Mixing
For what ordinals does $1+alpha=alpha$ hold? For what ordinals does $2alpha=alpha$ hold?
up vote
0
down vote
favorite
For what ordinals does $1+alpha=alpha$ hold?
It obviously does not hold for naturals, (though I don't know how to prove it).
My proposal is that it holds for any ordinal $geomega$.
Assume the contrary: $exists gamma ge omega $ such that $1+gammanegamma$. Let's divide gamma by omega: $gamma=omegatau+rho$. So,
$$1+omegatau+rhoneomegatau+rho$$
$$omegatau+rhoneomegatau+rho,$$
which is a contradiction.
For what ordinals does $2alpha=alpha $ hold?
My proposal is that it (1) holds for limit ordinals and (2) does not hold for non-limit ordinals.
Assume the contrary (1): $exists$ a limit ordinal $gamma$ such that $2gammanegamma$.
Let's divide gamma by omega: $gamma=omegatau+0$. So,
$$2(omegatau+0)neomegatau+0$$
$$(2omega)tauneomegatau$$
$$omegatauneomegatau,$$
which is a contradiction.
Assume the contrary (2): $exists$ a non-limit ordinal $gamma+k, k>0$ such that $2(gamma+k)=gamma+k$. So,
$$2gamma+2k=gamma+k$$
$$gamma+2k=gamma+k$$
$$2k=k,$$
which is a contradiction.
I'd like to know whether my proofs are correct or not, and if they are, whether they're rigourous enough. It would also be great to see alternative ways of proving these.
Thank you in advance.
proof-verification ordinals
asked May 14 at 9:57
fragileradius
237112
add a comment |
up vote
0
down vote
favorite
For what ordinals does $1+alpha=alpha$ hold?
It obviously does not hold for naturals, (though I don't know how to prove it).
My proposal is that it holds for any ordinal $geomega$.
Assume the contrary: $exists gamma ge omega $ such that $1+gammanegamma$. Let's divide gamma by omega: $gamma=omegatau+rho$. So,
$$1+omegatau+rhoneomegatau+rho$$
$$omegatau+rhoneomegatau+rho,$$
which is a contradiction.
For what ordinals does $2alpha=alpha $ hold?
My proposal is that it (1) holds for limit ordinals and (2) does not hold for non-limit ordinals.
Assume the contrary (1): $exists$ a limit ordinal $gamma$ such that $2gammanegamma$.
Let's divide gamma by omega: $gamma=omegatau+0$. So,
$$2(omegatau+0)neomegatau+0$$
$$(2omega)tauneomegatau$$
$$omegatauneomegatau,$$
which is a contradiction.
Assume the contrary (2): $exists$ a non-limit ordinal $gamma+k, k>0$ such that $2(gamma+k)=gamma+k$. So,
$$2gamma+2k=gamma+k$$
$$gamma+2k=gamma+k$$
$$2k=k,$$
which is a contradiction.
I'd like to know whether my proofs are correct or not, and if they are, whether they're rigourous enough. It would also be great to see alternative ways of proving these.
Thank you in advance.
proof-verification ordinals
asked May 14 at 9:57
fragileradius
237112
what did you do in the second line of the second question?
– Guillermo Mosse
May 30 at 18:36
@GuillermoMosse I am sorry, could you please specify the line? I can't notice anything that may need clarification. (Sorry!)
– fragileradius
May 31 at 20:10
You have an error when you go from $2gamma + 2k = gamma + k$ to $gamma + 2k = gamma + k$
– Guillermo Mosse
Jun 1 at 18:00
@GuillermoMosse Why? (Implied that $gamma$ is a limit ordinal and $k>0$ is a natural.)
– fragileradius
Jun 2 at 3:51
oh, then it's ok! sorry
– Guillermo Mosse
Jun 4 at 17:49
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For what ordinals does $1+alpha=alpha$ hold?
It obviously does not hold for naturals, (though I don't know how to prove it).
My proposal is that it holds for any ordinal $geomega$.
Assume the contrary: $exists gamma ge omega $ such that $1+gammanegamma$. Let's divide gamma by omega: $gamma=omegatau+rho$. So,
$$1+omegatau+rhoneomegatau+rho$$
$$omegatau+rhoneomegatau+rho,$$
which is a contradiction.
For what ordinals does $2alpha=alpha $ hold?
My proposal is that it (1) holds for limit ordinals and (2) does not hold for non-limit ordinals.
Assume the contrary (1): $exists$ a limit ordinal $gamma$ such that $2gammanegamma$.
Let's divide gamma by omega: $gamma=omegatau+0$. So,
$$2(omegatau+0)neomegatau+0$$
$$(2omega)tauneomegatau$$
$$omegatauneomegatau,$$
which is a contradiction.
