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.
elementary-set-theory ordinals
add a comment |
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.
elementary-set-theory ordinals
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
add a comment |
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.
elementary-set-theory ordinals
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
elementary-set-theory ordinals
asked Nov 19 at 12:22
Le Anh Dung
1,0011421
1,0011421
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
add a comment |
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
add a comment |
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.
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
add a comment |
1 Answer
1
active
oldest
votes
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.
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
add a comment |
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.
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
add a comment |
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.
"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.
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
add a comment |
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
1
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
add a comment |
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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