Mixing
If $x_n$ is a 0,1 sequence such that $text{p-lim}(x_n) = 1$, then any 0,1 sequence $y_n supset x_n$ has...
$begingroup$
I am working through this post from Terry Tao's blog.
Problem: Prove that if a binary (i.e. consisting of $0$ and $1$, or $top$ and $bot$) sequence $x_n$ is such that $text{p-lim}(x_n) = 1$, then any sequence $y_n$ containing $x_n$ has $text{p-lim}(y_n) = text{p-lim}(x_n) = 1$.
My attempt at a proof. We know the following facts regarding $text{p-lim}$:
- homomorphic: $text{p-lim}(1) = 1$, $text{p-lim}(Cx_n) = Ctext{p-lim}(x_n)$ (for some $C in {0, 1}$), $text{p-lim}(x_n oplus y_n) = text{p-lim}(x_n) oplus text{p-lim}(y_n)$ (where $oplus$ is the "or" operator), and $text{p-lim}(x_n otimes y_n) = text{p-lim}(x_n) otimes text{p-lim}(y_n)$ (where $otimes$ is the "and" operator)
- bounded: $text{p-inf}(x_n) leq text{p-lim}(x_n) leq text{p-sup}(x_n)$
- non-principality: deletion of a finite number of elements from $x_n$ should not change its $text{p-lim}$.
We do not know anything else about $text{p-lim}$, so a combination of these facts must be used to prove the required.
Note that $x_n$ will never be finite, since it is an indicator sequence which runs over all the natural numbers, hence $y_n$ will never be finite either. So non-principality is unlikely to be a useful tool. The supremum and infimum of every sequence of binary numbers which is not $1, 1, 1, 1, 1, 1 dots$ or $0, 0, 0, 0, 0, 0, dots$ is trivially $0$ and $1$ respectively, so it is unlikely that boundedness can be used as a tool.
Let us try to use the fact that $text{p-lim}$ is a homomorphism. Note that $y_n oplus x_n = y_n$, since $x_n subset y_n$ (in other words, $x_k = 1 Rightarrow y_k = 1$, while $x_k = 0 Rightarrow (y_k = 1 vee y_k = 0)$). Then:
begin{align*}
&quad;; x_n oplus y_n = y_n \
&Rightarrow text{p-lim}(x_n oplus y_n) = text{p-lim}(y_n) \
&Rightarrow text{p-lim}(x_n) oplus text{p-lim}(y_n) = text{p-lim}(y_n) \
&Rightarrow 1 oplus text{p-lim}(y_n) = text{p-lim}(y_n) \
&{text{1 is the zero of $oplus$}} \
&Rightarrow 1 = text{p-lim}(y_n)
end{align*}
Is this proof correct? The following assumptions, I think, will need confirmation:
- On the space of binary numbers, is it okay to assume that $oplus, otimes$ are the analogues of $+, times$, for the homomorphism property? I think it should be okay, because: 1) it makes things work out nicely (I think I can use similar logic to show that some of the other ultrafilter properties hold for the $text{p-lim}$ over the binary numbers), and connects nicely to classical "predicate calculus" stuff; 2) $oplus, otimes$ can also be thought of as "addition/multiplication $mathbb{Z}/2$")
limits proof-verification filters
asked Jan 27 at 6:47
user89user89
5831647
$endgroup$
add a comment |
$begingroup$
I am working through this post from Terry Tao's blog.
Problem: Prove that if a binary (i.e. consisting of $0$ and $1$, or $top$ and $bot$) sequence $x_n$ is such that $text{p-lim}(x_n) = 1$, then any sequence $y_n$ containing $x_n$ has $text{p-lim}(y_n) = text{p-lim}(x_n) = 1$.
My attempt at a proof. We know the following facts regarding $text{p-lim}$:
- homomorphic: $text{p-lim}(1) = 1$, $text{p-lim}(Cx_n) = Ctext{p-lim}(x_n)$ (for some $C in {0, 1}$), $text{p-lim}(x_n oplus y_n) = text{p-lim}(x_n) oplus text{p-lim}(y_n)$ (where $oplus$ is the "or" operator), and $text{p-lim}(x_n otimes y_n) = text{p-lim}(x_n) otimes text{p-lim}(y_n)$ (where $otimes$ is the "and" operator)
- bounded: $text{p-inf}(x_n) leq text{p-lim}(x_n) leq text{p-sup}(x_n)$
- non-principality: deletion of a finite number of elements from $x_n$ should not change its $text{p-lim}$.
