What is the meaning of $mathbbN^mathbbN$? [duplicate]

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  What's the meaning of a set to the power of another set?
  
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Here, $mathbbN$ is the set of natural numbers. I was reading why there is no bijection from $mathbbN^mathbbN$ to $mathbbN$, and I understood we took any subset from $mathbbN^mathbbN$ and proved it not bijective to $mathbbN$ by Cantor's Diagonalization argument.

What am I unable to understand is, what exactly is meant by $mathbbN^mathbbN$?

I read somewhere that it is recognized as the space of all sequences of natural numbers. I didn't understand why? What is the meaning of that notation?

elementary-set-theory

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edited Jul 26 at 22:05

Aloizio Macedo

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asked Jul 26 at 21:37

Avinash Bhawnani

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Jul 26 at 21:51

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is the infinite cartesian product $$mathbb N times mathbb N times mathbb N times cdots$$ This corresponds $1-1$ with a sequence of natural numbers.
  – Peter
  Jul 26 at 21:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Peter s/a sequence/all sequences of integers/
  – orlp
  Jul 26 at 21:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A^B$ is the set of all maps from $B$ to $A$.
  – Arnaud Mortier
  Jul 26 at 21:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For any set $X$, $X^mathbf N$ denotes the set of all maps from $mathbf N$ to $X$,i.e. the set of all infinite sequences of elements of $X$.
  – Bernard
  Jul 26 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @orlp Right, thanks for the correction !
  – Peter
  Jul 26 at 21:50
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

This question already has an answer here:

• 
  What's the meaning of a set to the power of another set?
  
  2 answers
  
  

Here, $mathbbN$ is the set of natural numbers. I was reading why there is no bijection from $mathbbN^mathbbN$ to $mathbbN$, and I understood we took any subset from $mathbbN^mathbbN$ and proved it not bijective to $mathbbN$ by Cantor's Diagonalization argument.

What am I unable to understand is, what exactly is meant by $mathbbN^mathbbN$?

I read somewhere that it is recognized as the space of all sequences of natural numbers. I didn't understand why? What is the meaning of that notation?

elementary-set-theory

share|cite|improve this question

edited Jul 26 at 22:05

Aloizio Macedo

22.5k23283

asked Jul 26 at 21:37

Avinash Bhawnani

33118

marked as duplicate by Aloizio Macedo, Xander Henderson, amWhy elementary-set-theory
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Jul 26 at 21:51

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is the infinite cartesian product $$mathbb N times mathbb N times mathbb N times cdots$$ This corresponds $1-1$ with a sequence of natural numbers.
  – Peter
  Jul 26 at 21:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Peter s/a sequence/all sequences of integers/
  – orlp
  Jul 26 at 21:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A^B$ is the set of all maps from $B$ to $A$.
  – Arnaud Mortier
  Jul 26 at 21:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For any set $X$, $X^mathbf N$ denotes the set of all maps from $mathbf N$ to $X$,i.e. the set of all infinite sequences of elements of $X$.
  – Bernard
  Jul 26 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @orlp Right, thanks for the correction !
  – Peter
  Jul 26 at 21:50
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

This question already has an answer here:

• 
  What's the meaning of a set to the power of another set?
  
  2 answers
  
  

Here, $mathbbN$ is the set of natural numbers. I was reading why there is no bijection from $mathbbN^mathbbN$ to $mathbbN$, and I understood we took any subset from $mathbbN^mathbbN$ and proved it not bijective to $mathbbN$ by Cantor's Diagonalization argument.

What am I unable to understand is, what exactly is meant by $mathbbN^mathbbN$?

I read somewhere that it is recognized as the space of all sequences of natural numbers. I didn't understand why? What is the meaning of that notation?

elementary-set-theory

share|cite|improve this question

edited Jul 26 at 22:05

Aloizio Macedo

22.5k23283

asked Jul 26 at 21:37

Avinash Bhawnani

33118

This question already has an answer here:

• 
  What's the meaning of a set to the power of another set?
  
  2 answers
  
  

Here, $mathbbN$ is the set of natural numbers. I was reading why there is no bijection from $mathbbN^mathbbN$ to $mathbbN$, and I understood we took any subset from $mathbbN^mathbbN$ and proved it not bijective to $mathbbN$ by Cantor's Diagonalization argument.

What am I unable to understand is, what exactly is meant by $mathbbN^mathbbN$?

I read somewhere that it is recognized as the space of all sequences of natural numbers. I didn't understand why? What is the meaning of that notation?

This question already has an answer here:

• 
  What's the meaning of a set to the power of another set?
  
