TEORIA AXIOMATICA DE CONJUNTOS Versi\u00b4 on Preliminar Renato A. Lewin Author address: Pontificia Universidad Cat \u00b4 olica de Chile, Facultad de Matem \u00b4 aticas, Casilla 306 - Correo 22, Santiago CHILE. e-mail:
[email protected]
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Indice CAPITULO 1. Introducci\u00b4on. Los Axiomas de Zermelo 5 Fraenkel 1. El Lenguaje Formalizado L 7 2. Los Axiomas de la Teor\u00b4\u0131a ZF . Conceptos 9 Fundamentales CAPITULO 2. Teor\u00b4\u0131a Elemental 1. Operaciones 2. Relaciones 3. Funciones 4. Relaciones de Equivalencia 5. Relaciones de Orden
17 17 23 30 40 44
CAPITULO3. Ordinales 57 1. N\u00b4 umeros Naturales 57 2. Ordinales 64 3. Inducci\u00b4onTrans\ufb01nita 70 4. Recursi\u00b4on 72 5. FuncionesNormales 75 6. Ordinales y Buenos Ordenes 78 7. Aritm\u00b4eticaOrdinal 80 8. La Jerarqu\u00b4\u0131a Acumulativa de Conjuntos 101 CAPITULO 4. El Axioma de Elecci\u00b4on 1. Equivalencias del Axioma de Elecci\u00b4on 2. Aplicaciones
105 105 111
CAPITULO5. Cardinales 1. De\ufb01niciones y Resultados B\u00b4asicos 2. Conjuntos Finitos y Conjuntos In\ufb01nitos 3. Aritm\u00b4eticaCardinal 4. Cardinales Regulares y Singulares 5. La Hip\u00b4otesis del Continuo
117 117 125 129 144 146
Bibliograf\u00b4\u0131a
151
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CAPITULO 1
Introducci´on. Los Axiomas de Zermelo Fraenkel Este libro trata sobre los conjuntos. Intuitivamente un conjunto es
Una teor´ıa axiom´atica es un modelo formal de una realidad que
Diversas teor´ıas axiom´aticas de conjuntos han logrado en mayor
umeros. Lo anterior parece condenar nuestro proyecto al fracaso, sin em-
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trabajo de E. Zermelo el primero en proponer una teor´ıa en los primeros a˜ nos de este siglo. Un conjunto est´a definido por los objetos que contiene. Nuestra umeros 1, 2, . . . , 99, le corresponde la propiedad “ser umero entero mayor que cero y menor que cien”. A la inversa, a toda n´ Temprano en el desarrollo de la teor´ıa de conjuntos se descubri´o
R = {x : x ∈ x}. La pregunta entonces es ¿pertenece R a R?, o en sımbolos, ´ ¿R ∈ R?. R ∈ R. En cualquier caso obtenemos la contradicci´on R ∈ R ↔ R ∈ R .
La paradoja de Russell (y otras) nos dice que el concepto de “pro-
Sin embargo, la noci´on de que a cada propiedad deber´ıa corresponon. Algunos matem´aticos no han
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una clase propia. Es decir, las clases propias son las extensiones de ultimas son la clase R definida En nuestra teor´ıa, ZF , no existen las clases propias, s´olo conjuntos.
La noci´on de “propiedad” no la hemos definido pero de lo anterior se desprende que es central en nuestro estudio. Vamoson a continuaci´ a definir este concepto. Como dijimos, una teor´ıa axiom´atica se desarrolla a partir de cier-
No es el prop´osito de este texto introducir al lector a la L´ogica
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3. Sımbolo ´ de pertenencia: ∈ 4. Sımbolo ´ de igualdad: = 5. Conectivos l´ogicos: ¬, ∨, ∧, →, ↔, es decir, los sımbolos ´ ha6. Cuantificadores: ∀, ∃, con su significado habitual. 7. Par´entesis: ( , ). Usados como signos deon. puntuaci´ Cualquier cadena finita formada por estos sımbolos ´ es una expresi´on Una f´ormula de L es una expresi´on de L construida como sigue: 1. X ∈ Y, X = Y son f´ormulas de L para cualquiera dos variables o constantes X e Y no necesariamente distintas. La primera se lee X pertenece a Y o bien Y contiene a X y la segunda X es igual a Y . Su significado intuitivo es el obvio. Estas se llamar´an f´ormulas at´omicas. 2. Si ϕ y ψ son f´ormulas de L , entonces tambi´en lo son (ϕ ∨ ψ), (ϕ ∧ ψ), (ϕ → ψ), (ϕ ↔ ψ). Estas f´ormulas corresponden respectivamente a la disyunci´on, conjunci´on, implicaci´on y equivalencia de ϕ y ψ . 3. Si ϕ es una f´ormula de L , entonces ¬ϕ tambi´en es una f´ormula de L . La f´ormula ¬ϕ corresponde a la negaci´on de ϕ . Tambi´en ∈ Y Xy /X = Y para escribir usaremos los sımbolos ´ auxiliares ¬(X ∈ Y ) y ¬(X = Y ), respectivamente. 4. Si ϕ es una f´ormula de L y x es una variable, ∀xϕ, ∃xϕ son f´ormulas de L . Estas se leen cualquier conjunto x verifica ϕ y existe (por lo menos) un conjunto que verifica ϕ , respectivamente. Su
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seamos capaces de determinar cu´al de las dos se cumple!). Una oraci´on
Como hemos dicho, s´olo aceptaremos f´ormulas de L para hablar de
x⊆ y
ssi ∀z (z ∈ x → z ∈ y ).
Entonces, como ya sabemos que x ⊆ y puede escribirse en el lenguaje L, un, en general usaremos expresiones del
2. Los Axiomas de la Teor´ıa ZF . Conceptos Fundamentales A1. Axioma de Extensionalidad: “Si todo elemento de X es un elemento de Y y todo elemento de Y es un elemento de X , entonces X es igual a Y ”. Dicho de otro modo, si dos conjuntos tienen los mismos elementos,
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En L escribimos
∃X ∀x x ∈ X. Observemos que, en particular, este axioma garantiza que existe al menos un conjunto. Lema 1.1. Existe un ´ conjunto que no contiene unico un ning´ ele-
mento.
Demostraci ´ on. Supongamos que existen dos conjuntos distintos
a y b ambos sin elementos. Por A1 ∃x((x ∈ a ∧ x ∈ b) ∨ (x ∈ b ∧ x ∈ a)), una contradicci´ on. Luego hay un unico ´ conjunto vac´ıo.
(´ Definici ´on 1.2. El unico) conjunto que no tiene elementos se
llama el conjunto vac´ıo y se le denota ∅ .
Obs´ervese que el sımbolo ´ ∅ no es la letra griega ϕ .
A3. Axioma de Separaci´on: “Si ϕ(x) es una f´ormula de L y X es un conjunto, entonces En L escribimos ∀X ∃Y ∀z (z ∈ Y ↔ (z ∈ X ∧ ϕ(x)). Este axioma nos dice que para cualquier propiedad (expresada por ϕ(x)) y cualquier conjunto A existe el subconjunto de A formado por los elementos que verifican esa propiedad. Obviamente este conjunto es unico. ´ Definici ´on 1.3. Si ϕ(x) es una f´ormula de L y A un conjunto,
el conjunto cuya existencia est´a garantizada por A3 se denotar´a con el sımbolo ´
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En este caso tenemos que si R ∈ R, entonces R∈ A
y
R ∈ R,
∈ R, lo cual es una contradicci´ on, luego R / lo que, a diferencia de antes, A. R / no es contradictorio, s´olo implica∈que Teorema 1.2. No existe el conjunto de todos los conjuntos. Demostraci ´ on. Supongamos que si existe y llamemoslo V . En-
tonces en virtud de A3 podemos construir el conjunto de Russell R = {x ∈ V : x ∈ x}, contradicci´ on.
Por ultimo, ´ cabe destacar que este no es propiamente un axioma sino m´as bien un esquema. En efecto, para cada f´ormula ϕ(x) de L tenemos un axioma distinto, o sea, hay una cantidad ilimitada de instancias de este axioma. A4. Axioma de Pares: “Dados dos conjuntos X e Y , existe un conjunto unicos cuyos ´ elementos son X e Y ”. Su expresi´on en L es
∀X ∀Y ∃Z ∀x(x ∈ Z ↔ (x = X ∨ x = Y )). Resulta claro por A1 que este conjunto unico.esLo´ denotamos
{X, Y }. y lo llamamos el par no–ordenado X, Y . El axioma A1 tambi´en garantiza la existencia del conjunto cuyo u ´ nico elemento es el conjunto X
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Un caso particular que merece una notaci´on especial es el siguiente. {X, Y }. (¿Por qu´e ?) Si X e Y son conjuntos, entonces existe Entonces x∈ {X, Y } ↔ (x ∈ X ∨ x ∈ Y ). Corresponde al conjunto de todos pertenecen {X, Y } se llama la uni´on de los X conjuntos e Y y se que le denota X ∪ Yya . sea a X o a Y (o a ambos). M´as generalmente, dados los conjuntos X1 , X2 , . . . , Xn de manera an´aloga al caso de la uni´on de dos conjuntos, definimos X1 ∪ X2 ∪ · · · ∪ Xn = {X1 , X2 , . . . , Xn }.
de tal manera que x ∈ X1 ∪ X2 ∪ · · · ∪ Xn ↔ x ∈ X1 ∨ x ∈ X2 ∨ · · · ∨ x ∈ Xn .
