Mixing
Equivalence on definition of Folner sequence
$begingroup$
$textbf{Definition 1.}$ Let ${F_n}_{n in mathbb{N}}$ be a sequence of finite subsets of $mathbb{Z}^d$. We say that ${F_n}_{n in mathbb{N}}$ is a Folner sequence if for every $v in mathbb{Z}^d$, $lim_{n rightarrow +infty} frac{|(F_n + v) Delta F_n|}{|F_n|} = 0$.
$textbf{Definition 2.}$ Let ${F_n}_{n in mathbb{N}}$ be a sequence of finite subsets of $mathbb{Z}^d$. We say that ${F_n}_{n in mathbb{N}}$ is a Folner sequence if $lim_{n rightarrow +infty} frac{|partial F_n|}{|F_n|} = 0$, where $partial F = {v in mathbb{Z}^d : v notin F, d(v, F) = 1}$.
What I did was the following:
$(1 implies 2)$ First of all, note that as ${F_n}_{n in mathbb{N}}$ is a Folner sequence in the sense of definition $1$, we have that $lim_{n rightarrow +infty} frac{|(F_n pm e_i) Delta F_n|}{|F_n|} = 0$, for all $i in {1, dots, d}$. Thus, $lim_{n rightarrow +infty} frac{|(F_n pm e_i) setminus F_n|}{|F_n|} = 0$, for all $i in {1, dots, d}$.
Now, note that $|partial F_n| leq |(F_n + e_1) setminus F_n| + dots + |(F_n + e_d) setminus F_n| + |(F_n - e_1) setminus F_n)| + dots + |(F_n - e_d) setminus F_n|$. The, by the remark made above, we have that
$lim_{n rightarrow +infty} frac{|partial F_n|}{|F_n|} = 0$.
Now, I have two questions:
1) Is it clear (and correct) the argument above?
2) I'm having some trouble to show that $2 implies 1$. Could someone help me?
Thanks in advance!
sequences-and-series group-theory finite-groups amenability
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 16:30
Pedro FilipiniPedro Filipini
233
$endgroup$
add a comment |
$begingroup$
$textbf{Definition 1.}$ Let ${F_n}_{n in mathbb{N}}$ be a sequence of finite subsets of $mathbb{Z}^d$. We say that ${F_n}_{n in mathbb{N}}$ is a Folner sequence if for every $v in mathbb{Z}^d$, $lim_{n rightarrow +infty} frac{|(F_n + v) Delta F_n|}{|F_n|} = 0$.
$textbf{Definition 2.}$ Let ${F_n}_{n in mathbb{N}}$ be a sequence of finite subsets of $mathbb{Z}^d$. We say that ${F_n}_{n in mathbb{N}}$ is a Folner sequence if $lim_{n rightarrow +infty} frac{|partial F_n|}{|F_n|} = 0$, where $partial F = {v in mathbb{Z}^d : v notin F, d(v, F) = 1}$.
What I did was the following:
$(1 implies 2)$ First of all, note that as ${F_n}_{n in mathbb{N}}$ is a Folner sequence in the sense of definition $1$, we have that $lim_{n rightarrow +infty} frac{|(F_n pm e_i) Delta F_n|}{|F_n|} = 0$, for all $i in {1, dots, d}$. Thus, $lim_{n rightarrow +infty} frac{|(F_n pm e_i) setminus F_n|}{|F_n|} = 0$, for all $i in {1, dots, d}$.
Now, note that $|partial F_n| leq |(F_n + e_1) setminus F_n| + dots + |(F_n + e_d) setminus F_n| + |(F_n - e_1) setminus F_n)| + dots + |(F_n - e_d) setminus F_n|$. The, by the remark made above, we have that
$lim_{n rightarrow +infty} frac{|partial F_n|}{|F_n|} = 0$.
Now, I have two questions:
1) Is it clear (and correct) the argument above?
2) I'm having some trouble to show that $2 implies 1$. Could someone help me?
Thanks in advance!
sequences-and-series group-theory finite-groups amenability
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 16:30
Pedro FilipiniPedro Filipini
233
$endgroup$
$begingroup$
Note that $d(v,F)=1$ implies that $vnot in F$ in definition 2.
