Let $α$ and $β$ be ordinals such that $α>1$ and $βneq0$. Let $γ$, $ζ$, and $η$ be ordinals st...












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Let $alpha$ and $beta$ be ordinals such that $alpha>1$ and $betaneq0$. Let $gamma$, $zeta$, and $eta$ be ordinals such that $gamma<beta$, $zeta<alpha$, and $eta<alpha^{gamma}$. I want to prove that $alpha^{gamma}cdotzeta+eta<alpha^{beta}$. I have already observed that $alpha^{gamma}cdotzeta<alpha^{beta}$. The hard part is showing that the equality still holds if we add $eta$. Any ideas?






set-theory ordinals







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    $begingroup$


    Let $alpha$ and $beta$ be ordinals such that $alpha>1$ and $betaneq0$. Let $gamma$, $zeta$, and $eta$ be ordinals such that $gamma<beta$, $zeta<alpha$, and $eta<alpha^{gamma}$. I want to prove that $alpha^{gamma}cdotzeta+eta<alpha^{beta}$. I have already observed that $alpha^{gamma}cdotzeta<alpha^{beta}$. The hard part is showing that the equality still holds if we add $eta$. Any ideas?






    set-theory ordinals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $alpha$ and $beta$ be ordinals such that $alpha>1$ and $betaneq0$. Let $gamma$, $zeta$, and $eta$ be ordinals such that $gamma<beta$, $zeta<alpha$, and $eta<alpha^{gamma}$. I want to prove that $alpha^{gamma}cdotzeta+eta<alpha^{beta}$. I have already observed that $alpha^{gamma}cdotzeta<alpha^{beta}$. The hard part is showing that the equality still holds if we add $eta$. Any ideas?






      set-theory ordinals







      share|cite|improve this question











      $endgroup$




      Let $alpha$ and $beta$ be ordinals such that $alpha>1$ and $betaneq0$. Let $gamma$, $zeta$, and $eta$ be ordinals such that $gamma<beta$, $zeta<alpha$, and $eta<alpha^{gamma}$. I want to prove that $alpha^{gamma}cdotzeta+eta<alpha^{beta}$. I have already observed that $alpha^{gamma}cdotzeta<alpha^{beta}$. The hard part is showing that the equality still holds if we add $eta$. Any ideas?






      set-theory ordinals




      set-theory ordinals






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      share|cite|improve this question













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      edited Jan 4 at 17:15







      Eigenfield

















      asked Jan 4 at 16:42









      EigenfieldEigenfield

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          2 Answers
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          $begingroup$

          Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.



          From the hypothesis, we know that:
          $$
          eta <alpha^gamma
          $$
          Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
          $$
          alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
          $$
          Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
          $$
          alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
          $$



          And thus we have shown what was needed.



          Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.






          share|cite|improve this answer









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            1












            $begingroup$

            Note that



            $$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$



            Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.






            share|cite|improve this answer









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              2 Answers
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              2 Answers
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              active

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              $begingroup$

              Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.



              From the hypothesis, we know that:
              $$
              eta <alpha^gamma
              $$
              Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
              $$
              alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
              $$
              Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
              $$
              alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
              $$



              And thus we have shown what was needed.



              Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.



                From the hypothesis, we know that:
                $$
                eta <alpha^gamma
                $$
                Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
                $$
                alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
                $$
                Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
                $$
                alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
                $$



                And thus we have shown what was needed.



                Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.



                  From the hypothesis, we know that:
                  $$
                  eta <alpha^gamma
                  $$
                  Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
                  $$
                  alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
                  $$
                  Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
                  $$
                  alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
                  $$



                  And thus we have shown what was needed.



                  Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.






                  share|cite|improve this answer









                  $endgroup$



                  Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.



                  From the hypothesis, we know that:
                  $$
                  eta <alpha^gamma
                  $$
                  Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
                  $$
                  alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
                  $$
                  Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
                  $$
                  alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
                  $$



                  And thus we have shown what was needed.



                  Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 17:17









                  ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro

                  661




                  661























                      1












                      $begingroup$

                      Note that



                      $$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$



                      Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Note that



                        $$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$



                        Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Note that



                          $$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$



                          Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.






                          share|cite|improve this answer









                          $endgroup$



                          Note that



                          $$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$



                          Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 4 at 17:08









                          Asaf Karagila♦Asaf Karagila

                          306k33438769




                          306k33438769






























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