Assume the contrary (2): $exists$ a non-limit ordinal $gamma+k, k>0$ such that $2(gamma+k)=gamma+k$. So,
$$2gamma+2k=gamma+k$$
$$gamma+2k=gamma+k$$
$$2k=k,$$
which is a contradiction.
I'd like to know whether my proofs are correct or not, and if they are, whether they're rigourous enough. It would also be great to see alternative ways of proving these.
Thank you in advance.
proof-verification ordinals
asked May 14 at 9:57
fragileradius
237112
For what ordinals does $1+alpha=alpha$ hold?
It obviously does not hold for naturals, (though I don't know how to prove it).
My proposal is that it holds for any ordinal $geomega$.
Assume the contrary: $exists gamma ge omega $ such that $1+gammanegamma$. Let's divide gamma by omega: $gamma=omegatau+rho$. So,
$$1+omegatau+rhoneomegatau+rho$$
$$omegatau+rhoneomegatau+rho,$$
which is a contradiction.
For what ordinals does $2alpha=alpha $ hold?
My proposal is that it (1) holds for limit ordinals and (2) does not hold for non-limit ordinals.
Assume the contrary (1): $exists$ a limit ordinal $gamma$ such that $2gammanegamma$.
Let's divide gamma by omega: $gamma=omegatau+0$. So,
$$2(omegatau+0)neomegatau+0$$
$$(2omega)tauneomegatau$$
$$omegatauneomegatau,$$
which is a contradiction.
Assume the contrary (2): $exists$ a non-limit ordinal $gamma+k, k>0$ such that $2(gamma+k)=gamma+k$. So,
$$2gamma+2k=gamma+k$$
$$gamma+2k=gamma+k$$
$$2k=k,$$
which is a contradiction.
I'd like to know whether my proofs are correct or not, and if they are, whether they're rigourous enough. It would also be great to see alternative ways of proving these.
Thank you in advance.
proof-verification ordinals
proof-verification ordinals
asked May 14 at 9:57
fragileradius
237112
asked May 14 at 9:57
fragileradius
237112
asked May 14 at 9:57
fragileradius
237112
asked May 14 at 9:57
fragileradius
237112
asked May 14 at 9:57
fragileradius
237112
237112
what did you do in the second line of the second question?
– Guillermo Mosse
May 30 at 18:36
@GuillermoMosse I am sorry, could you please specify the line? I can't notice anything that may need clarification. (Sorry!)
– fragileradius
May 31 at 20:10
You have an error when you go from $2gamma + 2k = gamma + k$ to $gamma + 2k = gamma + k$
– Guillermo Mosse
Jun 1 at 18:00
@GuillermoMosse Why? (Implied that $gamma$ is a limit ordinal and $k>0$ is a natural.)
– fragileradius
Jun 2 at 3:51
oh, then it's ok! sorry
– Guillermo Mosse
Jun 4 at 17:49
add a comment |
what did you do in the second line of the second question?
– Guillermo Mosse
May 30 at 18:36
@GuillermoMosse I am sorry, could you please specify the line? I can't notice anything that may need clarification. (Sorry!)
– fragileradius
May 31 at 20:10
You have an error when you go from $2gamma + 2k = gamma + k$ to $gamma + 2k = gamma + k$
– Guillermo Mosse
Jun 1 at 18:00
@GuillermoMosse Why? (Implied that $gamma$ is a limit ordinal and $k>0$ is a natural.)
– fragileradius
Jun 2 at 3:51
oh, then it's ok! sorry
– Guillermo Mosse
Jun 4 at 17:49
what did you do in the second line of the second question?
– Guillermo Mosse
May 30 at 18:36
what did you do in the second line of the second question?
– Guillermo Mosse
May 30 at 18:36
@GuillermoMosse I am sorry, could you please specify the line? I can't notice anything that may need clarification. (Sorry!)
– fragileradius
May 31 at 20:10
@GuillermoMosse I am sorry, could you please specify the line? I can't notice anything that may need clarification. (Sorry!)
– fragileradius
May 31 at 20:10
You have an error when you go from $2gamma + 2k = gamma + k$ to $gamma + 2k = gamma + k$
– Guillermo Mosse
Jun 1 at 18:00
You have an error when you go from $2gamma + 2k = gamma + k$ to $gamma + 2k = gamma + k$
– Guillermo Mosse
Jun 1 at 18:00
@GuillermoMosse Why? (Implied that $gamma$ is a limit ordinal and $k>0$ is a natural.)
– fragileradius
Jun 2 at 3:51
@GuillermoMosse Why? (Implied that $gamma$ is a limit ordinal and $k>0$ is a natural.)