El lector seguramente est´a familiarizado con el concepto de uni´on
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A6. Axioma del Conjunto Potencia: “Si X es un conjunto, entonces existe el conjunto de todos los subconjuntos de X ”. Esto es ∀X ∃Y ∀z (z ∈ Y ↔ z ⊆ X)) (En rigor deberiamos excribir
∀X ∃Y ∀z (z ∈ Y ↔ ∀u(u ∈ z → u ∈ X)) sin embargo, como la lectura de la f´ormula se complica bastante y Creemos que este axioma se explica por sı´ mismo. El (´ unico) conjunto cuya existencia garantiza este axioma se designa por P X y se llama el conjunto potencia de X . A7 Axioma de Regularidad: “Todo conjunto no vac´ıo contiene un elemento con el que no comparte ning´ un elemento.” En L escribimos
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A7 ya que para cualquier y ∈ A, digamos y = am un para alg´ m , am+1 ∈ am y am+1 ∈ X , o sea y ∩ X = ∅. El problema entonces se reduce a verificar que existe tal conjunto A. Sin embargo para poder hacerlo no bastan los axiomas que tenemos hasta ahora, necesitamos dos axiomas m´as. La demostraci´on deber´a posponerse hasta entonces. (Ver ejercicio 7.) Aunque la mayor parte de las matem´aticas puede desarrollarse sin el axioma de regularidad es m´as c´omodo contar con ´el. A8. Axioma del Conjunto Infinito: “Existe un conjunto que tiene infinitos elementos”. Para escribirlo en el lenguaje L debemos usar una expresi´on que no es muy transparente.
∃X(∅ ∈ X ∧ ∀y (y ∈ X → y ∪ {y } ∈ X ). Es claro que el conjunto ası´ formado es intuitivamente infinito, basta verificar que contiene a los siguientes conjuntos
∅, {∅}, {∅, {∅}}, {∅, {∅}, {∅, {∅}}}, . . .
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De hecho, este axioma parece m´as complicado de lo que es. La
x −→ P x
no es, como veremos formalmente m´as adelante, un objeto de nuestra
unico elemento.
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G¨odel en 1930) que no puede completarse, es decir, aunque agreguemos
Todos estos problemas requieren de conocimientos de L´ogica Mate-
Ejercicios. 1. Demuestre que el axioma de pares puede ser reemplazado por el axioma m´as d´ebil: “Dados dos conjuntos X e Y , existe un conjunto los contiene a ambos”. 2. Demuestre que el axioma de uniones puede ser reemplazado por el axioma m´as d´ebil: “Si X es un conjunto, entonces existe un conjunto que contiene a todos los elementos de los elementos de X ”. 3. Demuestre que el axioma del conjunto potencia puede ser reemplazado por el axioma m´as d´ebil: “Si X es un conjunto, entonces existe un conjunto que con-
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CAPITULO 2
Teor´ıa Elemental En este capıtulo ´ formalizaremos y profundizaremos las nociones de
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Teorema 2.1. No existe el “complemento”,un deconjunto. ning´ Demostraci ´ on. Sea a un conjunto. Si existiera su complemento
llam´emoslo a , entonces en virtud de A5, a ∪ a
ser´ıa un conjunto,
El siguiente teorema nos da las propiedades de estas tres operaciones. Teorema 2.2. (Algebra de Conjuntos).
Para todo conjunto
b
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iv) Si a ⊆ c y b ⊆ d, entonces a ∩ b ⊆ c ∩ d. vii) a ⊆ b si y s´olo si a ∪ b = b. viii) Si a ⊆ c y b ⊆ d, entonces a ∪ b ⊆ c ∪ d. Demostraci ´ on. Ejercicio.
Algunas propiedades del conjunto potencia de un conjunto son interesantes. Teorema 2.4 Para todo conjunto a, b
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3. Demostrar que : (a) (b) {{a, b, c}, {a, d, e}, {a, f }} = {a, b, c, d, e, f } . (c) {{a, b, c}{, a {} a,,d,para e}, {todo a, f }}conjunto = { a} . a . ( a ∩ b) . (d) ( a){∩ a}( =b)a = = 4. Probar que: (a) Si a ∩ c = ∅ , entonces a ∩ (b ∪ c) = a ∩ b . (b) Si a ∩ b = ∅ , entonces a − b = a . (c) Si a ∩ b = ∅ y a ∪ b = c , entonces a = c − b. (d) a ∩ (b − c) = (a ∩ b) − c . (e) (a ∪ b) − c = (a − c) ∪ (b − c) .
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(c) a + (b + c) = (a + b) + c (d) a ∩ (b + c) = (a ∩ b) + (a ∩ c) (e) a − b ⊆ a + b (f) a = b si y s´olo si a + b = ∅ (g) Si a + c = b + c , entonces a = b 9. Sean a = {x ∈ Z : x es divisible por 4}
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14. Si b ⊆ a y c ⊆ a , probar que
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2. Relaciones
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Lema 2.6. Si a ∈ A y b ∈ A,
a, b ∈ PP A.
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Es claro que tanto Dom R como Rec R son conjuntos (¿qu´e a-
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Esto implica que existe z tal que
x, z ∈ S y
z, y ∈ S ,
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embargo F nosiesy inyectiva. s´olo si ∃yEntonces (y ∈ a ∧ yexisten = F (x))d, e ∈ a , d = e y F (d) = F (e). si y s´olo si F (x) ∈ a.
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Debido a como valencia la importancia en ii) y iii). dePara ´este, demostrar le dedicaremos el rec´ıproco un capıtulo ´ de i),completo, es decir, si cuarto. el F es sobreyectiva, S´olo entonces entonces podremos existeanalizar G : b →este a talproblema. que F ◦ G = Ib , necesitamos el ´
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con (c)m´as Si : F0 de y−→ un G1N son elemento supongamos entonces Fno∗ G F es ◦ Ginyectiva = F ◦ H .para (b) F N −→ . , inyectivas, F es yinyectiva pero que sobreyectiva. toda (d) F funci´on Rec F−→ ∗ G F Rec F . )Probar × (Recque G)ni. Gsobreyectiva. =H . (c) : 2 N(× 3 −→ Z) :,= 6bF,(−→ F nocbiyecci´on. es inyectiva (d) F : Z −→ N , F no es inyectiva ni sobreyectiva. (e) F : Q −→ R , F no es inyectiva ni sobreyectiva. (f) F : R −→ Q , F es sobreyectiva pero no inyectiva.
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a que existe que b = una ⊆c b si 19. Probar Probar (b) a ⊆ que b implica ∅que sifunci´on yc as´olo .H :b c=−→ ∅ by tal a= ∅ .F = G ◦ H y s´olo si Rec F biyecci´ ⊆ Rec G. Probar que H es ´ 20. siProbar (c) Si existe que : una
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(b) F − G es inyectiva. (c) F ◦ G es inyectiva. (d) F
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1 des elementos de conjunto por consiguiente consecuencia ii)dey Rlos esz, sim´etrica y de ∈ R la. simetr´ıa Entonces siun y s´olo y transitividad siz, R xy−que ∈⊆Rqueramos R. y por de la lo relaci´on tanto, como (verRejerciidentificar areflexiva todos losla que dichas Por ejemplo, ⊆yR.∈ cios). iii) es RSin sim´etrica, esembargo, sobre x,incluimos z comparten ∈aRsiy y tambi´en en s´olo la definici´on si Ipropiedades. porque R, y como tradicionalR es a z, − 1 − 1 al iv) estudiar rectas plano podemos a todas las rectasde mente transitiva, Rsees lelas de define equivalencia x, ası ydel ´ y∈ para R. si Luego hacer y s´olo hincapi´e R siidentificar R ⊆en = R. una relaci´on ◦R ◦RR.que que son paralelas entreesta sı. ´ Si definimos la relaci´on R como todos los equivalencia verifica propiedad. pares de elementos que queremos identificar, R es lo que llamamos una relaci´on de equivalencia. Esta secci´on estudiar´a este concepto.
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v) Si x ∈ [y ]R , entonces [x]R = [y ]R . vi) Si x, y ∈ [z ]R , entonces [x]R = [y ]R .
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ceptos y de equivalencia. z,entonces yde∈relaci´on R por u1 ∩ transitividad, ii)de unos = dice ∅ luego queo todo u sea, de elemento u z partici´on, , por [y ]Rlo . del Luego tanto, acampo saber, [{ xx, ]R est´a toda z⊆ } [⊆y ]relaci´on uRen ., o 1 2 equivalencia 1y= 2∈ ⊆ [un Analogamente, alguna de equivalencia sea, clase R[y P que ] [sobre es y ]toda clase x]Rconjunto , luego es no [vac´ıa. xda ]R origen = [yiv) ]R .nos a una dice ´ que dos clases R transitiva. distintas son disjuntas. Estas tres propiedades definen otro importante concepto matem´atico, el de partici´on de un conjunto.
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sobre b. Sea c el conjunto { d, e ∈ a × a : F(d), F(e) ∈ R}. Probar que c es relaci´on de equivalencia sobre a .
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lencia; relaci´on determinar de equivalencia la partici´ondeterminada que determina porcada F yuna se denota de ellas, y describir por Rgeom´etricamente [F ]. los elementos de cada partici´on:
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nodos unidos por l´ıneas. Los nodos representan conjuntos pertenecientes campo de la relaci´on R y las l´ıneas al orden mismo, ası, ´ si a, b ∈ R, habr´a dos nodos unidos por una l´ınea (ver figura). Mientras m´as arriba aparezca un conjunto en el diagrama, m´as “grande” es.
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qu´e punto este es el ejemplo tıpico ´ de orden necesitamos primero una definici´on.