We do not know anything else about $text{p-lim}$, so a combination of these facts must be used to prove the required.
Note that $x_n$ will never be finite, since it is an indicator sequence which runs over all the natural numbers, hence $y_n$ will never be finite either. So non-principality is unlikely to be a useful tool. The supremum and infimum of every sequence of binary numbers which is not $1, 1, 1, 1, 1, 1 dots$ or $0, 0, 0, 0, 0, 0, dots$ is trivially $0$ and $1$ respectively, so it is unlikely that boundedness can be used as a tool.
Let us try to use the fact that $text{p-lim}$ is a homomorphism. Note that $y_n oplus x_n = y_n$, since $x_n subset y_n$ (in other words, $x_k = 1 Rightarrow y_k = 1$, while $x_k = 0 Rightarrow (y_k = 1 vee y_k = 0)$). Then:
begin{align*}
&quad;; x_n oplus y_n = y_n \
&Rightarrow text{p-lim}(x_n oplus y_n) = text{p-lim}(y_n) \
&Rightarrow text{p-lim}(x_n) oplus text{p-lim}(y_n) = text{p-lim}(y_n) \
&Rightarrow 1 oplus text{p-lim}(y_n) = text{p-lim}(y_n) \
&{text{1 is the zero of $oplus$}} \
&Rightarrow 1 = text{p-lim}(y_n)
end{align*}
Is this proof correct? The following assumptions, I think, will need confirmation:
- On the space of binary numbers, is it okay to assume that $oplus, otimes$ are the analogues of $+, times$, for the homomorphism property? I think it should be okay, because: 1) it makes things work out nicely (I think I can use similar logic to show that some of the other ultrafilter properties hold for the $text{p-lim}$ over the binary numbers), and connects nicely to classical "predicate calculus" stuff; 2) $oplus, otimes$ can also be thought of as "addition/multiplication $mathbb{Z}/2$")
limits proof-verification filters
asked Jan 27 at 6:47
user89user89
5831647
$endgroup$
$begingroup$
What do you mean by $(y_n)$ "containing" $(x_n)$? Sequences are functions, not sets.. It seems you mean $y_n le x_n$ for all $n$?
$endgroup$
– Henno Brandsma
Jan 30 at 5:34
$begingroup$
It seems that you mean by "$x_n subset y_n$" that $y_k le x_k$ for all $k$; is that true? In the subset interpretation Tao gives the latter means that the subset of $omega$ associated with $y$ is a subset of the one asssociated with $x$, so inclusion the other way around.... You seem confused.
$endgroup$
– Henno Brandsma
Jan 30 at 23:07
$begingroup$
@HennoBrandsma by $x_n$ being contained in $y_n$, I mean: $x_k = 1 Rightarrow y_k = 1$
$endgroup$
– user89
Jan 31 at 18:45
add a comment |
$begingroup$
I am working through this post from Terry Tao's blog.
Problem: Prove that if a binary (i.e. consisting of $0$ and $1$, or $top$ and $bot$) sequence $x_n$ is such that $text{p-lim}(x_n) = 1$, then any sequence $y_n$ containing $x_n$ has $text{p-lim}(y_n) = text{p-lim}(x_n) = 1$.
My attempt at a proof. We know the following facts regarding $text{p-lim}$:
- homomorphic: $text{p-lim}(1) = 1$, $text{p-lim}(Cx_n) = Ctext{p-lim}(x_n)$ (for some $C in {0, 1}$), $text{p-lim}(x_n oplus y_n) = text{p-lim}(x_n) oplus text{p-lim}(y_n)$ (where $oplus$ is the "or" operator), and $text{p-lim}(x_n otimes y_n) = text{p-lim}(x_n) otimes text{p-lim}(y_n)$ (where $otimes$ is the "and" operator)
- bounded: $text{p-inf}(x_n) leq text{p-lim}(x_n) leq text{p-sup}(x_n)$
- non-principality: deletion of a finite number of elements from $x_n$ should not change its $text{p-lim}$.
We do not know anything else about $text{p-lim}$, so a combination of these facts must be used to prove the required.
