Is this set perfect?












0












$begingroup$


Let $X$ be a compact and Hausdorff space and $f:Xto X$ is a homeomorphism.
Let for all $ineq j$ and $pin X$, we have $f^i(p)neq f^j(p)$. I need to know that whether $overline{{f^n(p)}_{ninmathbb{Z}}}$ is uncountable or not. I think that it is sufficient to $overline{{f^{n}(p)}_{ninmathbb{Z}}}$ is an infinite and closed without isolated point i.e. it is a perfect set. It is clear that it is infinite and close. Let $p$ be an isolated point, this implies that there is an open set $U$ such that $U={p}$, hence ${f^n(U)}_{ninmathbb{Z}}$ is an open cover for compact set $overline{{f^{n}(p)}_{ninmathbb{Z}}}$. Thus there is $m>0$ with $overline{{f^{n}(p)}_{ninmathbb{Z}}}subseteq cup_{n=0}^{m}f^n(U)$, that is a contradiction.



Is my proof true?
Please help me






general-topology dynamical-systems







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $X$ be a compact and Hausdorff space and $f:Xto X$ is a homeomorphism.
    Let for all $ineq j$ and $pin X$, we have $f^i(p)neq f^j(p)$. I need to know that whether $overline{{f^n(p)}_{ninmathbb{Z}}}$ is uncountable or not. I think that it is sufficient to $overline{{f^{n}(p)}_{ninmathbb{Z}}}$ is an infinite and closed without isolated point i.e. it is a perfect set. It is clear that it is infinite and close. Let $p$ be an isolated point, this implies that there is an open set $U$ such that $U={p}$, hence ${f^n(U)}_{ninmathbb{Z}}$ is an open cover for compact set $overline{{f^{n}(p)}_{ninmathbb{Z}}}$. Thus there is $m>0$ with $overline{{f^{n}(p)}_{ninmathbb{Z}}}subseteq cup_{n=0}^{m}f^n(U)$, that is a contradiction.



    Is my proof true?
    Please help me






    general-topology dynamical-systems







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X$ be a compact and Hausdorff space and $f:Xto X$ is a homeomorphism.
      Let for all $ineq j$ and $pin X$, we have $f^i(p)neq f^j(p)$. I need to know that whether $overline{{f^n(p)}_{ninmathbb{Z}}}$ is uncountable or not. I think that it is sufficient to $overline{{f^{n}(p)}_{ninmathbb{Z}}}$ is an infinite and closed without isolated point i.e. it is a perfect set. It is clear that it is infinite and close. Let $p$ be an isolated point, this implies that there is an open set $U$ such that $U={p}$, hence ${f^n(U)}_{ninmathbb{Z}}$ is an open cover for compact set $overline{{f^{n}(p)}_{ninmathbb{Z}}}$. Thus there is $m>0$ with $overline{{f^{n}(p)}_{ninmathbb{Z}}}subseteq cup_{n=0}^{m}f^n(U)$, that is a contradiction.



      Is my proof true?
      Please help me






      general-topology dynamical-systems







      share|cite|improve this question











      $endgroup$




      Let $X$ be a compact and Hausdorff space and $f:Xto X$ is a homeomorphism.
      Let for all $ineq j$ and $pin X$, we have $f^i(p)neq f^j(p)$. I need to know that whether $overline{{f^n(p)}_{ninmathbb{Z}}}$ is uncountable or not. I think that it is sufficient to $overline{{f^{n}(p)}_{ninmathbb{Z}}}$ is an infinite and closed without isolated point i.e. it is a perfect set. It is clear that it is infinite and close. Let $p$ be an isolated point, this implies that there is an open set $U$ such that $U={p}$, hence ${f^n(U)}_{ninmathbb{Z}}$ is an open cover for compact set $overline{{f^{n}(p)}_{ninmathbb{Z}}}$. Thus there is $m>0$ with $overline{{f^{n}(p)}_{ninmathbb{Z}}}subseteq cup_{n=0}^{m}f^n(U)$, that is a contradiction.