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pero no antisim´etricas (tal tipo de relaci´on se llama un preorden). Podemos obtener una relaci´on de equivalencia a partir de un preorden y ordenar las clases de una manera coherente con el preorden de la manera indicada en el siguiente teorema. Esta construcci´on es bastante com´
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probablemente familiares lector. que si un conjunto tiene ≤a[zes [los y ]ii) ]S . una Entonces minimales cota inferior x no ypara tienen y de yel X zpor y sicomo qu´e ∀xN´otese (x ser ∈ Xes el→ transitiva menor a≤ x). elemento x z , del Selementos supremo ≤ luego conjunto iii) a[xes ]o de el hecho, [supremo z ]S . ´estos este ni deson siquiera X ´si a tiene es laque menor existir. cota Ensuperior el ejemplo dede X . S´ınfimo, la iv) figura a esa ely ´ınfimo b son de minimales X si a es la mayor cota inferior.
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formulaci´on de . Obs´ervese delabstracci´on axioma que de=regularidad. ∅ que sDe ≤ F hecho s, donde el axioma s es el de son B veremos cota superior es una de ∅, B pues de si, ya la a conocida ∈ A no lo propiedad fuera, existir´ıa del orden x menor ∈regularde ∅ tal los elemento idad que de la A relaci´on . es imposible. { x, yComo ∈a: ∅ x ∈⊆yA },para cualquier conjunto que dice n´ x≤a lo que por hip´otesis tiene un supremo s , este es la menor de las cotas superiores, luego tiene que ser el menor elemento de A .
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tiene (b) (a) x Si un R2R menor yessiasim´etrica, yelemento s´olo si yluego entonces − x ∈ x8 ≤yRyyes o−yantisim´etrica. x≤ = x0. , o sea, ≤ es orden (c) (b) xIaR total. es sim´etrica si y s´oloysi antisim´etrica. y − x = 0. 3 y (d) x R4 y si y s´olo si y − x es un entero par. (e) x R6 y si y s´olo si y − x = 1.
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−1 es (g) unla Si R orden = Run {es sobre x, conexa, y “es b : estricto xyentonces + que y fina es si?R par es R} total, .es esun conexa. entonces R ∩ el (b conjunto × b) 13. que ¿(c) Es ∅ relaci´on orden m´as que” orden sobre tambi´en (h)todas (d) Si RP1= Rlo asim´etrica, : x + y de entonces es par es ∪ Iun m´ 14. de Sean las y{eses. Px, particiones particiones dea a . . ySexRdice que P1 antisim´etrica. es m´as fina que Cam R es 2 y (i) Si R es antisim´etrica, entonces R − ICam R es asim´etrica.
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para cada b en a .yProbar si) aes Sbuna : bsupremo ∈ a superior est´a de a de conbalguna en a cota si s´olo superior sique Fen (c tiene cota en ordenado a .(resp. Probar ∗ por S b ∈ aa con esalguna isomorfo a. que inclusi´on, todo inferior) subconjunto deentonces F bnoen vac´ıo a2b . : de cotaa inferior en
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tenga cotas Tiene a supremo superiores o ´ınfimo e inferiores ? Encontrar perotodos que no lostenga elementos menor minimalesniy mayor elemento maximales elemento. de a .
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de a es un buen orden si y s´olo si todo segmento inicial de a est´a un bien ordenado por (la restricci´on de) el orden de a .
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teor´ıa de conjuntos. est´a familiarizado forma para ordenar bastante lossencilla. conjuntos Debemos finitos. destacar Esseguramente esta segunda algunas propiedad caracter´ısticas lacon quelos de 2 = ElSlector 1 = {∅ , {∅}} n´ trataremos ´esta. En primer de generalizar para n´ 3lugar = 0 S 2es =un {∅conjuntos , {∅}, {∅,infinitos, {∅}}} empezaremos por construir dentro de nuestra teor´ıa el conjunto de los n´
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primero siguiente teorema, del que derivan las propiedades quese m´as naturales? que todoelconjunto La respuesta inductivo, de ω ser , contiene obvia, deahecho todos los 2 es = sı´en {pero 0,particular 1} dista nos interesan de los n´ necesita n´ el Axioma de Infinito construir el conjunto de los 3 = {para 0, 1, poder 2} n´
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entreEn cia. naturales este caso, define resulta unam´as noci´on evidente de orden queapropiada. P.I. simplemente En estricto afirma rigor, que la verdad ω es el esmenor todo loconjunto contrario, inductivo. los n´ Las dos maneras de enunciar el principio son equivalentes, (ver ejercicios.)
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∈ ssi n ∪ {m n}≤=n.Sn, o sea ϕ(Sn) se verifica. Luego, en iii) m < 0Sn de P.I., para todo iv) Si n virtud < m, entonces Sn ≤ m. n ∈ ω , 0 ≤ n.
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de P.I., lo llamaremos de k Inducci´on Completa. hip´otesis dePrincipio inducci´on ≤ n. Es decir k ∈ n ´o Debemos k = n. En recordar siempre que estek nuevo equivalente con P.I. Decualquier caso ∈ Sn, principio o sea, k ≤esSn . mostraremos a continuaci´
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ii) Supongamos ψ (n) se verifica. Entonces por (∗), ϕ(n) tambi´en, pero por el lema 3.3 iii), para todo n y < Sn → (y < n ∨ y = n),
luego ∀y (y < Sn → ϕ(y )), es decir ψ (Sn) es cierta. Esto completa nuestra inducci´on, o sea para todo n ∈ ω , se verifica, en particular, ψ (Sn), y como n < Sn, esto garantiza que se verifica ϕ(n). El principio de Inducci´on Completa tambi´en puede plantearse en terminos de la relaci´on de pertenencia. Teorema 3.6. Sea B ⊆ ω tal que
(∗ ) si {k ∈ ω : k < n} ⊆ B , entonces n ∈ B . Entonces B = ω .
Teorema 3.7. ≤ es un buen orden. Demostraci ´ on. Hemos visto que ≤ es un orden total. Debemos
demostrar que ≤ es bien fundado, es decir, que si X ⊆ ω, X = ∅, entonces X tiene un menor elemento. Para ello suponemos que X no tiene menor elemento y aplicamos el Principio de Inducci´on Completa 3.6 a X ∈ ω − X ) Dado n , si para todo k < n , k ∈ ω − X , entonces n ∈ ω − X porque, si no, n es el menor elemento de X , luego ω − X = ω , es decir, X = ∅ , lo que es una contradicci´on. Por lo tanto todo subconjunto no vac´ıo de ω tiene un menor elemento y ≤ es un buen orden. Volveremos sobre estas propiedades en un contexto m´as general en las pr´oximas secciones. Teorema 3.8. Si un subconjunto de ω tiene una cota superior, entonces tiene un m´aximo elemento. Demostraci ´ on. Sea X ⊆ ω, X = ∅, definimos Y = {x ∈ ω : x es cota superior de X }.
Por hip´otesis Y = ∅. Sea m el menor elemento de Y . Por definici´on m es el supremo de X . Debemos demostrar que m ∈ X . Si m = 0, entonces para todo x ∈ X, x ≤ 0, o sea X = ∅ o bien X = {0}. El primer caso no ocurre por hip´otesis y en el segundo, 0 es el m´aximo de X
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Ejercicios. 1. Demuestre la equivalencia de las dos maneras de enunciar el Prin2. Demuestre el Principio de Inducci´on Completa a partir del teorema 3.7 3. Demuestre la equivalencia de las dos maneras de enunciar el Prin4. Sea ai : i ∈ ω una familia de conjuntos tal que para todo i ∈ ω, ai ⊂ aSi . Probar que a0 ⊂ an para todo n ∈ ω − 1. 5. Sea ai : i ∈ ω una familia de conjuntos tal que para todo i ∈ ω , ai ⊆ aSi y i en ω . i∈ω ai ⊆ a0 . Probar que ai = a0 para todo 6. Probar que si n ∈ ω, Sn = n y que ω = ω . 7. Probar que si a ⊆ ω, a= ∅ ya = a , entonces a = ω . 8. Probar que si m y n est´an en ω y m= n , entonces: m si n ∈ m, (a) m ∪ n = n si m ∈ n. n si n ∈ m, (b) m ∩ n = m si m ∈ n. 9. Probar que si n ∈ m y n = 0 , entonces existe un mayor elemento en n . 10. Si a ⊆ b ⊆ ω son no vac´ıos, n es el menor elemento de a y m es el menor elemento de b , ¿cu´al es la relaci´on entre n y m ? Justificar y responder para los mayores elementos de a y de b si estos existen. 11. Probar que si n ∈ ω , entonces no existe k ∈ ω tal que n < k < Sn. 12. Probar que si n y m est´an en ω y n < m , entonces Sn < Sm. 13. Probar que no existe una funci´on F : ω −→ ω tal que para todo n ∈ ω, F (Sn) < F (n). 14. Probar que si ϕ(x) es una f´ormula con variable libre x y existe un k ∈ ω , tal que (a) ϕ(k ) se verifica y (b) para todo n ∈ ω tal que k ≤ n, si se verifica ϕ(n) entonces se verifica ϕ(Sn). Entonces ϕ( ) se verifica para todo ∈ ω k
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(b) para todo n ∈ ω tal que n > k , si se verifica ϕ(n) entonces se verifica ϕ(Sn). Entonces ϕ(n) se verifica para todo n ∈ ω − k . 2. Ordinales Los ordinales son conjuntos asociados con buenos ´ordenes, de hecho, son los ejemplos tıpicos ´ deultimos estos ´ en el sentido que todo buen orden es isomorfo aunalg´ ordinal. La definici´on de ordinal no hace sino extender las propiedades de los n´ umeros naturales estudiadas en
Definici ´on 3.6.
i) a es ∈–transitivo si
∀x∀y ((x ∈ y ∧ y ∈ a) → x ∈ a). ii) a es un ordinal si a es ∈–transitivo y todo elemento de a es ∈–
Notaci´on: Usaremos letras griegas min´ usculas para denotar ordinales. Nada en la definici´on indica que haya ordinales, ni siquiera que ex-
son ordinales.
umeros naturales
i) 0 es un ordinal. ii) Ord (x) → Ord (Sx). iii) ∀x(x ∈ ω → Ord (x)). iv) Ord (ω ). Teorema 3.9.
i) Ord (0) trivialmente ya que 0 = ∅. ii) Supongamos que x es ordinal. Entonces x es ∈–transitivo y Demostraci ´ on.