Note that $x_n$ will never be finite, since it is an indicator sequence which runs over all the natural numbers, hence $y_n$ will never be finite either. So non-principality is unlikely to be a useful tool. The supremum and infimum of every sequence of binary numbers which is not $1, 1, 1, 1, 1, 1 dots$ or $0, 0, 0, 0, 0, 0, dots$ is trivially $0$ and $1$ respectively, so it is unlikely that boundedness can be used as a tool.
Let us try to use the fact that $text{p-lim}$ is a homomorphism. Note that $y_n oplus x_n = y_n$, since $x_n subset y_n$ (in other words, $x_k = 1 Rightarrow y_k = 1$, while $x_k = 0 Rightarrow (y_k = 1 vee y_k = 0)$). Then:
begin{align*}
&quad;; x_n oplus y_n = y_n \
&Rightarrow text{p-lim}(x_n oplus y_n) = text{p-lim}(y_n) \
&Rightarrow text{p-lim}(x_n) oplus text{p-lim}(y_n) = text{p-lim}(y_n) \
&Rightarrow 1 oplus text{p-lim}(y_n) = text{p-lim}(y_n) \
&{text{1 is the zero of $oplus$}} \
&Rightarrow 1 = text{p-lim}(y_n)
end{align*}
Is this proof correct? The following assumptions, I think, will need confirmation:
- On the space of binary numbers, is it okay to assume that $oplus, otimes$ are the analogues of $+, times$, for the homomorphism property? I think it should be okay, because: 1) it makes things work out nicely (I think I can use similar logic to show that some of the other ultrafilter properties hold for the $text{p-lim}$ over the binary numbers), and connects nicely to classical "predicate calculus" stuff; 2) $oplus, otimes$ can also be thought of as "addition/multiplication $mathbb{Z}/2$")
limits proof-verification filters
asked Jan 27 at 6:47
user89user89
5831647
$endgroup$
I am working through this post from Terry Tao's blog.
Problem: Prove that if a binary (i.e. consisting of $0$ and $1$, or $top$ and $bot$) sequence $x_n$ is such that $text{p-lim}(x_n) = 1$, then any sequence $y_n$ containing $x_n$ has $text{p-lim}(y_n) = text{p-lim}(x_n) = 1$.
My attempt at a proof. We know the following facts regarding $text{p-lim}$:
- homomorphic: $text{p-lim}(1) = 1$, $text{p-lim}(Cx_n) = Ctext{p-lim}(x_n)$ (for some $C in {0, 1}$), $text{p-lim}(x_n oplus y_n) = text{p-lim}(x_n) oplus text{p-lim}(y_n)$ (where $oplus$ is the "or" operator), and $text{p-lim}(x_n otimes y_n) = text{p-lim}(x_n) otimes text{p-lim}(y_n)$ (where $otimes$ is the "and" operator)
- bounded: $text{p-inf}(x_n) leq text{p-lim}(x_n) leq text{p-sup}(x_n)$
- non-principality: deletion of a finite number of elements from $x_n$ should not change its $text{p-lim}$.
We do not know anything else about $text{p-lim}$, so a combination of these facts must be used to prove the required.
Note that $x_n$ will never be finite, since it is an indicator sequence which runs over all the natural numbers, hence $y_n$ will never be finite either. So non-principality is unlikely to be a useful tool. The supremum and infimum of every sequence of binary numbers which is not $1, 1, 1, 1, 1, 1 dots$ or $0, 0, 0, 0, 0, 0, dots$ is trivially $0$ and $1$ respectively, so it is unlikely that boundedness can be used as a tool.
