Jech Lemma 3.7: Why does this follow?
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I'm on Jech Chapter 3 (Cardinal Numbers) on the section on cofinalities. I don't understand why the implication in the red rectangle is true. If the aforementioned gamma-sequence was constant and every term equaled to alpha, why would the right hand limit in the red rectangle necessarily evaluate to gamma?
set-theory cardinals
edited Dec 24 '18 at 22:12
Andrés E. Caicedo
65.5k8158249
asked Dec 24 '18 at 20:48
Barycentric_BashBarycentric_Bash
42339
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add a comment |
$begingroup$
I'm on Jech Chapter 3 (Cardinal Numbers) on the section on cofinalities. I don't understand why the implication in the red rectangle is true. If the aforementioned gamma-sequence was constant and every term equaled to alpha, why would the right hand limit in the red rectangle necessarily evaluate to gamma?
set-theory cardinals
edited Dec 24 '18 at 22:12
Andrés E. Caicedo
65.5k8158249
asked Dec 24 '18 at 20:48
Barycentric_BashBarycentric_Bash
42339
$endgroup$
add a comment |
$begingroup$
I'm on Jech Chapter 3 (Cardinal Numbers) on the section on cofinalities. I don't understand why the implication in the red rectangle is true. If the aforementioned gamma-sequence was constant and every term equaled to alpha, why would the right hand limit in the red rectangle necessarily evaluate to gamma?
set-theory cardinals
edited Dec 24 '18 at 22:12
Andrés E. Caicedo
65.5k8158249
asked Dec 24 '18 at 20:48
Barycentric_BashBarycentric_Bash
42339
$endgroup$
I'm on Jech Chapter 3 (Cardinal Numbers) on the section on cofinalities. I don't understand why the implication in the red rectangle is true. If the aforementioned gamma-sequence was constant and every term equaled to alpha, why would the right hand limit in the red rectangle necessarily evaluate to gamma?
set-theory cardinals
set-theory cardinals
edited Dec 24 '18 at 22:12
Andrés E. Caicedo
65.5k8158249
asked Dec 24 '18 at 20:48
Barycentric_BashBarycentric_Bash
42339
edited Dec 24 '18 at 22:12
Andrés E. Caicedo
65.5k8158249
asked Dec 24 '18 at 20:48
Barycentric_BashBarycentric_Bash
42339
edited Dec 24 '18 at 22:12
Andrés E. Caicedo
65.5k8158249
edited Dec 24 '18 at 22:12
Andrés E. Caicedo
65.5k8158249
edited Dec 24 '18 at 22:12
Andrés E. Caicedo
65.5k8158249
65.5k8158249
asked Dec 24 '18 at 20:48
Barycentric_BashBarycentric_Bash
42339
asked Dec 24 '18 at 20:48
Barycentric_BashBarycentric_Bash
42339
asked Dec 24 '18 at 20:48
Barycentric_BashBarycentric_Bash
42339
42339
add a comment |
add a comment |
1 Answer
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The key is the phrase "in $alpha$" - that is, $beta_eta<alpha$ for each $eta<gamma$. This rules out the possibility that each $beta$ is equal to $alpha$, and in fact means that while the sequence of $beta$s is not necessarily increasing at each step (it's only guaranteed to be nondecreasing) it does grow unboundedly-in-$gamma$-often.
answered Dec 24 '18 at 20:57
Noah SchweberNoah Schweber
125k10150287
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The key is the phrase "in $alpha$" - that is, $beta_eta<alpha$ for each $eta<gamma$. This rules out the possibility that each $beta$ is equal to $alpha$, and in fact means that while the sequence of $beta$s is not necessarily increasing at each step (it's only guaranteed to be nondecreasing) it does grow unboundedly-in-$gamma$-often.
answered Dec 24 '18 at 20:57
Noah SchweberNoah Schweber
125k10150287
$endgroup$
add a comment |
$begingroup$
The key is the phrase "in $alpha$" - that is, $beta_eta<alpha$ for each $eta<gamma$. This rules out the possibility that each $beta$ is equal to $alpha$, and in fact means that while the sequence of $beta$s is not necessarily increasing at each step (it's only guaranteed to be nondecreasing) it does grow unboundedly-in-$gamma$-often.
answered Dec 24 '18 at 20:57
Noah SchweberNoah Schweber
125k10150287
$endgroup$
add a comment |
$begingroup$
The key is the phrase "in $alpha$" - that is, $beta_eta<alpha$ for each $eta<gamma$. This rules out the possibility that each $beta$ is equal to $alpha$, and in fact means that while the sequence of $beta$s is not necessarily increasing at each step (it's only guaranteed to be nondecreasing) it does grow unboundedly-in-$gamma$-often.
answered Dec 24 '18 at 20:57
Noah SchweberNoah Schweber
125k10150287
$endgroup$
The key is the phrase "in $alpha$" - that is, $beta_eta<alpha$ for each $eta<gamma$. This rules out the possibility that each $beta$ is equal to $alpha$, and in fact means that while the sequence of $beta$s is not necessarily increasing at each step (it's only guaranteed to be nondecreasing) it does grow unboundedly-in-$gamma$-often.
answered Dec 24 '18 at 20:57
Noah SchweberNoah Schweber
125k10150287
answered Dec 24 '18 at 20:57
Noah SchweberNoah Schweber
125k10150287
answered Dec 24 '18 at 20:57
Noah SchweberNoah Schweber
125k10150287
answered Dec 24 '18 at 20:57
Noah SchweberNoah Schweber
125k10150287
125k10150287
add a comment |
add a comment |
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