– fragileradius
Jun 2 at 3:51
oh, then it's ok! sorry
– Guillermo Mosse
Jun 4 at 17:49
oh, then it's ok! sorry
– Guillermo Mosse
Jun 4 at 17:49
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
It seems that those questions (124, and 125) are from A. Shen's "Basic Set Theory", which I've encountered recently.
You proofs help me about clarity in my approach, but I think the proof should be used the materials which are provided ealier in the book. It seems that you are using the result of Problem 126 (limit ordinals can be represented as $omegacdotalpha$), it's not wrong but if (like myself) studying from Shen's book, the problem 126 must be proved first.
So I try to modified your proof using "known" material provide in the book, which does NOT mean my proof is correct, but nevertheless I post as an answer instead of new question:
- Problem 1:
If $alpha<omega$, then the sets of order type $alpha$ are finite sets, and have finite cardinality. We have $left|Aright|<1+left|Aright|$, implies there is no one-to-one correspondence between the sets of order type $alpha$ and $1+alpha$.
Consider an ordinal $alpha$ that $geqomega$.
We have $1+omega=omega$.
Suppose that $forallbetaleft(omegaleqbeta<alpharight)rightarrowleft(1+beta=betaright)$.
If $alpha$ is a nonlimit ordinal, we have $alpha=beta+1$. Let $A,A^{prime}$ are sets have order types $alpha$ and $1+alpha$, and let $a$ is the largest element of A (and also, of $A^{prime}$). We have $Asetminusleft{ aright} ,A^{prime}setminusleft{ aright}$ have order type $beta$ and $1+beta$. By induction hypothesis, $Asetminusleft{ aright}$ is isomorphic to $A^{prime}setminusleft{ aright}$ by a map $f$. Let $fleft(aright)=a$, we have isomorphism between $A$ and $A^{prime}$.
Consider $A$ has order type $alpha$, we have $left{ bright} cup A$ has order type $1+alpha$.
Consider an initial segment $I$ of $left{ bright} cup A$, that $Ineqleft{ bright}$ , let $a$ (as $Ineqleft{ bright}$ , implies $aneq b$ (or $ain A$)) is the least element of $left(left{ bright} cup Aright)setminus I$, we have $forall xin I, x<a$. While $b$ is the least element of $left{ bright} cup A$, we have $forall xin Isetminusleft{ bright} ,x<a$, implies that $Isetminusleft{ bright}$ is an initial segment of $A$, and $Isetminusleft{ bright} =left[0,aright)$. (that part, I feel hard to explain clearly).
That is if an initial segment of $1+alpha$ has the form $1+beta$, where $beta$ is an initial segment of $alpha$.
If $alpha$ is a limit ordinal. Suppose that $1+alphaneqalpharightarrow1+alpha>alpha$, implies that $alpha$ is isomorphic to an initial segment $1+beta$ of $1+alpha$, where $beta$ is an initial segment of $alpha$. By induction hypothesis, $beta=1+beta$, which means $alpha$ is isomorphic to its initial segment $beta$. By Theorem 21 (of the book), it's either impossible or $beta=alpha$, but then contradicts to our hypothesis that $1+alphaneqalpha$.
So, we have for any $alphageqomega, 1+alpha=alpha$.
- Problem 2:
Suppose that $alpha$ is a nonlimit ordinal, then there's an ordinal $beta$, $beta+1=alpha$,
$$2cdotleft(beta+1right) =beta+1$$
$$2cdotbeta+2 =beta+1$$
Consider $A$ and $B$ have order types $alpha$ and $2cdotalpha$, so $A$ is (ordered) isomorphic to $B$. This means the greatest element of $A$ corresponds to the greatest element of $B$, implies that $2cdotbeta+1=beta$. But we have $betaleq2beta<2beta+1$, so there doesn't exist nonlimit ordinal $alpha$ that $2alpha=alpha$.
Suppose that $alpha$ is a limit ordinal, and $2alphaneqalpharightarrow2alpha>alpha$. As we know (p.89) that if $alpha<2alpha$, then there's unique representation of $alpha$, $alpha=2alpha_{1}+beta$, where $alpha_{1}<alpha, beta<2$.
If $beta=1$, it contradicts to hypothesis that $alpha$ is a limit ordinal.
If $beta=0, alpha=2alpha_{1}$, where $alpha_{1}$ is also limit ordinal. If $2alpha_{1}=alpha_{1}rightarrowalpha=alpha_{1}=2alpha_{1}=2alpha$.