Let us try to use the fact that $text{p-lim}$ is a homomorphism. Note that $y_n oplus x_n = y_n$, since $x_n subset y_n$ (in other words, $x_k = 1 Rightarrow y_k = 1$, while $x_k = 0 Rightarrow (y_k = 1 vee y_k = 0)$). Then:
begin{align*}
&quad;; x_n oplus y_n = y_n \
&Rightarrow text{p-lim}(x_n oplus y_n) = text{p-lim}(y_n) \
&Rightarrow text{p-lim}(x_n) oplus text{p-lim}(y_n) = text{p-lim}(y_n) \
&Rightarrow 1 oplus text{p-lim}(y_n) = text{p-lim}(y_n) \
&{text{1 is the zero of $oplus$}} \
&Rightarrow 1 = text{p-lim}(y_n)
end{align*}
Is this proof correct? The following assumptions, I think, will need confirmation:
- On the space of binary numbers, is it okay to assume that $oplus, otimes$ are the analogues of $+, times$, for the homomorphism property? I think it should be okay, because: 1) it makes things work out nicely (I think I can use similar logic to show that some of the other ultrafilter properties hold for the $text{p-lim}$ over the binary numbers), and connects nicely to classical "predicate calculus" stuff; 2) $oplus, otimes$ can also be thought of as "addition/multiplication $mathbb{Z}/2$")
limits proof-verification filters
limits proof-verification filters
asked Jan 27 at 6:47
user89user89
5831647
asked Jan 27 at 6:47
user89user89
5831647
edited Jan 31 at 18:46
user89
asked Jan 27 at 6:47
user89user89
5831647
asked Jan 27 at 6:47
user89user89
5831647
asked Jan 27 at 6:47
user89user89
5831647
5831647
$begingroup$
What do you mean by $(y_n)$ "containing" $(x_n)$? Sequences are functions, not sets.. It seems you mean $y_n le x_n$ for all $n$?
$endgroup$
– Henno Brandsma
Jan 30 at 5:34
$begingroup$
It seems that you mean by "$x_n subset y_n$" that $y_k le x_k$ for all $k$; is that true? In the subset interpretation Tao gives the latter means that the subset of $omega$ associated with $y$ is a subset of the one asssociated with $x$, so inclusion the other way around.... You seem confused.
$endgroup$
– Henno Brandsma
Jan 30 at 23:07
$begingroup$
@HennoBrandsma by $x_n$ being contained in $y_n$, I mean: $x_k = 1 Rightarrow y_k = 1$
$endgroup$
– user89
Jan 31 at 18:45
add a comment |
$begingroup$
What do you mean by $(y_n)$ "containing" $(x_n)$? Sequences are functions, not sets.. It seems you mean $y_n le x_n$ for all $n$?
$endgroup$
– Henno Brandsma
Jan 30 at 5:34
$begingroup$
It seems that you mean by "$x_n subset y_n$" that $y_k le x_k$ for all $k$; is that true? In the subset interpretation Tao gives the latter means that the subset of $omega$ associated with $y$ is a subset of the one asssociated with $x$, so inclusion the other way around.... You seem confused.
$endgroup$
– Henno Brandsma
Jan 30 at 23:07
$begingroup$
@HennoBrandsma by $x_n$ being contained in $y_n$, I mean: $x_k = 1 Rightarrow y_k = 1$
$endgroup$
– user89
Jan 31 at 18:45
$begingroup$
What do you mean by $(y_n)$ "containing" $(x_n)$? Sequences are functions, not sets.. It seems you mean $y_n le x_n$ for all $n$?
$endgroup$
– Henno Brandsma
Jan 30 at 5:34
$begingroup$
What do you mean by $(y_n)$ "containing" $(x_n)$? Sequences are functions, not sets.. It seems you mean $y_n le x_n$ for all $n$?
$endgroup$
– Henno Brandsma
Jan 30 at 5:34
$begingroup$
It seems that you mean by "$x_n subset y_n$" that $y_k le x_k$ for all $k$; is that true? In the subset interpretation Tao gives the latter means that the subset of $omega$ associated with $y$ is a subset of the one asssociated with $x$, so inclusion the other way around.... You seem confused.
$endgroup$
– Henno Brandsma
Jan 30 at 23:07
$begingroup$
It seems that you mean by "$x_n subset y_n$" that $y_k le x_k$ for all $k$; is that true? In the subset interpretation Tao gives the latter means that the subset of $omega$ associated with $y$ is a subset of the one asssociated with $x$, so inclusion the other way around.... You seem confused.
