A wrong counter example (on distance and compactness)












2












$begingroup$


Question: Let $E$ and $F$ be closed disjoint sets in $mathbb{R}^d$. Show that if one of the sets is compact, then necessarily $d(E, F) > 0$.



A proof was provided, and I understand it, but I can't figure out why the following 'counterexample' does not work:



Let $E = left{sum_{i=1}^n frac{1}{2^i} | nin mathbb{N} right} $ and $F = left{sum_{i=1}^n Big(frac{1}{2^i} + frac{1}{3^{i+1}}Big) | nin mathbb{N} right} $



Here, both $E$ and $F$ are compact, but $nexists epsilon > 0$ such that $d(E,F) > epsilon$.






real-analysis general-topology examples-counterexamples







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Question: Let $E$ and $F$ be closed disjoint sets in $mathbb{R}^d$. Show that if one of the sets is compact, then necessarily $d(E, F) > 0$.



    A proof was provided, and I understand it, but I can't figure out why the following 'counterexample' does not work:



    Let $E = left{sum_{i=1}^n frac{1}{2^i} | nin mathbb{N} right} $ and $F = left{sum_{i=1}^n Big(frac{1}{2^i} + frac{1}{3^{i+1}}Big) | nin mathbb{N} right} $



    Here, both $E$ and $F$ are compact, but $nexists epsilon > 0$ such that $d(E,F) > epsilon$.






    real-analysis general-topology examples-counterexamples







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Question: Let $E$ and $F$ be closed disjoint sets in $mathbb{R}^d$. Show that if one of the sets is compact, then necessarily $d(E, F) > 0$.



      A proof was provided, and I understand it, but I can't figure out why the following 'counterexample' does not work:



      Let $E = left{sum_{i=1}^n frac{1}{2^i} | nin mathbb{N} right} $ and $F = left{sum_{i=1}^n Big(frac{1}{2^i} + frac{1}{3^{i+1}}Big) | nin mathbb{N} right} $



      Here, both $E$ and $F$ are compact, but $nexists epsilon > 0$ such that $d(E,F) > epsilon$.






      real-analysis general-topology examples-counterexamples







      share|cite|improve this question











      $endgroup$




      Question: Let $E$ and $F$ be closed disjoint sets in $mathbb{R}^d$. Show that if one of the sets is compact, then necessarily $d(E, F) > 0$.



      A proof was provided, and I understand it, but I can't figure out why the following 'counterexample' does not work:



      Let $E = left{sum_{i=1}^n frac{1}{2^i} | nin mathbb{N} right} $ and $F = left{sum_{i=1}^n Big(frac{1}{2^i} + frac{1}{3^{i+1}}Big) | nin mathbb{N} right} $



      Here, both $E$ and $F$ are compact, but $nexists epsilon > 0$ such that $d(E,F) > epsilon$.






      real-analysis general-topology examples-counterexamples




      real-analysis general-topology examples-counterexamples






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 14:40









      Davide Giraudo

      126k16150261




      126k16150261










      asked Jan 3 at 5:46









      Sean LeeSean Lee

      369111




      369111






















          1 Answer
          1






          active

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          3












          $begingroup$

          Bot $E$ and $F$ are not compact nor closed, since they do not include their limit points
          $$
          sum_{i=1}^infty frac{1}{2^i};text{ and };sum_{i=1}^infty left(frac{1}{2^i}+frac{1}{3^{i+1}}right)
          $$
          therefore the theorem does not apply



          Note

          Following the comments, it should also be noted that $E$ and $F$ are not compact since not any of their open covers contains a finite subcovering: for example, think of the family of open balls
          $$
          mathcal{C}=left{mathring{B}big(x_i,10^{-i^2}big)right}_{iin mathbb{N}_+}
          $$
          where




          • $x_i=frac{1}{2^I}$ or $x_i=frac{1}{2^i}+frac{1}{3^{i+1}}$, $iin mathbb{N}_+$ and


          • $mathring{B}(x_o,r)={xinmathbb{R}^n|,|x_o-x|<r}$.



