Every ordinal $alpha>0$ can be expressed uniquely as $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot...











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Every ordinal $alpha>0$ can be expressed uniquely as



$$alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$$



where $beta_1>beta_2>cdots>beta_n$ and $k_1>0,k_2>0,cdots,k_n>0$ are finite.




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





My attempt:



Existence



Assume that $xi$ can be expressed as normal form for all $xi<alpha$.



Let $beta=max{xiinrm Ordmidomega^xilealpha}$. Then there is $delta$ and $rho<omega^beta$ such that $alpha=omega^betacdotdelta+rho$. Since $rho<omega^beta$, $rho<alpha$ and thus $delta>0$. I claim that $delta$ is finite. If not, $delta$ is infinite and thus $deltageomega$. Then $omega^{beta+1}=omega^betacdotomegaleomega^betacdotdeltalealpha$ and thus $omega^{beta+1}lealpha$. This contradicts the maximality of $beta$. Thus $delta$ is finite. Let $beta_1=beta$ and $k_1=delta$. If $rho=0$, then $alpha=omega^{beta_1}cdot k_1$. If $rho>0$, then $rho=omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ for some $beta_2>cdots>beta_n$ and finite $k_2,cdots,k_n>0$ by inductive hypothesis. We have $omega^{beta_2}lerho<omega^{beta_1}$, then $beta_2<beta_1$. As a result, $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ as desired.



We observed that $beta<gammaimpliesforall kinomega:omega^betacdot k<omega^gamma$. This is because $omega^betacdot k<omega^betacdotomega=omega^{beta+1}leomega^gamma$. It follows that if $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ is in normal form and $beta_1<gamma$, then $alpha<omega^gamma$.



Uniqueness



Assume that the normal form of $xi$ is unique for all $xi<alpha$.



Let $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n=omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$. The previous observation implies that $beta_1=gamma_1$. Let $delta=omega^{beta_1}=omega^{gamma_1}$, $rho=omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$, and $sigma=omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$. Then $alpha=deltacdot k_1+rho=deltacdot l_1+sigma$ where $rho<delta$ and $sigma<delta$. Thus $k_1=l_1$ and $rho=sigma$. By inductive hypothesis, the normal form of $rho$ is unique and thus $m=n$, $beta_2=gamma_2,cdots,beta_n=gamma_n, k_2=l_2,cdots,k_n=l_m$. It follows that the normal form of $alpha$ is unique.










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  • I think that you must to argue why $beta$ is a maximum.
    – Gödel
    Nov 19 at 15:18










  • Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
    – Le Anh Dung
    Nov 19 at 22:59

















up vote
2
down vote

favorite













Every ordinal $alpha>0$ can be expressed uniquely as



$$alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$$



where $beta_1>beta_2>cdots>beta_n$ and $k_1>0,k_2>0,cdots,k_n>0$ are finite.




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





My attempt:



Existence



Assume that $xi$ can be expressed as normal form for all $xi<alpha$.



Let $beta=max{xiinrm Ordmidomega^xilealpha}$. Then there is $delta$ and $rho<omega^beta$ such that $alpha=omega^betacdotdelta+rho$. Since $rho<omega^beta$, $rho<alpha$ and thus $delta>0$. I claim that $delta$ is finite. If not, $delta$ is infinite and thus $deltageomega$. Then $omega^{beta+1}=omega^betacdotomegaleomega^betacdotdeltalealpha$ and thus $omega^{beta+1}lealpha$. This contradicts the maximality of $beta$. Thus $delta$ is finite. Let $beta_1=beta$ and $k_1=delta$. If $rho=0$, then $alpha=omega^{beta_1}cdot k_1$. If $rho>0$, then $rho=omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ for some $beta_2>cdots>beta_n$ and finite $k_2,cdots,k_n>0$ by inductive hypothesis. We have $omega^{beta_2}lerho<omega^{beta_1}$, then $beta_2<beta_1$. As a result, $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ as desired.



We observed that $beta<gammaimpliesforall kinomega:omega^betacdot k<omega^gamma$. This is because $omega^betacdot k<omega^betacdotomega=omega^{beta+1}leomega^gamma$. It follows that if $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ is in normal form and $beta_1<gamma$, then $alpha<omega^gamma$.