$endgroup$
– Yanko
Jan 27 at 16:50
1
$begingroup$
1) it looks good to me. 2) you need to apply the definition multiple times. Each time move one step until you move "$v$" steps. (i.e. write $v=sum_{i=1}^d a_i e_i$ then move $a_i$ times in the $e_i$ direction for all $i$ each time the limit is zero so the sum of all limits will be zero).
$endgroup$
– Yanko
Jan 27 at 16:52
add a comment |
$begingroup$
$textbf{Definition 1.}$ Let ${F_n}_{n in mathbb{N}}$ be a sequence of finite subsets of $mathbb{Z}^d$. We say that ${F_n}_{n in mathbb{N}}$ is a Folner sequence if for every $v in mathbb{Z}^d$, $lim_{n rightarrow +infty} frac{|(F_n + v) Delta F_n|}{|F_n|} = 0$.
$textbf{Definition 2.}$ Let ${F_n}_{n in mathbb{N}}$ be a sequence of finite subsets of $mathbb{Z}^d$. We say that ${F_n}_{n in mathbb{N}}$ is a Folner sequence if $lim_{n rightarrow +infty} frac{|partial F_n|}{|F_n|} = 0$, where $partial F = {v in mathbb{Z}^d : v notin F, d(v, F) = 1}$.
What I did was the following:
$(1 implies 2)$ First of all, note that as ${F_n}_{n in mathbb{N}}$ is a Folner sequence in the sense of definition $1$, we have that $lim_{n rightarrow +infty} frac{|(F_n pm e_i) Delta F_n|}{|F_n|} = 0$, for all $i in {1, dots, d}$. Thus, $lim_{n rightarrow +infty} frac{|(F_n pm e_i) setminus F_n|}{|F_n|} = 0$, for all $i in {1, dots, d}$.
Now, note that $|partial F_n| leq |(F_n + e_1) setminus F_n| + dots + |(F_n + e_d) setminus F_n| + |(F_n - e_1) setminus F_n)| + dots + |(F_n - e_d) setminus F_n|$. The, by the remark made above, we have that
$lim_{n rightarrow +infty} frac{|partial F_n|}{|F_n|} = 0$.
Now, I have two questions:
1) Is it clear (and correct) the argument above?
2) I'm having some trouble to show that $2 implies 1$. Could someone help me?
Thanks in advance!
sequences-and-series group-theory finite-groups amenability
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 16:30
Pedro FilipiniPedro Filipini
233
$endgroup$
$textbf{Definition 1.}$ Let ${F_n}_{n in mathbb{N}}$ be a sequence of finite subsets of $mathbb{Z}^d$. We say that ${F_n}_{n in mathbb{N}}$ is a Folner sequence if for every $v in mathbb{Z}^d$, $lim_{n rightarrow +infty} frac{|(F_n + v) Delta F_n|}{|F_n|} = 0$.
$textbf{Definition 2.}$ Let ${F_n}_{n in mathbb{N}}$ be a sequence of finite subsets of $mathbb{Z}^d$. We say that ${F_n}_{n in mathbb{N}}$ is a Folner sequence if $lim_{n rightarrow +infty} frac{|partial F_n|}{|F_n|} = 0$, where $partial F = {v in mathbb{Z}^d : v notin F, d(v, F) = 1}$.
What I did was the following:
$(1 implies 2)$ First of all, note that as ${F_n}_{n in mathbb{N}}$ is a Folner sequence in the sense of definition $1$, we have that $lim_{n rightarrow +infty} frac{|(F_n pm e_i) Delta F_n|}{|F_n|} = 0$, for all $i in {1, dots, d}$. Thus, $lim_{n rightarrow +infty} frac{|(F_n pm e_i) setminus F_n|}{|F_n|} = 0$, for all $i in {1, dots, d}$.
Now, note that $|partial F_n| leq |(F_n + e_1) setminus F_n| + dots + |(F_n + e_d) setminus F_n| + |(F_n - e_1) setminus F_n)| + dots + |(F_n - e_d) setminus F_n|$. The, by the remark made above, we have that
$lim_{n rightarrow +infty} frac{|partial F_n|}{|F_n|} = 0$.
Now, I have two questions:
1) Is it clear (and correct) the argument above?
2) I'm having some trouble to show that $2 implies 1$. Could someone help me?