If $2alpha_{1}neqalpha_{1}$, and $alpha_{1}=2alpha_{2}$..., consider the set of all $alpha_{i}$, that $2alpha_{i}neqalpha_{i},alpha_{i}=2alpha_{i+1}$. The set must contain the least element $alpha_{j}$, that is $2alpha_{j}neqalpha_{j}$, apply the same argument above, it must $alpha_{j}=2alpha^{prime}rightarrow2alpha_{j}>2alpha^{prime}rightarrowalpha_{j}>alpha^{prime}$. Because $alpha_{j}$ is the least element of the set, we have $alpha^{prime}$ doesn't belong to the set, or $2alpha^{prime}=alpha^{prime}rightarrowalpha^{prime}=alpha_{j}$, which contradicts.
(I feel this part is over-complicated).
answered 16 hours ago
duqu
778
Busy with college and having no spare time to read and understand your answer, I still have to express my gratitude for praising my question with attention in a form of such a long post. Yes, that book was The One. (Да, у меня тоже книга Шеня).
– fragileradius
12 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It seems that those questions (124, and 125) are from A. Shen's "Basic Set Theory", which I've encountered recently.
You proofs help me about clarity in my approach, but I think the proof should be used the materials which are provided ealier in the book. It seems that you are using the result of Problem 126 (limit ordinals can be represented as $omegacdotalpha$), it's not wrong but if (like myself) studying from Shen's book, the problem 126 must be proved first.
So I try to modified your proof using "known" material provide in the book, which does NOT mean my proof is correct, but nevertheless I post as an answer instead of new question:
- Problem 1:
If $alpha<omega$, then the sets of order type $alpha$ are finite sets, and have finite cardinality. We have $left|Aright|<1+left|Aright|$, implies there is no one-to-one correspondence between the sets of order type $alpha$ and $1+alpha$.
Consider an ordinal $alpha$ that $geqomega$.
We have $1+omega=omega$.
Suppose that $forallbetaleft(omegaleqbeta<alpharight)rightarrowleft(1+beta=betaright)$.
If $alpha$ is a nonlimit ordinal, we have $alpha=beta+1$. Let $A,A^{prime}$ are sets have order types $alpha$ and $1+alpha$, and let $a$ is the largest element of A (and also, of $A^{prime}$). We have $Asetminusleft{ aright} ,A^{prime}setminusleft{ aright}$ have order type $beta$ and $1+beta$. By induction hypothesis, $Asetminusleft{ aright}$ is isomorphic to $A^{prime}setminusleft{ aright}$ by a map $f$. Let $fleft(aright)=a$, we have isomorphism between $A$ and $A^{prime}$.
Consider $A$ has order type $alpha$, we have $left{ bright} cup A$ has order type $1+alpha$.
Consider an initial segment $I$ of $left{ bright} cup A$, that $Ineqleft{ bright}$ , let $a$ (as $Ineqleft{ bright}$ , implies $aneq b$ (or $ain A$)) is the least element of $left(left{ bright} cup Aright)setminus I$, we have $forall xin I, x<a$. While $b$ is the least element of $left{ bright} cup A$, we have $forall xin Isetminusleft{ bright} ,x<a$, implies that $Isetminusleft{ bright}$ is an initial segment of $A$, and $Isetminusleft{ bright} =left[0,aright)$. (that part, I feel hard to explain clearly).
That is if an initial segment of $1+alpha$ has the form $1+beta$, where $beta$ is an initial segment of $alpha$.
If $alpha$ is a limit ordinal. Suppose that $1+alphaneqalpharightarrow1+alpha>alpha$, implies that $alpha$ is isomorphic to an initial segment $1+beta$ of $1+alpha$, where $beta$ is an initial segment of $alpha$. By induction hypothesis, $beta=1+beta$, which means $alpha$ is isomorphic to its initial segment $beta$. By Theorem 21 (of the book), it's either impossible or $beta=alpha$, but then contradicts to our hypothesis that $1+alphaneqalpha$.
So, we have for any $alphageqomega, 1+alpha=alpha$.
- Problem 2:
Suppose that $alpha$ is a nonlimit ordinal, then there's an ordinal $beta$, $beta+1=alpha$,
$$2cdotleft(beta+1right) =beta+1$$
$$2cdotbeta+2 =beta+1$$
Consider $A$ and $B$ have order types $alpha$ and $2cdotalpha$, so $A$ is (ordered) isomorphic to $B$. This means the greatest element of $A$ corresponds to the greatest element of $B$, implies that $2cdotbeta+1=beta$. But we have $betaleq2beta<2beta+1$, so there doesn't exist nonlimit ordinal $alpha$ that $2alpha=alpha$.
Suppose that $alpha$ is a limit ordinal, and $2alphaneqalpharightarrow2alpha>alpha$. As we know (p.89) that if $alpha<2alpha$, then there's unique representation of $alpha$, $alpha=2alpha_{1}+beta$, where $alpha_{1}<alpha, beta<2$.
