Prove that $|omega+omega|=|omegacdotomega|=|omega^omega|=|omega|$












3












$begingroup$



$|omega+omega|=|omegacdotomega|=|omega^omega|=|omega|$




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



We have:



$omega+omega={omega+alphamid alpha<omega}$



$omegacdotomega={omegacdotalpha mid alpha<omega}$



$omega^omega={omega^alpha mid alpha<omega}$



Thus we build bijections $f,g,h$ as follows:



$f:omega to omega+omega$ such that $f(alpha)=omega+alpha$



$g:omega to omegacdotomega$ such that $f(alpha)=omegacdotalpha$



$h:omega to omega^omega$ such that $f(alpha)=omega^alpha$



This completes the proof.






elementary-set-theory ordinals







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    These maps are not going to be surjective for the correct definitions..
    $endgroup$
    – Berci
    Dec 1 '18 at 8:51












  • $begingroup$
    You need Zorn's lemma for uncountable sets.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 1 '18 at 9:06










  • $begingroup$
    @Berci Please be more specific! I am unable to understand "the correct definitions".
    $endgroup$
    – Le Anh Dung
    Dec 1 '18 at 9:46










  • $begingroup$
    Is there some reason you are trying to explicitly construct bijections? You could simply use what you know about cardinals. Two of the equalities are simply sayling that $aleph_0+aleph_0=aleph_0cdotaleph_0=aleph_0$. And the third one follows from the fact that union of countably many countable sets is gains countable.
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 10:00






  • 3




    $begingroup$
    As previous comment mentioned, what you wrote for $omegacdotomega$ and $omega^omega$ differs from the usual definition of ordinal multiplication and ordinal power. Did you perhaps want to write unions of those sets, i.e., something like $bigcuplimits_{alpha<omega} omegacdotalpha$ and $bigcuplimits_{alpha<omega} omega^alpha$?
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 10:03
















3












$begingroup$



$|omega+omega|=|omegacdotomega|=|omega^omega|=|omega|$




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



We have:



$omega+omega={omega+alphamid alpha<omega}$



$omegacdotomega={omegacdotalpha mid alpha<omega}$



$omega^omega={omega^alpha mid alpha<omega}$



Thus we build bijections $f,g,h$ as follows:



$f:omega to omega+omega$ such that $f(alpha)=omega+alpha$



$g:omega to omegacdotomega$ such that $f(alpha)=omegacdotalpha$



$h:omega to omega^omega$ such that $f(alpha)=omega^alpha$



This completes the proof.






elementary-set-theory ordinals







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    These maps are not going to be surjective for the correct definitions..
    $endgroup$
    – Berci
    Dec 1 '18 at 8:51












  • $begingroup$
    You need Zorn's lemma for uncountable sets.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 1 '18 at 9:06










  • $begingroup$
    @Berci Please be more specific! I am unable to understand "the correct definitions".
    $endgroup$
    – Le Anh Dung
    Dec 1 '18 at 9:46










  • $begingroup$
    Is there some reason you are trying to explicitly construct bijections? You could simply use what you know about cardinals. Two of the equalities are simply sayling that $aleph_0+aleph_0=aleph_0cdotaleph_0=aleph_0$. And the third one follows from the fact that union of countably many countable sets is gains countable.
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 10:00






  • 3




    $begingroup$
    As previous comment mentioned, what you wrote for $omegacdotomega$ and $omega^omega$ differs from the usual definition of ordinal multiplication and ordinal power. Did you perhaps want to write unions of those sets, i.e., something like $bigcuplimits_{alpha<omega} omegacdotalpha$ and $bigcuplimits_{alpha<omega} omega^alpha$?
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 10:03














3












3








3





$begingroup$



$|omega+omega|=|omegacdotomega|=|omega^omega|=|omega|$




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



We have:



$omega+omega={omega+alphamid alpha<omega}$



$omegacdotomega={omegacdotalpha mid alpha<omega}$



$omega^omega={omega^alpha mid alpha<omega}$



Thus we build bijections $f,g,h$ as follows:



$f:omega to omega+omega$ such that $f(alpha)=omega+alpha$



$g:omega to omegacdotomega$ such that $f(alpha)=omegacdotalpha$



$h:omega to omega^omega$ such that $f(alpha)=omega^alpha$



This completes the proof.






