Introductory Cardinality Theory Alan Kaylor Cline

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1 Introductory Cardnalty Theory lan Kaylor Clne lthough by name the theory of set cardnalty may seem to be an offshoot of combnatorcs, the central nterest s actually nfnte sets. Combnatorcs deals wth fnte sets. s wll be seen, the tool employed for the majorty of the theory s establshng oneto-one correspondences between sets. One verson of cardnalty theory uses cardnal numbers - a type of number used to quantfy cardnaltes. We wll not use cardnal numbers here although the results are qute smlar. The ntal defnton of fnteness says no more than a set s fnte f we can count ts elements. Countng means establshng a one-to-one correspondence wth a set of consecutve ntegers begnnng wth one. Thus, f f s a functon mappng {, 2,..., n } oneto-one a set, we count as f (), f (2),..., f ( n ). The fact that f s one-to-one and ensures that each element of s counted (-ness) and that no element s counted more than once (one-to-one-ness). Defnton : set s fnte wth cardnalty n f t s empty or f there exsts a one-to-one functon mappng {,2,..., n }. set s nfnte f t s not fnte. Thus, snce the set of lower case Latn characters {a, b, c,, z} can be put nto one-to-one correspondence wth {,2,...,26}, that set s fnte. Let s prove that the set of natural numbers {0,,2,...} s nfnte. Notce to do ths we must show that for any n, no functon exsts mappng {,2,..., n } one-to-one. (It s not suffcent to show that some functon doesn t wor we must establsh that t s mpossble to have such a functon). s one mght expect, such arguments use proof-by-contradcton. Example : The set of natural numbers s nfnte. Proof: Suppose there exsts an n and a functon f mappng {,2,..., n} one-to-one. The set { f (), f (2),..., f ( n)} s fnte so t has a maxmum element. Let m max{ f (), f (2),..., f ( n)}. Snce m and f s, there must exst some {,2,..., n} such that f ( ) m. But then also f ( ) { f (), f (2),..., f ( n)} and we would have that m max{ f (), f (2),..., f ( n)} f ( ) f ( ) m, whch s a contradcton. We conclude that no such n and f exsts, so s nfnte. That wasn t hard but t wasn t much fun ether. It mght be nce to have a more drect method for provng a set s nfnte. Such s provded by ths alternatve defnton. one-to-one correspondence between sets and B mples that there s a one-to-one functon mappng B (and thus another functon, the nverse of the frst, mappng B ).

2 Defnton 2: set s nfnte f there exsts a one-to-one functon mappng a proper subset of. set s fnte f t s not nfnte. Whenever we see two defntons for the same concept, we should as are they logcally dentcal. The answer ths tme s not qute. Let me deal wth the easy part of the comparson of the two defntons frst and then I ll dscuss the trcy stuff. It s a smple matter to show that f a set s fnte accordng to Defnton then t must be fnte under Defnton 2. (If a set were fnte accordng to Defnton but nfnte under Defnton 2, we would end up wth a one-to-one correspondence between two fnte sets of dfferent cardnaltes. That can t happen.) By usng the contrapostve to ths, we see that f a set s nfnte accordng to Defnton 2 then t must be nfnte under Defnton. Ths s good because, snce we wll want to use Defnton for fnte sets and Defnton 2 for nfnte sets, we now that we are usng the stronger defntons (.e., the clams wll hold for ether defnton). Turnng now to the thorner queston of mplcaton n the other drecton, the answer s that wth the standard axoms of set theory 2, one cannot prove that a set nfnte under Defnton must also be nfnte under Defnton 2. In order to establsh the mplcaton we add the famous xom of Choce (loosely stated as Gven any collecton of nonempty sets, we can choose a member from each set n that collecton.). The xom of Choce allows us to construct the functon that Defnton 2 requres 3. To see the ease of usng Defnton 2 for nfnteness, let s reprove that the set of natural numbers s nfnte usng Defnton 2 nstead of. Example 2: The set of natural numbers s nfnte. Proof: Let denote the set of postve natural numbers and consder the functon f : defned by f ( n) n. (Ths s often called the successor functon.) Snce for n m, f ( n) n m f ( m), and f s one-to-one. But f s also snce for any n, n and f ( n ) n. Because proper subset of. but 0 ~, s a Notce that snce the defnton requres that the set be put nto one-to-one correspondence wth a proper subset, we must prove that the functon s both one-to-one and. The followng lemma wll allow us to cut some of the wor. It says that a one-to-one functon s also - and thus nvertble - f ts range s restrcted to exactly the mage of ts doman. 2 By "standard axoms of set theory", I am referrng to Zermelo Fraenel set theory (see J.M. Henle. n Outlne of Set Theory. Sprnger Verlag, 986). 3 Wthout gettng too deeply nto Gödel Consstency Theory, I ll add that, although the xom of Choce s not dervable from the standard set theoretc axoms, f those other axoms are consstent then the xom of Choce s consstent wth them. Lastly, one should not feel covered n shame because we use the xom of Choce. Mathematcans do t all of the tme wthout blnng. It does have some nterestng consequences however, such that a lne segment nch long can be cut nto a fnte number of peces and then glued bac together to form a segment one lght-year long.