If $beta=1$, it contradicts to hypothesis that $alpha$ is a limit ordinal.
If $beta=0, alpha=2alpha_{1}$, where $alpha_{1}$ is also limit ordinal. If $2alpha_{1}=alpha_{1}rightarrowalpha=alpha_{1}=2alpha_{1}=2alpha$.
If $2alpha_{1}neqalpha_{1}$, and $alpha_{1}=2alpha_{2}$..., consider the set of all $alpha_{i}$, that $2alpha_{i}neqalpha_{i},alpha_{i}=2alpha_{i+1}$. The set must contain the least element $alpha_{j}$, that is $2alpha_{j}neqalpha_{j}$, apply the same argument above, it must $alpha_{j}=2alpha^{prime}rightarrow2alpha_{j}>2alpha^{prime}rightarrowalpha_{j}>alpha^{prime}$. Because $alpha_{j}$ is the least element of the set, we have $alpha^{prime}$ doesn't belong to the set, or $2alpha^{prime}=alpha^{prime}rightarrowalpha^{prime}=alpha_{j}$, which contradicts.
(I feel this part is over-complicated).
answered 16 hours ago
duqu
778
Busy with college and having no spare time to read and understand your answer, I still have to express my gratitude for praising my question with attention in a form of such a long post. Yes, that book was The One. (Да, у меня тоже книга Шеня).
– fragileradius
12 hours ago
add a comment |
up vote
2
down vote
accepted
It seems that those questions (124, and 125) are from A. Shen's "Basic Set Theory", which I've encountered recently.
You proofs help me about clarity in my approach, but I think the proof should be used the materials which are provided ealier in the book. It seems that you are using the result of Problem 126 (limit ordinals can be represented as $omegacdotalpha$), it's not wrong but if (like myself) studying from Shen's book, the problem 126 must be proved first.
So I try to modified your proof using "known" material provide in the book, which does NOT mean my proof is correct, but nevertheless I post as an answer instead of new question:
- Problem 1:
If $alpha<omega$, then the sets of order type $alpha$ are finite sets, and have finite cardinality. We have $left|Aright|<1+left|Aright|$, implies there is no one-to-one correspondence between the sets of order type $alpha$ and $1+alpha$.
Consider an ordinal $alpha$ that $geqomega$.
We have $1+omega=omega$.
Suppose that $forallbetaleft(omegaleqbeta<alpharight)rightarrowleft(1+beta=betaright)$.
If $alpha$ is a nonlimit ordinal, we have $alpha=beta+1$. Let $A,A^{prime}$ are sets have order types $alpha$ and $1+alpha$, and let $a$ is the largest element of A (and also, of $A^{prime}$). We have $Asetminusleft{ aright} ,A^{prime}setminusleft{ aright}$ have order type $beta$ and $1+beta$. By induction hypothesis, $Asetminusleft{ aright}$ is isomorphic to $A^{prime}setminusleft{ aright}$ by a map $f$. Let $fleft(aright)=a$, we have isomorphism between $A$ and $A^{prime}$.
Consider $A$ has order type $alpha$, we have $left{ bright} cup A$ has order type $1+alpha$.
Consider an initial segment $I$ of $left{ bright} cup A$, that $Ineqleft{ bright}$ , let $a$ (as $Ineqleft{ bright}$ , implies $aneq b$ (or $ain A$)) is the least element of $left(left{ bright} cup Aright)setminus I$, we have $forall xin I, x<a$. While $b$ is the least element of $left{ bright} cup A$, we have $forall xin Isetminusleft{ bright} ,x<a$, implies that $Isetminusleft{ bright}$ is an initial segment of $A$, and $Isetminusleft{ bright} =left[0,aright)$. (that part, I feel hard to explain clearly).
That is if an initial segment of $1+alpha$ has the form $1+beta$, where $beta$ is an initial segment of $alpha$.
If $alpha$ is a limit ordinal. Suppose that $1+alphaneqalpharightarrow1+alpha>alpha$, implies that $alpha$ is isomorphic to an initial segment $1+beta$ of $1+alpha$, where $beta$ is an initial segment of $alpha$. By induction hypothesis, $beta=1+beta$, which means $alpha$ is isomorphic to its initial segment $beta$. By Theorem 21 (of the book), it's either impossible or $beta=alpha$, but then contradicts to our hypothesis that $1+alphaneqalpha$.
So, we have for any $alphageqomega, 1+alpha=alpha$.