Para ver que Sx es ∈–transitivo consideremos z , y tales que ∈ ∈ Sx
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iii) Consideramos la f´ormula ϕ(x) = Ord (x). Por i) y ii) y P.I., todo n´ umero natural es un ordinal. iv) El lema 3.3, ii) demuestra que ω es ∈–transitivo. Si n ∈ ω , iii) demuestra que n es ∈–transitivo. Luego Ord (ω ). Ver que no todo conjunto es un ordinal es f´acil. De hecho, {{0}} no es ni siquiera ∈–transitivo. Tambi´en es f´acil ver unque par ning´ ordenado a, b es un ordinal. Teorema 3.10. Si C es un conjunto de ordinales, entonces C
es un ordinal.
Demostraci ´ on. Para demostrar que C es ∈–transitivo, supong-
amos x ∈ y ∈ C . Entonces existe α ∈ C talque y ∈ α. Como es ∈–transitivo, x ∈ α y por lo tanto, x ∈ y ∈ α y esteultimo ´ x ∈ C, es decir, C es ∈–transitivo. Sea x ∈ C . Entonces existe α ∈ C tal que x ∈ α, entonces, x tambi´en es ∈–transitivo. Por lo tanto C es un ordinal. El siguiente teorema demuestra que todo ordinal est´a compuesto por ordinales. Teorema 3.11. Si α es un ordinal, entonces todos sus elementos
son ordinales.
x ∈ α. Entonces x es ∈– transitivo. Si y ∈ x, como α es ∈–transitivo, y ∈ α, luego y es ∈–transitivo, luego x es ordinal. Demostraci ´ on. Supongamos que
Los ordinales nos proporcionan un nuevo ejemplo de clase propia. Teorema 3.12. No existe el conjunto de todos los ordinales. Demostraci ´ on. Supongamos por el contrario que O es el conjunto
de todos los ordinales. Entonces es claro que O es ∈–transitivo y que todos sus elementos son ∈–transitivos, es decir, O es un ordinal, luego O ∈ O, contradicci´ on. β
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Demostraci ´ on. Supongamos por el contrario que existen ordi-
nales α y β tales que α ∈ β, α= β y β ∈ α. Sea
A = {x ∈ Sα ∪ Sβ : ∃y (y ∈ Sα ∪ Sβ ∧ x ∈ y ∧ x = y ∧ y ∈ x)}.
Nuestra suposici´on implica que α, β ∈ A. Luego A = ∅. B = {x ∈ Sα ∪ Sβ : a ∈ x ∧ a = x ∧ x ∈ a}
B= ∅ ya que a ∈ A. Por el axioma de regularidad nuevamente sea b ∈ B tal que B ∩ b = ∅. Como a ∈ B, a = b. La contradicci´ on se obtiene probando que a = b. Sea x ∈ a, como a es ordinal, x tambi´en lo es. Como a ∩ A = como x ∈ a ∈ Sα ∪ Sβ, x ∈ Sα ∪ Sβ , ya que ∅, x ∈ A. Porultimo ´ esteultimo, ´ al ser uni´on de dos ordinales, es tambi´en un ordinal y, por lo tanto, es ∈–transitivo. Entonces
∀y (y ∈ Sα ∪ Sβ → (x ∈ y ∨ x = y ∨ y ∈ x)) en particular como b ∈ Sα ∪ Sβ , x ∈ b ∨ x = b ∨ b ∈ x.
Si x = b, b ∈ a, luego b ∈ B . Si b ∈ x, como a es ordinal b ∈ a, luego b ∈ B . En ambos casos tenemos una contradicci´ on. unica posibilidad es x ∈ b, es decir, hemos dePor lo tanto la ´ mostrado que a ⊆ b. Supongamos ahora que x ∈ b. Como antes x ∈ Sα ∪ Sβ y x ∈ B ya que B ∩ b = ∅. Luego a∈ x ∨ a= x ∨ x∈ a.
Si a ∈ x ´o a = x, entonces a ∈ b ya que b es ordinal. Luego x ∈ a. Hemos demostrado que b ⊆ a. Por lo tanto a = b lo que contradice nuestra suposici´on, luego α∈ β ∨ α= β ∨ β∈ α.
El teorema 1.3 garantiza que s´olo una de las tres posibilidades anteriores se verifica.
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Demostraci ´ on. Si x ∈
A, entonces para todo α ∈ A, x ∈ α, luego x es ordinal, por lo tanto es ∈–transitivo. Supongamos x ∈ y ∈A. Entonces para todo α ∈ A, x ∈ y ∈ A es ∈–transitivo. Hemos demostrado que A es un A y ordinal. Por 3.13, para todo α ∈ A, A ∈ α ´o A = α ´o α ∈ A, ultima posibilidad implica que, en particular, α ∈ α. Las pero esta ´ ∈ A. dos primeras posibilidades implicanA que Teorema 3.15. Para cualquier α y β
i) ii) iii) iv) v)
α ∈ β ssi α ⊂ β . Si C ⊆ α, entonces C = α o´ C ∈ α. α = Sα. Si α ∈ β , entonces Sα ∈ β ´o Sα = β. α = S α o´ α = α.
i) ⇒ Por ∈–transitividad de β , si x ∈ α, entonces x ∈ β , o sea, α ⊆ β . Como α ∈ β − α, α= β . ⇐ Si α ⊂ β , entonces α = β y β ∈ α. Luego por el teorema 3.13, α ∈ β . ii) C es un ordinal por el teorema 3.10. Supongamos que α ∈C , entonces α ∈ x para alg´ un x ∈ C y como hemos supuesto que C ⊆ α, x ∈ α. O sea α ∈ x ∈ α, lo que contradice el teorema 1.3. Luego C ∈ α ´o C = α. iii) Si x ∈ Sα, x ∈ y para alg´ un y ∈ Sα. Entonces y ∈ α ´o y = α; en cualquier caso x ∈ α. Entonces x ∈ α ∈ Sα, luego x ∈ Sα. Demostraci ´ on.
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Definici ´on 3.7. Sea α un ordinal. Definimos la relaci´on ≤ en
α como sigue. Para x, y ∈ α x≤ y
si y s´olo si
x ∈ y ∨ x = y.
N´otese que, en rigor, para distintos ordinales α el orden reci´en
Tambi´en debemos notar que ´este coincide con el orden que definimos para losumeros n´ naturales. El siguiente teorema resume las principales propiedades de este orden. Teorema 3.16. Sea α un ordinal.
i) ≤ es un buen orden en α. ii) Si A ⊆ α y A= ∅, entonces A es el menor elemento de A. iii) 0 es el menor elemento de α . vi) No existe un ordinal β tal que α < β < Sα. A es el supremo de A . ix) α < β si y s´olo si Sα ≤ β . x) Sα = Sβ si y s´olo si α = β. xi) Sα < Sβ si y s´olo si α < β . Demostraci ´ on.
i) ≤ es obviamente reflexiva. Es antisim´etrica
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A es un ordinal y para x ∈ A viii) Supongamos A ⊆ α, entonces es claro que x ⊆A, o sea, por vii), x ≤ A, es decir, A es cota superior. Sea β otra cota superior de A . Entonces para todo A, x ∈ como x ∈ α ∈ A y β es cota superior de A , x ≤ α ≤ β , es decir, x ∈ β , o sea, A ⊆ β, por lo tanto ∪A es la menor de las cotas superiores de A . ix) Si α ∈ β , entonces por ∈–transitividad Sα ⊆ β , o sea, Sα ≤ β . Si Sα ≤ β , como α < Sα, α < β , por transitividad de < . x) Supongamos Sα = Sβ y α= β . Entonces como α ∈ Sα = Sβ, α ∈ Sβ . Pero entonces α ∈ β . An´alogamente β ∈ α lo que es imposible, luego, α = β . Es claro que si α = β, Sα = Sβ . xi) α < β si y s´olo si Sα ≤ β si y s´olo si Sα < Sβ por ix) y iv) respectivamente.
