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Is set of all surjections from $mathbbN cap m+1$ to finite subset of $mathbbN$ finite?

0

$begingroup$

Set of all surjections from $1,…,m$ to $1,…,n$ is finite.

But Set of all surjections from $mathbbN$ to $mathbbN$ is not finite.

So I thought this intermediate:

Is $$ $f$$ : mathbbN cap (m+1)(=1,…,m) to$ finite subset of $mathbbN$ $wedge$ f is surjection $ $ finite set?

Thanks in advance.

Edit: I had made a serious mistake.I added word, "finite subset".

elementary-set-theory

share|cite|improve this question

edited Mar 18 at 13:02

hina logic

asked Mar 18 at 11:44

hina logichina logic

63

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  There is no surjection from $1,2,...,m $ to $mathbb N$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 18 at 11:47
  

  
  
  
  

add a comment | 

0

$begingroup$

Set of all surjections from $1,…,m$ to $1,…,n$ is finite.

But Set of all surjections from $mathbbN$ to $mathbbN$ is not finite.

So I thought this intermediate:

Is $$ $f$$ : mathbbN cap (m+1)(=1,…,m) to$ finite subset of $mathbbN$ $wedge$ f is surjection $ $ finite set?

Thanks in advance.

Edit: I had made a serious mistake.I added word, "finite subset".

elementary-set-theory

share|cite|improve this question

edited Mar 18 at 13:02

hina logic

asked Mar 18 at 11:44

hina logichina logic

63

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  There is no surjection from $1,2,...,m $ to $mathbb N$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 18 at 11:47
  

  
  
  
  

add a comment | 

$begingroup$

Set of all surjections from $1,…,m$ to $1,…,n$ is finite.

But Set of all surjections from $mathbbN$ to $mathbbN$ is not finite.

So I thought this intermediate:

Is $$ $f$$ : mathbbN cap (m+1)(=1,…,m) to$ finite subset of $mathbbN$ $wedge$ f is surjection $ $ finite set?

Thanks in advance.

Edit: I had made a serious mistake.I added word, "finite subset".

elementary-set-theory

share|cite|improve this question

edited Mar 18 at 13:02

hina logic

asked Mar 18 at 11:44

hina logichina logic

63

$endgroup$

share|cite|improve this question

edited Mar 18 at 13:02

hina logic

asked Mar 18 at 11:44

hina logichina logic

63

share|cite|improve this question

edited Mar 18 at 13:02

hina logic

asked Mar 18 at 11:44

hina logichina logic

63

asked Mar 18 at 11:44

hina logichina logic

63

asked Mar 18 at 11:44

hina logichina logic

63

2 Answers
2

active

oldest

votes

4

$begingroup$

Yes, it's empty in fact:) Note the domain is finite, and the range is countably infinite.

share|cite|improve this answer

edited Mar 18 at 11:55

answered Mar 18 at 11:47

MariahMariah

2,1431718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CameronBuie right indeed! editing
  $endgroup$
  – Mariah
  Mar 18 at 11:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid this is complicated. $mathbbN$ is not range but codomain. But these are the same notation, hence I cannot fix it.
  $endgroup$
  – hina logic
  Mar 18 at 12:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hinalogic: "Surjection" means that you want the range to be equal to the codomain.
  $endgroup$
  – Henning Makholm
  Mar 18 at 12:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HenningMakholm Should I wrote "$1,…,m$ to subset of $mathbbN$"?
  $endgroup$
  – hina logic
  Mar 18 at 12:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @hinalogic: What do you intend for the word "surjection" to achieve in your question? A "surjection from $Y$ to a subset of $X$" is just the same as a "map from $Y$ to $X$" -- there is no point in speaking of surjections unless you care what the range is.
  $endgroup$
  – Henning Makholm
  Mar 18 at 12:16
  
  

  
  
  
  

 | 
show 5 more comments

0

$begingroup$

Since $mathbbN$ is infiite and $1,2,3,...,m$ is finite, surjections from the latter to the former are impossible. Otherwise, we would be claiming that $mathbbN$'s size is smaller or equal than $1,2,3,...,m$'s, which is absurd

share|cite|improve this answer

answered Mar 18 at 11:58

asdfasdf

1973

$endgroup$

add a comment | 

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4

$begingroup$

Yes, it's empty in fact:) Note the domain is finite, and the range is countably infinite.