- Problem 2:
Suppose that $alpha$ is a nonlimit ordinal, then there's an ordinal $beta$, $beta+1=alpha$,
$$2cdotleft(beta+1right) =beta+1$$
$$2cdotbeta+2 =beta+1$$
Consider $A$ and $B$ have order types $alpha$ and $2cdotalpha$, so $A$ is (ordered) isomorphic to $B$. This means the greatest element of $A$ corresponds to the greatest element of $B$, implies that $2cdotbeta+1=beta$. But we have $betaleq2beta<2beta+1$, so there doesn't exist nonlimit ordinal $alpha$ that $2alpha=alpha$.
Suppose that $alpha$ is a limit ordinal, and $2alphaneqalpharightarrow2alpha>alpha$. As we know (p.89) that if $alpha<2alpha$, then there's unique representation of $alpha$, $alpha=2alpha_{1}+beta$, where $alpha_{1}<alpha, beta<2$.
If $beta=1$, it contradicts to hypothesis that $alpha$ is a limit ordinal.
If $beta=0, alpha=2alpha_{1}$, where $alpha_{1}$ is also limit ordinal. If $2alpha_{1}=alpha_{1}rightarrowalpha=alpha_{1}=2alpha_{1}=2alpha$.
If $2alpha_{1}neqalpha_{1}$, and $alpha_{1}=2alpha_{2}$..., consider the set of all $alpha_{i}$, that $2alpha_{i}neqalpha_{i},alpha_{i}=2alpha_{i+1}$. The set must contain the least element $alpha_{j}$, that is $2alpha_{j}neqalpha_{j}$, apply the same argument above, it must $alpha_{j}=2alpha^{prime}rightarrow2alpha_{j}>2alpha^{prime}rightarrowalpha_{j}>alpha^{prime}$. Because $alpha_{j}$ is the least element of the set, we have $alpha^{prime}$ doesn't belong to the set, or $2alpha^{prime}=alpha^{prime}rightarrowalpha^{prime}=alpha_{j}$, which contradicts.
(I feel this part is over-complicated).
answered 16 hours ago
duqu
778
Busy with college and having no spare time to read and understand your answer, I still have to express my gratitude for praising my question with attention in a form of such a long post. Yes, that book was The One. (Да, у меня тоже книга Шеня).
– fragileradius
12 hours ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It seems that those questions (124, and 125) are from A. Shen's "Basic Set Theory", which I've encountered recently.
You proofs help me about clarity in my approach, but I think the proof should be used the materials which are provided ealier in the book. It seems that you are using the result of Problem 126 (limit ordinals can be represented as $omegacdotalpha$), it's not wrong but if (like myself) studying from Shen's book, the problem 126 must be proved first.
So I try to modified your proof using "known" material provide in the book, which does NOT mean my proof is correct, but nevertheless I post as an answer instead of new question:
- Problem 1:
If $alpha<omega$, then the sets of order type $alpha$ are finite sets, and have finite cardinality. We have $left|Aright|<1+left|Aright|$, implies there is no one-to-one correspondence between the sets of order type $alpha$ and $1+alpha$.
Consider an ordinal $alpha$ that $geqomega$.
We have $1+omega=omega$.
Suppose that $forallbetaleft(omegaleqbeta<alpharight)rightarrowleft(1+beta=betaright)$.
If $alpha$ is a nonlimit ordinal, we have $alpha=beta+1$. Let $A,A^{prime}$ are sets have order types $alpha$ and $1+alpha$, and let $a$ is the largest element of A (and also, of $A^{prime}$). We have $Asetminusleft{ aright} ,A^{prime}setminusleft{ aright}$ have order type $beta$ and $1+beta$. By induction hypothesis, $Asetminusleft{ aright}$ is isomorphic to $A^{prime}setminusleft{ aright}$ by a map $f$. Let $fleft(aright)=a$, we have isomorphism between $A$ and $A^{prime}$.
Consider $A$ has order type $alpha$, we have $left{ bright} cup A$ has order type $1+alpha$.
Consider an initial segment $I$ of $left{ bright} cup A$, that $Ineqleft{ bright}$ , let $a$ (as $Ineqleft{ bright}$ , implies $aneq b$ (or $ain A$)) is the least element of $left(left{ bright} cup Aright)setminus I$, we have $forall xin I, x<a$. While $b$ is the least element of $left{ bright} cup A$, we have $forall xin Isetminusleft{ bright} ,x<a$, implies that $Isetminusleft{ bright}$ is an initial segment of $A$, and $Isetminusleft{ bright} =left[0,aright)$. (that part, I feel hard to explain clearly).
That is if an initial segment of $1+alpha$ has the form $1+beta$, where $beta$ is an initial segment of $alpha$.
