Jech Lemma 3.10 (cofinalities)
$begingroup$
For some reason, this lemma remained elusive in my attempts to find it on the web. I couldn't find it in Hrbacek & Jech either.
I understand every part of the proof above except for the last sentence. How does beta = kappa follow? Many thanks.
soft question: reading this book is very hard for me :( the previous page took a solid 3 hours to fill in the blanks of the proofs, and the author sometimes assumes knowledge that was never stated. Any resources that are good supplements to Jech? very lost on how I should approach reading this book. any advice is helpful
soft-question set-theory cardinals
asked Dec 25 '18 at 5:59
Barycentric_BashBarycentric_Bash
42339
$endgroup$
add a comment |
$begingroup$
For some reason, this lemma remained elusive in my attempts to find it on the web. I couldn't find it in Hrbacek & Jech either.
I understand every part of the proof above except for the last sentence. How does beta = kappa follow? Many thanks.
soft question: reading this book is very hard for me :( the previous page took a solid 3 hours to fill in the blanks of the proofs, and the author sometimes assumes knowledge that was never stated. Any resources that are good supplements to Jech? very lost on how I should approach reading this book. any advice is helpful
soft-question set-theory cardinals
asked Dec 25 '18 at 5:59
Barycentric_BashBarycentric_Bash
42339
$endgroup$
1
$begingroup$
It's a difficult book. Yes. But that's not a bad thing. Working hard for something is good. We should all remember this from time to time.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:06
1
$begingroup$
Also, it's not fully clear to me what kind of set theory you are trying to learn, or what you already know or not know. But you can try karagila.org/wp-content/uploads/2016/01/ests-wh.pdf for some recommendations.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:07
$begingroup$
You can try the early chapters of Set Theory: An Introduction To Independence Proofs, by K. Kunen.... What is Jech's starting definition of singular cardinal? I'd like to comment or answer but I dk where Jech is starting from.
$endgroup$
– DanielWainfleet
Dec 26 '18 at 1:10
$begingroup$
@DanielWainfleet An infinite cardinal k is singular if cf k < k - Jech
$endgroup$
– Barycentric_Bash
Dec 26 '18 at 3:45
$begingroup$
Sorry, but I should have asked more precisely: What is Jech's starting def'n of cf$(k)?$
$endgroup$
– DanielWainfleet
Dec 27 '18 at 9:03
add a comment |
$begingroup$
For some reason, this lemma remained elusive in my attempts to find it on the web. I couldn't find it in Hrbacek & Jech either.
I understand every part of the proof above except for the last sentence. How does beta = kappa follow? Many thanks.
soft question: reading this book is very hard for me :( the previous page took a solid 3 hours to fill in the blanks of the proofs, and the author sometimes assumes knowledge that was never stated. Any resources that are good supplements to Jech? very lost on how I should approach reading this book. any advice is helpful
soft-question set-theory cardinals
asked Dec 25 '18 at 5:59
Barycentric_BashBarycentric_Bash
42339
$endgroup$
For some reason, this lemma remained elusive in my attempts to find it on the web. I couldn't find it in Hrbacek & Jech either.
I understand every part of the proof above except for the last sentence. How does beta = kappa follow? Many thanks.
soft question: reading this book is very hard for me :( the previous page took a solid 3 hours to fill in the blanks of the proofs, and the author sometimes assumes knowledge that was never stated. Any resources that are good supplements to Jech? very lost on how I should approach reading this book. any advice is helpful
soft-question set-theory cardinals
soft-question set-theory cardinals
asked Dec 25 '18 at 5:59
Barycentric_BashBarycentric_Bash
42339
asked Dec 25 '18 at 5:59
Barycentric_BashBarycentric_Bash
42339
asked Dec 25 '18 at 5:59
Barycentric_BashBarycentric_Bash
42339
asked Dec 25 '18 at 5:59
Barycentric_BashBarycentric_Bash
42339
asked Dec 25 '18 at 5:59
Barycentric_BashBarycentric_Bash
42339
42339
1
$begingroup$
It's a difficult book. Yes. But that's not a bad thing. Working hard for something is good. We should all remember this from time to time.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:06
1
$begingroup$
Also, it's not fully clear to me what kind of set theory you are trying to learn, or what you already know or not know. But you can try karagila.org/wp-content/uploads/2016/01/ests-wh.pdf for some recommendations.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:07
$begingroup$
You can try the early chapters of Set Theory: An Introduction To Independence Proofs, by K. Kunen.... What is Jech's starting definition of singular cardinal? I'd like to comment or answer but I dk where Jech is starting from.