Thanks in advance!
sequences-and-series group-theory finite-groups amenability
sequences-and-series group-theory finite-groups amenability
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 16:30
Pedro FilipiniPedro Filipini
233
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
asked Jan 27 at 16:30
Pedro FilipiniPedro Filipini
233
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
edited Jan 27 at 23:25
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Jan 27 at 16:30
Pedro FilipiniPedro Filipini
233
asked Jan 27 at 16:30
Pedro FilipiniPedro Filipini
233
asked Jan 27 at 16:30
Pedro FilipiniPedro Filipini
233
233
$begingroup$
Note that $d(v,F)=1$ implies that $vnot in F$ in definition 2.
$endgroup$
– Yanko
Jan 27 at 16:50
1
$begingroup$
1) it looks good to me. 2) you need to apply the definition multiple times. Each time move one step until you move "$v$" steps. (i.e. write $v=sum_{i=1}^d a_i e_i$ then move $a_i$ times in the $e_i$ direction for all $i$ each time the limit is zero so the sum of all limits will be zero).
$endgroup$
– Yanko
Jan 27 at 16:52
add a comment |
$begingroup$
Note that $d(v,F)=1$ implies that $vnot in F$ in definition 2.
$endgroup$
– Yanko
Jan 27 at 16:50
1
$begingroup$
1) it looks good to me. 2) you need to apply the definition multiple times. Each time move one step until you move "$v$" steps. (i.e. write $v=sum_{i=1}^d a_i e_i$ then move $a_i$ times in the $e_i$ direction for all $i$ each time the limit is zero so the sum of all limits will be zero).
$endgroup$
– Yanko
Jan 27 at 16:52
$begingroup$
Note that $d(v,F)=1$ implies that $vnot in F$ in definition 2.
$endgroup$
– Yanko
Jan 27 at 16:50
$begingroup$
Note that $d(v,F)=1$ implies that $vnot in F$ in definition 2.
$endgroup$
– Yanko
Jan 27 at 16:50
1
1
$begingroup$
1) it looks good to me. 2) you need to apply the definition multiple times. Each time move one step until you move "$v$" steps. (i.e. write $v=sum_{i=1}^d a_i e_i$ then move $a_i$ times in the $e_i$ direction for all $i$ each time the limit is zero so the sum of all limits will be zero).
$endgroup$
– Yanko
Jan 27 at 16:52
$begingroup$
1) it looks good to me. 2) you need to apply the definition multiple times. Each time move one step until you move "$v$" steps. (i.e. write $v=sum_{i=1}^d a_i e_i$ then move $a_i$ times in the $e_i$ direction for all $i$ each time the limit is zero so the sum of all limits will be zero).
$endgroup$
– Yanko
Jan 27 at 16:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Given $v$, pick a sequence $v_0=0,v_1,ldots, v_m=v$ such that $d(v_i,v_{i+1})=1$.
Now note that a point $xin(F_n+v)mathrel{Delta} F_n$ is either a point $xin F_n$ such that $x-vnotin F_n$ or vice versa. Thus in the sequence $x-v_0,x-v_1,ldots, x-v_m$, we encounter a switch form $in F_n$ to $notin F_n$ or vice versa. Thus to every point in $(F_n+v)mathrel{Delta} F_n$, we can associate an index $iin{0,ldots, m}$, a point in $delta F_n$, and an additional "bit" whether $xin F_n$ or $xin F_n+v$. We conclude
$$|(F_n+v)mathrel{Delta} F_n|le |delta F_n|cdot (m+1)cdot 2.$$
answered Jan 27 at 17:16
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Why does this switch form exist? Why couldn't it happen that it switches an odd number of times?
$endgroup$
– Pedro Filipini
Jan 27 at 18:44
add a comment |
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$begingroup$
Given $v$, pick a sequence $v_0=0,v_1,ldots, v_m=v$ such that $d(v_i,v_{i+1})=1$.
Now note that a point $xin(F_n+v)mathrel{Delta} F_n$ is either a point $xin F_n$ such that $x-vnotin F_n$ or vice versa. Thus in the sequence $x-v_0,x-v_1,ldots, x-v_m$, we encounter a switch form $in F_n$ to $notin F_n$ or vice versa. Thus to every point in $(F_n+v)mathrel{Delta} F_n$, we can associate an index $iin{0,ldots, m}$, a point in $delta F_n$, and an additional "bit" whether $xin F_n$ or $xin F_n+v$. We conclude
$$|(F_n+v)mathrel{Delta} F_n|le |delta F_n|cdot (m+1)cdot 2.$$
answered Jan 27 at 17:16
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Why does this switch form exist? Why couldn't it happen that it switches an odd number of times?