Uniqueness



Assume that the normal form of $xi$ is unique for all $xi<alpha$.



Let $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n=omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$. The previous observation implies that $beta_1=gamma_1$. Let $delta=omega^{beta_1}=omega^{gamma_1}$, $rho=omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$, and $sigma=omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$. Then $alpha=deltacdot k_1+rho=deltacdot l_1+sigma$ where $rho<delta$ and $sigma<delta$. Thus $k_1=l_1$ and $rho=sigma$. By inductive hypothesis, the normal form of $rho$ is unique and thus $m=n$, $beta_2=gamma_2,cdots,beta_n=gamma_n, k_2=l_2,cdots,k_n=l_m$. It follows that the normal form of $alpha$ is unique.










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  • I think that you must to argue why $beta$ is a maximum.
    – Gödel
    Nov 19 at 15:18










  • Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
    – Le Anh Dung
    Nov 19 at 22:59















up vote
2
down vote

favorite









up vote
2
down vote

favorite












Every ordinal $alpha>0$ can be expressed uniquely as



$$alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$$



where $beta_1>beta_2>cdots>beta_n$ and $k_1>0,k_2>0,cdots,k_n>0$ are finite.




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





My attempt:



Existence



Assume that $xi$ can be expressed as normal form for all $xi<alpha$.



Let $beta=max{xiinrm Ordmidomega^xilealpha}$. Then there is $delta$ and $rho<omega^beta$ such that $alpha=omega^betacdotdelta+rho$. Since $rho<omega^beta$, $rho<alpha$ and thus $delta>0$. I claim that $delta$ is finite. If not, $delta$ is infinite and thus $deltageomega$. Then $omega^{beta+1}=omega^betacdotomegaleomega^betacdotdeltalealpha$ and thus $omega^{beta+1}lealpha$. This contradicts the maximality of $beta$. Thus $delta$ is finite. Let $beta_1=beta$ and $k_1=delta$. If $rho=0$, then $alpha=omega^{beta_1}cdot k_1$. If $rho>0$, then $rho=omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ for some $beta_2>cdots>beta_n$ and finite $k_2,cdots,k_n>0$ by inductive hypothesis. We have $omega^{beta_2}lerho<omega^{beta_1}$, then $beta_2<beta_1$. As a result, $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ as desired.



We observed that $beta<gammaimpliesforall kinomega:omega^betacdot k<omega^gamma$. This is because $omega^betacdot k<omega^betacdotomega=omega^{beta+1}leomega^gamma$. It follows that if $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ is in normal form and $beta_1<gamma$, then $alpha<omega^gamma$.



Uniqueness



Assume that the normal form of $xi$ is unique for all $xi<alpha$.



Let $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n=omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$. The previous observation implies that $beta_1=gamma_1$. Let $delta=omega^{beta_1}=omega^{gamma_1}$, $rho=omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$, and $sigma=omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$. Then $alpha=deltacdot k_1+rho=deltacdot l_1+sigma$ where $rho<delta$ and $sigma<delta$. Thus $k_1=l_1$ and $rho=sigma$. By inductive hypothesis, the normal form of $rho$ is unique and thus $m=n$, $beta_2=gamma_2,cdots,beta_n=gamma_n, k_2=l_2,cdots,k_n=l_m$. It follows that the normal form of $alpha$ is unique.










share|cite|improve this question














Every ordinal $alpha>0$ can be expressed uniquely as



$$alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$$



where $beta_1>beta_2>cdots>beta_n$ and $k_1>0,k_2>0,cdots,k_n>0$ are finite.




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





My attempt:



Existence



Assume that $xi$ can be expressed as normal form for all $xi<alpha$.