$endgroup$
– DanielWainfleet
Dec 26 '18 at 1:10
$begingroup$
@DanielWainfleet An infinite cardinal k is singular if cf k < k - Jech
$endgroup$
– Barycentric_Bash
Dec 26 '18 at 3:45
$begingroup$
Sorry, but I should have asked more precisely: What is Jech's starting def'n of cf$(k)?$
$endgroup$
– DanielWainfleet
Dec 27 '18 at 9:03
add a comment |
1
$begingroup$
It's a difficult book. Yes. But that's not a bad thing. Working hard for something is good. We should all remember this from time to time.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:06
1
$begingroup$
Also, it's not fully clear to me what kind of set theory you are trying to learn, or what you already know or not know. But you can try karagila.org/wp-content/uploads/2016/01/ests-wh.pdf for some recommendations.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:07
$begingroup$
You can try the early chapters of Set Theory: An Introduction To Independence Proofs, by K. Kunen.... What is Jech's starting definition of singular cardinal? I'd like to comment or answer but I dk where Jech is starting from.
$endgroup$
– DanielWainfleet
Dec 26 '18 at 1:10
$begingroup$
@DanielWainfleet An infinite cardinal k is singular if cf k < k - Jech
$endgroup$
– Barycentric_Bash
Dec 26 '18 at 3:45
$begingroup$
Sorry, but I should have asked more precisely: What is Jech's starting def'n of cf$(k)?$
$endgroup$
– DanielWainfleet
Dec 27 '18 at 9:03
1
1
$begingroup$
It's a difficult book. Yes. But that's not a bad thing. Working hard for something is good. We should all remember this from time to time.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:06
$begingroup$
It's a difficult book. Yes. But that's not a bad thing. Working hard for something is good. We should all remember this from time to time.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:06
1
1
$begingroup$
Also, it's not fully clear to me what kind of set theory you are trying to learn, or what you already know or not know. But you can try karagila.org/wp-content/uploads/2016/01/ests-wh.pdf for some recommendations.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:07
$begingroup$
Also, it's not fully clear to me what kind of set theory you are trying to learn, or what you already know or not know. But you can try karagila.org/wp-content/uploads/2016/01/ests-wh.pdf for some recommendations.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:07
$begingroup$
You can try the early chapters of Set Theory: An Introduction To Independence Proofs, by K. Kunen.... What is Jech's starting definition of singular cardinal? I'd like to comment or answer but I dk where Jech is starting from.
$endgroup$
– DanielWainfleet
Dec 26 '18 at 1:10
$begingroup$
You can try the early chapters of Set Theory: An Introduction To Independence Proofs, by K. Kunen.... What is Jech's starting definition of singular cardinal? I'd like to comment or answer but I dk where Jech is starting from.
$endgroup$
– DanielWainfleet
Dec 26 '18 at 1:10
$begingroup$
@DanielWainfleet An infinite cardinal k is singular if cf k < k - Jech
$endgroup$
– Barycentric_Bash
Dec 26 '18 at 3:45
$begingroup$
@DanielWainfleet An infinite cardinal k is singular if cf k < k - Jech
$endgroup$
– Barycentric_Bash
Dec 26 '18 at 3:45
$begingroup$
Sorry, but I should have asked more precisely: What is Jech's starting def'n of cf$(k)?$
$endgroup$
– DanielWainfleet
Dec 27 '18 at 9:03
$begingroup$
Sorry, but I should have asked more precisely: What is Jech's starting def'n of cf$(k)?$
$endgroup$
– DanielWainfleet
Dec 27 '18 at 9:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The sentence before last shows there is a one-to-one mapping of $kappa$ into $lambdatimes beta,$ so $kappa le lambda cdot |beta|.$ Since $lambda < kappa,$ it follows that $|beta| ge kappa$ (otherwise we'd have $lambdacdot |beta| <kappa).$ Since $beta_xi < kappa$ for all $xi <lambda,$ $beta = sup_{xi < lambda}beta_xile kappa.$ We have $beta le kappa$ and $|beta|ge kappa,$ so it follows that $beta =kappa.$
As for books, the book by Enderton tends to get good reviews, and I've heard good things about Just and Weese as well (as well as Hrbacek/Jech that you mention). This is just what I've heard from others: I don't really know these first hand, other than Hrbacek and Jech which I've gone through bits and pieces of. The first book I learned from was Smullyan and Fitting, which I found quite pedagogical (although with a lot of annoying errors) but it does things a bit differently so might not serve the purpose of supplementing a more standard text. My second was Kunen (the older version), which was great, though the second chapter is rough going at first (it's probably actually best to skip it or read it selectively and then read it in full before the forcing chapters). Perhaps you can also browse the blurbs on the teach yourself logic guide (I think all the books I mentioned are written up there).
