Mixing
Why continuum function isn't strictly increasing?
$begingroup$
Is there any example that for cardinal numbers $kappa < lambda$, we have $2^kappa = 2^lambda$?
My guess is that it only depends on whether GCH holds. Is it true?
elementary-set-theory cardinals
asked Nov 26 '12 at 10:04
Metta World PeaceMetta World Peace
2,49242244
$endgroup$
add a comment |
$begingroup$
Is there any example that for cardinal numbers $kappa < lambda$, we have $2^kappa = 2^lambda$?
My guess is that it only depends on whether GCH holds. Is it true?
elementary-set-theory cardinals
asked Nov 26 '12 at 10:04
Metta World PeaceMetta World Peace
2,49242244
$endgroup$
$begingroup$
I just noticed the exact wording of the question was not answered by my answer previously. I added more.
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:02
$begingroup$
Thank you very much. I'm so amazed. It seems to me that you ponder upon these problems like crazy :)
$endgroup$
– Metta World Peace
Nov 28 '12 at 2:10
$begingroup$
I do (and I am). I mean, seriously. 4am. I am going to sleep! (I have to give a talk tomorrow afternoon, and I won't have time to doze off around noon.)
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:11
$begingroup$
What should I say?! I don't want to sound like Cee lo in a talent show. But what you have been doing do amaze and impress me。
$endgroup$
– Metta World Peace
Nov 28 '12 at 5:59
$begingroup$
See also math.stackexchange.com/questions/74477/… or math.stackexchange.com/questions/376509/… and other questions linked there.
$endgroup$
– Martin Sleziak
Jul 4 '14 at 11:41
add a comment |
$begingroup$
Is there any example that for cardinal numbers $kappa < lambda$, we have $2^kappa = 2^lambda$?
My guess is that it only depends on whether GCH holds. Is it true?
elementary-set-theory cardinals
asked Nov 26 '12 at 10:04
Metta World PeaceMetta World Peace
2,49242244
$endgroup$
Is there any example that for cardinal numbers $kappa < lambda$, we have $2^kappa = 2^lambda$?
My guess is that it only depends on whether GCH holds. Is it true?
elementary-set-theory cardinals
elementary-set-theory cardinals
asked Nov 26 '12 at 10:04
Metta World PeaceMetta World Peace
2,49242244
asked Nov 26 '12 at 10:04
Metta World PeaceMetta World Peace
2,49242244
asked Nov 26 '12 at 10:04
Metta World PeaceMetta World Peace
2,49242244
asked Nov 26 '12 at 10:04
Metta World PeaceMetta World Peace
2,49242244
asked Nov 26 '12 at 10:04
Metta World PeaceMetta World Peace
2,49242244
2,49242244
$begingroup$
I just noticed the exact wording of the question was not answered by my answer previously. I added more.
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:02
$begingroup$
Thank you very much. I'm so amazed. It seems to me that you ponder upon these problems like crazy :)
$endgroup$
– Metta World Peace
Nov 28 '12 at 2:10
$begingroup$
I do (and I am). I mean, seriously. 4am. I am going to sleep! (I have to give a talk tomorrow afternoon, and I won't have time to doze off around noon.)
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:11
$begingroup$
What should I say?! I don't want to sound like Cee lo in a talent show. But what you have been doing do amaze and impress me。
$endgroup$
– Metta World Peace
Nov 28 '12 at 5:59
$begingroup$
See also math.stackexchange.com/questions/74477/… or math.stackexchange.com/questions/376509/… and other questions linked there.
$endgroup$
– Martin Sleziak
Jul 4 '14 at 11:41
add a comment |
$begingroup$
I just noticed the exact wording of the question was not answered by my answer previously. I added more.
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:02
$begingroup$
Thank you very much. I'm so amazed. It seems to me that you ponder upon these problems like crazy :)
$endgroup$
– Metta World Peace
Nov 28 '12 at 2:10
$begingroup$
I do (and I am). I mean, seriously. 4am. I am going to sleep! (I have to give a talk tomorrow afternoon, and I won't have time to doze off around noon.)
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:11
$begingroup$
What should I say?! I don't want to sound like Cee lo in a talent show. But what you have been doing do amaze and impress me。
$endgroup$
– Metta World Peace
Nov 28 '12 at 5:59
$begingroup$
See also math.stackexchange.com/questions/74477/… or math.stackexchange.com/questions/376509/… and other questions linked there.
$endgroup$
– Martin Sleziak
Jul 4 '14 at 11:41
$begingroup$
I just noticed the exact wording of the question was not answered by my answer previously. I added more.