  2 answers
  
  

elementary-set-theory

share|cite|improve this question

edited Jul 26 at 22:05

Aloizio Macedo

22.5k23283

asked Jul 26 at 21:37

Avinash Bhawnani

33118

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 22:05

Aloizio Macedo

22.5k23283

edited Jul 26 at 22:05

Aloizio Macedo

22.5k23283

edited Jul 26 at 22:05

Aloizio Macedo

22.5k23283

22.5k23283

asked Jul 26 at 21:37

Avinash Bhawnani

33118

asked Jul 26 at 21:37

Avinash Bhawnani

33118

asked Jul 26 at 21:37

Avinash Bhawnani

33118

33118

marked as duplicate by Aloizio Macedo, Xander Henderson, amWhy elementary-set-theory
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Jul 26 at 21:51

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.

marked as duplicate by Aloizio Macedo, Xander Henderson, amWhy elementary-set-theory
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Jul 26 at 21:51

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is the infinite cartesian product $$mathbb N times mathbb N times mathbb N times cdots$$ This corresponds $1-1$ with a sequence of natural numbers.
  – Peter
  Jul 26 at 21:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Peter s/a sequence/all sequences of integers/
  – orlp
  Jul 26 at 21:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A^B$ is the set of all maps from $B$ to $A$.
  – Arnaud Mortier
  Jul 26 at 21:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For any set $X$, $X^mathbf N$ denotes the set of all maps from $mathbf N$ to $X$,i.e. the set of all infinite sequences of elements of $X$.
  – Bernard
  Jul 26 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @orlp Right, thanks for the correction !
  – Peter
  Jul 26 at 21:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is the infinite cartesian product $$mathbb N times mathbb N times mathbb N times cdots$$ This corresponds $1-1$ with a sequence of natural numbers.
  – Peter
  Jul 26 at 21:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Peter s/a sequence/all sequences of integers/
  – orlp
  Jul 26 at 21:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $A^B$ is the set of all maps from $B$ to $A$.
  – Arnaud Mortier
  Jul 26 at 21:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  For any set $X$, $X^mathbf N$ denotes the set of all maps from $mathbf N$ to $X$,i.e. the set of all infinite sequences of elements of $X$.
  – Bernard
  Jul 26 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @orlp Right, thanks for the correction !
  – Peter
  Jul 26 at 21:50
  

  
  
  
  

It is the infinite cartesian product $$mathbb N times mathbb N times mathbb N times cdots$$ This corresponds $1-1$ with a sequence of natural numbers.
– Peter
Jul 26 at 21:41

It is the infinite cartesian product $$mathbb N times mathbb N times mathbb N times cdots$$ This corresponds $1-1$ with a sequence of natural numbers.
– Peter
Jul 26 at 21:41

1

1

@Peter s/a sequence/all sequences of integers/
– orlp
Jul 26 at 21:41

@Peter s/a sequence/all sequences of integers/
– orlp
Jul 26 at 21:41

$A^B$ is the set of all maps from $B$ to $A$.
– Arnaud Mortier
Jul 26 at 21:42

$A^B$ is the set of all maps from $B$ to $A$.
– Arnaud Mortier
Jul 26 at 21:42

1

1

For any set $X$, $X^mathbf N$ denotes the set of all maps from $mathbf N$ to $X$,i.e. the set of all infinite sequences of elements of $X$.
– Bernard
Jul 26 at 21:44

For any set $X$, $X^mathbf N$ denotes the set of all maps from $mathbf N$ to $X$,i.e. the set of all infinite sequences of elements of $X$.
– Bernard
Jul 26 at 21:44

@orlp Right, thanks for the correction !
– Peter
Jul 26 at 21:50

@orlp Right, thanks for the correction !
– Peter
Jul 26 at 21:50

add a comment | 

2 Answers
2

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1
down vote



accepted

$A^B$ is the set of all functions from $B$ to $A$. When $A=B=mathbbN$, $mathbbN^mathbbN$ is the set of all functions $f:mathbbNtomathbbN$. Any such function can be viewed as a sequence $f(1),f(2),f(3),ldots$, so changing notation a bit with $a_n=f(n)$ we see that the set of all functions $mathbbN to mathbbN$ is the set of all sequences $a_1,a_2,ldots$ of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:41

Alon Amit

10.2k3765

add a comment | 

up vote
1
down vote

Let $kinmathbbN$. One could write an element $xinmathbbN^k$ as the $k$-tuple
$$x=(a_1,ldots,a_k),$$
which is as saying that for each $iin1,ldots,k$, there exists an entry $a_iinmathbbN$. Similarly, when we write $mathbbN^mathbbN$ we mean that if $xinmathbbN^mathbbN$, then for each $iinmathbbN$ there is an entry $a_iinmathbbN$. That translates to
$$x=(a_i)_iinmathbbN,$$
which, as you have said, is a sequence of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:44

