhas even finite order.

2. k(z) =(z) ifzhas odd finite order.

3. k(z) =h(z) ifzhas infinite order and z X

4. k(z) =(z) ifzhas infinite order and z Y

Then clearly k(k(z) = z for all z, and k maps X to Y and Y to X. hecne k,restricted to X, is a bijection from X to Y.

The statement of the next Lemma, which is essentially the axiom of choice, isthat we could have defined the order alternatively as follows, usiong surjective mapsinstead of injective maps.

For two nonempty sets, S Tif there is a surjective map from T to S. For theempty set we have Sfor all sets and S if and only ifS=.

Lemma 7 (Axiom of Choice) There exists a surjective map from T to S if andonly ifS is not empty and there exists an injective map fromS to T.

Proof: The if part follows by the following easy argument. Assume S is notempty, and assume there is an injective map ST. We need to prove that there isa surjective map TS. let f :STbe the injective map that is assumed to existand pick s Swhich exists since Sis not empty. Define a surjective map as follows:If t T is in the range ofT, define g(t) = s for the unique s such that f(s) = t. Ift T is not in the range off, define g(t) =s. Then this map is surjective, since forevery s Swe have s =g(f(s)).

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The only if part is follows from the axiom of choice. Given a surjective mapg : T S. Note that for every s S there exists t T with g(t) = s. That wecan choose one t for every s and build a function out of that with f(s) =t for thesechosen elements is precisely the axiom of choice. Clearly this map is injective, for

assume f(s) =f(s

) for two elements s, s

. then s= g(f(s) =g(f(s

) =s

. The axiom of choice is a noteworthy axiom of set theory. In is usually source

ofexistence proofs which are highly non-constructive. In general, the injective mapgiven by the above lemma is not constructive. In some cases, for example when T isfoinite, one can construct the injective map as we have done before. This is a casewhen the set theoretic axiom is not really needed and one gets away with other, lescontentious axioms.

Exercise 4 IfT is the power set ofS, thenS T.

Our final major goal is to prove:

Theorem 1 The order is a total order, that is for any two setsS andT we haveST orTS.

Definition 4 AssumeS is a partially ordered set.

1. Sis called totally ordered, if for anyx, y S we havex y ory x.

2. S is called well ordered, if it is totally ordered and every subset T of S has a

minimal element, that is an elementt T such that t s for everys T.

Lemma 8 Let S and T be two well ordered sets. Then one of the following twostatements is true

1. There is an injective monotone map S Tand an elementt Tsuch that therange of the map is the set of allt T witht < t.

2. There is an injective monotone map T S and an element s S such thatthe range of the map is the set of alls S withs < s.

Proof: Consider the setT0 of all t T such that there exists s Sand a orderpreserving isomorphism from {t T : t < t} to {t T : t < t}. T0 is not empty,since it contains the minimal element ofT. T We claim for given t T0 this elementsis unique. For assume not and we have two such elementss1ands2, assume w.l.o.g.s1 < s2 and two isomoprhisms f1 and f2. Let t

be the minimal element in T0 for

which f1(t

) = f2(t

) and let S

be the common range of f1 and f2 on the set ofelements smaller than t. But then clearly f1(t

) and f2(t) have to be the smallest

element in S\ S. A contradiction.Define a function f on T0 by this map to the unique s. It is order preserving and

bijective. IfT0 = Tthen we are done. IfT0 = T, choose t minimal in T\ T0. Notethat iffwas not surjective, then we could find minimal s in the complement of therange offand then note t T0 by associatiung it with this minimal element. Thissi a contradiction, hence fis surjective.

It then remains to prove

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Lemma 9 Every set can be endowed with a well order.

We give a proof relative to Zorns lemma below. Proof: Let Xbe a set. Consider theset of all structures (Y, ) where Y is a subset ofX and a well order ofY. Thisset is partially ordered by

(Y, )(Y, )

If either they are equal or Y Y and there exists y Y such that Y ={yY >y < y}and coincides with onY. One readily checks that this is a partial order.Given a totally ordered chain, one can take the union of allYin the change and definethe order in the natural way by extension. Then this union is an upper bound for thechain. By Zorns lemma there exists a maximal chain (Y, ). We show that Y =X.For if not, pick x Xand extende the well order on Y to a well order on Y {x}by setting y x for all y Y. Then this new well order contradicts maximality of(Y, )

An upper bound for a subset X in a partially ordered set is an element y suchtaht x y for all x X. A strict upper bound is an upper bound which in additionsatisfies x=y for all x X.

Lemma 10 (Zorns lemma) Let X be a nonempty partially ordered set with theproperty that every totally ordered subset ofXhas an upper bound. ThenX contains

at least one maximal element.

Proof: Assume to get a contradiction thatXhas no maximal element. Then everytotally ordered subset has a strict upper bound. Namely, it has an upper bound, andsince this upper bound is not maximal there is a strictly larger upper bound which is

then a strict upper bound for the set. Now this gives a contradiction to the followinglemma.

Lemma 11 Let X be a partially ordered set with ordering relation and let x besome element of it. Then there is a well ordered subset ofX containingx as minimal

element that has no strict upper bound.

Proof: For every set Y containing x and having a strict upper bound, we chooseone such upper bound and call it f(Y) (we need the axiom of choice). Call a wellordered subset Y ofXgood if if it contains xas minimal element and if for every

y Y we have f({y

Y : y < Y}) = y. Given any two good sets, there is aninjection of one into the other, say g : Y Y . We claim this injection is a restrictionof the identity map on X. For assume not, then let y be the minimal element forwhich g(y)= y. However, by goodness applied to the set{y Y : y < y} both inY and Y we conclude g(y) =y, a contradiction. Now consider the union of all goodsets. Then this set is well ordered and it contains no upper bound, or else we wouldfind a larger good set.

The following exercisie proves that D is countable.

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Exercise 5 LetXbe a countable set, each of its elements being a countable set again.Then

YXY is countable.

Lemma 12 The set of nonnegative real numbers has the same cardinality as the

power set ofN.

Proof: The identitfication of real numbers with Dedkind cust gives an injectivemap fromRto the power set ofD(which is equivalent toN) to R. To get an injectivemap from the power set ofN to R, we consider thes et of all sequences with values0 and 1 only and map it injectively to R via an

0 ann+n(0).

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Cardinality of Sets (UCLA)

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