What are the equivalence classes of this relation?












1












$begingroup$


Let $f colon X → X$ be an injective function.



For $y, z ∈ X$, define $y sim z$ to mean there
exists an integer $n ≥ 0$ such that either $f^n(z) = y$ or $f^n(y) = z$. (Here $f^0(z) = z$ for all $z ∈ X$.)



Prove this is defines an equivalence relation on $X$.



Give a list of the different possibilities for the
resulting equivalence classes



My attempt: Proving it was an equivalence relation was simple as it involved only using definitions of the function, but I have no idea what the equivalence classes must be.






relations equivalence-relations







share|cite|improve this question











$endgroup$












  • $begingroup$
    My guess is that you're being asked to describe the "shape" of the possible equivalence classes.
    $endgroup$
    – rogerl
    Dec 11 '18 at 21:21










  • $begingroup$
    I do not know how to even begin thinking about that, which is my main issue.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 21:24
















1












$begingroup$


Let $f colon X → X$ be an injective function.



For $y, z ∈ X$, define $y sim z$ to mean there
exists an integer $n ≥ 0$ such that either $f^n(z) = y$ or $f^n(y) = z$. (Here $f^0(z) = z$ for all $z ∈ X$.)



Prove this is defines an equivalence relation on $X$.



Give a list of the different possibilities for the
resulting equivalence classes



My attempt: Proving it was an equivalence relation was simple as it involved only using definitions of the function, but I have no idea what the equivalence classes must be.






relations equivalence-relations







share|cite|improve this question











$endgroup$












  • $begingroup$
    My guess is that you're being asked to describe the "shape" of the possible equivalence classes.
    $endgroup$
    – rogerl
    Dec 11 '18 at 21:21










  • $begingroup$
    I do not know how to even begin thinking about that, which is my main issue.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 21:24














1












1








1





$begingroup$


Let $f colon X → X$ be an injective function.



For $y, z ∈ X$, define $y sim z$ to mean there
exists an integer $n ≥ 0$ such that either $f^n(z) = y$ or $f^n(y) = z$. (Here $f^0(z) = z$ for all $z ∈ X$.)



Prove this is defines an equivalence relation on $X$.



Give a list of the different possibilities for the
resulting equivalence classes



My attempt: Proving it was an equivalence relation was simple as it involved only using definitions of the function, but I have no idea what the equivalence classes must be.






relations equivalence-relations







share|cite|improve this question











$endgroup$




Let $f colon X → X$ be an injective function.



For $y, z ∈ X$, define $y sim z$ to mean there
exists an integer $n ≥ 0$ such that either $f^n(z) = y$ or $f^n(y) = z$. (Here $f^0(z) = z$ for all $z ∈ X$.)



Prove this is defines an equivalence relation on $X$.



Give a list of the different possibilities for the
resulting equivalence classes



My attempt: Proving it was an equivalence relation was simple as it involved only using definitions of the function, but I have no idea what the equivalence classes must be.






relations equivalence-relations




relations equivalence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 23:11









egreg

185k1486206




185k1486206










asked Dec 11 '18 at 21:15









childishsadbinochildishsadbino

1148




1148












  • $begingroup$
    My guess is that you're being asked to describe the "shape" of the possible equivalence classes.
    $endgroup$
    – rogerl
    Dec 11 '18 at 21:21










  • $begingroup$
    I do not know how to even begin thinking about that, which is my main issue.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 21:24


















  • $begingroup$
    My guess is that you're being asked to describe the "shape" of the possible equivalence classes.
    $endgroup$
    – rogerl
    Dec 11 '18 at 21:21










  • $begingroup$
    I do not know how to even begin thinking about that, which is my main issue.
    $endgroup$
    – childishsadbino
    Dec 11 '18 at 21:24
















$begingroup$
My guess is that you're being asked to describe the "shape" of the possible equivalence classes.
$endgroup$
– rogerl
Dec 11 '18 at 21:21




$begingroup$
My guess is that you're being asked to describe the "shape" of the possible equivalence classes.
$endgroup$
– rogerl
Dec 11 '18 at 21:21












$begingroup$
I do not know how to even begin thinking about that, which is my main issue.
$endgroup$
– childishsadbino
Dec 11 '18 at 21:24




$begingroup$
I do not know how to even begin thinking about that, which is my main issue.
$endgroup$
– childishsadbino
Dec 11 '18 at 21:24










1 Answer
1






active

oldest

votes


















1












$begingroup$

The question is asking for various possibilities for the equivalence classes. So



1). If $x in X$ is a fixed point of the function i.e. $f(x)=x$, then $[x]={x}$. Due to injectivity, no other element will map to $x$, so it is the only member in it’s class.



