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

1,0751521




1,0751521








  • 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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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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

7,4742720




7,4742720








  • 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


















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