If $alpha$ is a limit ordinal. Suppose that $1+alphaneqalpharightarrow1+alpha>alpha$, implies that $alpha$ is isomorphic to an initial segment $1+beta$ of $1+alpha$, where $beta$ is an initial segment of $alpha$. By induction hypothesis, $beta=1+beta$, which means $alpha$ is isomorphic to its initial segment $beta$. By Theorem 21 (of the book), it's either impossible or $beta=alpha$, but then contradicts to our hypothesis that $1+alphaneqalpha$.
So, we have for any $alphageqomega, 1+alpha=alpha$.
- Problem 2:
Suppose that $alpha$ is a nonlimit ordinal, then there's an ordinal $beta$, $beta+1=alpha$,
$$2cdotleft(beta+1right) =beta+1$$
$$2cdotbeta+2 =beta+1$$
Consider $A$ and $B$ have order types $alpha$ and $2cdotalpha$, so $A$ is (ordered) isomorphic to $B$. This means the greatest element of $A$ corresponds to the greatest element of $B$, implies that $2cdotbeta+1=beta$. But we have $betaleq2beta<2beta+1$, so there doesn't exist nonlimit ordinal $alpha$ that $2alpha=alpha$.
Suppose that $alpha$ is a limit ordinal, and $2alphaneqalpharightarrow2alpha>alpha$. As we know (p.89) that if $alpha<2alpha$, then there's unique representation of $alpha$, $alpha=2alpha_{1}+beta$, where $alpha_{1}<alpha, beta<2$.
If $beta=1$, it contradicts to hypothesis that $alpha$ is a limit ordinal.
If $beta=0, alpha=2alpha_{1}$, where $alpha_{1}$ is also limit ordinal. If $2alpha_{1}=alpha_{1}rightarrowalpha=alpha_{1}=2alpha_{1}=2alpha$.
If $2alpha_{1}neqalpha_{1}$, and $alpha_{1}=2alpha_{2}$..., consider the set of all $alpha_{i}$, that $2alpha_{i}neqalpha_{i},alpha_{i}=2alpha_{i+1}$. The set must contain the least element $alpha_{j}$, that is $2alpha_{j}neqalpha_{j}$, apply the same argument above, it must $alpha_{j}=2alpha^{prime}rightarrow2alpha_{j}>2alpha^{prime}rightarrowalpha_{j}>alpha^{prime}$. Because $alpha_{j}$ is the least element of the set, we have $alpha^{prime}$ doesn't belong to the set, or $2alpha^{prime}=alpha^{prime}rightarrowalpha^{prime}=alpha_{j}$, which contradicts.
(I feel this part is over-complicated).
answered 16 hours ago
duqu
778
It seems that those questions (124, and 125) are from A. Shen's "Basic Set Theory", which I've encountered recently.
You proofs help me about clarity in my approach, but I think the proof should be used the materials which are provided ealier in the book. It seems that you are using the result of Problem 126 (limit ordinals can be represented as $omegacdotalpha$), it's not wrong but if (like myself) studying from Shen's book, the problem 126 must be proved first.
So I try to modified your proof using "known" material provide in the book, which does NOT mean my proof is correct, but nevertheless I post as an answer instead of new question:
- Problem 1:
If $alpha<omega$, then the sets of order type $alpha$ are finite sets, and have finite cardinality. We have $left|Aright|<1+left|Aright|$, implies there is no one-to-one correspondence between the sets of order type $alpha$ and $1+alpha$.
Consider an ordinal $alpha$ that $geqomega$.
We have $1+omega=omega$.
Suppose that $forallbetaleft(omegaleqbeta<alpharight)rightarrowleft(1+beta=betaright)$.
If $alpha$ is a nonlimit ordinal, we have $alpha=beta+1$. Let $A,A^{prime}$ are sets have order types $alpha$ and $1+alpha$, and let $a$ is the largest element of A (and also, of $A^{prime}$). We have $Asetminusleft{ aright} ,A^{prime}setminusleft{ aright}$ have order type $beta$ and $1+beta$. By induction hypothesis, $Asetminusleft{ aright}$ is isomorphic to $A^{prime}setminusleft{ aright}$ by a map $f$. Let $fleft(aright)=a$, we have isomorphism between $A$ and $A^{prime}$.
Consider $A$ has order type $alpha$, we have $left{ bright} cup A$ has order type $1+alpha$.
Consider an initial segment $I$ of $left{ bright} cup A$, that $Ineqleft{ bright}$ , let $a$ (as $Ineqleft{ bright}$ , implies $aneq b$ (or $ain A$)) is the least element of $left(left{ bright} cup Aright)setminus I$, we have $forall xin I, x<a$. While $b$ is the least element of $left{ bright} cup A$, we have $forall xin Isetminusleft{ bright} ,x<a$, implies that $Isetminusleft{ bright}$ is an initial segment of $A$, and $Isetminusleft{ bright} =left[0,aright)$. (that part, I feel hard to explain clearly).
