Continuous function on a compact metric space with no fixed point












3












$begingroup$



Let $(X,d)$ be a compact metric space, and suppose $f:X to X$ is a continuous map with no fixed point. Show that there is an $epsilon > 0$ so that $d(f(x),x) ge epsilon forall x in X$.




I considered a function $g:X to mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?






general-topology metric-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f: X to X$ for example.
    $endgroup$
    – Randall
    Dec 5 '18 at 4:30








  • 1




    $begingroup$
    If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
    $endgroup$
    – Aniruddha Deshmukh
    Dec 5 '18 at 4:35
















3












$begingroup$



Let $(X,d)$ be a compact metric space, and suppose $f:X to X$ is a continuous map with no fixed point. Show that there is an $epsilon > 0$ so that $d(f(x),x) ge epsilon forall x in X$.




I considered a function $g:X to mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?






general-topology metric-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f: X to X$ for example.
    $endgroup$
    – Randall
    Dec 5 '18 at 4:30








  • 1




    $begingroup$
    If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
    $endgroup$
    – Aniruddha Deshmukh
    Dec 5 '18 at 4:35














3












3








3





$begingroup$



Let $(X,d)$ be a compact metric space, and suppose $f:X to X$ is a continuous map with no fixed point. Show that there is an $epsilon > 0$ so that $d(f(x),x) ge epsilon forall x in X$.




I considered a function $g:X to mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?






general-topology metric-spaces







share|cite|improve this question











$endgroup$





Let $(X,d)$ be a compact metric space, and suppose $f:X to X$ is a continuous map with no fixed point. Show that there is an $epsilon > 0$ so that $d(f(x),x) ge epsilon forall x in X$.




I considered a function $g:X to mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?






general-topology metric-spaces




general-topology metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 4:41









gt6989b

34.6k22456




34.6k22456










asked Dec 5 '18 at 4:30









general1597general1597

313




313












  • $begingroup$
    $f: X to X$ for example.
    $endgroup$
    – Randall
    Dec 5 '18 at 4:30








  • 1




    $begingroup$
    If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
    $endgroup$
    – Aniruddha Deshmukh
    Dec 5 '18 at 4:35


















  • $begingroup$
    $f: X to X$ for example.
    $endgroup$
    – Randall
    Dec 5 '18 at 4:30








  • 1




    $begingroup$
    If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
    $endgroup$
    – Aniruddha Deshmukh
    Dec 5 '18 at 4:35
















$begingroup$
$f: X to X$ for example.
$endgroup$
– Randall
Dec 5 '18 at 4:30






$begingroup$
$f: X to X$ for example.
$endgroup$
– Randall
Dec 5 '18 at 4:30






1




1




$begingroup$
If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
$endgroup$
– Aniruddha Deshmukh
Dec 5 '18 at 4:35




$begingroup$
If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
$endgroup$
– Aniruddha Deshmukh
Dec 5 '18 at 4:35










2 Answers
2






active

oldest

votes


















2












$begingroup$

The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
    $endgroup$
    – DanielWainfleet
    Dec 5 '18 at 10:08



















1












$begingroup$

For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






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    active

    oldest

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    2












    $begingroup$

    The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
      $endgroup$
      – DanielWainfleet
      Dec 5 '18 at 10:08
















    2












    $begingroup$

    The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
      $endgroup$
      – DanielWainfleet
      Dec 5 '18 at 10:08














    2












    2








    2





    $begingroup$

    The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.






    share|cite|improve this answer











    $endgroup$



    The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 5 '18 at 5:26

























    answered Dec 5 '18 at 4:58









    mlerma54mlerma54

    1,177148




    1,177148








    • 1




      $begingroup$
      The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
      $endgroup$
      – DanielWainfleet
      Dec 5 '18 at 10:08














    • 1




      $begingroup$
      The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
      $endgroup$
      – DanielWainfleet
      Dec 5 '18 at 10:08








    1




    1




    $begingroup$
    The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
    $endgroup$
    – DanielWainfleet
    Dec 5 '18 at 10:08




    $begingroup$
    The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
    $endgroup$
    – DanielWainfleet
    Dec 5 '18 at 10:08











    1












    $begingroup$

    For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.






        share|cite|improve this answer









        $endgroup$



        For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 5:06









        zoidbergzoidberg

        1,080113




        1,080113






























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