Observaci´on. En viii) hay un peque˜ no abuso producido por el hecho de denotar = con el mismo sımbolo ´ ordenes distintos. En efecto, siAtenemos α, entonces ∪A no est´a en el campo de ≤, luego no puede ser una cota superior. Podemos sin embargo entender dicha proposici´on como
Ejercicios. 1. Probar que: (a) es un ordinal si y s´olo si
es un buen orden sobre
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7. Sea B un conjunto de ordinales . Probar que si α es un ordinal β ≤ α. Probar tambi´en y para todo β ∈ B , β ≤ α , entonces que lo anterior no es necesariamente cierto si no se requiere que α sea ordinal. 8. Probar que si α ∈ ω y β es el mayor elemento de α , entonces α = Sβ . ¿Es esto cierto si α es ordinal arbitrario? 3. Inducci´on Transfinita En esta secci´on generalizaremos el teorema de inducci´on de los n´ umeros naturales a los ordinales. De hecho, el primero es un caso particular del segundo. Teorema 3.17. (Principio de Inducci´on Transfinita)
α,
Si ϕ(x) es una f´ormula con una variable libre y para todo ordinal (∗ ) si para todo β < α, ϕ(β ) , entonces ϕ(α). Entonces para todo ordinal α, ϕ(α).
ning´ Demostraci ´ on. Observemos primero que como no existe un
β < 0, por ( ∗ ), ϕ(0) se verifica trivialmente. ordinal Supongamos que el teorema es falsoun para alg´ α. Entonces
β < γ, β ∈ C , es decir, para todo β < γ, ϕ(β ), pero entonces por la hip´otesis ( ∗ ), ϕ(γ ), pero como γ ∈ C , se llega a una contradicci´on. Luego para todo ordinal α ϕ(α).
N´otese que esta formulaci´on del teorema de inducci´on transfinita es similar a la del teorema de inducci´on completa umeros de los natn´
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iii) α es l´ımite si y s´olo α= 0 ∧ ∀β (β < α → ∃ γ (β < γ < α)). i) Si α = Sβ , entonces por el teorema 3.15 iii), α = Sβ = β < α, Reciprocamente, siα < α, S α ≤ α. Si suponemos que S α < α, α ∈ S α ∈ α,
Demostraci ´ on.
o sea, α ∈ α, una contradicci´ on. Luego α = Sα y como α es un ordinal, α es sucesor. ≤ α. Entonces por i), ii) Sabemos por el teorema 3.15 ii),αque α = α si y s´olo si α no es sucesor, luego α = 0 y α = α si y s´olo si α es l´ımite. iii) Por ii), α es l´ımite si y s´olo si α= 0, αα = y esta ultima ´ es equivalente a que para todo β ∈ α, existe γ ∈ α tal que β ∈ γ ∈ α. Teorema 3.19. Sea ϕ(x) una f´ormula con una variable libre tal
que i) ϕ(0), ii) para todo ordinal α , si ϕ(α) , entonces ϕ(Sα) y iii) para todo ordinal l´ımite α , si ϕ(β ) para todo β < α, entonces ϕ(α). Entonces para todo ordinal α, ϕ(α).
Demostraci ´ on. Para demostrar este teorema, demostraremos las
hip´otesis del teorema 3.17 a partir de i), ii) y iii). Sea entonces α un ordinal y supongamos que ϕ(β ) para todo β < α. Debemos demostrar que ϕ(α). Hay tres casos.
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1. Demuestre que el Principio de Inducci´on dado en el teorema 3.19 implica al Principio de Inducci´on del teorema 3.17. 2. Demuestre el Principio de Inducci´on 3.19 sin usar el teorema 3.17. 3. Demuestre el siguiente Principio de Inducci´on. Sean α un ordinal no nulo A ⊆ α tales que (a) 0 ∈ A, (b) si β ∈ A y Sβ < α, entonces Sβ ∈ A, (c) si β = β < α y para todo γ < β , γ ∈ A. Entonces A = α. 4. Recursi´on En esta secci´on estudiaremos un proceso, ´ıntimamente ligado a la En ´algebra elemental frecuentemente nos encontramos con sucesiones del tipo siguiente a0 = 1 an+1 = 2an + 1
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conjuntos cualquira. Este ultimo ´ caso es necesario ya que la operaci´on debe estar definida para todos los conjuntos y no s´olo para los ordinales. i) Sea y = G(x) una operaci´on unaria. Entonces existe una unica ´ operaci´on unaria y = F (x) tal que
Teorema 3.20.
∀α(Ord (α) → F (α) = G({F (β ) : β ∈ α}) ∧ ∀x(¬Ord (x) → F (x) = x) ii) Sean y = G(x), y = H (x) dos operaciones unarias, entonces existe una unica ´ operaci´on unaria y = F (x) tal que para todo ordinal α F (Sα) = G(F (α)) ∧ (α = α → F (α) = H {F (β ) : β ∈ α})
y si x no es ordinal, F (x) = x). Demostraci ´ on.
i) Demostraremos por inducci´on que para todo
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f ∗α0 α0 = = = =
{fα (β ) : β ∈ α0 } {fβ (β ) : β ∈ α0 } { gα ( β ) : β ∈ α 0 } gα∗ α0 , 0
0
0
luego fα0 (α0 ) = G(fα∗0 α0 ) = G(gα∗0 α0 ) = gα0 (α0 ),
Es decir fα0 = gα0 lo que contradice nuestra suposici´on. Por lo tanto, por el teorema 3.17, para todo ordinal α existe unaunica ´ funci´on fα que verifica a) y b). Podemos ahora definir nuestra operaci´on unaria y = F (x). F (x) =
fx (x) x
si x es ordinal, si x no es ordinal.
Por construcci´on, y = F (x) verifica la tesis del teorema. ii) Demostraremos por inducci´on que para todo α unica existe una ´ funci´on f tal que
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fβ tambi´en verifica a), b), c) y d). Luego por inducci´on, para todo ordinal α , existe una funci´on fα con las caracter´ısticas indicadas. La unicidad de fα es obvia y se deja como ejercicio. Podemos definir la operaci´on unaria F como en i).
Ejercicios. 1. Definimos la clausura transitiva de a, que denotamos T (a), como Indicaci´on: Definir recursivamente una funci´on F sobre ω de tal manera que F (0) = a F (n + 1 ) =
F (n)
y luego probar que T (a) = T (a) = a ∪ a ∪
ω F (n). Intuitivamente an∈∪ a··· ,
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1. La funci´on S es estrictamente creciente pero no continua, por ejemplo Sω= ω = Sω. 2. La funci´on es continua pero no estrictamente creciente, por ejemplo, ω < Sω , peroω = Sω. 3. Sea α un ordinal l´ımite. Definimos para β ∈ α f (β ) =
Sβ β
, si β es sucesor, , si no.
F´acilmente podemos ver que f es normal. Teorema 3.21. Sean A, B ordinales µ : A → B una funci´on.
Si µ es estrictamente creciente, entonces para todo α ∈ A, µ(α).
α≤
Demostraci ´ on. Por inducci´on. Teorema 3.22. Sean A, B ordinales µ : A → B una funci´on
continua tal que si Sβ ∈ A,
µ(β ) <µ(Sβ ). Entonces µ es normal.
Demostraci ´ on. Tenemos que demostrar que µ es estrictamente
creciente. Lo haremos por inducci´on usando la siguiente f´ormula
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δ < µ(γ ) < µ(α), ya que como µ es estrictamente creciente, µ(γ ) < µ(α). Luego por teorema 3.18 iii), µ(α) es l´ımite. Teorema 3.24. Si α es un ordinal, β < α y µ es no decreciente,
entonces
µ(γ ) = γ<α
Demostraci ´ on. Si γ < β
β ≤γ<α
µ(γ ).
µ(γ ) ≤ µ(β ), luego µ(γ ) ≤ µ(β )
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o sea
ν ◦ µ(α) =
ν ◦ µ(δ ),
δ<α
es decir ν ◦ µ es continua y por lo tanto ν ◦ µ es normal. Teorema 3.26. Sea α un ordinal, µ una funci´on normal con
un βalg´ ∈ α existen δ, γ ∈ α dominio α . Supongamos que para tal que µ(δ ) ≤ β < µ(γ ). Entonces existe unico un γ ´ ∈ α tal que µ(γ ) ≤ β < µ(Sγ ).
Demostraci ´ on. Sea ε el menor γ tal que β < µ(γ ). Este debe
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los posibles buenos ordenes. Volveremos sobre este tema en el pr´oximo capıtulo. ´ Teorema 3.27. Todo buen orden es isomorfo con (el orden de) un
ordinal.
Demostraci ´ on. Sea R un buen orden cuyo campo es A . Defi-
namos la operaci´on unaria F (α) =
R − menor elemento de A − F ∗ α si ´este es no vac´ıo, A si no.
(Para ser rigurosos, la operaci´on F debe estar definida para todo
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7. Aritm´etica Ordinal A continuaci´on definiremos la suma, el producto y la exponenciaci´on de ordinales. Las dos primeras operaciones tienen una interpretaci´on bastante un ordinal. β despu´es de los de α , gr´aficamente
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Formalizaremos ahora las ideas intuitivas sobre la suma de ordinales que dimos en la introducci´on de esta secci´on. Teorema 3.29. Dados α y β ordinales, definimos el orden R
cuyo campo es {0}× α ∪{1}× β como sigue:
x0 , y0 R x1 , y1 si y solamente si se verifica una de las siguientes condiciones 1. x0 = x1 = 0 y y0 ≤ y1 < α, 2. x0 = 0 , x1 = 1, 3. x0 = x1 = 1 y y0 ≤ y1 < β.
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iii) α < β si y s´olo si existe γ > 0 tal que α + γ = β. iv) Si α < β , entonces α + γ ≤ β + γ.
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– Si 0 = λ = ∪λ y para todo γ ∈ λ
γ ≤ α + γ . Entonces
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– 0 + Sα = S (0 + α) = Sα. – Si 0 = α = ∪α y para todo γ ∈ α, 0 + γ = γ ,
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Notemos que la suma de ordinales no es general conmutativa, en efecto,
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Teorema 3.34. Dados α y β ordinales, definimos el orden R
cuyo campo es α × β como sigue:
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= α β · (α γ · α ) = αβ · αSγ .
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debe usarse un argumento similar al empleado en la demostraci´on de vii). Esto completa la inducci´on.
