Complicated index models and Boolean algebras/ Shelah/ Unclear step in the proof
Here on the page $10$ in the $5$th line (the proof of lemma $1.10$), Shelah defines $n_*$ as $omega$: $$n_*=omega,$$
and then he continues:
be such that $n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
Does it make sense?
sequences-and-series set-theory boolean-algebra natural-numbers
asked Dec 5 '18 at 17:36
user122424user122424
1,1032616
add a comment |
Here on the page $10$ in the $5$th line (the proof of lemma $1.10$), Shelah defines $n_*$ as $omega$: $$n_*=omega,$$
and then he continues:
be such that $n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
Does it make sense?
sequences-and-series set-theory boolean-algebra natural-numbers
asked Dec 5 '18 at 17:36
user122424user122424
1,1032616
add a comment |
Here on the page $10$ in the $5$th line (the proof of lemma $1.10$), Shelah defines $n_*$ as $omega$: $$n_*=omega,$$
and then he continues:
be such that $n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
Does it make sense?
sequences-and-series set-theory boolean-algebra natural-numbers
asked Dec 5 '18 at 17:36
user122424user122424
1,1032616
Here on the page $10$ in the $5$th line (the proof of lemma $1.10$), Shelah defines $n_*$ as $omega$: $$n_*=omega,$$
and then he continues:
be such that $n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
Does it make sense?
sequences-and-series set-theory boolean-algebra natural-numbers
sequences-and-series set-theory boolean-algebra natural-numbers
asked Dec 5 '18 at 17:36
user122424user122424
1,1032616
asked Dec 5 '18 at 17:36
user122424user122424
1,1032616
edited Dec 5 '18 at 18:06
user122424
asked Dec 5 '18 at 17:36
user122424user122424
1,1032616
asked Dec 5 '18 at 17:36
user122424user122424
1,1032616
asked Dec 5 '18 at 17:36
user122424user122424
1,1032616
1,1032616
add a comment |
add a comment |
1 Answer
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The paper you are quoting does not say
$n_*geq{text{max}(n(0),...,n(m-1)}<omega.$
it says
$n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
and certainly if the middle $<omega$, it is going to be less than $n_ast=omega$.
I can't speak for the rest of the paper, but I don't see anything particularly odd about the line.
answered Dec 5 '18 at 17:49
rschwiebrschwieb
105k12100245
Fine, and what are $n$ and $m$ there?
– user122424
Dec 5 '18 at 17:54
1
@user122424 The $n(..)$'s are defined right after the last ${1.4}$ marker on page 9.
– Not Mike
Dec 5 '18 at 18:39
@NotMike But how is $n(...$ a function. It appears to me that it is a number $<omega$ ...But on the page $10$ 5th line they put arguments to it: $n(0),...,n(m-1)$
– user122424
Dec 6 '18 at 12:47
@user122424 by induction/recursion
– Not Mike
Jan 2 at 11:43
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
The paper you are quoting does not say
$n_*geq{text{max}(n(0),...,n(m-1)}<omega.$
it says
$n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
and certainly if the middle $<omega$, it is going to be less than $n_ast=omega$.
I can't speak for the rest of the paper, but I don't see anything particularly odd about the line.
answered Dec 5 '18 at 17:49
rschwiebrschwieb
105k12100245
Fine, and what are $n$ and $m$ there?
– user122424
Dec 5 '18 at 17:54
1
@user122424 The $n(..)$'s are defined right after the last ${1.4}$ marker on page 9.
– Not Mike
Dec 5 '18 at 18:39
@NotMike But how is $n(...$ a function. It appears to me that it is a number $<omega$ ...But on the page $10$ 5th line they put arguments to it: $n(0),...,n(m-1)$
– user122424
Dec 6 '18 at 12:47
@user122424 by induction/recursion
– Not Mike
Jan 2 at 11:43
add a comment |
The paper you are quoting does not say
$n_*geq{text{max}(n(0),...,n(m-1)}<omega.$
it says
$n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
and certainly if the middle $<omega$, it is going to be less than $n_ast=omega$.