That is if an initial segment of $1+alpha$ has the form $1+beta$, where $beta$ is an initial segment of $alpha$.
If $alpha$ is a limit ordinal. Suppose that $1+alphaneqalpharightarrow1+alpha>alpha$, implies that $alpha$ is isomorphic to an initial segment $1+beta$ of $1+alpha$, where $beta$ is an initial segment of $alpha$. By induction hypothesis, $beta=1+beta$, which means $alpha$ is isomorphic to its initial segment $beta$. By Theorem 21 (of the book), it's either impossible or $beta=alpha$, but then contradicts to our hypothesis that $1+alphaneqalpha$.
So, we have for any $alphageqomega, 1+alpha=alpha$.
- Problem 2:
Suppose that $alpha$ is a nonlimit ordinal, then there's an ordinal $beta$, $beta+1=alpha$,
$$2cdotleft(beta+1right) =beta+1$$
$$2cdotbeta+2 =beta+1$$
Consider $A$ and $B$ have order types $alpha$ and $2cdotalpha$, so $A$ is (ordered) isomorphic to $B$. This means the greatest element of $A$ corresponds to the greatest element of $B$, implies that $2cdotbeta+1=beta$. But we have $betaleq2beta<2beta+1$, so there doesn't exist nonlimit ordinal $alpha$ that $2alpha=alpha$.
Suppose that $alpha$ is a limit ordinal, and $2alphaneqalpharightarrow2alpha>alpha$. As we know (p.89) that if $alpha<2alpha$, then there's unique representation of $alpha$, $alpha=2alpha_{1}+beta$, where $alpha_{1}<alpha, beta<2$.
If $beta=1$, it contradicts to hypothesis that $alpha$ is a limit ordinal.
If $beta=0, alpha=2alpha_{1}$, where $alpha_{1}$ is also limit ordinal. If $2alpha_{1}=alpha_{1}rightarrowalpha=alpha_{1}=2alpha_{1}=2alpha$.
If $2alpha_{1}neqalpha_{1}$, and $alpha_{1}=2alpha_{2}$..., consider the set of all $alpha_{i}$, that $2alpha_{i}neqalpha_{i},alpha_{i}=2alpha_{i+1}$. The set must contain the least element $alpha_{j}$, that is $2alpha_{j}neqalpha_{j}$, apply the same argument above, it must $alpha_{j}=2alpha^{prime}rightarrow2alpha_{j}>2alpha^{prime}rightarrowalpha_{j}>alpha^{prime}$. Because $alpha_{j}$ is the least element of the set, we have $alpha^{prime}$ doesn't belong to the set, or $2alpha^{prime}=alpha^{prime}rightarrowalpha^{prime}=alpha_{j}$, which contradicts.
(I feel this part is over-complicated).
answered 16 hours ago
duqu
778
answered 16 hours ago
duqu
778
answered 16 hours ago
duqu
778
answered 16 hours ago
duqu
778
778
Busy with college and having no spare time to read and understand your answer, I still have to express my gratitude for praising my question with attention in a form of such a long post. Yes, that book was The One. (Да, у меня тоже книга Шеня).
– fragileradius
12 hours ago
add a comment |
Busy with college and having no spare time to read and understand your answer, I still have to express my gratitude for praising my question with attention in a form of such a long post. Yes, that book was The One. (Да, у меня тоже книга Шеня).
– fragileradius
12 hours ago
Busy with college and having no spare time to read and understand your answer, I still have to express my gratitude for praising my question with attention in a form of such a long post. Yes, that book was The One. (Да, у меня тоже книга Шеня).
– fragileradius
12 hours ago
Busy with college and having no spare time to read and understand your answer, I still have to express my gratitude for praising my question with attention in a form of such a long post. Yes, that book was The One. (Да, у меня тоже книга Шеня).
– fragileradius
12 hours ago
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2780529%2ffor-what-ordinals-does-1-alpha-alpha-hold-for-what-ordinals-does-2-alpha%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
what did you do in the second line of the second question?
– Guillermo Mosse
May 30 at 18:36
@GuillermoMosse I am sorry, could you please specify the line? I can't notice anything that may need clarification. (Sorry!)
– fragileradius
May 31 at 20:10
You have an error when you go from $2gamma + 2k = gamma + k$ to $gamma + 2k = gamma + k$
– Guillermo Mosse
Jun 1 at 18:00
@GuillermoMosse Why? (Implied that $gamma$ is a limit ordinal and $k>0$ is a natural.)
– fragileradius
Jun 2 at 3:51
oh, then it's ok! sorry
– Guillermo Mosse
Jun 4 at 17:49