$endgroup$
– Henno Brandsma
Jan 30 at 23:07
$begingroup$
@HennoBrandsma by $x_n$ being contained in $y_n$, I mean: $x_k = 1 Rightarrow y_k = 1$
$endgroup$
– user89
Jan 31 at 18:45
$begingroup$
@HennoBrandsma by $x_n$ being contained in $y_n$, I mean: $x_k = 1 Rightarrow y_k = 1$
$endgroup$
– user89
Jan 31 at 18:45
add a comment |
1 Answer
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You can go about it in two ways, which come down to the same thing. If we have a $p$-lim for sequences with values $0$ and $1$ we can identify a sequence $x=(x_n)$ with a subset $A(x) subseteq omega$ where $n in A(x)$ iff $x_n =1$ (using characteristic functions, in other words, or indicator sequences as Tao calls them). As $p$-$lim(x_n) in {0,1}$ as well, as he also shows, we have that a $p$-lim defines a non-principal ultrafilter $mathcal{F}$ on the powerset of $omega$: $A in mathcal{F}$ iff the indicator sequence $chi_A$ has $p$-$lim(chi_A)=1$. The monotonicity property you mention follows immediately from Tao's monotonicity principle: if two Boolean sequences $x,y$ obey: $forall k: x_k =1 implies y_k=1$, we see that $A(x) subseteq A(y)$ and as $p$-$lim(x)=1$ so $A(x) in mathcal{F}$ we have $A(y) in mathcal{F}$ by monotonicity of ultrafilters, and so $p$-$lim(y)=1$ as well.
To see the monotonicity of the ultrafilter (which Tao doesn't show) is to use the homomorphism properties. "$x$ is contained in $y$" (or $A(x) subseteq A(y)$) in your definition is equivalent to the identity $1-x+xy=1$ or $xy-x=x(y-1)=0$ (i.e. this holds for all $k$, as reals). This identity is preserved by $p$-$lim$ by its homomorphism property:
$$ptext{ -}lim(x)left(1-ptext{ -}lim(y)right) = 0$$
from which it follows that $ptext{ -}lim(x)=1$ implies $ptext{-}lim(y)=1$ which is what we needed.
answered Jan 31 at 23:18
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
$begingroup$
Can you comment on whether my proof approach was okay or not?
$endgroup$
– user89
Feb 1 at 3:47
$begingroup$
@user89 You have to prove separately that $oplus$ is preserved. I do think it's true, as it can be defined in terms of $+,cdot$ on the reals, when restricted to Boolean sequences, so do that. And you need to justify too that $x$ contained in $y$ iff $x oplus y = y$. I think the equivalence is not true.
$endgroup$
– Henno Brandsma
Feb 1 at 5:28
add a comment |
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$begingroup$
You can go about it in two ways, which come down to the same thing. If we have a $p$-lim for sequences with values $0$ and $1$ we can identify a sequence $x=(x_n)$ with a subset $A(x) subseteq omega$ where $n in A(x)$ iff $x_n =1$ (using characteristic functions, in other words, or indicator sequences as Tao calls them). As $p$-$lim(x_n) in {0,1}$ as well, as he also shows, we have that a $p$-lim defines a non-principal ultrafilter $mathcal{F}$ on the powerset of $omega$: $A in mathcal{F}$ iff the indicator sequence $chi_A$ has $p$-$lim(chi_A)=1$. The monotonicity property you mention follows immediately from Tao's monotonicity principle: if two Boolean sequences $x,y$ obey: $forall k: x_k =1 implies y_k=1$, we see that $A(x) subseteq A(y)$ and as $p$-$lim(x)=1$ so $A(x) in mathcal{F}$ we have $A(y) in mathcal{F}$ by monotonicity of ultrafilters, and so $p$-$lim(y)=1$ as well.
To see the monotonicity of the ultrafilter (which Tao doesn't show) is to use the homomorphism properties. "$x$ is contained in $y$" (or $A(x) subseteq A(y)$) in your definition is equivalent to the identity $1-x+xy=1$ or $xy-x=x(y-1)=0$ (i.e. this holds for all $k$, as reals). This identity is preserved by $p$-$lim$ by its homomorphism property:
$$ptext{ -}lim(x)left(1-ptext{ -}lim(y)right) = 0$$
from which it follows that $ptext{ -}lim(x)=1$ implies $ptext{-}lim(y)=1$ which is what we needed.
answered Jan 31 at 23:18
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
$begingroup$
Can you comment on whether my proof approach was okay or not?
$endgroup$
– user89
Feb 1 at 3:47
$begingroup$
@user89 You have to prove separately that $oplus$ is preserved. I do think it's true, as it can be defined in terms of $+,cdot$ on the reals, when restricted to Boolean sequences, so do that. And you need to justify too that $x$ contained in $y$ iff $x oplus y = y$. I think the equivalence is not true.