Let $beta=max{xiinrm Ordmidomega^xilealpha}$. Then there is $delta$ and $rho<omega^beta$ such that $alpha=omega^betacdotdelta+rho$. Since $rho<omega^beta$, $rho<alpha$ and thus $delta>0$. I claim that $delta$ is finite. If not, $delta$ is infinite and thus $deltageomega$. Then $omega^{beta+1}=omega^betacdotomegaleomega^betacdotdeltalealpha$ and thus $omega^{beta+1}lealpha$. This contradicts the maximality of $beta$. Thus $delta$ is finite. Let $beta_1=beta$ and $k_1=delta$. If $rho=0$, then $alpha=omega^{beta_1}cdot k_1$. If $rho>0$, then $rho=omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ for some $beta_2>cdots>beta_n$ and finite $k_2,cdots,k_n>0$ by inductive hypothesis. We have $omega^{beta_2}lerho<omega^{beta_1}$, then $beta_2<beta_1$. As a result, $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ as desired.



We observed that $beta<gammaimpliesforall kinomega:omega^betacdot k<omega^gamma$. This is because $omega^betacdot k<omega^betacdotomega=omega^{beta+1}leomega^gamma$. It follows that if $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$ is in normal form and $beta_1<gamma$, then $alpha<omega^gamma$.



Uniqueness



Assume that the normal form of $xi$ is unique for all $xi<alpha$.



Let $alpha=omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n=omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$. The previous observation implies that $beta_1=gamma_1$. Let $delta=omega^{beta_1}=omega^{gamma_1}$, $rho=omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n$, and $sigma=omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$. Then $alpha=deltacdot k_1+rho=deltacdot l_1+sigma$ where $rho<delta$ and $sigma<delta$. Thus $k_1=l_1$ and $rho=sigma$. By inductive hypothesis, the normal form of $rho$ is unique and thus $m=n$, $beta_2=gamma_2,cdots,beta_n=gamma_n, k_2=l_2,cdots,k_n=l_m$. It follows that the normal form of $alpha$ is unique.







elementary-set-theory ordinals






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asked Nov 19 at 12:22









Le Anh Dung

1,0011421




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  • I think that you must to argue why $beta$ is a maximum.
    – Gödel
    Nov 19 at 15:18










  • Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
    – Le Anh Dung
    Nov 19 at 22:59




















  • I think that you must to argue why $beta$ is a maximum.
    – Gödel
    Nov 19 at 15:18










  • Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
    – Le Anh Dung
    Nov 19 at 22:59


















I think that you must to argue why $beta$ is a maximum.
– Gödel
Nov 19 at 15:18




I think that you must to argue why $beta$ is a maximum.
– Gödel
Nov 19 at 15:18












Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
– Le Anh Dung
Nov 19 at 22:59






Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
– Le Anh Dung
Nov 19 at 22:59












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










"The previous observation implies that $beta_1 = gamma_1$." I don't see that. For the existence part, I don't see any errors.



Also, this looks similar to the proof in section 3 of the excelent article "A short introduction to Ordinal Notations", by Simmons. It's freely available online; you should check it out. (Are you doing a thesis on ordinal notations?)



I like Simmon's way of writing the proof because it feels more like an algorithm :-). However, it's basically the same as yours.






share|cite|improve this answer





















  • Hi! Assume the contrary that $beta_1 neq gamma_1$. WLOG, we assume $beta_1 < gamma_1$, then $omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n<omega^{beta_1}cdot k_1+omega^{beta_1}cdot k_2+cdots+omega^{beta_1}cdot k_n=omega^{beta_1}cdot(k_1+k_2+cdots+k_n)$ $<omega^{gamma_1}<omega^{gamma_1}cdot l_1<omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$ and thus $alpha<alpha$. This is clearly a contradiction.
    – Le Anh Dung
    Nov 23 at 12:41












  • What I'm saying is that maybe you should clarify that. Also, are you doing a thesis on ordinals? I just finished mine. Feel free to chat.
    – Guillermo Mosse
    Nov 23 at 13:36










  • Thank you for your suggestion! I'm just learning basic stuff about set theory ;)
    – Le Anh Dung
    Nov 23 at 14:08






  • 1




    A nice book is Introduction to Set Theory, by Hrbace.k
    – Guillermo Mosse
    Nov 23 at 14:34











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










"The previous observation implies that $beta_1 = gamma_1$." I don't see that. For the existence part, I don't see any errors.