answered Dec 25 '18 at 7:29
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
sorry, but why is (kappa)(kappa) = kappa? The book gives (aleph)^2=aleph but we haven't proven that every set can be well ordered so, I don't see how the above statement follows (from what has been proven so far).
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:01
$begingroup$
also, thanks I'll definitely check out smullyan and fitting as I've searched around and seen some great reviews
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:02
1
$begingroup$
@Barycentric_Bash I'm pretty sure axiom of choice is being assumed here. (Or that we're implicitly talking about well-orderable cardinals.)
$endgroup$
– spaceisdarkgreen
Dec 25 '18 at 8:08
$begingroup$
@Barycentric_Bash: Indeed this part is after Jech focuses on the $aleph$ cardinals, so choice is not needed.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:24
add a comment |
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$begingroup$
The sentence before last shows there is a one-to-one mapping of $kappa$ into $lambdatimes beta,$ so $kappa le lambda cdot |beta|.$ Since $lambda < kappa,$ it follows that $|beta| ge kappa$ (otherwise we'd have $lambdacdot |beta| <kappa).$ Since $beta_xi < kappa$ for all $xi <lambda,$ $beta = sup_{xi < lambda}beta_xile kappa.$ We have $beta le kappa$ and $|beta|ge kappa,$ so it follows that $beta =kappa.$
As for books, the book by Enderton tends to get good reviews, and I've heard good things about Just and Weese as well (as well as Hrbacek/Jech that you mention). This is just what I've heard from others: I don't really know these first hand, other than Hrbacek and Jech which I've gone through bits and pieces of. The first book I learned from was Smullyan and Fitting, which I found quite pedagogical (although with a lot of annoying errors) but it does things a bit differently so might not serve the purpose of supplementing a more standard text. My second was Kunen (the older version), which was great, though the second chapter is rough going at first (it's probably actually best to skip it or read it selectively and then read it in full before the forcing chapters). Perhaps you can also browse the blurbs on the teach yourself logic guide (I think all the books I mentioned are written up there).
answered Dec 25 '18 at 7:29
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
sorry, but why is (kappa)(kappa) = kappa? The book gives (aleph)^2=aleph but we haven't proven that every set can be well ordered so, I don't see how the above statement follows (from what has been proven so far).
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:01
$begingroup$
also, thanks I'll definitely check out smullyan and fitting as I've searched around and seen some great reviews
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:02
1
$begingroup$
@Barycentric_Bash I'm pretty sure axiom of choice is being assumed here. (Or that we're implicitly talking about well-orderable cardinals.)
$endgroup$
– spaceisdarkgreen
Dec 25 '18 at 8:08
$begingroup$
@Barycentric_Bash: Indeed this part is after Jech focuses on the $aleph$ cardinals, so choice is not needed.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:24
add a comment |
$begingroup$
The sentence before last shows there is a one-to-one mapping of $kappa$ into $lambdatimes beta,$ so $kappa le lambda cdot |beta|.$ Since $lambda < kappa,$ it follows that $|beta| ge kappa$ (otherwise we'd have $lambdacdot |beta| <kappa).$ Since $beta_xi < kappa$ for all $xi <lambda,$ $beta = sup_{xi < lambda}beta_xile kappa.$ We have $beta le kappa$ and $|beta|ge kappa,$ so it follows that $beta =kappa.$
As for books, the book by Enderton tends to get good reviews, and I've heard good things about Just and Weese as well (as well as Hrbacek/Jech that you mention). This is just what I've heard from others: I don't really know these first hand, other than Hrbacek and Jech which I've gone through bits and pieces of. The first book I learned from was Smullyan and Fitting, which I found quite pedagogical (although with a lot of annoying errors) but it does things a bit differently so might not serve the purpose of supplementing a more standard text. My second was Kunen (the older version), which was great, though the second chapter is rough going at first (it's probably actually best to skip it or read it selectively and then read it in full before the forcing chapters). Perhaps you can also browse the blurbs on the teach yourself logic guide (I think all the books I mentioned are written up there).
