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How many permutations does a countable set have? [duplicate]

2

0

$begingroup$

This question already has an answer here:

• 
  Is symmetric group on natural numbers countable?
  
  8 answers
  
  

I have just come across the Riemann Rearrangement theorem and, from what I understand, it shows that any real can be written as a permutation of a conditionally convergent series. The problem I have is that permutations of an infinite series are countably infinite, so in theory one could list all possible permutations of the series which would be equivalent to listing all real numbers. But reals are uncountable.

This seems like a contradiction to me, where am I wrong?

permutations real-numbers

share|cite|improve this question

edited Apr 8 at 12:38

Asaf Karagila♦

308k33441775

asked Apr 8 at 10:53

Lorenzo Lorenzo

184

New contributor

Lorenzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

marked as duplicate by Asaf Karagila♦ Apr 8 at 12:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While the motivation for this question is the rearrangement theorem, the title should reflect the actual question instead.
  $endgroup$
  – Asaf Karagila♦
  Apr 8 at 12:38
  

  
  
  
  

add a comment | 

2

0

$begingroup$

This question already has an answer here:

• 
  Is symmetric group on natural numbers countable?
  
  8 answers
  
  

I have just come across the Riemann Rearrangement theorem and, from what I understand, it shows that any real can be written as a permutation of a conditionally convergent series. The problem I have is that permutations of an infinite series are countably infinite, so in theory one could list all possible permutations of the series which would be equivalent to listing all real numbers. But reals are uncountable.

This seems like a contradiction to me, where am I wrong?

permutations real-numbers

share|cite|improve this question

edited Apr 8 at 12:38

Asaf Karagila♦

308k33441775

asked Apr 8 at 10:53

Lorenzo Lorenzo

184

New contributor

Lorenzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

marked as duplicate by Asaf Karagila♦ Apr 8 at 12:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While the motivation for this question is the rearrangement theorem, the title should reflect the actual question instead.
  $endgroup$
  – Asaf Karagila♦
  Apr 8 at 12:38
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Is symmetric group on natural numbers countable?
  
  8 answers
  
  

I have just come across the Riemann Rearrangement theorem and, from what I understand, it shows that any real can be written as a permutation of a conditionally convergent series. The problem I have is that permutations of an infinite series are countably infinite, so in theory one could list all possible permutations of the series which would be equivalent to listing all real numbers. But reals are uncountable.

This seems like a contradiction to me, where am I wrong?

permutations real-numbers

share|cite|improve this question

edited Apr 8 at 12:38

Asaf Karagila♦

308k33441775

asked Apr 8 at 10:53

Lorenzo Lorenzo

184

New contributor

Lorenzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Apr 8 at 12:38

Asaf Karagila♦

308k33441775

asked Apr 8 at 10:53

Lorenzo Lorenzo

184

New contributor

Lorenzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 8 at 12:38

Asaf Karagila♦

308k33441775

asked Apr 8 at 10:53

Lorenzo Lorenzo

184

New contributor

Lorenzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 8 at 12:38

Asaf Karagila♦

308k33441775

edited Apr 8 at 12:38

Asaf Karagila♦

308k33441775

asked Apr 8 at 10:53

Lorenzo Lorenzo

184

New contributor

Lorenzo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 8 at 10:53

Lorenzo Lorenzo

184

1 Answer
1

active

oldest

votes

6

$begingroup$

Permutations of an infinite series are not countably infinite, so there is no contradiction.

The set of permutations of an infinite series is the set of all bijections from $mathbb N$ to $mathbb N$, which is an uncountably infinite set. For one, the Riemann theorem you state can be used to prove that it is uncountable, or, more set-theoretically, a variant of the diagonal argument can be used, as in this post

share|cite|improve this answer

answered Apr 8 at 10:57

5xum5xum

92.6k394162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok that makes sense, but I thought one could list all permutations by having permutations of different 'lengths' n (that permutate only the first n terms, where all permutations of length n are finite) and then listing each one after the other ie all permutations of length 1, all permutations of length 2 etc.
  $endgroup$
  – Lorenzo
  Apr 8 at 11:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Lorenzo There is no "length" to speak of here. Each permutation is a permutation of "infinite" length, if you will. It is a rearangement of the entire set $mathbb N$.
  $endgroup$
  – 5xum
  Apr 8 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just saying length n as in all other terms beyond n are not changed eg. (13245678...) would be of length 3
  $endgroup$
  – Lorenzo
  Apr 8 at 11:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lorenzo For example, when, in your listing, to you list the permutation that maps $1$ to $2$, $2$ to $1$, $3$ to $4$, $4$ to $3$, $2k-1$ to $2k$ and $2k$ to $2k-1$?. There is no $n$ at which all terms beyond $n$ are unchanged! Every integer gets moved either one up or one down.
  $endgroup$
  – 5xum
  Apr 8 at 11:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok I get it now, thanks
  $endgroup$
  – Lorenzo
  Apr 8 at 11:20
  

  
  
  
  

 | 
show 1 more comment

6

$begingroup$

Permutations of an infinite series are not countably infinite, so there is no contradiction.