3 Lemma: If f : B then f maps one-to-one B f( ) and thus s nvertable. Consder now Defnton 2 but omt the necessty for the functon to be : Defnton 2': set s nfnte f there exsts a one-to-one functon mappng nto a proper subset of. set s fnte f t s not nfnte. Obvously f a set s nfnte under Defnton 2 t wll be nfnte under defnton 2'. From the lemma, however, the opposte s also true. To see ths, suppose s nfnte under Defnton 2' and so a functon f mappng nto a proper subset ' of. exsts. Let f ( ) and notce that ' so must also be a proper subset of. We conclude that s also nfnte under Defnton 2. By usng the contrapostve we may show that Defntons 2 and 2' for fnte sets are also equvalent. Snce Defnton 2' saves some wor, we wll use t. More generally we wll use Defnton 2' for nfnte sets and Defnton for fnte sets. The set of natural numbers has been proved to be nfnte usng both Defnton and Defnton 2. The proof usng Defnton 2' s the same as that usng Defnton 2 except that the second to last sentence (showng that the mappng s ) could be elmnated. We state t as our frst theorem and then prove the real nterval [0,] also s nfnte.. Theorem : The set N of natural numbers s nfnte. Theorem 2: The real nterval [0,] s nfnte. Proof: Consder the functon f ( x) x / 2 defned on [0,]. Clearly f maps the nterval nto [0,/ 2], a proper subset of [[0,]. Snce for x y, f ( x) x / 2 y / 2 f ( y), f s one-toone. Thus [0,] s nfnte. The substance of the next theorem may seem obvous: f a set s nfnte and has addtonal elements added, t stll s nfnte. Theorem 3: superset of an nfnte set s nfnte. Proof: Let be an nfnte set and assume ˆ. We see to show that  s nfnte as well. By Defnton 2', we now there exsts an f : ' for some '. (We wll

4 use the symbol to ndcate proper subsets.). Defne a new functon g, an "extenson" of f to all of Â, as follows f( a) f a ga ( ). Frst we must show that g s oneto-one on Â. To that end, consder dstnct elements a ˆ a f a ˆ ~, a2. Ether a, a2, a ˆ, a2 ~, or one element s n each of and ˆ ~ (and, wthout loss of generalty, we assume a and a ˆ 2 ~ ). If a, a2, then g( a ) f ( a ) f ( a2) g( a2), snce f s one-to-one. If a ˆ, a2 ~, then g( a ) a a2 g( a2). Fnally, f a and a ˆ 2 ~, then g( a) f ( a) ', so g( a ), but g( a ˆ 2) a2 ~. Snce g( a ) and g( a ˆ 2) ~, g( a) g( a2). We have shown that g s one-to-one. Fnally, snce ', there exsts some element a ~ '. We want to show that for no a ˆ s g( a) a. To that end, suppose g( a) a. If a, then g( a) f ( a) ', so g( a) a. If a ˆ ~, then g( a) a ˆ ~, so g( a) a. We have a contradcton n ether case, so we now there exsts no a ˆ so that g( a) a. We conclude that g maps  nto a proper subset of tself and thus  s nfnte. n easy corollary follows from the fact that f s a subset of Â, then  s a superset of. If  were fnte yet nfnte, we would have a contradcton of the theorem. So, f  s fnte so must be fnte as well. Corollary: subset of a fnte set s fnte. We use Theorem three to show that t s easy to establsh that a very strange set s nfnte by showng that t has an nfnte subset. Example 3: The set of ratos of ntegers to odd numbers s nfnte. Proof: ll natural numbers can be expressed as ratos of themselves to. Thus the set of natural numbers s a subset of ths set. The set of natural numbers s nfnte, therefore ths set s nfnte. The next theorem emphaszes that nfnte cardnalty s preserved by one-to-one mappngs. Theorem 4: Let be nfnte and f : B, then B s nfnte.

