symmetry in natural numbers Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Grouping natural numbers into arithmetic progressionSymmetry groups of discrete functionsIs there a “natural” / “categorical” definition of the “parity” of a permutation?Demonstrate the isomorphism $textPerm(X) to S_n$ where $X$ is a finite set.Determining the symmetry group of an infinite horizontal line.Permutations on the set of natural numbers.Lexicographic Rank of First $n$ Natural Numbers PermutationDoes the concept of permutation make sense for a set indexed by the real numbers?Rotational symmetry and translation symmetry on a infinitely large plane.The Distribution of Weird Numbers in the Natural Numbers
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symmetry in natural numbers
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Grouping natural numbers into arithmetic progressionSymmetry groups of discrete functionsIs there a “natural” / “categorical” definition of the “parity” of a permutation?Demonstrate the isomorphism $textPerm(X) to S_n$ where $X$ is a finite set.Determining the symmetry group of an infinite horizontal line.Permutations on the set of natural numbers.Lexicographic Rank of First $n$ Natural Numbers PermutationDoes the concept of permutation make sense for a set indexed by the real numbers?Rotational symmetry and translation symmetry on a infinitely large plane.The Distribution of Weird Numbers in the Natural Numbers
$begingroup$
Given a finite set $E$ we can associate a group of symmetries or permutation $S_n$ where $n=|E|$ is the cardinal of $E$. My question is what if the set is infinite or more precisely countable ? Is there any symmetries in the set of natural numbers $mathbbN$ and by extension in countable sets ?
number-theory permutations symmetry
$endgroup$
add a comment |
$begingroup$
Given a finite set $E$ we can associate a group of symmetries or permutation $S_n$ where $n=|E|$ is the cardinal of $E$. My question is what if the set is infinite or more precisely countable ? Is there any symmetries in the set of natural numbers $mathbbN$ and by extension in countable sets ?
number-theory permutations symmetry
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Mar 28 at 12:45
add a comment |
$begingroup$
Given a finite set $E$ we can associate a group of symmetries or permutation $S_n$ where $n=|E|$ is the cardinal of $E$. My question is what if the set is infinite or more precisely countable ? Is there any symmetries in the set of natural numbers $mathbbN$ and by extension in countable sets ?
number-theory permutations symmetry
$endgroup$
Given a finite set $E$ we can associate a group of symmetries or permutation $S_n$ where $n=|E|$ is the cardinal of $E$. My question is what if the set is infinite or more precisely countable ? Is there any symmetries in the set of natural numbers $mathbbN$ and by extension in countable sets ?
number-theory permutations symmetry
number-theory permutations symmetry
edited Mar 27 at 20:40
Bernard
124k742117
124k742117
asked Mar 27 at 20:39
HassanBHassanB
62
62
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Mar 28 at 12:45
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Mar 28 at 12:45
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Mar 28 at 12:45
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Mar 28 at 12:45
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The objective of my question was trying to exhibit analytically some permutations of the set of natural numbers $mathbbN$. Although there is no consensus on what is a permutation of a countable set as @lulu had mentioned. The following is based on the discussion and particularly on the idea given by @TheSilverDoe. He (She) is the one who gave the following example of a bijection $f: mathbbN^* longrightarrow mathbbN^*$ defined by:
$$f(n)=begincases
n+1 & mboxif $n$ is odd \
n-1 & mboxif $n$ is even
endcases$$
To which I will be referring as $f_1$. From there I defined $f_2$ and $f_3$ as follows:
$$f_2(n)=begincases
n+2 & mboxif & nequiv 1 mod 3 \
n & mboxif & nequiv 2 mod 3 \
n-1 & mboxif & nequiv 0 mod 3
endcases$$
And,
$$f_3(n)=begincases
n+3 & mboxif & nequiv 1 mod 4 \
n+1 & mboxif & nequiv 2 mod 4 \
n-1 & mboxif & nequiv 3 mod 4 \
n-3 & mboxif & nequiv 0 mod 4
endcases$$
One can see that we can define a sequence of bijections, symmetries or "permutations" $f_k : mathbbN^* longrightarrow mathbbN^*$ where the pattern is now clear.