elementary-set-theory ordinals







share|cite|improve this question









$endgroup$





$|omega+omega|=|omegacdotomega|=|omega^omega|=|omega|$




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



We have:



$omega+omega={omega+alphamid alpha<omega}$



$omegacdotomega={omegacdotalpha mid alpha<omega}$



$omega^omega={omega^alpha mid alpha<omega}$



Thus we build bijections $f,g,h$ as follows:



$f:omega to omega+omega$ such that $f(alpha)=omega+alpha$



$g:omega to omegacdotomega$ such that $f(alpha)=omegacdotalpha$



$h:omega to omega^omega$ such that $f(alpha)=omega^alpha$



This completes the proof.






elementary-set-theory ordinals




elementary-set-theory ordinals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 8:42









Le Anh DungLe Anh Dung

1,1111521




1,1111521








  • 3




    $begingroup$
    These maps are not going to be surjective for the correct definitions..
    $endgroup$
    – Berci
    Dec 1 '18 at 8:51












  • $begingroup$
    You need Zorn's lemma for uncountable sets.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 1 '18 at 9:06










  • $begingroup$
    @Berci Please be more specific! I am unable to understand "the correct definitions".
    $endgroup$
    – Le Anh Dung
    Dec 1 '18 at 9:46










  • $begingroup$
    Is there some reason you are trying to explicitly construct bijections? You could simply use what you know about cardinals. Two of the equalities are simply sayling that $aleph_0+aleph_0=aleph_0cdotaleph_0=aleph_0$. And the third one follows from the fact that union of countably many countable sets is gains countable.
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 10:00






  • 3




    $begingroup$
    As previous comment mentioned, what you wrote for $omegacdotomega$ and $omega^omega$ differs from the usual definition of ordinal multiplication and ordinal power. Did you perhaps want to write unions of those sets, i.e., something like $bigcuplimits_{alpha<omega} omegacdotalpha$ and $bigcuplimits_{alpha<omega} omega^alpha$?
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 10:03














  • 3




    $begingroup$
    These maps are not going to be surjective for the correct definitions..
    $endgroup$
    – Berci
    Dec 1 '18 at 8:51












  • $begingroup$
    You need Zorn's lemma for uncountable sets.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 1 '18 at 9:06










  • $begingroup$
    @Berci Please be more specific! I am unable to understand "the correct definitions".
    $endgroup$
    – Le Anh Dung
    Dec 1 '18 at 9:46










  • $begingroup$
    Is there some reason you are trying to explicitly construct bijections? You could simply use what you know about cardinals. Two of the equalities are simply sayling that $aleph_0+aleph_0=aleph_0cdotaleph_0=aleph_0$. And the third one follows from the fact that union of countably many countable sets is gains countable.
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 10:00






  • 3




    $begingroup$
    As previous comment mentioned, what you wrote for $omegacdotomega$ and $omega^omega$ differs from the usual definition of ordinal multiplication and ordinal power. Did you perhaps want to write unions of those sets, i.e., something like $bigcuplimits_{alpha<omega} omegacdotalpha$ and $bigcuplimits_{alpha<omega} omega^alpha$?
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 10:03








3




3




$begingroup$
These maps are not going to be surjective for the correct definitions..
$endgroup$
– Berci
Dec 1 '18 at 8:51






$begingroup$
These maps are not going to be surjective for the correct definitions..
$endgroup$
– Berci
Dec 1 '18 at 8:51














$begingroup$
You need Zorn's lemma for uncountable sets.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 1 '18 at 9:06




$begingroup$
You need Zorn's lemma for uncountable sets.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 1 '18 at 9:06












$begingroup$
@Berci Please be more specific! I am unable to understand "the correct definitions".
$endgroup$
– Le Anh Dung
Dec 1 '18 at 9:46




$begingroup$
@Berci Please be more specific! I am unable to understand "the correct definitions".
$endgroup$
– Le Anh Dung
Dec 1 '18 at 9:46












$begingroup$
Is there some reason you are trying to explicitly construct bijections? You could simply use what you know about cardinals. Two of the equalities are simply sayling that $aleph_0+aleph_0=aleph_0cdotaleph_0=aleph_0$. And the third one follows from the fact that union of countably many countable sets is gains countable.
$endgroup$
– Martin Sleziak
Dec 1 '18 at 10:00




