Which sets are “clompact”?

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This is an exercise taken verbatim from Abbott's Understanding Analysis:




Let’s call a set clompact if it has the property that every
closed cover (i.e., a cover consisting of closed sets) admits a finite subcover.
Describe all of the clompact subsets of $mathbf R$.




I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover
$$ 0cupbigcup_1^inftyleft[frac1n+1,frac1nright] $$
for $[0,1]$ does not have a finite subcover. Similar constructions of closed covers show that $[a,b]$, $(a,b]$, $[b,a)$ and $(a,b)$ are not clompact as well. In addition, $mathbf R$ itself and any unbounded interval is also not clompact.



Is anyone able to help in solving this problem? Any assistance is appreciated.










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    This is an exercise taken verbatim from Abbott's Understanding Analysis:




    Let’s call a set clompact if it has the property that every
    closed cover (i.e., a cover consisting of closed sets) admits a finite subcover.
    Describe all of the clompact subsets of $mathbf R$.




    I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover
    $$ 0cupbigcup_1^inftyleft[frac1n+1,frac1nright] $$
    for $[0,1]$ does not have a finite subcover. Similar constructions of closed covers show that $[a,b]$, $(a,b]$, $[b,a)$ and $(a,b)$ are not clompact as well. In addition, $mathbf R$ itself and any unbounded interval is also not clompact.



    Is anyone able to help in solving this problem? Any assistance is appreciated.










    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      This is an exercise taken verbatim from Abbott's Understanding Analysis:




      Let’s call a set clompact if it has the property that every
      closed cover (i.e., a cover consisting of closed sets) admits a finite subcover.
      Describe all of the clompact subsets of $mathbf R$.




      I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover
      $$ 0cupbigcup_1^inftyleft[frac1n+1,frac1nright] $$
      for $[0,1]$ does not have a finite subcover. Similar constructions of closed covers show that $[a,b]$, $(a,b]$, $[b,a)$ and $(a,b)$ are not clompact as well. In addition, $mathbf R$ itself and any unbounded interval is also not clompact.



      Is anyone able to help in solving this problem? Any assistance is appreciated.










      share|cite|improve this question













      This is an exercise taken verbatim from Abbott's Understanding Analysis:




      Let’s call a set clompact if it has the property that every
      closed cover (i.e., a cover consisting of closed sets) admits a finite subcover.
      Describe all of the clompact subsets of $mathbf R$.




      I am unable to fully resolve the problem. So far, I have been able to see that singleton sets are always clompact, because the single element must be in at least one set belonging to the closed cover, and that one set is a sufficient finite subcover. The null set is also clompact for obvious reasons. I know that every non-singleton interval (regardless of if they are open, closed, or half-open) is not clompact. As an example, the closed cover
      $$ 0cupbigcup_1^inftyleft[frac1n+1,frac1nright] $$
      for $[0,1]$ does not have a finite subcover. Similar constructions of closed covers show that $[a,b]$, $(a,b]$, $[b,a)$ and $(a,b)$ are not clompact as well. In addition, $mathbf R$ itself and any unbounded interval is also not clompact.



      Is anyone able to help in solving this problem? Any assistance is appreciated.







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      asked 3 hours ago









      YiFan

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          Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.






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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes








            up vote
            6
            down vote



            accepted










            Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.






            share|cite|improve this answer
























              up vote
              6
              down vote



              accepted










              Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.






              share|cite|improve this answer






















                up vote
                6
                down vote



                accepted







                up vote
                6
                down vote



                accepted






                Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.






                share|cite|improve this answer












                Hint: Singletons are closed, so if $Ksubsetmathbb R$, then $x:xin K$ is a closed cover of $K$.







                share|cite|improve this answer












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                answered 3 hours ago









                Aweygan

                12.7k21441




                12.7k21441



























                     

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