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:02
$begingroup$
I just noticed the exact wording of the question was not answered by my answer previously. I added more.
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:02
$begingroup$
Thank you very much. I'm so amazed. It seems to me that you ponder upon these problems like crazy :)
$endgroup$
– Metta World Peace
Nov 28 '12 at 2:10
$begingroup$
Thank you very much. I'm so amazed. It seems to me that you ponder upon these problems like crazy :)
$endgroup$
– Metta World Peace
Nov 28 '12 at 2:10
$begingroup$
I do (and I am). I mean, seriously. 4am. I am going to sleep! (I have to give a talk tomorrow afternoon, and I won't have time to doze off around noon.)
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:11
$begingroup$
I do (and I am). I mean, seriously. 4am. I am going to sleep! (I have to give a talk tomorrow afternoon, and I won't have time to doze off around noon.)
$endgroup$
– Asaf Karagila♦
Nov 28 '12 at 2:11
$begingroup$
What should I say?! I don't want to sound like Cee lo in a talent show. But what you have been doing do amaze and impress me。
$endgroup$
– Metta World Peace
Nov 28 '12 at 5:59
$begingroup$
What should I say?! I don't want to sound like Cee lo in a talent show. But what you have been doing do amaze and impress me。
$endgroup$
– Metta World Peace
Nov 28 '12 at 5:59
$begingroup$
See also math.stackexchange.com/questions/74477/… or math.stackexchange.com/questions/376509/… and other questions linked there.
$endgroup$
– Martin Sleziak
Jul 4 '14 at 11:41
$begingroup$
See also math.stackexchange.com/questions/74477/… or math.stackexchange.com/questions/376509/… and other questions linked there.
$endgroup$
– Martin Sleziak
Jul 4 '14 at 11:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is independent of ZFC. It is consistent that there are no such cardinals, for example if GCH holds. Note that $lambdaleqkappaimplies2^lambdaleq2^kappa$, so it is enough to show that the continuum function is injective.
However it is consistent that $2^{aleph_0}=2^{aleph_1}=aleph_3$.
There is not much we can say about the continuum function in ZFC. This is a dire consequence from Easton's theorem.
Easton theorem tells us that if $F$ is a function whose domain is the regular cardinals and:
- $kappa<lambdaimplies F(kappa)leq F(lambda)$,
- $operatorname{cf}(kappa)<operatorname{cf}(F(kappa))$
Then there is a forcing extension which does not collapse cardinals and for every regular $kappa$, $2^kappa=F(kappa)$ in the extension.
Assume GCH holds and take the function $F(kappa)=kappa^{++}$. We can show that in the extension where $F$ describes the continuum function we have $2^{kappa}=kappa^{++}$ for regular cardinals, and $F(mu)=mu^+$ for singular $mu$. This means that GCH fails for all regular cardinals, but $2^lambda=2^kappaifflambda=kappa$. So the injectivity of the continuum function holds, while GCH fails.
(If one is not in the mood for a class-forcing, which can be a bit complicated, one can simply start with GCH and set $2^{aleph_n}=aleph_{n+2}$ for $n<omega$, and GCH to hold otherwise instead.)
edited Jul 4 '14 at 11:31
Martin Sleziak
44.9k10121274
answered Nov 26 '12 at 10:14
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
$begingroup$
Thank you. It sounds fairly interesting.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:27
$begingroup$
You're welcome. You should search for the term Easton on this site, it has been mentioned and explained before (I am writing from a cellphone, so I am being brief).
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 10:33
$begingroup$
Good advise. Although it is probably too advanced for me, I'll have a look.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:39
$begingroup$
Now that I am by a computer again, I added a bit.
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 11:39
1
$begingroup$
@AsafKaragila I wasn't even thinking in terms of consistency, just combinatorics. I do not agree that it is just the very basics (nothing very basic about: If $aleph_omega$ is strong limit, then $2^{aleph_omega}<aleph_{omega_4}$, for instance) but yes, it does not tell us what the value of the continuum is, or the value of $2^{aleph_{17}}$, even as a function of the previous values, or any of that. So most likely yes, you are using the phrase in a different fashion that I am. (I would still shake my head silently if I were in the audience of a talk where the phrase is mentioned.)
$endgroup$
– Andrés E. Caicedo
Jul 5 '14 at 14:41
|
show 5 more comments
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1 Answer
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$begingroup$
This is independent of ZFC. It is consistent that there are no such cardinals, for example if GCH holds. Note that $lambdaleqkappaimplies2^lambdaleq2^kappa$, so it is enough to show that the continuum function is injective.
However it is consistent that $2^{aleph_0}=2^{aleph_1}=aleph_3$.