answered Dec 25 '18 at 7:29
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
sorry, but why is (kappa)(kappa) = kappa? The book gives (aleph)^2=aleph but we haven't proven that every set can be well ordered so, I don't see how the above statement follows (from what has been proven so far).
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:01
$begingroup$
also, thanks I'll definitely check out smullyan and fitting as I've searched around and seen some great reviews
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:02
1
$begingroup$
@Barycentric_Bash I'm pretty sure axiom of choice is being assumed here. (Or that we're implicitly talking about well-orderable cardinals.)
$endgroup$
– spaceisdarkgreen
Dec 25 '18 at 8:08
$begingroup$
@Barycentric_Bash: Indeed this part is after Jech focuses on the $aleph$ cardinals, so choice is not needed.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:24
add a comment |
$begingroup$
The sentence before last shows there is a one-to-one mapping of $kappa$ into $lambdatimes beta,$ so $kappa le lambda cdot |beta|.$ Since $lambda < kappa,$ it follows that $|beta| ge kappa$ (otherwise we'd have $lambdacdot |beta| <kappa).$ Since $beta_xi < kappa$ for all $xi <lambda,$ $beta = sup_{xi < lambda}beta_xile kappa.$ We have $beta le kappa$ and $|beta|ge kappa,$ so it follows that $beta =kappa.$
As for books, the book by Enderton tends to get good reviews, and I've heard good things about Just and Weese as well (as well as Hrbacek/Jech that you mention). This is just what I've heard from others: I don't really know these first hand, other than Hrbacek and Jech which I've gone through bits and pieces of. The first book I learned from was Smullyan and Fitting, which I found quite pedagogical (although with a lot of annoying errors) but it does things a bit differently so might not serve the purpose of supplementing a more standard text. My second was Kunen (the older version), which was great, though the second chapter is rough going at first (it's probably actually best to skip it or read it selectively and then read it in full before the forcing chapters). Perhaps you can also browse the blurbs on the teach yourself logic guide (I think all the books I mentioned are written up there).
answered Dec 25 '18 at 7:29
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
The sentence before last shows there is a one-to-one mapping of $kappa$ into $lambdatimes beta,$ so $kappa le lambda cdot |beta|.$ Since $lambda < kappa,$ it follows that $|beta| ge kappa$ (otherwise we'd have $lambdacdot |beta| <kappa).$ Since $beta_xi < kappa$ for all $xi <lambda,$ $beta = sup_{xi < lambda}beta_xile kappa.$ We have $beta le kappa$ and $|beta|ge kappa,$ so it follows that $beta =kappa.$
As for books, the book by Enderton tends to get good reviews, and I've heard good things about Just and Weese as well (as well as Hrbacek/Jech that you mention). This is just what I've heard from others: I don't really know these first hand, other than Hrbacek and Jech which I've gone through bits and pieces of. The first book I learned from was Smullyan and Fitting, which I found quite pedagogical (although with a lot of annoying errors) but it does things a bit differently so might not serve the purpose of supplementing a more standard text. My second was Kunen (the older version), which was great, though the second chapter is rough going at first (it's probably actually best to skip it or read it selectively and then read it in full before the forcing chapters). Perhaps you can also browse the blurbs on the teach yourself logic guide (I think all the books I mentioned are written up there).
answered Dec 25 '18 at 7:29
spaceisdarkgreenspaceisdarkgreen
33.5k21753
edited Dec 25 '18 at 20:13
answered Dec 25 '18 at 7:29
spaceisdarkgreenspaceisdarkgreen
33.5k21753
answered Dec 25 '18 at 7:29
spaceisdarkgreenspaceisdarkgreen
33.5k21753
answered Dec 25 '18 at 7:29
spaceisdarkgreenspaceisdarkgreen
33.5k21753
33.5k21753
$begingroup$
sorry, but why is (kappa)(kappa) = kappa? The book gives (aleph)^2=aleph but we haven't proven that every set can be well ordered so, I don't see how the above statement follows (from what has been proven so far).