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3. Demuestre (a) Si α ≤ tambi´en β , entonces 3.35, γ + iii), α≤ iv),γ v) + βy. ix). 4. Demuestre (b) Si γ + el α< teorema γ + β , entonces 3.41. α < β. (c) Si α + γ < β + γ , entonces α < β .
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⊆ Sα es par ,o pero los una respectivos casos. dei∈ordinales Clasificar (que β, (a)γtal )que mente Si =ordinal α {(α δ+ crecientes :iβ∈αδ= esω ordinal , entonces (impar αi )yi∈no yen < α (·δ ββi )< ω , }tales .sucesi´on al que que pertenece lim el i) αβ existe i= ∈ αγ αα i+ los (lim αi )ii∈γ que en pares ordinal. (b)siguientes Si Una operaci´on
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ordinales, probar creciente. Probar que que para si α estodo ordinal ordinal sucesor λ seentonces tiene λ +existe limi∈α αi ,= pero limsi λ+ esαordinal l´ımite no necesariamente no necesariamente se i∈α (α i ) , pero existelim tiene tali∈l´ımite. α αi + λ = limi∈α (αi + λ). Probar tambi´en que
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entonces si para todo β ∈ α se tiene F(β ) ≤ G(β ) , ello implica que
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primo αm , con los que se (c) todo (b) ω 22 ·+3m ω+·<2 1 n+ =existe 1 (ω 2 +un 1) ordinal ·3 2 3 2 2 :+=ω2 · (7ω+ + (c) tiene (d) ω ·+32 6 1) 3 2 (d) ω · 7 + ω · 5 + 3 = 3 · (ω 2 + 1) · 5 · (ω + 1) · 7 (e) ω ω + ω + 1 = (ω + 1) · (ω ω + 1)
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(b) lim •n∈εωSαn=· ω E (εα ); n ordinal l´ımite, entonces εα = limβ ∈α εβ , (c) lim•n∈siω ω α · es (d) limβ ∈ω2 β · ω
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teor´ıa tos 2. que Siyes debe α de ∈ βgran entenderse , entonces importancia s´olo Vα ∈ V en como una manera m´asde avanzado abreviarde unlos concepto funβ . el estudio damentos correspondiente 3. Para de todo la teor´ıa α a, laVf´ormula es conjuntos. ∈–transitivo. Si bien estos temas caen fuera α de del alcance de este libro, esta construcci´on, llamada la jerarqu´ıa acu-
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pares, potencias productos cartesianos y dem´as construcciones que hemos estudiado en los capıtulos ´ precedentes. Esto implica que todos los conjuntos que aparecen en la pr´actica, pertenecen a V . ¿ Existir´a alg´
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3. Demuestre α es el mayor ordinal que pertenece a Vα . 4. Demuestre el axioma de regularidad a partir del teorema 3.43.
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a no una es de en categor´ıa absoluto aparte intuitivo debido que se aconjunto las pueda N´otese consecuencias tambi´en que que deel neamente equivalencia un elemento cuyo de cada campo esimportantes A y hacer. cuyas que pertenece clases de a equivalencia A . Esto no ´el axioma seson desprenden. no ninguna Estudiaremos ideaP de varias construir formulaciones tal funci´on. diferentes de ´este representa losda ning´ elementos de . Elcomo principio de Zermelo nos dice que y enseguida algunas aplicaciones. podemos escogerde unsus elemento de cada clase de equivalencia, es decir un sistema de representantes de las clases de equivalencia.
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subconjunto de A totalmente ordenado por R tiene una cota superior en A , entonces A tiene un elemento maximal.
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no vac´ıos, luego existe una funci´on de elecci´on F para ´el. Notemos que ´esta asigna a cada subconjunto no vac´ıo de A un elemento de si mismo.
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el S -menor (i.e. A − H ∗elemento β = ∅). de Supongamos {a ∈ A : F (0) por
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una <-cadena. Como F ∗ α ⊆ A, por el Lema de Zorn, existe c ∈ A que es <-cota superior de F ∗ α, es decir
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sea g : αα−→ Rque tal isomorfismo. por es generalidad ordinal talque existe R y una { β,funci´on γ : Entonces β ≤ inyectiva, γ < α} , esser fdecir, : Aisomorfismo, −→elBorden . Sea de ag∈α A, inyectiva y g : αEntonces −→ Cam R ⊆ B , es decir α ∈ Γ. y definamos son isomorfos.
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del2. axioma La uni´on de elecci´on. enumerable A continuaci´on de conjuntos daremos enumerables una breve es enumerable. lista con 5. Todo anillo con unidad tiene un ideal maximal. algunas 3. Todo Existe de ´estas. un conjunto n´ extenderse a un orden total. 6. orden parcial de puede 7. El teorema de Hahn-Banach. 8. El teorema de completud para la l´ogica de primer orden. 9. Toda ´algebra de Boole es isomorfa a un campo de conjuntos.
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plicidad podemos que elementos de linealmente C son todosindependisjuntos. por lo tanto v0 , . . .suponer , vn−1 ∈ B y como ´este es n−1 los Si C = {Ai : i ∈ ω }, sea diente,
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de Zermelo, existe ⊆ [0 tal´este que B us´o contiene capıtulo mos como ´ usado el anterior 5) yelsea axioma B sin =darse de Bsubconjunto +elecci´on. ri cuenta = {x + rde :Bxque ∈ embargo B estaban },,1) donde usando la suma se un esprincipio cuando i un i Sin exactamente representante de cada clase de equivalencia. escogimos especial. m´odulo 1 i.e. lasun funciones biyectivas fi , especificamente,
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una funci´on de elecci´on. (Indicaci´on: Use inducci´on sobre el n´
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de elementos del ejemplo queanterior tiene undemuestra conjunto. Hasta que cualquier el momento intervalo sabemos abierto c´omo es π π contar equinumeroso los elementoscon de los (−conjuntos finitos. En este capıtulo ´ for, ). 2 2 malizaremos estos conceptos intuitivos y los generalizaremos a conjuntos infinitos. La mayor parte de los teoremas de este capıtulo ´ dependen † del axioma de elecci´on. Indicaremos ´estos mediante el sımbolo ´ .
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generalizar la tal idea anterior a conjuntos infinitos, sin embargo, dos ´el. trica todos la de funci´on los De yaquellos conjuntos transitiva, hecho que que aparece finitos afirmaci´on pero son no equivalentes aen los esla una infinitos es demostraci´on ni relaci´on m´as bajo es ni suponer la dentro menos, “relaci´on”anterior. del el teorema de elAxioma principio nuestra 5.3, de teor´ıa del Elecci´on lahay que buen Una nos dificultades, primera que dado un conjunto infinito cualquiera, no orden, yaescoger posible dice y que que es su idea decir, campo α unes yla representante α definir equivalente es + 1la son clase laes cardinalidad siempre de con detodos cada el axioma equinumerosos. una los deconjuntos. de un delas conjunto elecci´on. “clases Naturalmente, como de En equivalencia”, segundo su “claselupodemos que exista alg´ gar, sisaber, de a nos equivalencia” esrestringimos f´acil elgarantizar menor ver que bajo ordinal asiun la existe conjunto relaci´on que un pertenece ordinal particular, de equinumerosidad, equinumeroso a ella. ´esta Necesitamos es una con sin relaci´on un embargo, antes conjunto de un es infinito equivalencia. claro par de que dado, resultados. la clase ´este deno equivalencia ser´a nuncade ´ cualquier conjunto no vac´ıo es una clase propia, por lo tanto no existen dentro de nuestra teor´ıa,luego no podemos usarlas como “cardinales”.
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iii) |α| ≤ α. iv) Si α ∼ A, entonces |A| ≤ α.
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entonces Si x∈Dn−1f, n: − x∈ii) An , Hay x∈An−1 . inyectiva = B0. y entonces x∈ una funci´on A1 −→ y por lo que anteriormente, (An−→ An−iii) An−= o hay una vimos funci´on sobreyectivax∈ g h:∗B AD .n−2 ), 1 −D 1 ∅ −2 −
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iii) ⇒ i)† Supongamos A = 0 y sea g : B −→ A sobreyectiva. Entonces −1∗ g A induce una partici´on de B , a saber, {g −1∗ {a} : a ∈ A}. Por el axioma de elecci´on, escogemos un sistema de representantes para esta
|A| ∼ |f ∗ A| y f ∗ A ⊆ B, luego por 5.5, |A| = |f ∗ A| ≤ |B |. Observemos que el axioma de elecci´on se usa solamente en la deun ejemplo de cardinal. Los siguientes dos teoremas remedian esta situaci´on. Teorema 5.7. Si n ∈ ω , entonces n es un cardinal. Demostraci ´ on. Por inducci´on.
Es claro que 0 es un cardinal. Supongamos que n es un cardinal y n + 1 no lo es. Entonces existe m < n + 1 tal que m ∼ n + 1, es claro que m = 0. Sea f : n + 1 −→ m una biyecci´ on. Podemos suponer que f(n) = m − 1, on, luego, por hip´otesis
Teorema 5.8. ω es cardinal. Demostraci ´ on. Sabemos por 5.3, iii), que |ω | ≤ ω . Si |ω | <
ω, |ω | = m para alg´ un n´ umero natural m. Pero m ⊆ m + 1 ⊆ ω, luego
|ω | = m = | m | ≤ | m + 1 | = m + 1 ≤ | ω | luego m = m + 1, contradicci´ on. Luego |ω | = ω . Teorema 5.9. Cantor
Para todo A,
|A| < |P A|.