5 Proof: Let be an nfnte set and f : B. If we set B' f ( ), then from the lemma above f maps one-to-one B '. The dea of the proof s to show that B ' s nfnte. Snce B s a superset of B ', Theorem 3 wll guarantee that B s nfnte as well. We now that snce s nfnte there s some functon g mappng one-to-one nto some proper subset ' of tself. Furthermore, snce f B : ', we have f : B' so h f g f : B' B' s defned. Snce h s the composton of one-to-one functons, t s one-to-one. Now consder some element a ~ '. (We now such an element exsts snce ' s a proper subset of.) Let b f ( a). If we can show that there s no b B' so that h() b b, then h wll have been shown to have mapped B ' one-to-one nto a proper subset of tself - and thus be nfnte. To that end, suppose there s such a b B' wth h() b b. That says f ( g( f ( b))) b, so g( f ( b)) f ( b). But snce b f ( a), we have maps nto g( f ( b)) a ', so we would have. Ths s a contradcton because a ~ ' but g g f ( ( b)) both beng an element of ' and beng outsde of '. We conclude that no such element b B' so that h() b b exsts and h maps B ' one-to-one nto a proper subset of tself. B ' s then shown to be nfnte and B, a superset of B ', s also nfnte. In Example 3, we saw that Theorem 3 smplfed showng sets were nfnte when we could fnd nfnte subsets. Theorem 4 extends that smplfcaton but no longer must we fnd nfnte sube sets we may establsh nfnteness by fndng one-to-one mappngs from nown nfnte sets. Example 4: The set of ponts n the plane W x y x y 2 4 {(, ) ( 3) 6} s nfnte. Proof: Consder the mappng f :[0,] W defned by f ( t ) ( t,3). Notce that ths actually does map nto W snce for 0 t, t (3 3) t 6. To show f s one-to-one, consder dstnct st, [0,], f ( s) ( s,3) ( t,3) f ( t ). Thus f s one-to-one and by Theorem 4, W s nfnte. s was stated ntally, the theory of cardnalty deals wth one-to-one correspondences between sets. We wll now refne the concept of nfnte set by dstngushng those sets that can be put nto one-to-one correspondence wth the natural numbers from those that cannot. Defnton 3: set s countably nfnte f there exsts a one-to-one functon mappng. set s countable f t s fnte or countably nfnte. set s uncountably nfnte f t s not countable. Let's show that, the set of ntegers s countable nfnte. Example 5: The set of ntegers s countably nfnte.

6 n /2 f ns even Proof: Consder f : defned by f( n). We see to ( n ) / 2 f nsodd show f maps one-to-one. Frst we wll show that f s one-to-one. Consder dstnct nm., Ether both of n and m are even, both are odd, or one s even and one s odd. If n and m are even, f ( n) n/ 2 m/ 2 f ( m). If n and m are odd, f ( n) ( n )/2 ( m )/2 f ( m). Lastly, f n s even and m s odd then f ( n) 0 f ( m). We conclude that f s one-to-one. Lastly we wll show that f s. Consder any. If 0 then 2 and s even so f (2 ) 2 / 2. If 0 then ( 2 ) and s odd so f ( ( 2 )) ( ( 2 ) )/2. We conclude that f s both one-to-one and and thus s countably nfnte. We could show that sets such as the even ntegers, the odd ntegers, and the powers of two are all countably nfnte. legtmate queston then s "re there any uncountably nfnte sets?". The followng theorem shows that there are. The proof uses the classc dagonalzaton argument of Georg Cantor. It s an proof by contradcton: we assume that the real nterval [0,] s countably nfnte, attempt to "count" all of them, fnd that at least one s mssng, and get a contradcton. Theorem 5: The real nterval [0,] s uncountably nfnte. Proof: Theorem 2 guarantees that the nterval s nfnte. To prove that t s uncountably nfnte, let us assume that t s countably nfnte. Thus there exsts g : [0,]. If we can show that there s a number n [0,] that s not equal to g( ) for any N, then g s not and we have a contradcton. We then may conclude that [0,] s uncountably nfnte. To ths end, consder the decmal expansons of g(0), g (),.... For N, let g( ) be expressed as. dd2... d.... Some real numbers may have two dfferent decmal expansons: one termnatng n zeros, the other n nnes. If there s the opton, we wll choose the expanson termnatng n nnes. Notce that 0 tself wll then be the only number that termnates n zeros and that every number n [0,] has a unque such decmal expanson. f d Now consder constructng the number e. e e2... e... defned by e 2 f d for = 0,, 2,. Frst recognze that e s a real number and e [0, ] (n fact e [ / 9, 2 / 9 ]), and yet, e cannot equal g( ) for any N. Suppose e g( ), then the decmal expansons of the two must agree n every poston, but n fact they dffer n the st decmal dgt. Snce for no N s e g( ), g s not. We conclude that no such g exsts and that [0,] s uncountably nfnte. The name "dagonalzaton" s suggested n the constructon of the number e n the proof. If one were to mae a column of the decmal expansons of g(0), g(),.., then e s created by alterng the entres on the dagonal of ths table.