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:01
$begingroup$
also, thanks I'll definitely check out smullyan and fitting as I've searched around and seen some great reviews
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:02
1
$begingroup$
@Barycentric_Bash I'm pretty sure axiom of choice is being assumed here. (Or that we're implicitly talking about well-orderable cardinals.)
$endgroup$
– spaceisdarkgreen
Dec 25 '18 at 8:08
$begingroup$
@Barycentric_Bash: Indeed this part is after Jech focuses on the $aleph$ cardinals, so choice is not needed.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:24
add a comment |
$begingroup$
sorry, but why is (kappa)(kappa) = kappa? The book gives (aleph)^2=aleph but we haven't proven that every set can be well ordered so, I don't see how the above statement follows (from what has been proven so far).
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:01
$begingroup$
also, thanks I'll definitely check out smullyan and fitting as I've searched around and seen some great reviews
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:02
1
$begingroup$
@Barycentric_Bash I'm pretty sure axiom of choice is being assumed here. (Or that we're implicitly talking about well-orderable cardinals.)
$endgroup$
– spaceisdarkgreen
Dec 25 '18 at 8:08
$begingroup$
@Barycentric_Bash: Indeed this part is after Jech focuses on the $aleph$ cardinals, so choice is not needed.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:24
$begingroup$
sorry, but why is (kappa)(kappa) = kappa? The book gives (aleph)^2=aleph but we haven't proven that every set can be well ordered so, I don't see how the above statement follows (from what has been proven so far).
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:01
$begingroup$
sorry, but why is (kappa)(kappa) = kappa? The book gives (aleph)^2=aleph but we haven't proven that every set can be well ordered so, I don't see how the above statement follows (from what has been proven so far).
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:01
$begingroup$
also, thanks I'll definitely check out smullyan and fitting as I've searched around and seen some great reviews
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:02
$begingroup$
also, thanks I'll definitely check out smullyan and fitting as I've searched around and seen some great reviews
$endgroup$
– Barycentric_Bash
Dec 25 '18 at 8:02
1
1
$begingroup$
@Barycentric_Bash I'm pretty sure axiom of choice is being assumed here. (Or that we're implicitly talking about well-orderable cardinals.)
$endgroup$
– spaceisdarkgreen
Dec 25 '18 at 8:08
$begingroup$
@Barycentric_Bash I'm pretty sure axiom of choice is being assumed here. (Or that we're implicitly talking about well-orderable cardinals.)
$endgroup$
– spaceisdarkgreen
Dec 25 '18 at 8:08
$begingroup$
@Barycentric_Bash: Indeed this part is after Jech focuses on the $aleph$ cardinals, so choice is not needed.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:24
$begingroup$
@Barycentric_Bash: Indeed this part is after Jech focuses on the $aleph$ cardinals, so choice is not needed.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:24
add a comment |
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1
$begingroup$
It's a difficult book. Yes. But that's not a bad thing. Working hard for something is good. We should all remember this from time to time.
$endgroup$
– Asaf Karagila♦
Dec 25 '18 at 8:06
1
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Also, it's not fully clear to me what kind of set theory you are trying to learn, or what you already know or not know. But you can try karagila.org/wp-content/uploads/2016/01/ests-wh.pdf for some recommendations.
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– Asaf Karagila♦
Dec 25 '18 at 8:07
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You can try the early chapters of Set Theory: An Introduction To Independence Proofs, by K. Kunen.... What is Jech's starting definition of singular cardinal? I'd like to comment or answer but I dk where Jech is starting from.
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– DanielWainfleet
Dec 26 '18 at 1:10
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@DanielWainfleet An infinite cardinal k is singular if cf k < k - Jech
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– Barycentric_Bash
Dec 26 '18 at 3:45
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Sorry, but I should have asked more precisely: What is Jech's starting def'n of cf$(k)?$
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– DanielWainfleet
Dec 27 '18 at 9:03