Demostraci ´ on. Es claro que |A| ≤ |P A| ya que la asignaci´on
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B = {x ∈ A : x ∈ f (x)}.
Como B ⊆ A y f es sobreyectiva, existe a ∈ A tal que B = f (a). Entonces si a ∈ B, por definici´on a ∈ f (a), luego a ∈ B . Si a ∈ B , entonces a ∈ f (a), o sea, a ∈ B , es decir, a ∈ B si y s´olo si a ∈ B,
contradicci´on, luego tal funci´on sobreyectiva no puede=existir y |A| |P A|. Corolario 5.10. Dado un ordinal α , existe un cardinal m > α. m = |P α|. Demostraci ´ on. Basta considerar
Corolario 5.11. No existe el conjunto de todos los cardinales. Demostraci ´ on. Si existiera el conjunto C de todos los cardinales,
por el teorema reci´en demostrado, el conjunto
{ α ∈ n : α < n} n
∈C
contendr´ıa a todos los ordinales y por lo tanto ´estos constituir´ıan tambi´en un conjunto. Teorema 5.12. Si Γ es un conjunto de cardinales, entonces Γ
es un cardinal.
Demostraci ´ on. Supongamos Γ no es un cardinal, entonces,
Γ| < n∈Γ
|
Γ
(recordemos queΓ es un ordinal). Luego, existe
| pero
n
Γ| ∈ n ∈ Γ ,
⊆ Γy |
Γ| < n = |n| ≤ | Γ| .
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ℵ0 = ω ℵα+1 = (ℵα )+ ℵλ =
α<λ
ℵα , si λ es l´ımite.
Ejercicios. 1. Demuestre que la relaci´on “ser equinumeroso” es refleja sim´etrica y transitiva. 2. Probar que: (a) PPP 0 ∼ 4. (b) Si A ∼ B , entonces C A ∼ C B . (c) Si B ∩ C = ∅, entonces B ∪C A ∼ B A × C A. (d) C (B A) ∼ B ×C A. (e) C (A × B ) ∼ C A × C B . (f) Si R1 es el conjunto de todos los ´ordenes estrictos sobre X y R2 es el conjunto de todos los ´ordenes sobre X , entonces R1 ∼ R2 . (g) Probar que si E es el conjunto de todas las clases de equiv3. Sea A un conjunto. Probar que P A ∼ A 2. 4. Sea {Ij : j ∈ J } una partici´on de I . Probar que i∈I
Ai ∼
j ∈J
(
i ∈I j
Ai ).
5. Si A1 ∼ A2 y B1 ∼ B2 y A1 ∩ B1 = A2 ∩ B2 = ∅, probar que A1 ∪ B1 ∼ A2 ∪ B2 . 6. Probar que si A ∼ C y B ∼ D , entonces A × B ∼ C × D. 9. Probar que si A ∼ B (B − {b}). 10. Probar que si A ∼ B (A − C ) ∼ (B − D ). 11. Sean (Ai )i∈I y (Bi )i∈I disjuntos dos a dos.
y a ∈ A y b ∈ B , entonces (A − {a}) ∼ y C ∼ D y C ⊆ A y D ⊆ B , entonces dos familias, cada una de conjuntos Si A B para todo i ∈ I , probar que
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13. Demuestre las partes del teorema 5.3 que no fueron demostradas en el texto. 14. Haga la inducci´on mencionada en la demostraci´on del teorema 5.4. 15. Demuestre directamente, sin usar el axioma de elecci´on, la equivalencia de las dos primeras afirmaciones del teorema 5.6. 16. Demuestre directamente, sin usar el axioma de elecci´on, la implicaci´on i) ⇒ iii) del teorema 5.6. umeros 17. Probar que para intervalos de n´ reales se tiene (a, b) ∼ [a, b]. 18. Probar que ω y | ω 2| son cardinales distintos. 19. Probar que |Q| = |Q[x]| = ω , donde Q[x] es el conjunto de polinomios en x con co eficientes en Q. 20. Sea R el conjunto de todas las raıces ´ reales de polinomios de Q[x]. Probar que |R| = ω. 21. Si B ⊆ A y ω ⊆ |B | < |A|, cual es la cardinalidad de A − B ?. 22. Probar que α es un cardinal tal que ω < α si y s´olo si α es ordinal l´ımite. 23. Para α y β ordinales, probar o dar contraejemplo de: (c) |α + β | = |α| + |β |. (d) |α · β | = |α| · |β |. (e) Si α < β , entonces |α| < |β |. 24. Sean A , B y C conjuntos. Probar que: (a) Si B = ∅, entonces |A| ≤ | B A|. 25. Probar que para todo α , ℵα es un cardinal. 26. Probar que: (a) ℵα = ℵβ si y s´olo si α = β . (b) ℵα < ℵβ si y s´olo si α < β . 27. Probar que para todo cardinal α existe un ordinal β tal que α < ℵβ . 2. Conjuntos Finitos y Conjuntos Infinitos En esta secci´on daremos un marco formal a la nociones intuitivas
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iii) A es enumerable si |A| = ℵ0 . umeros Una consecuencia inmediata de esta definici´on es que los n´ umero natural. Ası´ por ejemplo |{x}| = 1 , |{x, y }| = 2 , si x = y, etc. Teorema 5.13. Si A ⊆ B o existe una funci´on sobreyectiva f :
B −→ A, entonces: i) Si A es infinito, B es infinito, ii) † Si B es finito, A es finito.
Demostraci ´ on. Si A ⊆ B , use el teorema 5.5.
Si existe una funci´on sobreyectiva f : B −→ A, entonces como on g : |A| −→ A, tenemos que existe una biyecci´ g ◦ f : |A| −→ B
es una funci´on sobreyectiva, luego por el teorema 5.6, |A| ≤ |B|. (Aquı´ es donde hemos usado el axioma de elecci´on.) La conclusi´on del teorema se sigue de esta afirmaci´on. Teorema 5.14. Si A es finito y B ⊂ A, entonces |B | < |A|. Demostraci ´ on. Sea b ∈ A − B y f : A −→ m una biyecci´on.
Podemos suponer que f(b) = m − 1 pues si f(b) = m − 1 podemos f B : B −→ m − 1 es inyectiva y por el teorema 5.6,
|B | ≤ m − 1 < m = | A |. Recu´erdese que esta parte del teorema 5.6 no requiere del axioma de elecci´on. El teorema anterior caracteriza a los conjuntos finitos, de hecho, se
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⇐) Por el teorema anterior. ⇒) Como A es infinito, |A| ≥ ℵ0 = ω , luego existe una funci´on inyectiva f : ω −→ A. Definamos h : A −→ A − {f (0)} como sigue Demostraci ´ on. †
h(x) =
f (f −1 (x) + 1) si x ∈ f ∗ ω, si x ∈ f ∗ ω. x
es claro que h es biyectiva. Teorema 5.16. Si A y B son finitos, |A| = |B | y f : A −→ B ,
entonces f es inyectiva si y s´olo si f es sobreyectiva.
⇒) Supongamos f es inyectiva. Si f no es sobreyectiva, entonces como f ∗ A ⊆ B , por 5.14
Demostraci ´ on.
|A | = | f ∗ A | < | B |, contradicci´on. ⇐) Supongamos que f es sobreyectiva. Entonces f induce la siguiente partici´on sobre A :
{f −1∗ {b} : b ∈ B } Podemos seleccionar un representante de cada elemento de la partici´on y definir g : B −→ A b −→ el representante de f −1∗ {b} seleccionado.
Obs´ervese que como B es finito, la partici´on indicada es finita y por lo
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3. Probar que, si A y B son finitos, entonces A ∪ B , A × B y A B son finitos. Si A o B son enumerables ¿es alguno de los conjuntos mencionados infinito no numerable?. 4. Probar que A es infinito si y s´olo si existe un subconjunto enumerable de A . 5. Probar que si A es infinito y B no es vac´ıo, entonces A × B es infinito. 6. Probar que si A es infinito y B ⊆ A es finito, entonces A − B es infinito. 7. Probar que A es infinito si y s´olo si para todo n ∈ ω existe B ⊆ A con B ∼ n. 8. Sean A infinito y B enumerable. Probar que A ∼ (A ∪ B ). 9. Probar que si A es enumerable y x ∈ A, entonces A − {x} es enumerable. Probar que A es infinito si y s´olo si A ∼ (A − {x}). 10. Probar que si P A es infinito, entonces A es infinito. 11. Probar que todo subconjunto de un conjunto enumerable es finito o enumerable. 12. Probar que si A y B son enumerables, entonces A × B es enumerable. Probar lo mismo si B es finito no vac´ıo. 13. Sea (Ai )i∈ω una familia de conjuntos enumerables. Probar que 14. Probar que si A es enumerable, entonces existe B ⊆ A enumeri∈ω Ai es enumerable. able tal que A − B es enumerable. 15. Probar que si n ∈ ω , entonces ω n es enumerable (exponenciaci´on ordinal). 16. Probar que si I = ω , entonces i los subconjuntos finitos de un 17. Probar que el conjunto de todos i∈I ω es enumerable. conjunto enumerable es enumerable. 18. Probar que si A es infinito, entonces A es enumerable si y s´olo
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(c) Diremos que a es progresi´on aritm´etica si existe d ∈ ω tal umeros naturales
es enumerable. 21. Probar que una partici´on de un conjunto enumerable tiene un 22. Sea ϕ una f´ormula con variable libre x . Probar que si A es
23. Probar que A es finito si y s´olo si A pertenece a todo conjunto K tal que ∅ ∈ K y para todo x ∈ A y todo B ⊆ A tal que B ∈ K , se tenga B ∪ {x} ∈ K . 24. Probar que A es finito si y s´olo si A pertenece a todo conjunto K tal que ∅ ∈ K , y x ∈ A implica {x} ∈ K y, si B ∈ K y C ∈ K , entonces B ∪ C ∈ K . 25. Probar que A es finito si y s´olo si Punico A es conjunto el ´ K 26. Probar que, si A es enumerable, entonces existe una familia B de conjuntos tal que: (a) B es enumerable, (b) si C ∈ B , entonces C es enumerable, (c) Si C y D est´an en B , y C= D , entonces C ∩ D = ∅ , y (d) B = A. Probar el rec´ıproco usando el Axioma de Elecci´on. 3. Aritm´etica Cardinal
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Si I = 2, escribimos mi
=
m0
+c
m1
i ∈2
M´as generalmente si i = n, escribimos mi
=
m0
+c · · · +c
mn−1 .
i ∈n
La idea de la definici´on es que la suma de cardinales representa la
m0
+c
m1
= |{(0, α) : α ∈
m0 }
como
{(1, α) : α ∈ m1 }|
|{(0, α) : α ∈ m0 }| = m0 y |{(1, α) : α ∈ m1 }| = m1 , la finalidad de tomar estos conjuntos de pares ordenados es simplemente
Teorema 5.17.
