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functions and set of numbers

Onto and everywhere defined for functions having same cardinalityNotation: Concatenate two functions (piecewise) / Concatenate two vectors, lists or tuplesFunctions and Set TheorySet theory and functionsCartesian Products and FunctionsProve this theorem about infinity setShould the invertible functions be bijective?Characterise certain set functions which could be “reduced” in terms of argumentsProof verification of set equalities about functions and union/intersectionFunctions: Induced Set functions

0

1

$begingroup$

Denote $f$ : $Nto N$ a function

for any $n$ $in$ $N$ we will mark $A_n = {0,1,2,3,4...,n}$ and $A(-1) = emptyset$.

I need to prove that $f$ is one to one function if and only if $f[A_n] neq f[A_m]$ for every $m,n in N cup {-1}$ and $mneq n$.

how can I approach this function at first? f.e what is the image of $f[A_5]$?

then the proof that it's one to one...

Thanks.

functions

share|cite|improve this question

edited Mar 15 at 8:18

Max

9211319

asked Mar 15 at 7:44

OO1OO1

174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Groups are a specific thing in mathematics and not the thing you are looking at here. The word for the thing you have here is a "set".
  $endgroup$
  – Tobias Kildetoft
  Mar 15 at 7:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Tobias, I have edited the title.
  $endgroup$
  – OO1
  Mar 15 at 8:04
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Denote $f$ : $Nto N$ a function

for any $n$ $in$ $N$ we will mark $A_n = {0,1,2,3,4...,n}$ and $A(-1) = emptyset$.

I need to prove that $f$ is one to one function if and only if $f[A_n] neq f[A_m]$ for every $m,n in N cup {-1}$ and $mneq n$.

how can I approach this function at first? f.e what is the image of $f[A_5]$?

then the proof that it's one to one...

Thanks.

functions

share|cite|improve this question

edited Mar 15 at 8:18

Max

9211319

asked Mar 15 at 7:44

OO1OO1

174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Groups are a specific thing in mathematics and not the thing you are looking at here. The word for the thing you have here is a "set".
  $endgroup$
  – Tobias Kildetoft
  Mar 15 at 7:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Tobias, I have edited the title.
  $endgroup$
  – OO1
  Mar 15 at 8:04
  

  
  
  
  

add a comment | 

$begingroup$

Denote $f$ : $Nto N$ a function

for any $n$ $in$ $N$ we will mark $A_n = {0,1,2,3,4...,n}$ and $A(-1) = emptyset$.

I need to prove that $f$ is one to one function if and only if $f[A_n] neq f[A_m]$ for every $m,n in N cup {-1}$ and $mneq n$.

how can I approach this function at first? f.e what is the image of $f[A_5]$?

then the proof that it's one to one...

Thanks.

functions

share|cite|improve this question

edited Mar 15 at 8:18

Max

9211319

asked Mar 15 at 7:44

OO1OO1

174

$endgroup$

share|cite|improve this question

edited Mar 15 at 8:18

Max

9211319

asked Mar 15 at 7:44

OO1OO1

174

share|cite|improve this question

edited Mar 15 at 8:18

Max

9211319

asked Mar 15 at 7:44

OO1OO1

174

edited Mar 15 at 8:18

Max

9211319

edited Mar 15 at 8:18

Max

9211319

asked Mar 15 at 7:44

OO1OO1

174

asked Mar 15 at 7:44

OO1OO1

174

1 Answer
1

active

oldest

votes

0

$begingroup$

We have
$$
A_{-1}subset A_0subset A_1subset A_2subsetcdots
$$
By definition of the term "function", and the notion of a function applied to a set, this means that we have
$$
f(A_{-1})subseteq f(A_{0})subseteq f(A_{1})subseteq f(A_{2})subseteqcdots
$$
Now, let's see what it would mean if we had equality anywhere in this chain. Say $f(A_{k-1}) = f(A_{k})$. That means that $f(k)in f(A_{k-1})$. This implies that there is some $iin A_{k-1}$ such that $f(k) = f(i)$. So $f$ cannot be one-to-one.

On the other hand, let's say $f$ is not one-to-one. Let $i<j$ be two natural numbers such that $f(i) = f(j)$. Then $f(j) = f(i)in f(A_{j-1})$, which means that $f(A_j) = f(A_{j-1})$.

share|cite|improve this answer

answered Mar 15 at 8:04

ArthurArthur

120k7120203

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the comment, but I am still struggling to understand how function work on sets? for example is $f[A_5]$ = {$0,1,2,3,4,5$}? i mean this function is like the identity function? When you refer to $f(i)$ does it mean a number in a set? for example, $f(0)$ in any group would be number $0$?
  $endgroup$
  – OO1
  Mar 15 at 8:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OO1 $f(A_5) = {f(0), f(1), f(2), f(3), f(4), f(5)}$. And when I refer to $f(i)$, I mean $f(i)$ as normal. For instance, $f$ could be the function "add $3$", and in that case, $f(A_5) = {3, 4, 5, 6, 7, 8}$ and $f(i) = i+3$. Or $f$ could be the function "constantly $8$", in which case $f(A_5) = {8}$ (duplicates are redundant, and therefore removed) and $f(i) = 8$. Or $f$ could be something entirely different.
  $endgroup$
  – Arthur
  Mar 15 at 8:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's really helpful, thanks Arthur.
  $endgroup$
  – OO1
  Mar 15 at 18:38
  

  
  
  
  

add a comment | 

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0

$begingroup$

We have
$$
A_{-1}subset A_0subset A_1subset A_2subsetcdots
$$
By definition of the term "function", and the notion of a function applied to a set, this means that we have
$$
f(A_{-1})subseteq f(A_{0})subseteq f(A_{1})subseteq f(A_{2})subseteqcdots
$$
Now, let's see what it would mean if we had equality anywhere in this chain. Say $f(A_{k-1}) = f(A_{k})$. That means that $f(k)in f(A_{k-1})$. This implies that there is some $iin A_{k-1}$ such that $f(k) = f(i)$. So $f$ cannot be one-to-one.