share|cite|improve this answer

edited Mar 18 at 11:55

answered Mar 18 at 11:47

MariahMariah

2,1431718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CameronBuie right indeed! editing
  $endgroup$
  – Mariah
  Mar 18 at 11:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid this is complicated. $mathbbN$ is not range but codomain. But these are the same notation, hence I cannot fix it.
  $endgroup$
  – hina logic
  Mar 18 at 12:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hinalogic: "Surjection" means that you want the range to be equal to the codomain.
  $endgroup$
  – Henning Makholm
  Mar 18 at 12:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HenningMakholm Should I wrote "$1,…,m$ to subset of $mathbbN$"?
  $endgroup$
  – hina logic
  Mar 18 at 12:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @hinalogic: What do you intend for the word "surjection" to achieve in your question? A "surjection from $Y$ to a subset of $X$" is just the same as a "map from $Y$ to $X$" -- there is no point in speaking of surjections unless you care what the range is.
  $endgroup$
  – Henning Makholm
  Mar 18 at 12:16
  
  

  
  
  
  

 | 
show 5 more comments

4

$begingroup$

Yes, it's empty in fact:) Note the domain is finite, and the range is countably infinite.

share|cite|improve this answer

edited Mar 18 at 11:55

answered Mar 18 at 11:47

MariahMariah

2,1431718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CameronBuie right indeed! editing
  $endgroup$
  – Mariah
  Mar 18 at 11:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm afraid this is complicated. $mathbbN$ is not range but codomain. But these are the same notation, hence I cannot fix it.
  $endgroup$
  – hina logic
  Mar 18 at 12:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hinalogic: "Surjection" means that you want the range to be equal to the codomain.
  $endgroup$
  – Henning Makholm
  Mar 18 at 12:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HenningMakholm Should I wrote "$1,…,m$ to subset of $mathbbN$"?
  $endgroup$
  – hina logic
  Mar 18 at 12:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @hinalogic: What do you intend for the word "surjection" to achieve in your question? A "surjection from $Y$ to a subset of $X$" is just the same as a "map from $Y$ to $X$" -- there is no point in speaking of surjections unless you care what the range is.
  $endgroup$
  – Henning Makholm
  Mar 18 at 12:16
  
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

Yes, it's empty in fact:) Note the domain is finite, and the range is countably infinite.

share|cite|improve this answer

edited Mar 18 at 11:55

answered Mar 18 at 11:47

MariahMariah

2,1431718

$endgroup$

share|cite|improve this answer

edited Mar 18 at 11:55

answered Mar 18 at 11:47

MariahMariah

2,1431718

answered Mar 18 at 11:47

MariahMariah

2,1431718

answered Mar 18 at 11:47

MariahMariah

2,1431718

0

$begingroup$

Since $mathbbN$ is infiite and $1,2,3,...,m$ is finite, surjections from the latter to the former are impossible. Otherwise, we would be claiming that $mathbbN$'s size is smaller or equal than $1,2,3,...,m$'s, which is absurd

share|cite|improve this answer

answered Mar 18 at 11:58

asdfasdf

1973

$endgroup$

add a comment | 

0

$begingroup$

Since $mathbbN$ is infiite and $1,2,3,...,m$ is finite, surjections from the latter to the former are impossible. Otherwise, we would be claiming that $mathbbN$'s size is smaller or equal than $1,2,3,...,m$'s, which is absurd

share|cite|improve this answer

answered Mar 18 at 11:58

asdfasdf

1973

$endgroup$

add a comment | 

$begingroup$

Since $mathbbN$ is infiite and $1,2,3,...,m$ is finite, surjections from the latter to the former are impossible. Otherwise, we would be claiming that $mathbbN$'s size is smaller or equal than $1,2,3,...,m$'s, which is absurd

share|cite|improve this answer

answered Mar 18 at 11:58

asdfasdf

1973

$endgroup$

share|cite|improve this answer

answered Mar 18 at 11:58

asdfasdf

1973

answered Mar 18 at 11:58

asdfasdf

1973

answered Mar 18 at 11:58

asdfasdf

1973