Also, this looks similar to the proof in section 3 of the excelent article "A short introduction to Ordinal Notations", by Simmons. It's freely available online; you should check it out. (Are you doing a thesis on ordinal notations?)



I like Simmon's way of writing the proof because it feels more like an algorithm :-). However, it's basically the same as yours.






share|cite|improve this answer





















  • Hi! Assume the contrary that $beta_1 neq gamma_1$. WLOG, we assume $beta_1 < gamma_1$, then $omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n<omega^{beta_1}cdot k_1+omega^{beta_1}cdot k_2+cdots+omega^{beta_1}cdot k_n=omega^{beta_1}cdot(k_1+k_2+cdots+k_n)$ $<omega^{gamma_1}<omega^{gamma_1}cdot l_1<omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$ and thus $alpha<alpha$. This is clearly a contradiction.
    – Le Anh Dung
    Nov 23 at 12:41












  • What I'm saying is that maybe you should clarify that. Also, are you doing a thesis on ordinals? I just finished mine. Feel free to chat.
    – Guillermo Mosse
    Nov 23 at 13:36










  • Thank you for your suggestion! I'm just learning basic stuff about set theory ;)
    – Le Anh Dung
    Nov 23 at 14:08






  • 1




    A nice book is Introduction to Set Theory, by Hrbace.k
    – Guillermo Mosse
    Nov 23 at 14:34















up vote
0
down vote



accepted










"The previous observation implies that $beta_1 = gamma_1$." I don't see that. For the existence part, I don't see any errors.



Also, this looks similar to the proof in section 3 of the excelent article "A short introduction to Ordinal Notations", by Simmons. It's freely available online; you should check it out. (Are you doing a thesis on ordinal notations?)



I like Simmon's way of writing the proof because it feels more like an algorithm :-). However, it's basically the same as yours.






share|cite|improve this answer





















  • Hi! Assume the contrary that $beta_1 neq gamma_1$. WLOG, we assume $beta_1 < gamma_1$, then $omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n<omega^{beta_1}cdot k_1+omega^{beta_1}cdot k_2+cdots+omega^{beta_1}cdot k_n=omega^{beta_1}cdot(k_1+k_2+cdots+k_n)$ $<omega^{gamma_1}<omega^{gamma_1}cdot l_1<omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$ and thus $alpha<alpha$. This is clearly a contradiction.
    – Le Anh Dung
    Nov 23 at 12:41












  • What I'm saying is that maybe you should clarify that. Also, are you doing a thesis on ordinals? I just finished mine. Feel free to chat.
    – Guillermo Mosse
    Nov 23 at 13:36










  • Thank you for your suggestion! I'm just learning basic stuff about set theory ;)
    – Le Anh Dung
    Nov 23 at 14:08






  • 1




    A nice book is Introduction to Set Theory, by Hrbace.k
    – Guillermo Mosse
    Nov 23 at 14:34













up vote
0
down vote



accepted







up vote
0
down vote



accepted






"The previous observation implies that $beta_1 = gamma_1$." I don't see that. For the existence part, I don't see any errors.



Also, this looks similar to the proof in section 3 of the excelent article "A short introduction to Ordinal Notations", by Simmons. It's freely available online; you should check it out. (Are you doing a thesis on ordinal notations?)



I like Simmon's way of writing the proof because it feels more like an algorithm :-). However, it's basically the same as yours.






share|cite|improve this answer












"The previous observation implies that $beta_1 = gamma_1$." I don't see that. For the existence part, I don't see any errors.



Also, this looks similar to the proof in section 3 of the excelent article "A short introduction to Ordinal Notations", by Simmons. It's freely available online; you should check it out. (Are you doing a thesis on ordinal notations?)