      Is my proof true?
      Please help me






      general-topology dynamical-systems




      general-topology dynamical-systems






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 9:29









      Jean Marie

      31.5k42355




      31.5k42355










      asked Dec 15 '18 at 9:05









      user479859user479859

      1037




      1037






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          There are several problems with your proof:




          1. You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.

          2. Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:28












          • $begingroup$
            I've deleted my third remark.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 9:30










          • $begingroup$
            Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:31










          • $begingroup$
            If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:32












          • $begingroup$
            What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
            $endgroup$
            – user479859
            Dec 15 '18 at 9:35














          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          There are several problems with your proof:




          1. You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.

          2. Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:28












          • $begingroup$
            I've deleted my third remark.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 9:30










          • $begingroup$
            Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:31










          • $begingroup$
            If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:32












          • $begingroup$
            What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
            $endgroup$
            – user479859
            Dec 15 '18 at 9:35


















          0












          $begingroup$

          There are several problems with your proof:




          1. You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.

          2. Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:28












          • $begingroup$
            I've deleted my third remark.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 9:30










          • $begingroup$
            Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:31










          • $begingroup$
            If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:32












          • $begingroup$
            What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
            $endgroup$
            – user479859
            Dec 15 '18 at 9:35
















          0












          0








          0





          $begingroup$

          There are several problems with your proof:




          1. You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.

          2. Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.






          share|cite|improve this answer











          $endgroup$



          There are several problems with your proof:




          1. You wrote “Let $p$ be an isolated point”, but $p$ here already has another meaning.

          2. Then you write “this implies that there is an open set $U$ such that $U={p}$”. This is wrong. What it implies is that there is an open subset $U$ of $X$ such that $overline{{f^n(p),|,ninmathbb{Z}}}cap U={p}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '18 at 9:30

























          answered Dec 15 '18 at 9:20









          José Carlos SantosJosé Carlos Santos

          174k23133242




          174k23133242












          • $begingroup$
            @ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:28












          • $begingroup$
            I've deleted my third remark.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 9:30










          • $begingroup$
            Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:31










          • $begingroup$
            If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:32












          • $begingroup$
            What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
            $endgroup$
            – user479859
            Dec 15 '18 at 9:35




















          • $begingroup$
            @ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:28












          • $begingroup$
            I've deleted my third remark.
            $endgroup$
            – José Carlos Santos
            Dec 15 '18 at 9:30










          • $begingroup$
            Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:31










          • $begingroup$
            If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
            $endgroup$
            – user479859
            Dec 15 '18 at 9:32












          • $begingroup$
            What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
            $endgroup$
            – user479859
            Dec 15 '18 at 9:35


















          $begingroup$
          @ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
          $endgroup$
          – user479859
          Dec 15 '18 at 9:28






          $begingroup$
          @ José Carlos Santos. Thanks. I think that $overline{{f^n(p)}_{ninmathbb{Z}}}$ is an uncountable set. In above proof $f$ is a homeomorphism, hence it is an open map and $f(overline{U})= overline{f(U)}$.
          $endgroup$
          – user479859
          Dec 15 '18 at 9:28














          $begingroup$
          I've deleted my third remark.
          $endgroup$
          – José Carlos Santos
          Dec 15 '18 at 9:30




          $begingroup$
          I've deleted my third remark.
          $endgroup$
          – José Carlos Santos
          Dec 15 '18 at 9:30












          $begingroup$
          Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
          $endgroup$
          – user479859
          Dec 15 '18 at 9:31




          $begingroup$
          Let $overline{{f^n(p)| ninmathbb{Z}}}cap U={p}$.
          $endgroup$
          – user479859
          Dec 15 '18 at 9:31












          $begingroup$
          If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
          $endgroup$
          – user479859
          Dec 15 '18 at 9:32






          $begingroup$
          If $overline{f^n(p)| ninmathbb{Z}}}cap U={ f^k(p)}$, in above proof we work with $f^k(p)$.
          $endgroup$
          – user479859
          Dec 15 '18 at 9:32














          $begingroup$
          What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
          $endgroup$
          – user479859
          Dec 15 '18 at 9:35






          $begingroup$
          What the means of "$ overline{{f^n(p)|ninmathbb{Z}}}$ has an isolated point"?
          $endgroup$
          – user479859
          Dec 15 '18 at 9:35




















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