A question about the proof of...











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I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.



Below is the verbatim proof by @user642796.




By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.



The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
alpha^{beta + gamma}
&= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
&= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
&= alpha^beta cdot sup { alpha^delta : delta < gamma }
&&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
\
&=alpha ^beta cdot alpha^gamma &&text{(by definition)}
end{align}$$



(The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)






My questions:




  1. I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve



$$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
= &alpha^beta cdot sup { alpha^delta : delta < gamma }
left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$





  1. The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.


Thank you so much!










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This question has an open bounty worth +50
reputation from Le Anh Dung ending in 6 days.


This question has not received enough attention.


It will be great if someone helps me fill in the blank of the proof.




















    up vote
    2
    down vote

    favorite












    I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.



    Below is the verbatim proof by @user642796.




    By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.



    The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
    alpha^{beta + gamma}
    &= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
    &= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
    &= alpha^beta cdot sup { alpha^delta : delta < gamma }
    &&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
    \
    &=alpha ^beta cdot alpha^gamma &&text{(by definition)}
    end{align}$$



    (The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)






    My questions:




    1. I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve



    $$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
    = &alpha^beta cdot sup { alpha^delta : delta < gamma }
    left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$





    1. The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.


    Thank you so much!










    share|cite|improve this question

















    This question has an open bounty worth +50
    reputation from Le Anh Dung ending in 6 days.


    This question has not received enough attention.


    It will be great if someone helps me fill in the blank of the proof.


















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.



      Below is the verbatim proof by @user642796.




      By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.



      The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
      alpha^{beta + gamma}
      &= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
      &= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
      &= alpha^beta cdot sup { alpha^delta : delta < gamma }
      &&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
      \
      &=alpha ^beta cdot alpha^gamma &&text{(by definition)}
      end{align}$$



      (The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)






      My questions:




      1. I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve



      $$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
      = &alpha^beta cdot sup { alpha^delta : delta < gamma }
      left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$





      1. The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.


      Thank you so much!










      share|cite|improve this question















      I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.



      Below is the verbatim proof by @user642796.




      By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.



      The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
      alpha^{beta + gamma}
      &= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
      &= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
      &= alpha^beta cdot sup { alpha^delta : delta < gamma }
      &&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
      \
      &=alpha ^beta cdot alpha^gamma &&text{(by definition)}
      end{align}$$



      (The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)






      My questions:




      1. I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve



      $$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
      = &alpha^beta cdot sup { alpha^delta : delta < gamma }
      left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$





      1. The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.


      Thank you so much!







      proof-explanation ordinals






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      edited Nov 8 at 1:29

























      asked Nov 8 at 1:20









      Le Anh Dung

      1,2021421




      1,2021421






      This question has an open bounty worth +50
      reputation from Le Anh Dung ending in 6 days.


      This question has not received enough attention.


      It will be great if someone helps me fill in the blank of the proof.








      This question has an open bounty worth +50
      reputation from Le Anh Dung ending in 6 days.


      This question has not received enough attention.


      It will be great if someone helps me fill in the blank of the proof.





























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