If $f$ is a permutation and $f=fcirc fcirccdots$, then $f$ is the identity function on its domain












0












$begingroup$



Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




Here is my take on it. Let $f:Uto U$ be bijective, hence, a permutation on $U$. Then, because $f=fcirc fcirc fcdots$, then $fcirc f= fcirc f circ f circcdots= f$, so $fcirc f=f$, and because it is bijective we can take its inverse giving us $f^{-1}circ fcirc f=f^{-1}circ fiff f=I$. Could anyone confirm this? Thanks.






sequences-and-series functions proof-verification permutations







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$endgroup$








  • 2




    $begingroup$
    I don't understand nothing. What's $fcirc f circ...$? this makes no sense. How many times do you compose?
    $endgroup$
    – Yanko
    Dec 16 '18 at 18:31










  • $begingroup$
    countably infinite times.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 18:32






  • 4




    $begingroup$
    Now I'm even more confused. How do you compose infinitely many times? Do you take limits? what's $fcirc f circ ... (x) = ?$
    $endgroup$
    – Yanko
    Dec 16 '18 at 18:33










  • $begingroup$
    Apparently, just $x$
    $endgroup$
    – Garmekain
    Dec 16 '18 at 18:34








  • 3




    $begingroup$
    If we have the set ${mbox{on},mbox{off}}$ and $f(mbox{on})=mbox{off}$ and $f(mbox{off})=mbox{on}$ then what is $(fcirc fcirc...)(mbox{on})$? Like if we toggle a light switch infinitely many times, does it end up on or off? :P
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 18:36
















0












$begingroup$



Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




Here is my take on it. Let $f:Uto U$ be bijective, hence, a permutation on $U$. Then, because $f=fcirc fcirc fcdots$, then $fcirc f= fcirc f circ f circcdots= f$, so $fcirc f=f$, and because it is bijective we can take its inverse giving us $f^{-1}circ fcirc f=f^{-1}circ fiff f=I$. Could anyone confirm this? Thanks.






sequences-and-series functions proof-verification permutations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I don't understand nothing. What's $fcirc f circ...$? this makes no sense. How many times do you compose?
    $endgroup$
    – Yanko
    Dec 16 '18 at 18:31










  • $begingroup$
    countably infinite times.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 18:32






  • 4




    $begingroup$
    Now I'm even more confused. How do you compose infinitely many times? Do you take limits? what's $fcirc f circ ... (x) = ?$
    $endgroup$
    – Yanko
    Dec 16 '18 at 18:33










  • $begingroup$
    Apparently, just $x$
    $endgroup$
    – Garmekain
    Dec 16 '18 at 18:34








  • 3




    $begingroup$
    If we have the set ${mbox{on},mbox{off}}$ and $f(mbox{on})=mbox{off}$ and $f(mbox{off})=mbox{on}$ then what is $(fcirc fcirc...)(mbox{on})$? Like if we toggle a light switch infinitely many times, does it end up on or off? :P
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 18:36














0












0








0


1



$begingroup$



Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




Here is my take on it. Let $f:Uto U$ be bijective, hence, a permutation on $U$. Then, because $f=fcirc fcirc fcdots$, then $fcirc f= fcirc f circ f circcdots= f$, so $fcirc f=f$, and because it is bijective we can take its inverse giving us $f^{-1}circ fcirc f=f^{-1}circ fiff f=I$. Could anyone confirm this? Thanks.






sequences-and-series functions proof-verification permutations







share|cite|improve this question











$endgroup$





Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




Here is my take on it. Let $f:Uto U$ be bijective, hence, a permutation on $U$. Then, because $f=fcirc fcirc fcdots$, then $fcirc f= fcirc f circ f circcdots= f$, so $fcirc f=f$, and because it is bijective we can take its inverse giving us $f^{-1}circ fcirc f=f^{-1}circ fiff f=I$. Could anyone confirm this? Thanks.






sequences-and-series functions proof-verification permutations




sequences-and-series functions proof-verification permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 1:35