$begingroup$
Is there some reason you are trying to explicitly construct bijections? You could simply use what you know about cardinals. Two of the equalities are simply sayling that $aleph_0+aleph_0=aleph_0cdotaleph_0=aleph_0$. And the third one follows from the fact that union of countably many countable sets is gains countable.
$endgroup$
– Martin Sleziak
Dec 1 '18 at 10:00




3




3




$begingroup$
As previous comment mentioned, what you wrote for $omegacdotomega$ and $omega^omega$ differs from the usual definition of ordinal multiplication and ordinal power. Did you perhaps want to write unions of those sets, i.e., something like $bigcuplimits_{alpha<omega} omegacdotalpha$ and $bigcuplimits_{alpha<omega} omega^alpha$?
$endgroup$
– Martin Sleziak
Dec 1 '18 at 10:03




$begingroup$
As previous comment mentioned, what you wrote for $omegacdotomega$ and $omega^omega$ differs from the usual definition of ordinal multiplication and ordinal power. Did you perhaps want to write unions of those sets, i.e., something like $bigcuplimits_{alpha<omega} omegacdotalpha$ and $bigcuplimits_{alpha<omega} omega^alpha$?
$endgroup$
– Martin Sleziak
Dec 1 '18 at 10:03










1 Answer
1






active

oldest

votes


















1












$begingroup$

Since $omega$ × $omega$ equinumerous to N×N, it is countable.

By induction for all n in N, $omega^n$ is countable.



$omega^{omega}$ = sup$_n$ $omega^n$ = $cup_n omega^n$ is a countable union of countable sets equinumerous to N×N, hence countable.



Finally as $omega$ + $omega$ < $omega$ × $omega,$ it too is countable.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    How is this helpful to verify the presented proof?
    $endgroup$
    – Asaf Karagila♦
    Dec 1 '18 at 10:10











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021131%2fprove-that-omega-omega-omega-cdot-omega-omega-omega-omega%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Since $omega$ × $omega$ equinumerous to N×N, it is countable.

By induction for all n in N, $omega^n$ is countable.



$omega^{omega}$ = sup$_n$ $omega^n$ = $cup_n omega^n$ is a countable union of countable sets equinumerous to N×N, hence countable.



Finally as $omega$ + $omega$ < $omega$ × $omega,$ it too is countable.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    How is this helpful to verify the presented proof?
    $endgroup$
    – Asaf Karagila♦
    Dec 1 '18 at 10:10
















1












$begingroup$

Since $omega$ × $omega$ equinumerous to N×N, it is countable.

By induction for all n in N, $omega^n$ is countable.



$omega^{omega}$ = sup$_n$ $omega^n$ = $cup_n omega^n$ is a countable union of countable sets equinumerous to N×N, hence countable.



Finally as $omega$ + $omega$ < $omega$ × $omega,$ it too is countable.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    How is this helpful to verify the presented proof?
    $endgroup$
    – Asaf Karagila♦
    Dec 1 '18 at 10:10














1












1








1





$begingroup$

Since $omega$ × $omega$ equinumerous to N×N, it is countable.

By induction for all n in N, $omega^n$ is countable.



$omega^{omega}$ = sup$_n$ $omega^n$ = $cup_n omega^n$ is a countable union of countable sets equinumerous to N×N, hence countable.



Finally as $omega$ + $omega$ < $omega$ × $omega,$ it too is countable.






share|cite|improve this answer









$endgroup$



Since $omega$ × $omega$ equinumerous to N×N, it is countable.

By induction for all n in N, $omega^n$ is countable.



$omega^{omega}$ = sup$_n$ $omega^n$ = $cup_n omega^n$ is a countable union of countable sets equinumerous to N×N, hence countable.



Finally as $omega$ + $omega$ < $omega$ × $omega,$ it too is countable.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 10:08









William ElliotWilliam Elliot

7,6322720




7,6322720








  • 2




    $begingroup$
    How is this helpful to verify the presented proof?
    $endgroup$
    – Asaf Karagila♦
    Dec 1 '18 at 10:10














  • 2




    $begingroup$
    How is this helpful to verify the presented proof?
    $endgroup$
    – Asaf Karagila♦
    Dec 1 '18 at 10:10








2




2




$begingroup$
How is this helpful to verify the presented proof?
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 10:10




$begingroup$
How is this helpful to verify the presented proof?
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 10:10


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021131%2fprove-that-omega-omega-omega-cdot-omega-omega-omega-omega%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
Read more