$$f_k(n)=begincasesbegincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack2 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack2+1,ldots,k
endcases & mboxif $k$ is even \
begincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack-12 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack-12+1,ldots,k
endcases & mboxif $k$ is odd
endcases$$
One can notice that if $k$ is odd $f_k$ has no fixed points and if it is even, then $f_k$ has infinitely many fixed points. A closer look at the above definition of $f_k$, shows that the set $mathbbN^*$ was partitioned into a union of disjoints sets. Namely $mathbbN^*=cup_m in mathbbN^*E_m,k$, where $E_m,k=m(k+1)-k,m(k+1)-k+1,ldots,m(k+1)$ with the same cardinality of each set, $|E_m,k|=k+1$ for any index $m$. And the restriction of the $f_k$ to $E_m,k$ noted $f_k_$ is just a flip of that set, which is one permutation. As examples written in Cauchy's two line representation
$$f_4_=
beginpmatrix
6 & 7 & 8 & 9 & 10 \
10 & 9 & 8 & 7 & 6
endpmatrix
qquad
f_5_=
beginpmatrix
13 & 14 & 15 & 16 & 17 & 18 \
18 & 17 & 16 & 15 & 14 & 13
endpmatrix$$
The above can be generalized to different partition of the set of natural numbers where the sets of that partition do not have the same cardinality. This suggest that we actually can make sense for a permutation of an infinite countable set. Therefore I drop the following definition.
Definition (Permutation in countable set):
Let $E$ be a countable set, and $(E_n)_n=1,ldots , infty$ is a disjoint partition of E composed by finite sets that is $$ E=cup_n in mathbbNE_n qquad mbox, qquad cap_n in mathbbNE_n = emptyset qquad mboxand qquad |E_n| < infty$$
A bijection $sigma : E longrightarrow E$ is a permuation of $E$ if and only if the following two conditions are satisfied $$ sigma_(E_n)=E_n qquad mboxand qquad sigma_ in S_E_n$$ where $S_E_n$ is the group of permutations of order $|E_n|$.
Let $p$ denote one partition of a countable set $E$ satisfying the conditions of the above definition, the set of permutations $sigma$ as defined which we denote $Sigma_p(E)$ with the composition operation is a group.
Thank you.
$endgroup$
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
The objective of my question was trying to exhibit analytically some permutations of the set of natural numbers $mathbbN$. Although there is no consensus on what is a permutation of a countable set as @lulu had mentioned. The following is based on the discussion and particularly on the idea given by @TheSilverDoe. He (She) is the one who gave the following example of a bijection $f: mathbbN^* longrightarrow mathbbN^*$ defined by:
$$f(n)=begincases
n+1 & mboxif $n$ is odd \
n-1 & mboxif $n$ is even
endcases$$
To which I will be referring as $f_1$. From there I defined $f_2$ and $f_3$ as follows:
$$f_2(n)=begincases
n+2 & mboxif & nequiv 1 mod 3 \
n & mboxif & nequiv 2 mod 3 \
n-1 & mboxif & nequiv 0 mod 3
endcases$$
And,
$$f_3(n)=begincases
n+3 & mboxif & nequiv 1 mod 4 \
n+1 & mboxif & nequiv 2 mod 4 \
n-1 & mboxif & nequiv 3 mod 4 \
n-3 & mboxif & nequiv 0 mod 4
endcases$$
One can see that we can define a sequence of bijections, symmetries or "permutations" $f_k : mathbbN^* longrightarrow mathbbN^*$ where the pattern is now clear.