Batominovski

33.1k33293




33.1k33293










asked Dec 16 '18 at 18:29









GarmekainGarmekain

1,502720




1,502720








  • 2




    $begingroup$
    I don't understand nothing. What's $fcirc f circ...$? this makes no sense. How many times do you compose?
    $endgroup$
    – Yanko
    Dec 16 '18 at 18:31










  • $begingroup$
    countably infinite times.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 18:32






  • 4




    $begingroup$
    Now I'm even more confused. How do you compose infinitely many times? Do you take limits? what's $fcirc f circ ... (x) = ?$
    $endgroup$
    – Yanko
    Dec 16 '18 at 18:33










  • $begingroup$
    Apparently, just $x$
    $endgroup$
    – Garmekain
    Dec 16 '18 at 18:34








  • 3




    $begingroup$
    If we have the set ${mbox{on},mbox{off}}$ and $f(mbox{on})=mbox{off}$ and $f(mbox{off})=mbox{on}$ then what is $(fcirc fcirc...)(mbox{on})$? Like if we toggle a light switch infinitely many times, does it end up on or off? :P
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 18:36














  • 2




    $begingroup$
    I don't understand nothing. What's $fcirc f circ...$? this makes no sense. How many times do you compose?
    $endgroup$
    – Yanko
    Dec 16 '18 at 18:31










  • $begingroup$
    countably infinite times.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 18:32






  • 4




    $begingroup$
    Now I'm even more confused. How do you compose infinitely many times? Do you take limits? what's $fcirc f circ ... (x) = ?$
    $endgroup$
    – Yanko
    Dec 16 '18 at 18:33










  • $begingroup$
    Apparently, just $x$
    $endgroup$
    – Garmekain
    Dec 16 '18 at 18:34








  • 3




    $begingroup$
    If we have the set ${mbox{on},mbox{off}}$ and $f(mbox{on})=mbox{off}$ and $f(mbox{off})=mbox{on}$ then what is $(fcirc fcirc...)(mbox{on})$? Like if we toggle a light switch infinitely many times, does it end up on or off? :P
    $endgroup$
    – SmileyCraft
    Dec 16 '18 at 18:36








2




2




$begingroup$
I don't understand nothing. What's $fcirc f circ...$? this makes no sense. How many times do you compose?
$endgroup$
– Yanko
Dec 16 '18 at 18:31




$begingroup$
I don't understand nothing. What's $fcirc f circ...$? this makes no sense. How many times do you compose?
$endgroup$
– Yanko
Dec 16 '18 at 18:31












$begingroup$
countably infinite times.
$endgroup$
– Garmekain
Dec 16 '18 at 18:32




$begingroup$
countably infinite times.
$endgroup$
– Garmekain
Dec 16 '18 at 18:32




4




4




$begingroup$
Now I'm even more confused. How do you compose infinitely many times? Do you take limits? what's $fcirc f circ ... (x) = ?$
$endgroup$
– Yanko
Dec 16 '18 at 18:33




$begingroup$
Now I'm even more confused. How do you compose infinitely many times? Do you take limits? what's $fcirc f circ ... (x) = ?$
$endgroup$
– Yanko
Dec 16 '18 at 18:33












$begingroup$
Apparently, just $x$
$endgroup$
– Garmekain
Dec 16 '18 at 18:34






$begingroup$
Apparently, just $x$
$endgroup$
– Garmekain
Dec 16 '18 at 18:34






3




3




$begingroup$
If we have the set ${mbox{on},mbox{off}}$ and $f(mbox{on})=mbox{off}$ and $f(mbox{off})=mbox{on}$ then what is $(fcirc fcirc...)(mbox{on})$? Like if we toggle a light switch infinitely many times, does it end up on or off? :P
$endgroup$
– SmileyCraft
Dec 16 '18 at 18:36




$begingroup$
If we have the set ${mbox{on},mbox{off}}$ and $f(mbox{on})=mbox{off}$ and $f(mbox{off})=mbox{on}$ then what is $(fcirc fcirc...)(mbox{on})$? Like if we toggle a light switch infinitely many times, does it end up on or off? :P
$endgroup$
– SmileyCraft
Dec 16 '18 at 18:36










1 Answer
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$begingroup$

Let me try to rewrite the question and please tell me if this is what the question means. If this is not the case, Ill delete the answer.




Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




If this is what the question means, then you are kind of correct. This is because $f^infty=fcirc f^infty$. If $f=f^infty$, then $f=fcirc f^infty$. Since $f$ is a bijection, $f^infty$ must be the identity function on $U$. But as $f=f^infty$, $f$ is also the identity function on $U$.




However, if $U$ is equipped with a topology so that it is possible to discuss limits, then this is also a valid interpretation of the problem. The proof remains the same. My first interpretation is actually a consequence of the second interpretation. That is, in the first interpretation, $U$ is equipped with the discrete topology.




Let $f$ be a permutation on a topological space $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ converges to a unique limit, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.







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    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    3












    $begingroup$

    Let me try to rewrite the question and please tell me if this is what the question means. If this is not the case, Ill delete the answer.




    Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




    If this is what the question means, then you are kind of correct. This is because $f^infty=fcirc f^infty$. If $f=f^infty$, then $f=fcirc f^infty$. Since $f$ is a bijection, $f^infty$ must be the identity function on $U$. But as $f=f^infty$, $f$ is also the identity function on $U$.




    However, if $U$ is equipped with a topology so that it is possible to discuss limits, then this is also a valid interpretation of the problem. The proof remains the same. My first interpretation is actually a consequence of the second interpretation. That is, in the first interpretation, $U$ is equipped with the discrete topology.




    Let $f$ be a permutation on a topological space $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ converges to a unique limit, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.







    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Let me try to rewrite the question and please tell me if this is what the question means. If this is not the case, Ill delete the answer.




      Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




      If this is what the question means, then you are kind of correct. This is because $f^infty=fcirc f^infty$. If $f=f^infty$, then $f=fcirc f^infty$. Since $f$ is a bijection, $f^infty$ must be the identity function on $U$. But as $f=f^infty$, $f$ is also the identity function on $U$.




      However, if $U$ is equipped with a topology so that it is possible to discuss limits, then this is also a valid interpretation of the problem. The proof remains the same. My first interpretation is actually a consequence of the second interpretation. That is, in the first interpretation, $U$ is equipped with the discrete topology.




      Let $f$ be a permutation on a topological space $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ converges to a unique limit, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.







      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Let me try to rewrite the question and please tell me if this is what the question means. If this is not the case, Ill delete the answer.




        Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




        If this is what the question means, then you are kind of correct. This is because $f^infty=fcirc f^infty$. If $f=f^infty$, then $f=fcirc f^infty$. Since $f$ is a bijection, $f^infty$ must be the identity function on $U$. But as $f=f^infty$, $f$ is also the identity function on $U$.




        However, if $U$ is equipped with a topology so that it is possible to discuss limits, then this is also a valid interpretation of the problem. The proof remains the same. My first interpretation is actually a consequence of the second interpretation. That is, in the first interpretation, $U$ is equipped with the discrete topology.




        Let $f$ be a permutation on a topological space $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ converges to a unique limit, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.







        share|cite|improve this answer











        $endgroup$



        Let me try to rewrite the question and please tell me if this is what the question means. If this is not the case, Ill delete the answer.




        Let $f$ be a permutation on a set $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ stabilizes, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.




        If this is what the question means, then you are kind of correct. This is because $f^infty=fcirc f^infty$. If $f=f^infty$, then $f=fcirc f^infty$. Since $f$ is a bijection, $f^infty$ must be the identity function on $U$. But as $f=f^infty$, $f$ is also the identity function on $U$.




        However, if $U$ is equipped with a topology so that it is possible to discuss limits, then this is also a valid interpretation of the problem. The proof remains the same. My first interpretation is actually a consequence of the second interpretation. That is, in the first interpretation, $U$ is equipped with the discrete topology.




        Let $f$ be a permutation on a topological space $U$. Suppose that for every $xin U$, the sequence $$x,f(x),fcirc f(x), fcirc fcirc f(x), ldots$$ converges to a unique limit, and let this value of $x$ be called $f^infty(x)$. Suppose that $f=f^infty$. Then, $f$ is the identity function on $U$.








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 20:31

























        answered Dec 16 '18 at 20:22







        user614671





































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