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Let $A$ be a finite set. Prove that there exists an $f:n to A$ that is onto $A$ for some $n in omega$

1

$begingroup$

Let $A$ be a finite set. Prove that there exists an $f:n to A$ that is onto $A$ for some $n in omega$.

Here by finite we mean: A set $X$ is finite iff there is a one-to-one function $f:X→n$ for some natural number $n$.

I can see that $A$ being finite implies the existence of some one-to-one function $g:Ato N$ for some $Nin omega$. and thus, its inverse function will be $g^-1:Nto A$, with $g$ onto as needed. But is it this simple? Is it always the case that such an inverse exists, or do we need to prove by induction on $n$?

elementary-set-theory

share|cite|improve this question

asked Mar 18 at 22:13

George W KushGeorge W Kush

256

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say "$n$" I think you mean "$1, 2, ...., n$." It may also help to define "$omega$" and "natural number" (does it include 0? Is the empty set finite?)
  $endgroup$
  – Michael
  Mar 18 at 22:23
  
  

  
  
  
  


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  @Michael it's rather $0,1, dots, n-1$.
  $endgroup$
  – Berci
  Mar 18 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Berci :?? Is it the case Berci = George W Kush?
  $endgroup$
  – Michael
  Mar 18 at 22:26
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  With this definition, yes, it's that simple..
  $endgroup$
  – Berci
  Mar 18 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In set theory, natural numbers are inductively encoded as $n=0,1, dots, n-1$, starting with $0:=emptyset, 1:=emptyset$, and so on.
  $endgroup$
  – Berci
  Mar 18 at 22:29
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $A$ be a finite set. Prove that there exists an $f:n to A$ that is onto $A$ for some $n in omega$.

Here by finite we mean: A set $X$ is finite iff there is a one-to-one function $f:X→n$ for some natural number $n$.

I can see that $A$ being finite implies the existence of some one-to-one function $g:Ato N$ for some $Nin omega$. and thus, its inverse function will be $g^-1:Nto A$, with $g$ onto as needed. But is it this simple? Is it always the case that such an inverse exists, or do we need to prove by induction on $n$?

elementary-set-theory

share|cite|improve this question

asked Mar 18 at 22:13

George W KushGeorge W Kush

256

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say "$n$" I think you mean "$1, 2, ...., n$." It may also help to define "$omega$" and "natural number" (does it include 0? Is the empty set finite?)
  $endgroup$
  – Michael
  Mar 18 at 22:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael it's rather $0,1, dots, n-1$.
  $endgroup$
  – Berci
  Mar 18 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Berci :?? Is it the case Berci = George W Kush?
  $endgroup$
  – Michael
  Mar 18 at 22:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  With this definition, yes, it's that simple..
  $endgroup$
  – Berci
  Mar 18 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In set theory, natural numbers are inductively encoded as $n=0,1, dots, n-1$, starting with $0:=emptyset, 1:=emptyset$, and so on.
  $endgroup$
  – Berci
  Mar 18 at 22:29
  

  
  
  
  

add a comment | 

$begingroup$

Let $A$ be a finite set. Prove that there exists an $f:n to A$ that is onto $A$ for some $n in omega$.

Here by finite we mean: A set $X$ is finite iff there is a one-to-one function $f:X→n$ for some natural number $n$.

I can see that $A$ being finite implies the existence of some one-to-one function $g:Ato N$ for some $Nin omega$. and thus, its inverse function will be $g^-1:Nto A$, with $g$ onto as needed. But is it this simple? Is it always the case that such an inverse exists, or do we need to prove by induction on $n$?

elementary-set-theory

share|cite|improve this question

asked Mar 18 at 22:13

George W KushGeorge W Kush

256

$endgroup$

share|cite|improve this question

asked Mar 18 at 22:13

George W KushGeorge W Kush

256

share|cite|improve this question

asked Mar 18 at 22:13

George W KushGeorge W Kush

256

asked Mar 18 at 22:13

George W KushGeorge W Kush

256

asked Mar 18 at 22:13

George W KushGeorge W Kush

256