There is not much we can say about the continuum function in ZFC. This is a dire consequence from Easton's theorem.
Easton theorem tells us that if $F$ is a function whose domain is the regular cardinals and:
- $kappa<lambdaimplies F(kappa)leq F(lambda)$,
- $operatorname{cf}(kappa)<operatorname{cf}(F(kappa))$
Then there is a forcing extension which does not collapse cardinals and for every regular $kappa$, $2^kappa=F(kappa)$ in the extension.
Assume GCH holds and take the function $F(kappa)=kappa^{++}$. We can show that in the extension where $F$ describes the continuum function we have $2^{kappa}=kappa^{++}$ for regular cardinals, and $F(mu)=mu^+$ for singular $mu$. This means that GCH fails for all regular cardinals, but $2^lambda=2^kappaifflambda=kappa$. So the injectivity of the continuum function holds, while GCH fails.
(If one is not in the mood for a class-forcing, which can be a bit complicated, one can simply start with GCH and set $2^{aleph_n}=aleph_{n+2}$ for $n<omega$, and GCH to hold otherwise instead.)
edited Jul 4 '14 at 11:31
Martin Sleziak
44.9k10121274
answered Nov 26 '12 at 10:14
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
$begingroup$
Thank you. It sounds fairly interesting.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:27
$begingroup$
You're welcome. You should search for the term Easton on this site, it has been mentioned and explained before (I am writing from a cellphone, so I am being brief).
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 10:33
$begingroup$
Good advise. Although it is probably too advanced for me, I'll have a look.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:39
$begingroup$
Now that I am by a computer again, I added a bit.
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 11:39
1
$begingroup$
@AsafKaragila I wasn't even thinking in terms of consistency, just combinatorics. I do not agree that it is just the very basics (nothing very basic about: If $aleph_omega$ is strong limit, then $2^{aleph_omega}<aleph_{omega_4}$, for instance) but yes, it does not tell us what the value of the continuum is, or the value of $2^{aleph_{17}}$, even as a function of the previous values, or any of that. So most likely yes, you are using the phrase in a different fashion that I am. (I would still shake my head silently if I were in the audience of a talk where the phrase is mentioned.)
$endgroup$
– Andrés E. Caicedo
Jul 5 '14 at 14:41
|
show 5 more comments
$begingroup$
This is independent of ZFC. It is consistent that there are no such cardinals, for example if GCH holds. Note that $lambdaleqkappaimplies2^lambdaleq2^kappa$, so it is enough to show that the continuum function is injective.
However it is consistent that $2^{aleph_0}=2^{aleph_1}=aleph_3$.
There is not much we can say about the continuum function in ZFC. This is a dire consequence from Easton's theorem.
Easton theorem tells us that if $F$ is a function whose domain is the regular cardinals and:
- $kappa<lambdaimplies F(kappa)leq F(lambda)$,
- $operatorname{cf}(kappa)<operatorname{cf}(F(kappa))$
Then there is a forcing extension which does not collapse cardinals and for every regular $kappa$, $2^kappa=F(kappa)$ in the extension.
Assume GCH holds and take the function $F(kappa)=kappa^{++}$. We can show that in the extension where $F$ describes the continuum function we have $2^{kappa}=kappa^{++}$ for regular cardinals, and $F(mu)=mu^+$ for singular $mu$. This means that GCH fails for all regular cardinals, but $2^lambda=2^kappaifflambda=kappa$. So the injectivity of the continuum function holds, while GCH fails.
(If one is not in the mood for a class-forcing, which can be a bit complicated, one can simply start with GCH and set $2^{aleph_n}=aleph_{n+2}$ for $n<omega$, and GCH to hold otherwise instead.)
edited Jul 4 '14 at 11:31
Martin Sleziak
44.9k10121274
answered Nov 26 '12 at 10:14
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
$begingroup$
Thank you. It sounds fairly interesting.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:27
$begingroup$
You're welcome. You should search for the term Easton on this site, it has been mentioned and explained before (I am writing from a cellphone, so I am being brief).
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 10:33
$begingroup$
Good advise. Although it is probably too advanced for me, I'll have a look.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:39
$begingroup$
Now that I am by a computer again, I added a bit.
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 11:39
1
$begingroup$
@AsafKaragila I wasn't even thinking in terms of consistency, just combinatorics. I do not agree that it is just the very basics (nothing very basic about: If $aleph_omega$ is strong limit, then $2^{aleph_omega}<aleph_{omega_4}$, for instance) but yes, it does not tell us what the value of the continuum is, or the value of $2^{aleph_{17}}$, even as a function of the previous values, or any of that. So most likely yes, you are using the phrase in a different fashion that I am. (I would still shake my head silently if I were in the audience of a talk where the phrase is mentioned.)
