A wrong counter example (on distance and compactness)












2














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







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    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



























      2












      2








      2







      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















      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






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      edited yesterday









      Davide Giraudo

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          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



















          • 2




            But a compact subset of a metric space is closed and bounded.
            – Thomas Shelby
            yesterday






          • 2




            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            – Kavi Rama Murthy
            yesterday








          • 1




            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.
            – Daniele Tampieri
            yesterday










          • @KaviRamaMurthy: you're right. I have corrected the answer.
            – Daniele Tampieri
            yesterday











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          active

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          3














          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



















          • 2




            But a compact subset of a metric space is closed and bounded.
            – Thomas Shelby
            yesterday






          • 2




            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            – Kavi Rama Murthy
            yesterday








          • 1




            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.
            – Daniele Tampieri
            yesterday










          • @KaviRamaMurthy: you're right. I have corrected the answer.
            – Daniele Tampieri
            yesterday
















          3














          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



















          • 2




            But a compact subset of a metric space is closed and bounded.
            – Thomas Shelby
            yesterday






          • 2




            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            – Kavi Rama Murthy
            yesterday








          • 1




            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.
            – Daniele Tampieri
            yesterday










          • @KaviRamaMurthy: you're right. I have corrected the answer.
            – Daniele Tampieri
            yesterday














          3












          3








          3






          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














          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 yesterday

























          answered yesterday









          Daniele Tampieri

          1,8301619




          1,8301619








          • 2




            But a compact subset of a metric space is closed and bounded.
            – Thomas Shelby
            yesterday






          • 2




            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            – Kavi Rama Murthy
            yesterday








          • 1




            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.
            – Daniele Tampieri
            yesterday










          • @KaviRamaMurthy: you're right. I have corrected the answer.
            – Daniele Tampieri
            yesterday














          • 2




            But a compact subset of a metric space is closed and bounded.
            – Thomas Shelby
            yesterday






          • 2




            I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
            – Kavi Rama Murthy
            yesterday








          • 1




            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.
            – Daniele Tampieri
            yesterday










          • @KaviRamaMurthy: you're right. I have corrected the answer.
            – Daniele Tampieri
            yesterday








          2




          2




          But a compact subset of a metric space is closed and bounded.
          – Thomas Shelby
          yesterday




          But a compact subset of a metric space is closed and bounded.
          – Thomas Shelby
          yesterday




          2




          2




          I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
          – Kavi Rama Murthy
          yesterday






          I think @Daniele Tampeire meant to say $E$ and $F$ are no compact because they are not closed.
          – Kavi Rama Murthy
          yesterday






          1




          1




          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.
          – Daniele Tampieri
          yesterday




          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.
          – Daniele Tampieri
          yesterday












          @KaviRamaMurthy: you're right. I have corrected the answer.
          – Daniele Tampieri
          yesterday




          @KaviRamaMurthy: you're right. I have corrected the answer.
          – Daniele Tampieri
          yesterday


















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