If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal












1












$begingroup$



If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal.




My attempt:



Assume the contrary that $alpha$ is not a limit ordinal. Then $alpha$ is a successor ordinal and thus $alpha=beta+1$ for a unique $beta$.



It follows that $beta bigcup{beta}=aleph_alpha$. Thus $|aleph_alpha|=|beta|$ or $aleph_alpha=|beta|$. This means $aleph_alpha$ is equipotent to a smaller ordrinal. This is clearly a contradiction.





My proof is quite short and I wonder if it contains any logical flaw/gap. Thank you for your help!










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$endgroup$

















    1












    $begingroup$



    If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal.




    My attempt:



    Assume the contrary that $alpha$ is not a limit ordinal. Then $alpha$ is a successor ordinal and thus $alpha=beta+1$ for a unique $beta$.



    It follows that $beta bigcup{beta}=aleph_alpha$. Thus $|aleph_alpha|=|beta|$ or $aleph_alpha=|beta|$. This means $aleph_alpha$ is equipotent to a smaller ordrinal. This is clearly a contradiction.





    My proof is quite short and I wonder if it contains any logical flaw/gap. Thank you for your help!










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal.




      My attempt:



      Assume the contrary that $alpha$ is not a limit ordinal. Then $alpha$ is a successor ordinal and thus $alpha=beta+1$ for a unique $beta$.



      It follows that $beta bigcup{beta}=aleph_alpha$. Thus $|aleph_alpha|=|beta|$ or $aleph_alpha=|beta|$. This means $aleph_alpha$ is equipotent to a smaller ordrinal. This is clearly a contradiction.





      My proof is quite short and I wonder if it contains any logical flaw/gap. Thank you for your help!










      share|cite|improve this question









      $endgroup$





      If $aleph_alpha=alpha$, then $alpha$ is a limit ordinal.




      My attempt:



      Assume the contrary that $alpha$ is not a limit ordinal. Then $alpha$ is a successor ordinal and thus $alpha=beta+1$ for a unique $beta$.



      It follows that $beta bigcup{beta}=aleph_alpha$. Thus $|aleph_alpha|=|beta|$ or $aleph_alpha=|beta|$. This means $aleph_alpha$ is equipotent to a smaller ordrinal. This is clearly a contradiction.





      My proof is quite short and I wonder if it contains any logical flaw/gap. Thank you for your help!







      proof-verification elementary-set-theory ordinals






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      asked Dec 9 '18 at 1:24









      Le Anh DungLe Anh Dung

      1,4141621




      1,4141621






















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          $begingroup$

          Yes your argument follows. In general, any infinite cardinal is a limit ordinal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh i forget that. Many thanks!
            $endgroup$
            – Le Anh Dung
            Dec 9 '18 at 1:29











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          0












          $begingroup$

          Yes your argument follows. In general, any infinite cardinal is a limit ordinal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh i forget that. Many thanks!
            $endgroup$
            – Le Anh Dung
            Dec 9 '18 at 1:29
















          0












          $begingroup$

          Yes your argument follows. In general, any infinite cardinal is a limit ordinal.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Oh i forget that. Many thanks!
            $endgroup$
            – Le Anh Dung
            Dec 9 '18 at 1:29














          0












          0








          0





          $begingroup$

          Yes your argument follows. In general, any infinite cardinal is a limit ordinal.






          share|cite|improve this answer









          $endgroup$



          Yes your argument follows. In general, any infinite cardinal is a limit ordinal.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 1:28









          Alberto TakaseAlberto Takase

          2,380719




          2,380719












          • $begingroup$
            Oh i forget that. Many thanks!
            $endgroup$
            – Le Anh Dung
            Dec 9 '18 at 1:29


















          • $begingroup$
            Oh i forget that. Many thanks!
            $endgroup$
            – Le Anh Dung
            Dec 9 '18 at 1:29
















          $begingroup$
          Oh i forget that. Many thanks!
          $endgroup$
          – Le Anh Dung
          Dec 9 '18 at 1:29




          $begingroup$
          Oh i forget that. Many thanks!
          $endgroup$
          – Le Anh Dung
          Dec 9 '18 at 1:29


















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