Let $n,kinomega$. Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$
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Let $n,kinomega$. Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: $kinomegaimplies k+omega=omega$.
Proof: By definition, $k+omega=sup_{ninomega}(k+n)$ and $omega=sup_{ninomega}(n)$. It is clear that ${k+n mid ninomega} subseteq {n mid ninomega}$ and that $forall ninomega, exists n'inomega:nle k+n'$. The result is then followed.
We proceed to prove our main theorem by induction on $n$.
The statement is trivially true for $n=1$.
Assume that $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$.
Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{n+1text{ times}}$
$=underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}+(omega+k)=(omegacdot n+k)+(omega+k)$
$=omegacdot n+(k+(omega+k))=omegacdot n+((k+omega)+k)overset{mathrm{Lemma}}{=}omegacdot n+(omega+k)=(omegacdot n+omega)+k=omegacdot (n+1)+k$.
This completes the proof.
elementary-set-theory ordinals
asked Dec 1 at 14:03
Le Anh Dung
9701421
add a comment |
up vote
1
down vote
favorite
Let $n,kinomega$. Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: $kinomegaimplies k+omega=omega$.
Proof: By definition, $k+omega=sup_{ninomega}(k+n)$ and $omega=sup_{ninomega}(n)$. It is clear that ${k+n mid ninomega} subseteq {n mid ninomega}$ and that $forall ninomega, exists n'inomega:nle k+n'$. The result is then followed.
We proceed to prove our main theorem by induction on $n$.
The statement is trivially true for $n=1$.
Assume that $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$.
Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{n+1text{ times}}$
$=underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}+(omega+k)=(omegacdot n+k)+(omega+k)$
$=omegacdot n+(k+(omega+k))=omegacdot n+((k+omega)+k)overset{mathrm{Lemma}}{=}omegacdot n+(omega+k)=(omegacdot n+omega)+k=omegacdot (n+1)+k$.
This completes the proof.
elementary-set-theory ordinals
asked Dec 1 at 14:03
Le Anh Dung
9701421
1
Have you already verified associativity?
– Andrés E. Caicedo
Dec 1 at 16:30
Hi @AndrésE.Caicedo, I have proved the associativity of ordinal addition.
– Le Anh Dung
Dec 2 at 0:56
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $n,kinomega$. Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: $kinomegaimplies k+omega=omega$.
Proof: By definition, $k+omega=sup_{ninomega}(k+n)$ and $omega=sup_{ninomega}(n)$. It is clear that ${k+n mid ninomega} subseteq {n mid ninomega}$ and that $forall ninomega, exists n'inomega:nle k+n'$. The result is then followed.
We proceed to prove our main theorem by induction on $n$.
The statement is trivially true for $n=1$.
Assume that $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$.
Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{n+1text{ times}}$
$=underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}+(omega+k)=(omegacdot n+k)+(omega+k)$
$=omegacdot n+(k+(omega+k))=omegacdot n+((k+omega)+k)overset{mathrm{Lemma}}{=}omegacdot n+(omega+k)=(omegacdot n+omega)+k=omegacdot (n+1)+k$.
This completes the proof.
elementary-set-theory ordinals
asked Dec 1 at 14:03
Le Anh Dung
9701421
Let $n,kinomega$. Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: $kinomegaimplies k+omega=omega$.
Proof: By definition, $k+omega=sup_{ninomega}(k+n)$ and $omega=sup_{ninomega}(n)$. It is clear that ${k+n mid ninomega} subseteq {n mid ninomega}$ and that $forall ninomega, exists n'inomega:nle k+n'$. The result is then followed.
We proceed to prove our main theorem by induction on $n$.
The statement is trivially true for $n=1$.
Assume that $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}=omegacdot n+k$.
Then $underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{n+1text{ times}}$
$=underbrace{(omega+k)+(omega+k)+ldots+(omega+k)}_{ntext{ times}}+(omega+k)=(omegacdot n+k)+(omega+k)$
$=omegacdot n+(k+(omega+k))=omegacdot n+((k+omega)+k)overset{mathrm{Lemma}}{=}omegacdot n+(omega+k)=(omegacdot n+omega)+k=omegacdot (n+1)+k$.
This completes the proof.
elementary-set-theory ordinals
elementary-set-theory ordinals
asked Dec 1 at 14:03
Le Anh Dung
9701421
asked Dec 1 at 14:03
Le Anh Dung
9701421
asked Dec 1 at 14:03
Le Anh Dung
9701421
asked Dec 1 at 14:03
Le Anh Dung
9701421
asked Dec 1 at 14:03
Le Anh Dung
9701421
9701421
1
Have you already verified associativity?
– Andrés E. Caicedo
Dec 1 at 16:30
Hi @AndrésE.Caicedo, I have proved the associativity of ordinal addition.
– Le Anh Dung
Dec 2 at 0:56
add a comment |
1
Have you already verified associativity?
– Andrés E. Caicedo
Dec 1 at 16:30
Hi @AndrésE.Caicedo, I have proved the associativity of ordinal addition.
– Le Anh Dung
Dec 2 at 0:56
1
1
Have you already verified associativity?
– Andrés E. Caicedo
Dec 1 at 16:30
Have you already verified associativity?
– Andrés E. Caicedo
Dec 1 at 16:30
Hi @AndrésE.Caicedo, I have proved the associativity of ordinal addition.
– Le Anh Dung
Dec 2 at 0:56
Hi @AndrésE.Caicedo, I have proved the associativity of ordinal addition.
– Le Anh Dung
Dec 2 at 0:56
add a comment |
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1
Have you already verified associativity?
– Andrés E. Caicedo
Dec 1 at 16:30
Hi @AndrésE.Caicedo, I have proved the associativity of ordinal addition.
– Le Anh Dung
Dec 2 at 0:56