$endgroup$
– Pedro Filipini
Jan 27 at 18:44
add a comment |
$begingroup$
Given $v$, pick a sequence $v_0=0,v_1,ldots, v_m=v$ such that $d(v_i,v_{i+1})=1$.
Now note that a point $xin(F_n+v)mathrel{Delta} F_n$ is either a point $xin F_n$ such that $x-vnotin F_n$ or vice versa. Thus in the sequence $x-v_0,x-v_1,ldots, x-v_m$, we encounter a switch form $in F_n$ to $notin F_n$ or vice versa. Thus to every point in $(F_n+v)mathrel{Delta} F_n$, we can associate an index $iin{0,ldots, m}$, a point in $delta F_n$, and an additional "bit" whether $xin F_n$ or $xin F_n+v$. We conclude
$$|(F_n+v)mathrel{Delta} F_n|le |delta F_n|cdot (m+1)cdot 2.$$
answered Jan 27 at 17:16
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
$begingroup$
Why does this switch form exist? Why couldn't it happen that it switches an odd number of times?
$endgroup$
– Pedro Filipini
Jan 27 at 18:44
add a comment |
$begingroup$
Given $v$, pick a sequence $v_0=0,v_1,ldots, v_m=v$ such that $d(v_i,v_{i+1})=1$.
Now note that a point $xin(F_n+v)mathrel{Delta} F_n$ is either a point $xin F_n$ such that $x-vnotin F_n$ or vice versa. Thus in the sequence $x-v_0,x-v_1,ldots, x-v_m$, we encounter a switch form $in F_n$ to $notin F_n$ or vice versa. Thus to every point in $(F_n+v)mathrel{Delta} F_n$, we can associate an index $iin{0,ldots, m}$, a point in $delta F_n$, and an additional "bit" whether $xin F_n$ or $xin F_n+v$. We conclude
$$|(F_n+v)mathrel{Delta} F_n|le |delta F_n|cdot (m+1)cdot 2.$$
answered Jan 27 at 17:16
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
Given $v$, pick a sequence $v_0=0,v_1,ldots, v_m=v$ such that $d(v_i,v_{i+1})=1$.
Now note that a point $xin(F_n+v)mathrel{Delta} F_n$ is either a point $xin F_n$ such that $x-vnotin F_n$ or vice versa. Thus in the sequence $x-v_0,x-v_1,ldots, x-v_m$, we encounter a switch form $in F_n$ to $notin F_n$ or vice versa. Thus to every point in $(F_n+v)mathrel{Delta} F_n$, we can associate an index $iin{0,ldots, m}$, a point in $delta F_n$, and an additional "bit" whether $xin F_n$ or $xin F_n+v$. We conclude
$$|(F_n+v)mathrel{Delta} F_n|le |delta F_n|cdot (m+1)cdot 2.$$
answered Jan 27 at 17:16
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 17:16
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 17:16
Hagen von EitzenHagen von Eitzen
283k23272507
answered Jan 27 at 17:16
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
$begingroup$
Why does this switch form exist? Why couldn't it happen that it switches an odd number of times?
$endgroup$
– Pedro Filipini
Jan 27 at 18:44
add a comment |
$begingroup$
Why does this switch form exist? Why couldn't it happen that it switches an odd number of times?
$endgroup$
– Pedro Filipini
Jan 27 at 18:44
$begingroup$
Why does this switch form exist? Why couldn't it happen that it switches an odd number of times?
$endgroup$
– Pedro Filipini
Jan 27 at 18:44
$begingroup$
Why does this switch form exist? Why couldn't it happen that it switches an odd number of times?
$endgroup$
– Pedro Filipini
Jan 27 at 18:44
add a comment |
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$begingroup$
Note that $d(v,F)=1$ implies that $vnot in F$ in definition 2.
$endgroup$
– Yanko
Jan 27 at 16:50
1
$begingroup$
1) it looks good to me. 2) you need to apply the definition multiple times. Each time move one step until you move "$v$" steps. (i.e. write $v=sum_{i=1}^d a_i e_i$ then move $a_i$ times in the $e_i$ direction for all $i$ each time the limit is zero so the sum of all limits will be zero).
$endgroup$
– Yanko
Jan 27 at 16:52