$$f_k(n)=begincasesbegincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack2 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack2+1,ldots,k
endcases & mboxif $k$ is even \
begincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack-12 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack-12+1,ldots,k
endcases & mboxif $k$ is odd
endcases$$
One can notice that if $k$ is odd $f_k$ has no fixed points and if it is even, then $f_k$ has infinitely many fixed points. A closer look at the above definition of $f_k$, shows that the set $mathbbN^*$ was partitioned into a union of disjoints sets. Namely $mathbbN^*=cup_m in mathbbN^*E_m,k$, where $E_m,k=m(k+1)-k,m(k+1)-k+1,ldots,m(k+1)$ with the same cardinality of each set, $|E_m,k|=k+1$ for any index $m$. And the restriction of the $f_k$ to $E_m,k$ noted $f_k_$ is just a flip of that set, which is one permutation. As examples written in Cauchy's two line representation
$$f_4_=
beginpmatrix
6 & 7 & 8 & 9 & 10 \
10 & 9 & 8 & 7 & 6
endpmatrix
qquad
f_5_=
beginpmatrix
13 & 14 & 15 & 16 & 17 & 18 \
18 & 17 & 16 & 15 & 14 & 13
endpmatrix$$
The above can be generalized to different partition of the set of natural numbers where the sets of that partition do not have the same cardinality. This suggest that we actually can make sense for a permutation of an infinite countable set. Therefore I drop the following definition.
Definition (Permutation in countable set):
Let $E$ be a countable set, and $(E_n)_n=1,ldots , infty$ is a disjoint partition of E composed by finite sets that is $$ E=cup_n in mathbbNE_n qquad mbox, qquad cap_n in mathbbNE_n = emptyset qquad mboxand qquad |E_n| < infty$$
A bijection $sigma : E longrightarrow E$ is a permuation of $E$ if and only if the following two conditions are satisfied $$ sigma_(E_n)=E_n qquad mboxand qquad sigma_ in S_E_n$$ where $S_E_n$ is the group of permutations of order $|E_n|$.
Let $p$ denote one partition of a countable set $E$ satisfying the conditions of the above definition, the set of permutations $sigma$ as defined which we denote $Sigma_p(E)$ with the composition operation is a group.
Thank you.
$endgroup$
add a comment |
$begingroup$
The objective of my question was trying to exhibit analytically some permutations of the set of natural numbers $mathbbN$. Although there is no consensus on what is a permutation of a countable set as @lulu had mentioned. The following is based on the discussion and particularly on the idea given by @TheSilverDoe. He (She) is the one who gave the following example of a bijection $f: mathbbN^* longrightarrow mathbbN^*$ defined by:
$$f(n)=begincases
n+1 & mboxif $n$ is odd \
n-1 & mboxif $n$ is even
endcases$$
To which I will be referring as $f_1$. From there I defined $f_2$ and $f_3$ as follows:
$$f_2(n)=begincases
n+2 & mboxif & nequiv 1 mod 3 \
n & mboxif & nequiv 2 mod 3 \
n-1 & mboxif & nequiv 0 mod 3
endcases$$
And,
$$f_3(n)=begincases
n+3 & mboxif & nequiv 1 mod 4 \
n+1 & mboxif & nequiv 2 mod 4 \
n-1 & mboxif & nequiv 3 mod 4 \
n-3 & mboxif & nequiv 0 mod 4
endcases$$
One can see that we can define a sequence of bijections, symmetries or "permutations" $f_k : mathbbN^* longrightarrow mathbbN^*$ where the pattern is now clear.
$$f_k(n)=begincasesbegincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack2 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack2+1,ldots,k
endcases & mboxif $k$ is even \
begincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack-12 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack-12+1,ldots,k
endcases & mboxif $k$ is odd
endcases$$
One can notice that if $k$ is odd $f_k$ has no fixed points and if it is even, then $f_k$ has infinitely many fixed points. A closer look at the above definition of $f_k$, shows that the set $mathbbN^*$ was partitioned into a union of disjoints sets. Namely $mathbbN^*=cup_m in mathbbN^*E_m,k$, where $E_m,k=m(k+1)-k,m(k+1)-k+1,ldots,m(k+1)$ with the same cardinality of each set, $|E_m,k|=k+1$ for any index $m$. And the restriction of the $f_k$ to $E_m,k$ noted $f_k_$ is just a flip of that set, which is one permutation. As examples written in Cauchy's two line representation
$$f_4_=
beginpmatrix
6 & 7 & 8 & 9 & 10 \
10 & 9 & 8 & 7 & 6
endpmatrix
qquad
f_5_=
beginpmatrix
13 & 14 & 15 & 16 & 17 & 18 \
18 & 17 & 16 & 15 & 14 & 13
endpmatrix$$
The above can be generalized to different partition of the set of natural numbers where the sets of that partition do not have the same cardinality. This suggest that we actually can make sense for a permutation of an infinite countable set. Therefore I drop the following definition.