7 To come are several theorems that can be summarzed as "the unon of a countable collecton of countable sets s countable." For that purpose the followng theorem s very helpful. Theorem 6: If there exsts a functon f : then s countable. Proof: If s fnte we are done. ssume then that s nfnte. Consder an array of the elements of nduced by the functon f a0a a where a f ( ). We wll defne a new functon g : and show that ths functon s both one-to-one and. To that end, defne g(0) a0 and then remove all copes of elements of the array equal to a 0. Defne next g () equal to the leadng element on the remanng array and then remove all copes of t. In general, defne g () as the leadng element of the array after all copes of g(0), g(),, g( ) have been removed. Snce s nfnte wll not end. Thus for every, g () s defned. We need to show ths mappng s both one-to-one and. To that end consder to propertes of g : For any element a j n the array (and therefore n ), there wll be some value of, so that g() a. In fact, the value of j. j By constructon, the value of g () must be dstnct from g(0), g(),, g( ), snce all copes of those elements were removed pror to the defnton of g (). The frst property guarantees that g s and the second property guarantees that t s one-to-one. We have establshed that ether the set s fnte or t s countably nfnte. Ths Theorem has an mmedate consequence. Theorem 7: subset of a countable set s countable. Proof: ssume set  s countable and ˆ. If s empty then t s obvously fnte. Henceforth, we assume s nonempty. If  s fnte then so wll be because of Theorem 3. If  s countably nfnte then there exsts a functon f ˆ. Choose : any fxed element a and defne a new functon g: as follows: f ( n) f f ( n) gn ( ) a f f ( n) Snce f s, for any element a there exsts an n such that f ( n) a. But then gn ( ) s also equal to a snce g( n) f ( n) f f ( n). We conclude that g s and from Theorem 6, s countable. Corollary: superset of an uncountably nfnte set s uncountably nfnte. Proof: ssume set s uncountably nfnte and ˆ. If  were countable then so would. by Theorem 7. That s a contradcton so  must be uncountably nfnte.

8 Theorem 8: The unon of a fnte collecton of fnte sets s fnte. Proof: For n, let {, 2,, n } be a collecton of fnte sets. We see to prove a stronger verson of the theorem that the cardnalty of the unon s less than or equal to the sum of the cardnaltes of the sets. To ths end, for n let n #( ). We proceed by nducton. For n, #( ) #( ). ssume now that all unons of n sets: n n #( ) #( ). We see to show that n n #( ) #( ). But then n n n n n. #( ) #( ) #( ) #( ) #( ) #( ) #( ) n n n Theorem 9: The unon of a countably nfnte collecton of fnte sets s countable. Proof: Let the fnte sets be,,..., defne n to be the cardnalty of 0, for, and let. Snce for, each s fnte, the elements can be ordered n 0 n some form aa2 a, where n s the cardnalty of 0. Consder an array of all of the elements of stretched out as a a a a a a a a a. We wll defne a functon f : 2 n0 2 n 2 n and then prove that ths functon s. To defne the functon we use the array: f () a l f a l s the th element of the array (startng the countng from 0.) Snce every element of the array (hence every element of ) s the mage under f for some, f s. From Theorem 4, s countable. Example 5: The set of ratonal numbers s countably nfnte. Proof:.For, defne { p / q p, q p q q 0}. Notce that each s fnte (n fact, havng exactly (2) 2 elements) and gven any ratonal number, ths wll be contaned n p/ q s ratonals equals for max{ p, q }. Thus the set of and ths set s countable by Theorem 9. Snce t contans the natural numbers as a subset, t s nfnte. We conclude that t s countably nfnte. The last of these theorems allows us to unon a countably nfnte collecton of countably nfnte sets and stll the result s countably nfnte. Theorem 0: The unon of a countably nfnte collecton of countably nfnte sets s countably nfnte. Proof: Let the countably nfnte sets be,,..., and let. Snce for, each 0 s countably nfnte, there s a functon f 0 :. Thus, the elements can

9 be ordered n the form a0a a, where a f(). Form now a new collecton of sets 0 2 { B0, B, } defned as follows: for B { a, a, a 2,, a, a0}. From Theorem 9, s countable. Snce the nfnte set 0, s nfnte, and therefore t s countably nfnte. Theorem : Let be uncountably nfnte and nfnte. f : B, then B s uncountably Proof: From the lemma above, f B, where B f ( ). By Theorem 4, B s : nfnte. Suppose B s countably nfnte. Then there exsts a functon Consder the functon h: defned as h f g g B. :. Ths functon s both one-to-one and snce f and g are. We would then have that s countably nfnte but that would be a contradcton. We conclude that B s uncountably nfnte.