          No one of its finite subfamilies covers the whole $E$ nor $F$.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            But a compact subset of a metric space is closed and bounded.
            $endgroup$
            – Thomas Shelby
            Jan 3 at 6:14






          • 2




            $begingroup$
            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 6:20








          • 1




            $begingroup$
            You're right @ThomasShelby: as Mr. Kavi Rama Murty pointed out, $E$ and $F$ are not compact because they are not closed: each finite (sub-)covering should be made of open sets in order to have compactness, an this is not the case.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:27










          • $begingroup$
            @KaviRamaMurthy: you're right. I have corrected the answer.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:28











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

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          3












          $begingroup$

          Bot $E$ and $F$ are not compact nor closed, since they do not include their limit points
          $$
          sum_{i=1}^infty frac{1}{2^i};text{ and };sum_{i=1}^infty left(frac{1}{2^i}+frac{1}{3^{i+1}}right)
          $$
          therefore the theorem does not apply



          Note

          Following the comments, it should also be noted that $E$ and $F$ are not compact since not any of their open covers contains a finite subcovering: for example, think of the family of open balls
          $$
          mathcal{C}=left{mathring{B}big(x_i,10^{-i^2}big)right}_{iin mathbb{N}_+}
          $$
          where




          • $x_i=frac{1}{2^I}$ or $x_i=frac{1}{2^i}+frac{1}{3^{i+1}}$, $iin mathbb{N}_+$ and


          • $mathring{B}(x_o,r)={xinmathbb{R}^n|,|x_o-x|<r}$.



          No one of its finite subfamilies covers the whole $E$ nor $F$.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            But a compact subset of a metric space is closed and bounded.
            $endgroup$
            – Thomas Shelby
            Jan 3 at 6:14






          • 2




            $begingroup$
            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 6:20








          • 1




            $begingroup$
            You're right @ThomasShelby: as Mr. Kavi Rama Murty pointed out, $E$ and $F$ are not compact because they are not closed: each finite (sub-)covering should be made of open sets in order to have compactness, an this is not the case.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:27










          • $begingroup$
            @KaviRamaMurthy: you're right. I have corrected the answer.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:28
















          3












          $begingroup$

          Bot $E$ and $F$ are not compact nor closed, since they do not include their limit points
          $$
          sum_{i=1}^infty frac{1}{2^i};text{ and };sum_{i=1}^infty left(frac{1}{2^i}+frac{1}{3^{i+1}}right)
          $$
          therefore the theorem does not apply



          Note

          Following the comments, it should also be noted that $E$ and $F$ are not compact since not any of their open covers contains a finite subcovering: for example, think of the family of open balls
          $$
          mathcal{C}=left{mathring{B}big(x_i,10^{-i^2}big)right}_{iin mathbb{N}_+}
          $$
          where




          • $x_i=frac{1}{2^I}$ or $x_i=frac{1}{2^i}+frac{1}{3^{i+1}}$, $iin mathbb{N}_+$ and


          • $mathring{B}(x_o,r)={xinmathbb{R}^n|,|x_o-x|<r}$.



          No one of its finite subfamilies covers the whole $E$ nor $F$.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            But a compact subset of a metric space is closed and bounded.
            $endgroup$
            – Thomas Shelby
            Jan 3 at 6:14






          • 2




            $begingroup$
            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 6:20








          • 1




            $begingroup$
            You're right @ThomasShelby: as Mr. Kavi Rama Murty pointed out, $E$ and $F$ are not compact because they are not closed: each finite (sub-)covering should be made of open sets in order to have compactness, an this is not the case.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:27










          • $begingroup$
            @KaviRamaMurthy: you're right. I have corrected the answer.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:28














          3












          3








          3





          $begingroup$

          Bot $E$ and $F$ are not compact nor closed, since they do not include their limit points
          $$
          sum_{i=1}^infty frac{1}{2^i};text{ and };sum_{i=1}^infty left(frac{1}{2^i}+frac{1}{3^{i+1}}right)
          $$
          therefore the theorem does not apply



          Note

          Following the comments, it should also be noted that $E$ and $F$ are not compact since not any of their open covers contains a finite subcovering: for example, think of the family of open balls
          $$
          mathcal{C}=left{mathring{B}big(x_i,10^{-i^2}big)right}_{iin mathbb{N}_+}
          $$
          where




          • $x_i=frac{1}{2^I}$ or $x_i=frac{1}{2^i}+frac{1}{3^{i+1}}$, $iin mathbb{N}_+$ and


          • $mathring{B}(x_o,r)={xinmathbb{R}^n|,|x_o-x|<r}$.