Definition (Permutation in countable set):
Let $E$ be a countable set, and $(E_n)_n=1,ldots , infty$ is a disjoint partition of E composed by finite sets that is $$ E=cup_n in mathbbNE_n qquad mbox, qquad cap_n in mathbbNE_n = emptyset qquad mboxand qquad |E_n| < infty$$
A bijection $sigma : E longrightarrow E$ is a permuation of $E$ if and only if the following two conditions are satisfied $$ sigma_(E_n)=E_n qquad mboxand qquad sigma_ in S_E_n$$ where $S_E_n$ is the group of permutations of order $|E_n|$.
Let $p$ denote one partition of a countable set $E$ satisfying the conditions of the above definition, the set of permutations $sigma$ as defined which we denote $Sigma_p(E)$ with the composition operation is a group.
Thank you.
$endgroup$
add a comment |
$begingroup$
The objective of my question was trying to exhibit analytically some permutations of the set of natural numbers $mathbbN$. Although there is no consensus on what is a permutation of a countable set as @lulu had mentioned. The following is based on the discussion and particularly on the idea given by @TheSilverDoe. He (She) is the one who gave the following example of a bijection $f: mathbbN^* longrightarrow mathbbN^*$ defined by:
$$f(n)=begincases
n+1 & mboxif $n$ is odd \
n-1 & mboxif $n$ is even
endcases$$
To which I will be referring as $f_1$. From there I defined $f_2$ and $f_3$ as follows:
$$f_2(n)=begincases
n+2 & mboxif & nequiv 1 mod 3 \
n & mboxif & nequiv 2 mod 3 \
n-1 & mboxif & nequiv 0 mod 3
endcases$$
And,
$$f_3(n)=begincases
n+3 & mboxif & nequiv 1 mod 4 \
n+1 & mboxif & nequiv 2 mod 4 \
n-1 & mboxif & nequiv 3 mod 4 \
n-3 & mboxif & nequiv 0 mod 4
endcases$$
One can see that we can define a sequence of bijections, symmetries or "permutations" $f_k : mathbbN^* longrightarrow mathbbN^*$ where the pattern is now clear.
$$f_k(n)=begincasesbegincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack2 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack2+1,ldots,k
endcases & mboxif $k$ is even \
begincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack-12 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack-12+1,ldots,k
endcases & mboxif $k$ is odd
endcases$$
One can notice that if $k$ is odd $f_k$ has no fixed points and if it is even, then $f_k$ has infinitely many fixed points. A closer look at the above definition of $f_k$, shows that the set $mathbbN^*$ was partitioned into a union of disjoints sets. Namely $mathbbN^*=cup_m in mathbbN^*E_m,k$, where $E_m,k=m(k+1)-k,m(k+1)-k+1,ldots,m(k+1)$ with the same cardinality of each set, $|E_m,k|=k+1$ for any index $m$. And the restriction of the $f_k$ to $E_m,k$ noted $f_k_$ is just a flip of that set, which is one permutation. As examples written in Cauchy's two line representation
$$f_4_=
beginpmatrix
6 & 7 & 8 & 9 & 10 \
10 & 9 & 8 & 7 & 6
endpmatrix
qquad
f_5_=
beginpmatrix
13 & 14 & 15 & 16 & 17 & 18 \
18 & 17 & 16 & 15 & 14 & 13
endpmatrix$$
The above can be generalized to different partition of the set of natural numbers where the sets of that partition do not have the same cardinality. This suggest that we actually can make sense for a permutation of an infinite countable set. Therefore I drop the following definition.
Definition (Permutation in countable set):
Let $E$ be a countable set, and $(E_n)_n=1,ldots , infty$ is a disjoint partition of E composed by finite sets that is $$ E=cup_n in mathbbNE_n qquad mbox, qquad cap_n in mathbbNE_n = emptyset qquad mboxand qquad |E_n| < infty$$
A bijection $sigma : E longrightarrow E$ is a permuation of $E$ if and only if the following two conditions are satisfied $$ sigma_(E_n)=E_n qquad mboxand qquad sigma_ in S_E_n$$ where $S_E_n$ is the group of permutations of order $|E_n|$.