I can't speak for the rest of the paper, but I don't see anything particularly odd about the line.
answered Dec 5 '18 at 17:49
rschwiebrschwieb
105k12100245
Fine, and what are $n$ and $m$ there?
– user122424
Dec 5 '18 at 17:54
1
@user122424 The $n(..)$'s are defined right after the last ${1.4}$ marker on page 9.
– Not Mike
Dec 5 '18 at 18:39
@NotMike But how is $n(...$ a function. It appears to me that it is a number $<omega$ ...But on the page $10$ 5th line they put arguments to it: $n(0),...,n(m-1)$
– user122424
Dec 6 '18 at 12:47
@user122424 by induction/recursion
– Not Mike
Jan 2 at 11:43
add a comment |
The paper you are quoting does not say
$n_*geq{text{max}(n(0),...,n(m-1)}<omega.$
it says
$n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
and certainly if the middle $<omega$, it is going to be less than $n_ast=omega$.
I can't speak for the rest of the paper, but I don't see anything particularly odd about the line.
answered Dec 5 '18 at 17:49
rschwiebrschwieb
105k12100245
The paper you are quoting does not say
$n_*geq{text{max}(n(0),...,n(m-1)}<omega.$
it says
$n_*geqtext{max}{n(0),...,n(m-1)}<omega.$
and certainly if the middle $<omega$, it is going to be less than $n_ast=omega$.
I can't speak for the rest of the paper, but I don't see anything particularly odd about the line.
answered Dec 5 '18 at 17:49
rschwiebrschwieb
105k12100245
answered Dec 5 '18 at 17:49
rschwiebrschwieb
105k12100245
answered Dec 5 '18 at 17:49
rschwiebrschwieb
105k12100245
answered Dec 5 '18 at 17:49
rschwiebrschwieb
105k12100245
105k12100245
Fine, and what are $n$ and $m$ there?
– user122424
Dec 5 '18 at 17:54
1
@user122424 The $n(..)$'s are defined right after the last ${1.4}$ marker on page 9.
– Not Mike
Dec 5 '18 at 18:39
@NotMike But how is $n(...$ a function. It appears to me that it is a number $<omega$ ...But on the page $10$ 5th line they put arguments to it: $n(0),...,n(m-1)$
– user122424
Dec 6 '18 at 12:47
@user122424 by induction/recursion
– Not Mike
Jan 2 at 11:43
add a comment |
Fine, and what are $n$ and $m$ there?
– user122424
Dec 5 '18 at 17:54
1
@user122424 The $n(..)$'s are defined right after the last ${1.4}$ marker on page 9.
– Not Mike
Dec 5 '18 at 18:39
@NotMike But how is $n(...$ a function. It appears to me that it is a number $<omega$ ...But on the page $10$ 5th line they put arguments to it: $n(0),...,n(m-1)$
– user122424
Dec 6 '18 at 12:47
@user122424 by induction/recursion
– Not Mike
Jan 2 at 11:43
Fine, and what are $n$ and $m$ there?
– user122424
Dec 5 '18 at 17:54
Fine, and what are $n$ and $m$ there?
– user122424
Dec 5 '18 at 17:54
1
1
@user122424 The $n(..)$'s are defined right after the last ${1.4}$ marker on page 9.
– Not Mike
Dec 5 '18 at 18:39
@user122424 The $n(..)$'s are defined right after the last ${1.4}$ marker on page 9.
– Not Mike
Dec 5 '18 at 18:39
@NotMike But how is $n(...$ a function. It appears to me that it is a number $<omega$ ...But on the page $10$ 5th line they put arguments to it: $n(0),...,n(m-1)$
– user122424
Dec 6 '18 at 12:47
@NotMike But how is $n(...$ a function. It appears to me that it is a number $<omega$ ...But on the page $10$ 5th line they put arguments to it: $n(0),...,n(m-1)$
– user122424
Dec 6 '18 at 12:47
@user122424 by induction/recursion
– Not Mike
Jan 2 at 11:43
@user122424 by induction/recursion
– Not Mike
Jan 2 at 11:43
add a comment |
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