†
Si Ai : i ∈ I y Bi : i ∈ I son familias de conjuntos disjuntos dos a dos tales que para todo i ∈ I Ai ∼ Bi , entonces i∈I
Ai ∼
i∈I
Bi ∼
i ∈I
|A i |
Demostraci ´ on. Para i ∈ I , sea fi : Ai −→ Bi una biyecci´ on.
Definimos
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En particular mi
≤
i∈I
mi . i ∈I
Demostraci ´ on. Si |Ai | = mi para cada i ∈ I , sea fi : mi −→ A
biyectiva y definamos: g:
i∈I
{ i, α : α ∈ mi } −→ Ai −→ fi (a).
i, α
Es f´acil ver que g es sobreyectiva luego por 5.6, i | ≤| i∈I A i ∈I | A i | . Algunas propiedades elementales de la suma de cardinales est´an resumidas en el pr´oximo teorema. Teorema 5.19.
ii)
mi
i)
= 0.
i ∈I
0 = 0.
i∈0
iii) Si I ⊆ J , entonces iv) Si
mi
v)
mi i∈I
≤
mi . i ∈J
≤ ni , para i ∈ I , entonces
i∈I
mi ≤
ni . i ∈I
1 = m. β<
m
Demostraci ´ on. i) y ii) son obvias a partir de la definici´on.
iii) Basta notar que
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es una biyecci´ on. Teorema 5.20. (Conmutatividad Generalizada)
Si mi : i ∈ I es una familia de cardinales y σ es una permutaci´on de I (i.e. σ : I −→ I es una biyecci´on), entonces mi i ∈I
= i ∈I
mσ(i)
Demostraci ´ on. Consid´erese
f:
i ∈I
{ i, α : α ∈ mi } −→ { i, α : α ∈ mσ(i) } i ∈I
i, α
−→
σ −1 (i), α .
Es claro que f es biyectiva, por ejemplo, para verificar que f es : α ∈ mσ(i) } es decir k ∈ I y sobreyectiva, sea k, β i∈ ∈I { i, α β ∈ mσ(k) . Sea j = σ (k ), entonces
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Pero
Aij : i ∈ I es tambi´en una familia de conjuntos disjuntos a pares luego | Aij | = | ( Aij )|. j ∈J
i ∈I
j ∈J
i ∈I j ∈J
Para terminar basta con notar que Aij = i,j ∈I ×J
i ∈I j ∈J
Aij ,
por asociatividad de la uni´on. Juntando los cuatro resultados anteriores, mij = mij . ( i,j ∈I ×J
i ∈I
j ∈J
Para completar estos resultados sobre la suma cardinal veremos algunos teoremas sobre cardinales finitos. Teorema 5.22. Para todo parumeros de n´ naturales m, n,
m +c n = m + n
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Corolario 5.23. Si A y B son finitos, entoncesBAtambi´en
lo es.
Demostraci ´ on. Por 5.17, |A∪B | ≤ |A|+c |B | = |A|+|B | < ω .
3.2. Producto de Cardinales. : i ∈ I una familia de cardinales. El producto cardinal de los mi es
Definici ´on 5.6. Sea
mi
Xi∈I mi = |
mi | i ∈I
Si I = 2, Xi∈I mi se escribe m0 ×c m1 . En general si I = n, Xi∈I mi = m0 ×c m1 ×c · · · ×c mn−1 . Como en el caso de la suma, el producto de cardinales no depende de
Teorema 5.24.
†
Si |Ai | = |Bi | para todo i ∈ I , entonces
|
i ∈I
Ai | = |
i ∈I
Bi |.
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El pr´oximo teorema resume algunas propiedades elementales de el producto cardinal. Teorema 5.26.
Xi∈I mi = 0. ii) Xi∈∅ mi = 1. iii) Xi∈I 1 = 1. m = |I| × m. iv) i ∈I
v) Si
mi
i) Si
mi
= 0 para alg´ un i ∈ I , entonces
≤ ni , para todo i ∈ I , entonces Xi∈I mi ≤ Xi∈I ni . i) En este caso
Demostraci ´ on.
ii) i∈I mi = ∅. iii) i∈∅ mi = {0} iv) N´otese que i∈I 1 = {f }, donde f (i) = 0, para todo i ∈ I . I ×m=
i ∈I
{ i, α : α ∈ m},
luego por definici´on, m i∈I
= | i ∈I
{ i, α : α ∈ m}|
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Demostraci ´ on. Por el teorema 5.24
Xi∈I (Xj ∈J mij ) = | Xi,j
∈I ×J mij
i ∈I
(
mij )| j ∈J
mij |
= | i,j ∈I ×J
Definimos F :
mij −→ i,j ∈ FI(×fJ )( i)(j ) =
f ( i∈i,J (j )j.∈J mij )
Es claro que F es biyecci´ on. Teorema 5.29. (Distributividad)
Si para todo i ∈ Ini es un cardinal, n ×c
ni i∈I
m × c ni .
= i ∈I
Demostraci ´ on. Notemos primero que m × c ni
=|
{ i, α, β : α ∈ m, β ∈ ni }|,
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Demostraci ´ on. Supongamos que esto es falso. Sea m el menor
cardinal tal que m ×c m= m. Definiremos un orden sobre el producto cartesiano m ×c m y demostraremos que ´este es un buen orden. Para α, β, γ, δ ∈ m definimos
α, β γ, δ si y s´olo si 1. α = γ y β = δ ´o 2. α ∪ β < γ ∪ δ ´o 3. α ∪ β = γ ∪ δ y α<γ 4. α ∪ β = γ ∪ δ y α= γ
´o y
β ≤ δ.
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f (ε) f (m) δ,δ y f (ε) = δ. Por lo tanto f m : m −→ δ × δ es inyectiva, luego, m ≤ | δ × δ | = |δ | × c |δ |.
Como δ < m, o bien, δ es finito, o bien |δ |×c |δ | = |δ |. En el primer caso m es finito. En el segundo m ≤ |δ| y δ < m lo que contradice el que m sea cardinal. Esto concluye la demostraci´on.
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Demostraci ´ on.
|
i ∈I
Ai | ≤
≤
i ∈I
|A i | m
i ∈I
= |I |×c m ≤ m ×c m = m.
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Para cada i ∈ I definimos hi : { i, α : α ∈ mi } −→ ni i, α −→ h( i, α )(i) .
Es claro que hi es inyectiva porque h lo es. Entonces
|hi∗{ i, α : α ∈ mi }| ≤ |{ i, α : α ∈ mi }| = mi <
ni
Podemos usar nuevamente el principio multiplicativo y encontrar
∈
i∈I
(ni − k ∗i { i, α : α ∈ mi }).
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3.3. Exponenciaci´on de Cardinales. Para terminar con esta secci´on definiremos la operaci´on de exponenciaci´on de cardinales. Debe-
un m´as una notaci´on de suyo algo
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Demostraci ´ on. i) - iv) Son obvias.
v) Basta notar que
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Demostraci ´ on. Asociamos a cada conjunto de A su funci´on car-
acter´ıstica:
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9. Probar que si m ≤ n, entonces m ×c p ≤ n ×c p. 10. (a) Probar que si m ≤ n, entonces ≤ nm . p
p
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Obs´ervese que si α = β + 1 es un sucesor, la funci´on f : 1 −→ α
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Teorema 5 44
†
Todo cardinal sucesor infi
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dominiode pertenencia Axioma dePares una 8 funci´on 11 30 Axioma par recorrido no-ordenado de de Extensionalidad una11 funci´on 30 9 subconjunto singleton campo de11 una funci´on 9 30 Axioma de Uniones composici´on de funciones 11 30 uni´on 11 intersecci´on 12
relaci´on funci´on sobreyectiva asociada a una 31partici´on 42 relaci´on funci´on identidad antisim´etrica 33 44 relaci´on restricci´on conexa de una 45funci´on 33 orden relaci´on parcial reflexiva 45 40 orden relaci´on total sim´etrica 45 40 relaci´on relaci´onasim´etrica transitiva 4046
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conjuntode Principio infinito Inducci´on 126 Transfinita 70 ordinal sucesor conjunto enumerable 70 126 ordinalcardinal suma 70 129 clausura transitiva Teorema de Zermelo 75 139 funci´on144 cofinal no–decreciente 75 funci´on estrictamente cofinalidad 144 creciente 75