Let $p$ denote one partition of a countable set $E$ satisfying the conditions of the above definition, the set of permutations $sigma$ as defined which we denote $Sigma_p(E)$ with the composition operation is a group.
Thank you.
$endgroup$
The objective of my question was trying to exhibit analytically some permutations of the set of natural numbers $mathbbN$. Although there is no consensus on what is a permutation of a countable set as @lulu had mentioned. The following is based on the discussion and particularly on the idea given by @TheSilverDoe. He (She) is the one who gave the following example of a bijection $f: mathbbN^* longrightarrow mathbbN^*$ defined by:
$$f(n)=begincases
n+1 & mboxif $n$ is odd \
n-1 & mboxif $n$ is even
endcases$$
To which I will be referring as $f_1$. From there I defined $f_2$ and $f_3$ as follows:
$$f_2(n)=begincases
n+2 & mboxif & nequiv 1 mod 3 \
n & mboxif & nequiv 2 mod 3 \
n-1 & mboxif & nequiv 0 mod 3
endcases$$
And,
$$f_3(n)=begincases
n+3 & mboxif & nequiv 1 mod 4 \
n+1 & mboxif & nequiv 2 mod 4 \
n-1 & mboxif & nequiv 3 mod 4 \
n-3 & mboxif & nequiv 0 mod 4
endcases$$
One can see that we can define a sequence of bijections, symmetries or "permutations" $f_k : mathbbN^* longrightarrow mathbbN^*$ where the pattern is now clear.
$$f_k(n)=begincasesbegincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack2 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack2+1,ldots,k
endcases & mboxif $k$ is even \
begincases
n+k-2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=0,1,ldots,frack-12 \
n-k+2i & mboxif & nequiv i+1 mod (k+1) & mboxfor & i=frack-12+1,ldots,k
endcases & mboxif $k$ is odd
endcases$$
One can notice that if $k$ is odd $f_k$ has no fixed points and if it is even, then $f_k$ has infinitely many fixed points. A closer look at the above definition of $f_k$, shows that the set $mathbbN^*$ was partitioned into a union of disjoints sets. Namely $mathbbN^*=cup_m in mathbbN^*E_m,k$, where $E_m,k=m(k+1)-k,m(k+1)-k+1,ldots,m(k+1)$ with the same cardinality of each set, $|E_m,k|=k+1$ for any index $m$. And the restriction of the $f_k$ to $E_m,k$ noted $f_k_$ is just a flip of that set, which is one permutation. As examples written in Cauchy's two line representation
$$f_4_=
beginpmatrix
6 & 7 & 8 & 9 & 10 \
10 & 9 & 8 & 7 & 6
endpmatrix
qquad
f_5_=
beginpmatrix
13 & 14 & 15 & 16 & 17 & 18 \
18 & 17 & 16 & 15 & 14 & 13
endpmatrix$$
The above can be generalized to different partition of the set of natural numbers where the sets of that partition do not have the same cardinality. This suggest that we actually can make sense for a permutation of an infinite countable set. Therefore I drop the following definition.
Definition (Permutation in countable set):
Let $E$ be a countable set, and $(E_n)_n=1,ldots , infty$ is a disjoint partition of E composed by finite sets that is $$ E=cup_n in mathbbNE_n qquad mbox, qquad cap_n in mathbbNE_n = emptyset qquad mboxand qquad |E_n| < infty$$
A bijection $sigma : E longrightarrow E$ is a permuation of $E$ if and only if the following two conditions are satisfied $$ sigma_(E_n)=E_n qquad mboxand qquad sigma_ in S_E_n$$ where $S_E_n$ is the group of permutations of order $|E_n|$.
Let $p$ denote one partition of a countable set $E$ satisfying the conditions of the above definition, the set of permutations $sigma$ as defined which we denote $Sigma_p(E)$ with the composition operation is a group.
Thank you.
answered Mar 29 at 15:41
HassanBHassanB
62
62
add a comment |
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
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$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Mar 28 at 12:45