$endgroup$
– Andrés E. Caicedo
Jul 5 '14 at 14:41
|
show 5 more comments
$begingroup$
This is independent of ZFC. It is consistent that there are no such cardinals, for example if GCH holds. Note that $lambdaleqkappaimplies2^lambdaleq2^kappa$, so it is enough to show that the continuum function is injective.
However it is consistent that $2^{aleph_0}=2^{aleph_1}=aleph_3$.
There is not much we can say about the continuum function in ZFC. This is a dire consequence from Easton's theorem.
Easton theorem tells us that if $F$ is a function whose domain is the regular cardinals and:
- $kappa<lambdaimplies F(kappa)leq F(lambda)$,
- $operatorname{cf}(kappa)<operatorname{cf}(F(kappa))$
Then there is a forcing extension which does not collapse cardinals and for every regular $kappa$, $2^kappa=F(kappa)$ in the extension.
Assume GCH holds and take the function $F(kappa)=kappa^{++}$. We can show that in the extension where $F$ describes the continuum function we have $2^{kappa}=kappa^{++}$ for regular cardinals, and $F(mu)=mu^+$ for singular $mu$. This means that GCH fails for all regular cardinals, but $2^lambda=2^kappaifflambda=kappa$. So the injectivity of the continuum function holds, while GCH fails.
(If one is not in the mood for a class-forcing, which can be a bit complicated, one can simply start with GCH and set $2^{aleph_n}=aleph_{n+2}$ for $n<omega$, and GCH to hold otherwise instead.)
edited Jul 4 '14 at 11:31
Martin Sleziak
44.9k10121274
answered Nov 26 '12 at 10:14
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
This is independent of ZFC. It is consistent that there are no such cardinals, for example if GCH holds. Note that $lambdaleqkappaimplies2^lambdaleq2^kappa$, so it is enough to show that the continuum function is injective.
However it is consistent that $2^{aleph_0}=2^{aleph_1}=aleph_3$.
There is not much we can say about the continuum function in ZFC. This is a dire consequence from Easton's theorem.
Easton theorem tells us that if $F$ is a function whose domain is the regular cardinals and:
- $kappa<lambdaimplies F(kappa)leq F(lambda)$,
- $operatorname{cf}(kappa)<operatorname{cf}(F(kappa))$
Then there is a forcing extension which does not collapse cardinals and for every regular $kappa$, $2^kappa=F(kappa)$ in the extension.
Assume GCH holds and take the function $F(kappa)=kappa^{++}$. We can show that in the extension where $F$ describes the continuum function we have $2^{kappa}=kappa^{++}$ for regular cardinals, and $F(mu)=mu^+$ for singular $mu$. This means that GCH fails for all regular cardinals, but $2^lambda=2^kappaifflambda=kappa$. So the injectivity of the continuum function holds, while GCH fails.
(If one is not in the mood for a class-forcing, which can be a bit complicated, one can simply start with GCH and set $2^{aleph_n}=aleph_{n+2}$ for $n<omega$, and GCH to hold otherwise instead.)
edited Jul 4 '14 at 11:31
Martin Sleziak
44.9k10121274
answered Nov 26 '12 at 10:14
Asaf Karagila♦Asaf Karagila
306k33438769
edited Jul 4 '14 at 11:31
Martin Sleziak
44.9k10121274
edited Jul 4 '14 at 11:31
Martin Sleziak
44.9k10121274
edited Jul 4 '14 at 11:31
Martin Sleziak
44.9k10121274
44.9k10121274
answered Nov 26 '12 at 10:14
Asaf Karagila♦Asaf Karagila
306k33438769
answered Nov 26 '12 at 10:14
Asaf Karagila♦Asaf Karagila
306k33438769
answered Nov 26 '12 at 10:14
Asaf Karagila♦Asaf Karagila
306k33438769
306k33438769
$begingroup$
Thank you. It sounds fairly interesting.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:27
$begingroup$
You're welcome. You should search for the term Easton on this site, it has been mentioned and explained before (I am writing from a cellphone, so I am being brief).
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 10:33
$begingroup$
Good advise. Although it is probably too advanced for me, I'll have a look.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:39
$begingroup$
Now that I am by a computer again, I added a bit.