Aquerman Kuczmenda

214

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

$A^B$ is the set of all functions from $B$ to $A$. When $A=B=mathbbN$, $mathbbN^mathbbN$ is the set of all functions $f:mathbbNtomathbbN$. Any such function can be viewed as a sequence $f(1),f(2),f(3),ldots$, so changing notation a bit with $a_n=f(n)$ we see that the set of all functions $mathbbN to mathbbN$ is the set of all sequences $a_1,a_2,ldots$ of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:41

Alon Amit

10.2k3765

add a comment | 

up vote
1
down vote



accepted

$A^B$ is the set of all functions from $B$ to $A$. When $A=B=mathbbN$, $mathbbN^mathbbN$ is the set of all functions $f:mathbbNtomathbbN$. Any such function can be viewed as a sequence $f(1),f(2),f(3),ldots$, so changing notation a bit with $a_n=f(n)$ we see that the set of all functions $mathbbN to mathbbN$ is the set of all sequences $a_1,a_2,ldots$ of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:41

Alon Amit

10.2k3765

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

$A^B$ is the set of all functions from $B$ to $A$. When $A=B=mathbbN$, $mathbbN^mathbbN$ is the set of all functions $f:mathbbNtomathbbN$. Any such function can be viewed as a sequence $f(1),f(2),f(3),ldots$, so changing notation a bit with $a_n=f(n)$ we see that the set of all functions $mathbbN to mathbbN$ is the set of all sequences $a_1,a_2,ldots$ of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:41

Alon Amit

10.2k3765

$A^B$ is the set of all functions from $B$ to $A$. When $A=B=mathbbN$, $mathbbN^mathbbN$ is the set of all functions $f:mathbbNtomathbbN$. Any such function can be viewed as a sequence $f(1),f(2),f(3),ldots$, so changing notation a bit with $a_n=f(n)$ we see that the set of all functions $mathbbN to mathbbN$ is the set of all sequences $a_1,a_2,ldots$ of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:41

Alon Amit

10.2k3765

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 21:41

Alon Amit

10.2k3765

answered Jul 26 at 21:41

Alon Amit

10.2k3765

answered Jul 26 at 21:41

Alon Amit

10.2k3765

10.2k3765

add a comment | 

add a comment | 

up vote
1
down vote

Let $kinmathbbN$. One could write an element $xinmathbbN^k$ as the $k$-tuple
$$x=(a_1,ldots,a_k),$$
which is as saying that for each $iin1,ldots,k$, there exists an entry $a_iinmathbbN$. Similarly, when we write $mathbbN^mathbbN$ we mean that if $xinmathbbN^mathbbN$, then for each $iinmathbbN$ there is an entry $a_iinmathbbN$. That translates to
$$x=(a_i)_iinmathbbN,$$
which, as you have said, is a sequence of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:44

Aquerman Kuczmenda

214

add a comment | 

up vote
1
down vote

Let $kinmathbbN$. One could write an element $xinmathbbN^k$ as the $k$-tuple
$$x=(a_1,ldots,a_k),$$
which is as saying that for each $iin1,ldots,k$, there exists an entry $a_iinmathbbN$. Similarly, when we write $mathbbN^mathbbN$ we mean that if $xinmathbbN^mathbbN$, then for each $iinmathbbN$ there is an entry $a_iinmathbbN$. That translates to
$$x=(a_i)_iinmathbbN,$$
which, as you have said, is a sequence of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:44

Aquerman Kuczmenda

214

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $kinmathbbN$. One could write an element $xinmathbbN^k$ as the $k$-tuple
$$x=(a_1,ldots,a_k),$$
which is as saying that for each $iin1,ldots,k$, there exists an entry $a_iinmathbbN$. Similarly, when we write $mathbbN^mathbbN$ we mean that if $xinmathbbN^mathbbN$, then for each $iinmathbbN$ there is an entry $a_iinmathbbN$. That translates to
$$x=(a_i)_iinmathbbN,$$
which, as you have said, is a sequence of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:44

Aquerman Kuczmenda

214

Let $kinmathbbN$. One could write an element $xinmathbbN^k$ as the $k$-tuple
$$x=(a_1,ldots,a_k),$$
which is as saying that for each $iin1,ldots,k$, there exists an entry $a_iinmathbbN$. Similarly, when we write $mathbbN^mathbbN$ we mean that if $xinmathbbN^mathbbN$, then for each $iinmathbbN$ there is an entry $a_iinmathbbN$. That translates to
$$x=(a_i)_iinmathbbN,$$
which, as you have said, is a sequence of natural numbers.

share|cite|improve this answer

answered Jul 26 at 21:44

Aquerman Kuczmenda

214

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 21:44

Aquerman Kuczmenda

214

answered Jul 26 at 21:44

Aquerman Kuczmenda

214

answered Jul 26 at 21:44

Aquerman Kuczmenda

214

214

add a comment | 

add a comment |