I like Simmon's way of writing the proof because it feels more like an algorithm :-). However, it's basically the same as yours.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 16:19









Guillermo Mosse

867316




867316












  • Hi! Assume the contrary that $beta_1 neq gamma_1$. WLOG, we assume $beta_1 < gamma_1$, then $omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n<omega^{beta_1}cdot k_1+omega^{beta_1}cdot k_2+cdots+omega^{beta_1}cdot k_n=omega^{beta_1}cdot(k_1+k_2+cdots+k_n)$ $<omega^{gamma_1}<omega^{gamma_1}cdot l_1<omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$ and thus $alpha<alpha$. This is clearly a contradiction.
    – Le Anh Dung
    Nov 23 at 12:41












  • What I'm saying is that maybe you should clarify that. Also, are you doing a thesis on ordinals? I just finished mine. Feel free to chat.
    – Guillermo Mosse
    Nov 23 at 13:36










  • Thank you for your suggestion! I'm just learning basic stuff about set theory ;)
    – Le Anh Dung
    Nov 23 at 14:08






  • 1




    A nice book is Introduction to Set Theory, by Hrbace.k
    – Guillermo Mosse
    Nov 23 at 14:34


















  • Hi! Assume the contrary that $beta_1 neq gamma_1$. WLOG, we assume $beta_1 < gamma_1$, then $omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n<omega^{beta_1}cdot k_1+omega^{beta_1}cdot k_2+cdots+omega^{beta_1}cdot k_n=omega^{beta_1}cdot(k_1+k_2+cdots+k_n)$ $<omega^{gamma_1}<omega^{gamma_1}cdot l_1<omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$ and thus $alpha<alpha$. This is clearly a contradiction.
    – Le Anh Dung
    Nov 23 at 12:41












  • What I'm saying is that maybe you should clarify that. Also, are you doing a thesis on ordinals? I just finished mine. Feel free to chat.
    – Guillermo Mosse
    Nov 23 at 13:36










  • Thank you for your suggestion! I'm just learning basic stuff about set theory ;)
    – Le Anh Dung
    Nov 23 at 14:08






  • 1




    A nice book is Introduction to Set Theory, by Hrbace.k
    – Guillermo Mosse
    Nov 23 at 14:34
















Hi! Assume the contrary that $beta_1 neq gamma_1$. WLOG, we assume $beta_1 < gamma_1$, then $omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n<omega^{beta_1}cdot k_1+omega^{beta_1}cdot k_2+cdots+omega^{beta_1}cdot k_n=omega^{beta_1}cdot(k_1+k_2+cdots+k_n)$ $<omega^{gamma_1}<omega^{gamma_1}cdot l_1<omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$ and thus $alpha<alpha$. This is clearly a contradiction.
– Le Anh Dung
Nov 23 at 12:41






Hi! Assume the contrary that $beta_1 neq gamma_1$. WLOG, we assume $beta_1 < gamma_1$, then $omega^{beta_1}cdot k_1+omega^{beta_2}cdot k_2+cdots+omega^{beta_n}cdot k_n<omega^{beta_1}cdot k_1+omega^{beta_1}cdot k_2+cdots+omega^{beta_1}cdot k_n=omega^{beta_1}cdot(k_1+k_2+cdots+k_n)$ $<omega^{gamma_1}<omega^{gamma_1}cdot l_1<omega^{gamma_1}cdot l_1+omega^{gamma_2}cdot l_2+cdots+omega^{gamma_m}cdot l_m$ and thus $alpha<alpha$. This is clearly a contradiction.
– Le Anh Dung
Nov 23 at 12:41














What I'm saying is that maybe you should clarify that. Also, are you doing a thesis on ordinals? I just finished mine. Feel free to chat.
– Guillermo Mosse
Nov 23 at 13:36




What I'm saying is that maybe you should clarify that. Also, are you doing a thesis on ordinals? I just finished mine. Feel free to chat.
– Guillermo Mosse
Nov 23 at 13:36












Thank you for your suggestion! I'm just learning basic stuff about set theory ;)
– Le Anh Dung
Nov 23 at 14:08




Thank you for your suggestion! I'm just learning basic stuff about set theory ;)
– Le Anh Dung
Nov 23 at 14:08




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1




A nice book is Introduction to Set Theory, by Hrbace.k
– Guillermo Mosse
Nov 23 at 14:34




A nice book is Introduction to Set Theory, by Hrbace.k
– Guillermo Mosse
Nov 23 at 14:34


















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