2). If there exists a finite sequence $x_1, x_2, ldots ,x_k$, such that $f(x_i)=x_{i+1}$ and $f(x_k)=x_1$. Then $[x_1]={x_1, x_2, ldots ,x_k }$.



3). Other possibility is to have infinite orbits.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The "infinite orbit" case might also split into the case ${ ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{Z}$, or the case ${ x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{N}$ but $x_0$ is not in the image of $f$.
    $endgroup$
    – Daniel Schepler
    Dec 11 '18 at 23:51










  • $begingroup$
    I understand 1 and 2, but could you please explain what you mean by infinite orbits?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:34






  • 1




    $begingroup$
    @childishsadbino Here is an example that can help. Consider $f(x_{i})=x_{i+2}$. Then $f(x_1)=x_3, f(x_3)=x_5, ldots$ and $f(x_2)=x_4, f(x_4)=x_6, ldots$. Thus $[x_1]={x_1,x_3,x_5, ldots}$ and $[x_2]={x_2,x_4,x_6, ldots}$. So they are infinite orbits.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 4:51












  • $begingroup$
    Are these the only possibilities?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:54












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The question is asking for various possibilities for the equivalence classes. So



1). If $x in X$ is a fixed point of the function i.e. $f(x)=x$, then $[x]={x}$. Due to injectivity, no other element will map to $x$, so it is the only member in it’s class.



2). If there exists a finite sequence $x_1, x_2, ldots ,x_k$, such that $f(x_i)=x_{i+1}$ and $f(x_k)=x_1$. Then $[x_1]={x_1, x_2, ldots ,x_k }$.



3). Other possibility is to have infinite orbits.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The "infinite orbit" case might also split into the case ${ ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{Z}$, or the case ${ x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{N}$ but $x_0$ is not in the image of $f$.
    $endgroup$
    – Daniel Schepler
    Dec 11 '18 at 23:51










  • $begingroup$
    I understand 1 and 2, but could you please explain what you mean by infinite orbits?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:34






  • 1




    $begingroup$
    @childishsadbino Here is an example that can help. Consider $f(x_{i})=x_{i+2}$. Then $f(x_1)=x_3, f(x_3)=x_5, ldots$ and $f(x_2)=x_4, f(x_4)=x_6, ldots$. Thus $[x_1]={x_1,x_3,x_5, ldots}$ and $[x_2]={x_2,x_4,x_6, ldots}$. So they are infinite orbits.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 4:51












  • $begingroup$
    Are these the only possibilities?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:54
















1












$begingroup$

The question is asking for various possibilities for the equivalence classes. So



1). If $x in X$ is a fixed point of the function i.e. $f(x)=x$, then $[x]={x}$. Due to injectivity, no other element will map to $x$, so it is the only member in it’s class.



2). If there exists a finite sequence $x_1, x_2, ldots ,x_k$, such that $f(x_i)=x_{i+1}$ and $f(x_k)=x_1$. Then $[x_1]={x_1, x_2, ldots ,x_k }$.



3). Other possibility is to have infinite orbits.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The "infinite orbit" case might also split into the case ${ ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{Z}$, or the case ${ x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{N}$ but $x_0$ is not in the image of $f$.
    $endgroup$
    – Daniel Schepler
    Dec 11 '18 at 23:51










  • $begingroup$
    I understand 1 and 2, but could you please explain what you mean by infinite orbits?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:34






  • 1




    $begingroup$
    @childishsadbino Here is an example that can help. Consider $f(x_{i})=x_{i+2}$. Then $f(x_1)=x_3, f(x_3)=x_5, ldots$ and $f(x_2)=x_4, f(x_4)=x_6, ldots$. Thus $[x_1]={x_1,x_3,x_5, ldots}$ and $[x_2]={x_2,x_4,x_6, ldots}$. So they are infinite orbits.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 4:51












  • $begingroup$
    Are these the only possibilities?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:54














1












1








1





$begingroup$

The question is asking for various possibilities for the equivalence classes. So



1). If $x in X$ is a fixed point of the function i.e. $f(x)=x$, then $[x]={x}$. Due to injectivity, no other element will map to $x$, so it is the only member in it’s class.



2). If there exists a finite sequence $x_1, x_2, ldots ,x_k$, such that $f(x_i)=x_{i+1}$ and $f(x_k)=x_1$. Then $[x_1]={x_1, x_2, ldots ,x_k }$.



3). Other possibility is to have infinite orbits.






share|cite|improve this answer











$endgroup$



The question is asking for various possibilities for the equivalence classes. So



1). If $x in X$ is a fixed point of the function i.e. $f(x)=x$, then $[x]={x}$. Due to injectivity, no other element will map to $x$, so it is the only member in it’s class.