$endgroup$
– Henno Brandsma
Feb 1 at 5:28
add a comment |
$begingroup$
You can go about it in two ways, which come down to the same thing. If we have a $p$-lim for sequences with values $0$ and $1$ we can identify a sequence $x=(x_n)$ with a subset $A(x) subseteq omega$ where $n in A(x)$ iff $x_n =1$ (using characteristic functions, in other words, or indicator sequences as Tao calls them). As $p$-$lim(x_n) in {0,1}$ as well, as he also shows, we have that a $p$-lim defines a non-principal ultrafilter $mathcal{F}$ on the powerset of $omega$: $A in mathcal{F}$ iff the indicator sequence $chi_A$ has $p$-$lim(chi_A)=1$. The monotonicity property you mention follows immediately from Tao's monotonicity principle: if two Boolean sequences $x,y$ obey: $forall k: x_k =1 implies y_k=1$, we see that $A(x) subseteq A(y)$ and as $p$-$lim(x)=1$ so $A(x) in mathcal{F}$ we have $A(y) in mathcal{F}$ by monotonicity of ultrafilters, and so $p$-$lim(y)=1$ as well.
To see the monotonicity of the ultrafilter (which Tao doesn't show) is to use the homomorphism properties. "$x$ is contained in $y$" (or $A(x) subseteq A(y)$) in your definition is equivalent to the identity $1-x+xy=1$ or $xy-x=x(y-1)=0$ (i.e. this holds for all $k$, as reals). This identity is preserved by $p$-$lim$ by its homomorphism property:
$$ptext{ -}lim(x)left(1-ptext{ -}lim(y)right) = 0$$
from which it follows that $ptext{ -}lim(x)=1$ implies $ptext{-}lim(y)=1$ which is what we needed.
answered Jan 31 at 23:18
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
$begingroup$
Can you comment on whether my proof approach was okay or not?
$endgroup$
– user89
Feb 1 at 3:47
$begingroup$
@user89 You have to prove separately that $oplus$ is preserved. I do think it's true, as it can be defined in terms of $+,cdot$ on the reals, when restricted to Boolean sequences, so do that. And you need to justify too that $x$ contained in $y$ iff $x oplus y = y$. I think the equivalence is not true.
$endgroup$
– Henno Brandsma
Feb 1 at 5:28
add a comment |
$begingroup$
You can go about it in two ways, which come down to the same thing. If we have a $p$-lim for sequences with values $0$ and $1$ we can identify a sequence $x=(x_n)$ with a subset $A(x) subseteq omega$ where $n in A(x)$ iff $x_n =1$ (using characteristic functions, in other words, or indicator sequences as Tao calls them). As $p$-$lim(x_n) in {0,1}$ as well, as he also shows, we have that a $p$-lim defines a non-principal ultrafilter $mathcal{F}$ on the powerset of $omega$: $A in mathcal{F}$ iff the indicator sequence $chi_A$ has $p$-$lim(chi_A)=1$. The monotonicity property you mention follows immediately from Tao's monotonicity principle: if two Boolean sequences $x,y$ obey: $forall k: x_k =1 implies y_k=1$, we see that $A(x) subseteq A(y)$ and as $p$-$lim(x)=1$ so $A(x) in mathcal{F}$ we have $A(y) in mathcal{F}$ by monotonicity of ultrafilters, and so $p$-$lim(y)=1$ as well.
To see the monotonicity of the ultrafilter (which Tao doesn't show) is to use the homomorphism properties. "$x$ is contained in $y$" (or $A(x) subseteq A(y)$) in your definition is equivalent to the identity $1-x+xy=1$ or $xy-x=x(y-1)=0$ (i.e. this holds for all $k$, as reals). This identity is preserved by $p$-$lim$ by its homomorphism property:
$$ptext{ -}lim(x)left(1-ptext{ -}lim(y)right) = 0$$
from which it follows that $ptext{ -}lim(x)=1$ implies $ptext{-}lim(y)=1$ which is what we needed.
answered Jan 31 at 23:18
Henno BrandsmaHenno Brandsma
114k348123
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You can go about it in two ways, which come down to the same thing. If we have a $p$-lim for sequences with values $0$ and $1$ we can identify a sequence $x=(x_n)$ with a subset $A(x) subseteq omega$ where $n in A(x)$ iff $x_n =1$ (using characteristic functions, in other words, or indicator sequences as Tao calls them). As $p$-$lim(x_n) in {0,1}$ as well, as he also shows, we have that a $p$-lim defines a non-principal ultrafilter $mathcal{F}$ on the powerset of $omega$: $A in mathcal{F}$ iff the indicator sequence $chi_A$ has $p$-$lim(chi_A)=1$. The monotonicity property you mention follows immediately from Tao's monotonicity principle: if two Boolean sequences $x,y$ obey: $forall k: x_k =1 implies y_k=1$, we see that $A(x) subseteq A(y)$ and as $p$-$lim(x)=1$ so $A(x) in mathcal{F}$ we have $A(y) in mathcal{F}$ by monotonicity of ultrafilters, and so $p$-$lim(y)=1$ as well.
