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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
asked 23 hours ago
Le Anh Dung
8921421
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
asked 23 hours ago
Le Anh Dung
8921421
I think that you must to argue why $beta$ is a maximum.
– Gödel
20 hours ago
Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
– Le Anh Dung
13 hours ago
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
asked 23 hours ago
Le Anh Dung
8921421
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 23 hours ago
Le Anh Dung
8921421
asked 23 hours ago
Le Anh Dung
8921421
asked 23 hours ago
Le Anh Dung
8921421
asked 23 hours ago
Le Anh Dung
8921421
asked 23 hours ago
Le Anh Dung
8921421
8921421
I think that you must to argue why $beta$ is a maximum.
– Gödel
20 hours ago
Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
– Le Anh Dung
13 hours ago
add a comment |
I think that you must to argue why $beta$ is a maximum.
– Gödel
20 hours ago
Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
– Le Anh Dung
13 hours ago
I think that you must to argue why $beta$ is a maximum.
– Gödel
20 hours ago
I think that you must to argue why $beta$ is a maximum.
– Gödel
20 hours ago
Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
– Le Anh Dung
13 hours ago
Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
– Le Anh Dung
13 hours ago
add a comment |
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I think that you must to argue why $beta$ is a maximum.
– Gödel
20 hours ago
Hi @Gödel! I proved that such $beta$ exists before and take that for granted.
– Le Anh Dung
13 hours ago