2). If there exists a finite sequence $x_1, x_2, ldots ,x_k$, such that $f(x_i)=x_{i+1}$ and $f(x_k)=x_1$. Then $[x_1]={x_1, x_2, ldots ,x_k }$.



3). Other possibility is to have infinite orbits.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 22:48

























answered Dec 11 '18 at 21:42









Anurag AAnurag A

26.4k12351




26.4k12351












  • $begingroup$
    The "infinite orbit" case might also split into the case ${ ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{Z}$, or the case ${ x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{N}$ but $x_0$ is not in the image of $f$.
    $endgroup$
    – Daniel Schepler
    Dec 11 '18 at 23:51










  • $begingroup$
    I understand 1 and 2, but could you please explain what you mean by infinite orbits?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:34






  • 1




    $begingroup$
    @childishsadbino Here is an example that can help. Consider $f(x_{i})=x_{i+2}$. Then $f(x_1)=x_3, f(x_3)=x_5, ldots$ and $f(x_2)=x_4, f(x_4)=x_6, ldots$. Thus $[x_1]={x_1,x_3,x_5, ldots}$ and $[x_2]={x_2,x_4,x_6, ldots}$. So they are infinite orbits.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 4:51












  • $begingroup$
    Are these the only possibilities?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:54


















  • $begingroup$
    The "infinite orbit" case might also split into the case ${ ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{Z}$, or the case ${ x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{N}$ but $x_0$ is not in the image of $f$.
    $endgroup$
    – Daniel Schepler
    Dec 11 '18 at 23:51










  • $begingroup$
    I understand 1 and 2, but could you please explain what you mean by infinite orbits?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:34






  • 1




    $begingroup$
    @childishsadbino Here is an example that can help. Consider $f(x_{i})=x_{i+2}$. Then $f(x_1)=x_3, f(x_3)=x_5, ldots$ and $f(x_2)=x_4, f(x_4)=x_6, ldots$. Thus $[x_1]={x_1,x_3,x_5, ldots}$ and $[x_2]={x_2,x_4,x_6, ldots}$. So they are infinite orbits.
    $endgroup$
    – Anurag A
    Dec 12 '18 at 4:51












  • $begingroup$
    Are these the only possibilities?
    $endgroup$
    – childishsadbino
    Dec 12 '18 at 4:54
















$begingroup$
The "infinite orbit" case might also split into the case ${ ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{Z}$, or the case ${ x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{N}$ but $x_0$ is not in the image of $f$.
$endgroup$
– Daniel Schepler
Dec 11 '18 at 23:51




$begingroup$
The "infinite orbit" case might also split into the case ${ ldots, x_{-2}, x_{-1}, x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{Z}$, or the case ${ x_0, x_1, x_2, ldots }$ where $x_{n+1} = f(x_n)$ for all $n in mathbb{N}$ but $x_0$ is not in the image of $f$.
$endgroup$
– Daniel Schepler
Dec 11 '18 at 23:51












$begingroup$
I understand 1 and 2, but could you please explain what you mean by infinite orbits?
$endgroup$
– childishsadbino
Dec 12 '18 at 4:34




$begingroup$
I understand 1 and 2, but could you please explain what you mean by infinite orbits?
$endgroup$
– childishsadbino
Dec 12 '18 at 4:34




1




1




$begingroup$
@childishsadbino Here is an example that can help. Consider $f(x_{i})=x_{i+2}$. Then $f(x_1)=x_3, f(x_3)=x_5, ldots$ and $f(x_2)=x_4, f(x_4)=x_6, ldots$. Thus $[x_1]={x_1,x_3,x_5, ldots}$ and $[x_2]={x_2,x_4,x_6, ldots}$. So they are infinite orbits.
$endgroup$
– Anurag A
Dec 12 '18 at 4:51






$begingroup$
@childishsadbino Here is an example that can help. Consider $f(x_{i})=x_{i+2}$. Then $f(x_1)=x_3, f(x_3)=x_5, ldots$ and $f(x_2)=x_4, f(x_4)=x_6, ldots$. Thus $[x_1]={x_1,x_3,x_5, ldots}$ and $[x_2]={x_2,x_4,x_6, ldots}$. So they are infinite orbits.
$endgroup$
– Anurag A
Dec 12 '18 at 4:51














$begingroup$
Are these the only possibilities?
$endgroup$
– childishsadbino
Dec 12 '18 at 4:54




$begingroup$
Are these the only possibilities?
$endgroup$
– childishsadbino
Dec 12 '18 at 4:54


















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