          No one of its finite subfamilies covers the whole $E$ nor $F$.






          share|cite|improve this answer











          $endgroup$



          Bot $E$ and $F$ are not compact nor closed, since they do not include their limit points
          $$
          sum_{i=1}^infty frac{1}{2^i};text{ and };sum_{i=1}^infty left(frac{1}{2^i}+frac{1}{3^{i+1}}right)
          $$
          therefore the theorem does not apply



          Note

          Following the comments, it should also be noted that $E$ and $F$ are not compact since not any of their open covers contains a finite subcovering: for example, think of the family of open balls
          $$
          mathcal{C}=left{mathring{B}big(x_i,10^{-i^2}big)right}_{iin mathbb{N}_+}
          $$
          where




          • $x_i=frac{1}{2^I}$ or $x_i=frac{1}{2^i}+frac{1}{3^{i+1}}$, $iin mathbb{N}_+$ and


          • $mathring{B}(x_o,r)={xinmathbb{R}^n|,|x_o-x|<r}$.



          No one of its finite subfamilies covers the whole $E$ nor $F$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 3 at 11:07

























          answered Jan 3 at 5:54









          Daniele TampieriDaniele Tampieri

          2,1271621




          2,1271621








          • 2




            $begingroup$
            But a compact subset of a metric space is closed and bounded.
            $endgroup$
            – Thomas Shelby
            Jan 3 at 6:14






          • 2




            $begingroup$
            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 6:20








          • 1




            $begingroup$
            You're right @ThomasShelby: as Mr. Kavi Rama Murty pointed out, $E$ and $F$ are not compact because they are not closed: each finite (sub-)covering should be made of open sets in order to have compactness, an this is not the case.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:27










          • $begingroup$
            @KaviRamaMurthy: you're right. I have corrected the answer.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:28














          • 2




            $begingroup$
            But a compact subset of a metric space is closed and bounded.
            $endgroup$
            – Thomas Shelby
            Jan 3 at 6:14






          • 2




            $begingroup$
            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            $endgroup$
            – Kavi Rama Murthy
            Jan 3 at 6:20








          • 1




            $begingroup$
            You're right @ThomasShelby: as Mr. Kavi Rama Murty pointed out, $E$ and $F$ are not compact because they are not closed: each finite (sub-)covering should be made of open sets in order to have compactness, an this is not the case.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:27










          • $begingroup$
            @KaviRamaMurthy: you're right. I have corrected the answer.
            $endgroup$
            – Daniele Tampieri
            Jan 3 at 6:28








          2




          2




          $begingroup$
          But a compact subset of a metric space is closed and bounded.
          $endgroup$
          – Thomas Shelby
          Jan 3 at 6:14




          $begingroup$
          But a compact subset of a metric space is closed and bounded.
          $endgroup$
          – Thomas Shelby
          Jan 3 at 6:14




          2




          2




          $begingroup$
          I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
          $endgroup$
          – Kavi Rama Murthy
          Jan 3 at 6:20






          $begingroup$
          I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
          $endgroup$
          – Kavi Rama Murthy
          Jan 3 at 6:20






          1




          1




          $begingroup$
          You're right @ThomasShelby: as Mr. Kavi Rama Murty pointed out, $E$ and $F$ are not compact because they are not closed: each finite (sub-)covering should be made of open sets in order to have compactness, an this is not the case.
          $endgroup$
          – Daniele Tampieri
          Jan 3 at 6:27




          $begingroup$
          You're right @ThomasShelby: as Mr. Kavi Rama Murty pointed out, $E$ and $F$ are not compact because they are not closed: each finite (sub-)covering should be made of open sets in order to have compactness, an this is not the case.
          $endgroup$
          – Daniele Tampieri
          Jan 3 at 6:27












          $begingroup$
          @KaviRamaMurthy: you're right. I have corrected the answer.
          $endgroup$
          – Daniele Tampieri
          Jan 3 at 6:28




          $begingroup$
          @KaviRamaMurthy: you're right. I have corrected the answer.
          $endgroup$
          – Daniele Tampieri
          Jan 3 at 6:28


















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