The set of permutations of an infinite series is the set of all bijections from $mathbb N$ to $mathbb N$, which is an uncountably infinite set. For one, the Riemann theorem you state can be used to prove that it is uncountable, or, more set-theoretically, a variant of the diagonal argument can be used, as in this post

share|cite|improve this answer

answered Apr 8 at 10:57

5xum5xum

92.6k394162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok that makes sense, but I thought one could list all permutations by having permutations of different 'lengths' n (that permutate only the first n terms, where all permutations of length n are finite) and then listing each one after the other ie all permutations of length 1, all permutations of length 2 etc.
  $endgroup$
  – Lorenzo
  Apr 8 at 11:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Lorenzo There is no "length" to speak of here. Each permutation is a permutation of "infinite" length, if you will. It is a rearangement of the entire set $mathbb N$.
  $endgroup$
  – 5xum
  Apr 8 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just saying length n as in all other terms beyond n are not changed eg. (13245678...) would be of length 3
  $endgroup$
  – Lorenzo
  Apr 8 at 11:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lorenzo For example, when, in your listing, to you list the permutation that maps $1$ to $2$, $2$ to $1$, $3$ to $4$, $4$ to $3$, $2k-1$ to $2k$ and $2k$ to $2k-1$?. There is no $n$ at which all terms beyond $n$ are unchanged! Every integer gets moved either one up or one down.
  $endgroup$
  – 5xum
  Apr 8 at 11:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok I get it now, thanks
  $endgroup$
  – Lorenzo
  Apr 8 at 11:20
  

  
  
  
  

 | 
show 1 more comment

6

$begingroup$

Permutations of an infinite series are not countably infinite, so there is no contradiction.

The set of permutations of an infinite series is the set of all bijections from $mathbb N$ to $mathbb N$, which is an uncountably infinite set. For one, the Riemann theorem you state can be used to prove that it is uncountable, or, more set-theoretically, a variant of the diagonal argument can be used, as in this post

share|cite|improve this answer

answered Apr 8 at 10:57

5xum5xum

92.6k394162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok that makes sense, but I thought one could list all permutations by having permutations of different 'lengths' n (that permutate only the first n terms, where all permutations of length n are finite) and then listing each one after the other ie all permutations of length 1, all permutations of length 2 etc.
  $endgroup$
  – Lorenzo
  Apr 8 at 11:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Lorenzo There is no "length" to speak of here. Each permutation is a permutation of "infinite" length, if you will. It is a rearangement of the entire set $mathbb N$.
  $endgroup$
  – 5xum
  Apr 8 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just saying length n as in all other terms beyond n are not changed eg. (13245678...) would be of length 3
  $endgroup$
  – Lorenzo
  Apr 8 at 11:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lorenzo For example, when, in your listing, to you list the permutation that maps $1$ to $2$, $2$ to $1$, $3$ to $4$, $4$ to $3$, $2k-1$ to $2k$ and $2k$ to $2k-1$?. There is no $n$ at which all terms beyond $n$ are unchanged! Every integer gets moved either one up or one down.
  $endgroup$
  – 5xum
  Apr 8 at 11:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok I get it now, thanks
  $endgroup$
  – Lorenzo
  Apr 8 at 11:20
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Permutations of an infinite series are not countably infinite, so there is no contradiction.

The set of permutations of an infinite series is the set of all bijections from $mathbb N$ to $mathbb N$, which is an uncountably infinite set. For one, the Riemann theorem you state can be used to prove that it is uncountable, or, more set-theoretically, a variant of the diagonal argument can be used, as in this post

share|cite|improve this answer

answered Apr 8 at 10:57

5xum5xum

92.6k394162

$endgroup$

share|cite|improve this answer

answered Apr 8 at 10:57

5xum5xum

92.6k394162

answered Apr 8 at 10:57

5xum5xum

92.6k394162

answered Apr 8 at 10:57

5xum5xum

92.6k394162