To see the monotonicity of the ultrafilter (which Tao doesn't show) is to use the homomorphism properties. "$x$ is contained in $y$" (or $A(x) subseteq A(y)$) in your definition is equivalent to the identity $1-x+xy=1$ or $xy-x=x(y-1)=0$ (i.e. this holds for all $k$, as reals). This identity is preserved by $p$-$lim$ by its homomorphism property:
$$ptext{ -}lim(x)left(1-ptext{ -}lim(y)right) = 0$$
from which it follows that $ptext{ -}lim(x)=1$ implies $ptext{-}lim(y)=1$ which is what we needed.
answered Jan 31 at 23:18
Henno BrandsmaHenno Brandsma
114k348123
edited Jan 31 at 23:41
answered Jan 31 at 23:18
Henno BrandsmaHenno Brandsma
114k348123
answered Jan 31 at 23:18
Henno BrandsmaHenno Brandsma
114k348123
answered Jan 31 at 23:18
Henno BrandsmaHenno Brandsma
114k348123
114k348123
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Can you comment on whether my proof approach was okay or not?
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– user89
Feb 1 at 3:47
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@user89 You have to prove separately that $oplus$ is preserved. I do think it's true, as it can be defined in terms of $+,cdot$ on the reals, when restricted to Boolean sequences, so do that. And you need to justify too that $x$ contained in $y$ iff $x oplus y = y$. I think the equivalence is not true.
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– Henno Brandsma
Feb 1 at 5:28
add a comment |
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Can you comment on whether my proof approach was okay or not?
$endgroup$
– user89
Feb 1 at 3:47
$begingroup$
@user89 You have to prove separately that $oplus$ is preserved. I do think it's true, as it can be defined in terms of $+,cdot$ on the reals, when restricted to Boolean sequences, so do that. And you need to justify too that $x$ contained in $y$ iff $x oplus y = y$. I think the equivalence is not true.
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– Henno Brandsma
Feb 1 at 5:28
$begingroup$
Can you comment on whether my proof approach was okay or not?
$endgroup$
– user89
Feb 1 at 3:47
$begingroup$
Can you comment on whether my proof approach was okay or not?
$endgroup$
– user89
Feb 1 at 3:47
$begingroup$
@user89 You have to prove separately that $oplus$ is preserved. I do think it's true, as it can be defined in terms of $+,cdot$ on the reals, when restricted to Boolean sequences, so do that. And you need to justify too that $x$ contained in $y$ iff $x oplus y = y$. I think the equivalence is not true.
$endgroup$
– Henno Brandsma
Feb 1 at 5:28
$begingroup$
@user89 You have to prove separately that $oplus$ is preserved. I do think it's true, as it can be defined in terms of $+,cdot$ on the reals, when restricted to Boolean sequences, so do that. And you need to justify too that $x$ contained in $y$ iff $x oplus y = y$. I think the equivalence is not true.
$endgroup$
– Henno Brandsma
Feb 1 at 5:28
add a comment |
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$begingroup$
What do you mean by $(y_n)$ "containing" $(x_n)$? Sequences are functions, not sets.. It seems you mean $y_n le x_n$ for all $n$?
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– Henno Brandsma
Jan 30 at 5:34
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It seems that you mean by "$x_n subset y_n$" that $y_k le x_k$ for all $k$; is that true? In the subset interpretation Tao gives the latter means that the subset of $omega$ associated with $y$ is a subset of the one asssociated with $x$, so inclusion the other way around.... You seem confused.
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– Henno Brandsma
Jan 30 at 23:07
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@HennoBrandsma by $x_n$ being contained in $y_n$, I mean: $x_k = 1 Rightarrow y_k = 1$
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– user89
Jan 31 at 18:45