On the other hand, let's say $f$ is not one-to-one. Let $i<j$ be two natural numbers such that $f(i) = f(j)$. Then $f(j) = f(i)in f(A_{j-1})$, which means that $f(A_j) = f(A_{j-1})$.

share|cite|improve this answer

answered Mar 15 at 8:04

ArthurArthur

120k7120203

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the comment, but I am still struggling to understand how function work on sets? for example is $f[A_5]$ = {$0,1,2,3,4,5$}? i mean this function is like the identity function? When you refer to $f(i)$ does it mean a number in a set? for example, $f(0)$ in any group would be number $0$?
  $endgroup$
  – OO1
  Mar 15 at 8:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OO1 $f(A_5) = {f(0), f(1), f(2), f(3), f(4), f(5)}$. And when I refer to $f(i)$, I mean $f(i)$ as normal. For instance, $f$ could be the function "add $3$", and in that case, $f(A_5) = {3, 4, 5, 6, 7, 8}$ and $f(i) = i+3$. Or $f$ could be the function "constantly $8$", in which case $f(A_5) = {8}$ (duplicates are redundant, and therefore removed) and $f(i) = 8$. Or $f$ could be something entirely different.
  $endgroup$
  – Arthur
  Mar 15 at 8:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's really helpful, thanks Arthur.
  $endgroup$
  – OO1
  Mar 15 at 18:38
  

  
  
  
  

add a comment | 

0

$begingroup$

We have
$$
A_{-1}subset A_0subset A_1subset A_2subsetcdots
$$
By definition of the term "function", and the notion of a function applied to a set, this means that we have
$$
f(A_{-1})subseteq f(A_{0})subseteq f(A_{1})subseteq f(A_{2})subseteqcdots
$$
Now, let's see what it would mean if we had equality anywhere in this chain. Say $f(A_{k-1}) = f(A_{k})$. That means that $f(k)in f(A_{k-1})$. This implies that there is some $iin A_{k-1}$ such that $f(k) = f(i)$. So $f$ cannot be one-to-one.

On the other hand, let's say $f$ is not one-to-one. Let $i<j$ be two natural numbers such that $f(i) = f(j)$. Then $f(j) = f(i)in f(A_{j-1})$, which means that $f(A_j) = f(A_{j-1})$.

share|cite|improve this answer

answered Mar 15 at 8:04

ArthurArthur

120k7120203

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the comment, but I am still struggling to understand how function work on sets? for example is $f[A_5]$ = {$0,1,2,3,4,5$}? i mean this function is like the identity function? When you refer to $f(i)$ does it mean a number in a set? for example, $f(0)$ in any group would be number $0$?
  $endgroup$
  – OO1
  Mar 15 at 8:10
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OO1 $f(A_5) = {f(0), f(1), f(2), f(3), f(4), f(5)}$. And when I refer to $f(i)$, I mean $f(i)$ as normal. For instance, $f$ could be the function "add $3$", and in that case, $f(A_5) = {3, 4, 5, 6, 7, 8}$ and $f(i) = i+3$. Or $f$ could be the function "constantly $8$", in which case $f(A_5) = {8}$ (duplicates are redundant, and therefore removed) and $f(i) = 8$. Or $f$ could be something entirely different.
  $endgroup$
  – Arthur
  Mar 15 at 8:42
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's really helpful, thanks Arthur.
  $endgroup$
  – OO1
  Mar 15 at 18:38
  

  
  
  
  

add a comment | 

$begingroup$

We have
$$
A_{-1}subset A_0subset A_1subset A_2subsetcdots
$$
By definition of the term "function", and the notion of a function applied to a set, this means that we have
$$
f(A_{-1})subseteq f(A_{0})subseteq f(A_{1})subseteq f(A_{2})subseteqcdots
$$
Now, let's see what it would mean if we had equality anywhere in this chain. Say $f(A_{k-1}) = f(A_{k})$. That means that $f(k)in f(A_{k-1})$. This implies that there is some $iin A_{k-1}$ such that $f(k) = f(i)$. So $f$ cannot be one-to-one.

On the other hand, let's say $f$ is not one-to-one. Let $i<j$ be two natural numbers such that $f(i) = f(j)$. Then $f(j) = f(i)in f(A_{j-1})$, which means that $f(A_j) = f(A_{j-1})$.

share|cite|improve this answer

answered Mar 15 at 8:04

ArthurArthur

120k7120203

$endgroup$

share|cite|improve this answer

answered Mar 15 at 8:04

ArthurArthur

120k7120203

answered Mar 15 at 8:04

ArthurArthur

120k7120203

answered Mar 15 at 8:04

ArthurArthur

120k7120203