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 11:39
1
$begingroup$
@AsafKaragila I wasn't even thinking in terms of consistency, just combinatorics. I do not agree that it is just the very basics (nothing very basic about: If $aleph_omega$ is strong limit, then $2^{aleph_omega}<aleph_{omega_4}$, for instance) but yes, it does not tell us what the value of the continuum is, or the value of $2^{aleph_{17}}$, even as a function of the previous values, or any of that. So most likely yes, you are using the phrase in a different fashion that I am. (I would still shake my head silently if I were in the audience of a talk where the phrase is mentioned.)
$endgroup$
– Andrés E. Caicedo
Jul 5 '14 at 14:41
|
show 5 more comments
$begingroup$
Thank you. It sounds fairly interesting.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:27
$begingroup$
You're welcome. You should search for the term Easton on this site, it has been mentioned and explained before (I am writing from a cellphone, so I am being brief).
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 10:33
$begingroup$
Good advise. Although it is probably too advanced for me, I'll have a look.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:39
$begingroup$
Now that I am by a computer again, I added a bit.
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 11:39
1
$begingroup$
@AsafKaragila I wasn't even thinking in terms of consistency, just combinatorics. I do not agree that it is just the very basics (nothing very basic about: If $aleph_omega$ is strong limit, then $2^{aleph_omega}<aleph_{omega_4}$, for instance) but yes, it does not tell us what the value of the continuum is, or the value of $2^{aleph_{17}}$, even as a function of the previous values, or any of that. So most likely yes, you are using the phrase in a different fashion that I am. (I would still shake my head silently if I were in the audience of a talk where the phrase is mentioned.)
$endgroup$
– Andrés E. Caicedo
Jul 5 '14 at 14:41
$begingroup$
Thank you. It sounds fairly interesting.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:27
$begingroup$
Thank you. It sounds fairly interesting.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:27
$begingroup$
You're welcome. You should search for the term Easton on this site, it has been mentioned and explained before (I am writing from a cellphone, so I am being brief).
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 10:33
$begingroup$
You're welcome. You should search for the term Easton on this site, it has been mentioned and explained before (I am writing from a cellphone, so I am being brief).
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 10:33
$begingroup$
Good advise. Although it is probably too advanced for me, I'll have a look.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:39
$begingroup$
Good advise. Although it is probably too advanced for me, I'll have a look.
$endgroup$
– Metta World Peace
Nov 26 '12 at 10:39
$begingroup$
Now that I am by a computer again, I added a bit.
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 11:39
$begingroup$
Now that I am by a computer again, I added a bit.
$endgroup$
– Asaf Karagila♦
Nov 26 '12 at 11:39
1
1
$begingroup$
@AsafKaragila I wasn't even thinking in terms of consistency, just combinatorics. I do not agree that it is just the very basics (nothing very basic about: If $aleph_omega$ is strong limit, then $2^{aleph_omega}<aleph_{omega_4}$, for instance) but yes, it does not tell us what the value of the continuum is, or the value of $2^{aleph_{17}}$, even as a function of the previous values, or any of that. So most likely yes, you are using the phrase in a different fashion that I am. (I would still shake my head silently if I were in the audience of a talk where the phrase is mentioned.)
$endgroup$
– Andrés E. Caicedo
Jul 5 '14 at 14:41
$begingroup$
@AsafKaragila I wasn't even thinking in terms of consistency, just combinatorics. I do not agree that it is just the very basics (nothing very basic about: If $aleph_omega$ is strong limit, then $2^{aleph_omega}<aleph_{omega_4}$, for instance) but yes, it does not tell us what the value of the continuum is, or the value of $2^{aleph_{17}}$, even as a function of the previous values, or any of that. So most likely yes, you are using the phrase in a different fashion that I am. (I would still shake my head silently if I were in the audience of a talk where the phrase is mentioned.)
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– Andrés E. Caicedo
Jul 5 '14 at 14:41
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$begingroup$
I just noticed the exact wording of the question was not answered by my answer previously. I added more.
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– Asaf Karagila♦
Nov 28 '12 at 2:02
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Thank you very much. I'm so amazed. It seems to me that you ponder upon these problems like crazy :)
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– Metta World Peace
Nov 28 '12 at 2:10
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I do (and I am). I mean, seriously. 4am. I am going to sleep! (I have to give a talk tomorrow afternoon, and I won't have time to doze off around noon.)
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– Asaf Karagila♦
Nov 28 '12 at 2:11
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What should I say?! I don't want to sound like Cee lo in a talent show. But what you have been doing do amaze and impress me。
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– Metta World Peace
Nov 28 '12 at 5:59
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See also math.stackexchange.com/questions/74477/… or math.stackexchange.com/questions/376509/… and other questions linked there.
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– Martin Sleziak
Jul 4 '14 at 11:41