Find the supremeum and infimum, and is the set compact?

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up vote
3
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A)
$S=(1,2) cup (2,3]$



Here the Supremum would be 3, since it's the least upper bound of the set. Which 3 would be in the set.



The infimum would be 1, since it's the greatest lower bound, but it's not included in the set.



For the compact I know it's supposed to be closed and bounded so looking at this set, there is a finite subcover covering [3], but when looking at the open set (1,2), there doesn't exist a finite subcover covering (1). Would this be a non compact case?






real-analysis general-topology compactness supremum-and-infimum







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  • 2




    An open cover must cover the entire set.
    – copper.hat
    22 hours ago










  • You got the supermum right, you got the infimum right, and you came up with a good intuition (yes the set is non-compact). But your reasoning is not formal. I would stick to the fact that a compact needs to be closed and bounded, is this closed?
    – Yanko
    22 hours ago











  • Thank you very much, i get it now!
    – Killercamin
    21 hours ago














up vote
3
down vote

favorite












A)
$S=(1,2) cup (2,3]$



Here the Supremum would be 3, since it's the least upper bound of the set. Which 3 would be in the set.



The infimum would be 1, since it's the greatest lower bound, but it's not included in the set.



For the compact I know it's supposed to be closed and bounded so looking at this set, there is a finite subcover covering [3], but when looking at the open set (1,2), there doesn't exist a finite subcover covering (1). Would this be a non compact case?






real-analysis general-topology compactness supremum-and-infimum







share|cite|improve this question



















  • 2




    An open cover must cover the entire set.
    – copper.hat
    22 hours ago










  • You got the supermum right, you got the infimum right, and you came up with a good intuition (yes the set is non-compact). But your reasoning is not formal. I would stick to the fact that a compact needs to be closed and bounded, is this closed?
    – Yanko
    22 hours ago











  • Thank you very much, i get it now!
    – Killercamin
    21 hours ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











A)
$S=(1,2) cup (2,3]$



Here the Supremum would be 3, since it's the least upper bound of the set. Which 3 would be in the set.



The infimum would be 1, since it's the greatest lower bound, but it's not included in the set.



For the compact I know it's supposed to be closed and bounded so looking at this set, there is a finite subcover covering [3], but when looking at the open set (1,2), there doesn't exist a finite subcover covering (1). Would this be a non compact case?






real-analysis general-topology compactness supremum-and-infimum







share|cite|improve this question















A)
$S=(1,2) cup (2,3]$



Here the Supremum would be 3, since it's the least upper bound of the set. Which 3 would be in the set.



The infimum would be 1, since it's the greatest lower bound, but it's not included in the set.



For the compact I know it's supposed to be closed and bounded so looking at this set, there is a finite subcover covering [3], but when looking at the open set (1,2), there doesn't exist a finite subcover covering (1). Would this be a non compact case?






real-analysis general-topology compactness supremum-and-infimum




real-analysis general-topology compactness supremum-and-infimum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 22 hours ago









Marios Gretsas

8,28611337




8,28611337










asked 22 hours ago









Killercamin

403216




403216







  • 2




    An open cover must cover the entire set.
    – copper.hat
    22 hours ago










  • You got the supermum right, you got the infimum right, and you came up with a good intuition (yes the set is non-compact). But your reasoning is not formal. I would stick to the fact that a compact needs to be closed and bounded, is this closed?
    – Yanko
    22 hours ago











  • Thank you very much, i get it now!
    – Killercamin
    21 hours ago












  • 2




    An open cover must cover the entire set.
    – copper.hat
    22 hours ago










  • You got the supermum right, you got the infimum right, and you came up with a good intuition (yes the set is non-compact). But your reasoning is not formal. I would stick to the fact that a compact needs to be closed and bounded, is this closed?
    – Yanko
    22 hours ago











  • Thank you very much, i get it now!
    – Killercamin
    21 hours ago







2




2




An open cover must cover the entire set.
– copper.hat
22 hours ago




An open cover must cover the entire set.
– copper.hat
22 hours ago












You got the supermum right, you got the infimum right, and you came up with a good intuition (yes the set is non-compact). But your reasoning is not formal. I would stick to the fact that a compact needs to be closed and bounded, is this closed?
– Yanko
22 hours ago





You got the supermum right, you got the infimum right, and you came up with a good intuition (yes the set is non-compact). But your reasoning is not formal. I would stick to the fact that a compact needs to be closed and bounded, is this closed?
– Yanko
22 hours ago













Thank you very much, i get it now!
– Killercamin
21 hours ago




Thank you very much, i get it now!
– Killercamin
21 hours ago










3 Answers
3






active

oldest

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up vote
5
down vote



accepted










The set is bounded but not closed. For example, the collection $(1,2) cup (2+1 over n,4)$ is an open cover that has no finite subcover.






share|cite|improve this answer



























    up vote
    1
    down vote













    The set is not compact because the sequence $x_n=frac1n+1+1 $ which lies in $S,forall n in BbbN$, does not have a convergent subsequence in $S$



    $x_n to 1 $ thus every subsequence of $x_n$ converges to $1$






    share|cite|improve this answer



























      up vote
      1
      down vote













      The logic about the supremum and infimum is correct.



      This is non-compact, more precisely because compact sets are to be closed, and this set is not closed : $2$ is a limit point of the set(why?), and $2$ is not part of the set. Note that you need not consider the logic of open covers if you are looking at closedness and boundedness.






      share|cite|improve this answer




















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        5
        down vote



        accepted










        The set is bounded but not closed. For example, the collection $(1,2) cup (2+1 over n,4)$ is an open cover that has no finite subcover.






        share|cite|improve this answer
























          up vote
          5
          down vote



          accepted










          The set is bounded but not closed. For example, the collection $(1,2) cup (2+1 over n,4)$ is an open cover that has no finite subcover.






          share|cite|improve this answer






















            up vote
            5
            down vote



            accepted







            up vote
            5
            down vote



            accepted






            The set is bounded but not closed. For example, the collection $(1,2) cup (2+1 over n,4)$ is an open cover that has no finite subcover.






            share|cite|improve this answer












            The set is bounded but not closed. For example, the collection $(1,2) cup (2+1 over n,4)$ is an open cover that has no finite subcover.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 22 hours ago









            copper.hat

            123k557156




            123k557156




















                up vote
                1
                down vote













                The set is not compact because the sequence $x_n=frac1n+1+1 $ which lies in $S,forall n in BbbN$, does not have a convergent subsequence in $S$



                $x_n to 1 $ thus every subsequence of $x_n$ converges to $1$






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  The set is not compact because the sequence $x_n=frac1n+1+1 $ which lies in $S,forall n in BbbN$, does not have a convergent subsequence in $S$



                  $x_n to 1 $ thus every subsequence of $x_n$ converges to $1$






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    The set is not compact because the sequence $x_n=frac1n+1+1 $ which lies in $S,forall n in BbbN$, does not have a convergent subsequence in $S$



                    $x_n to 1 $ thus every subsequence of $x_n$ converges to $1$






                    share|cite|improve this answer












                    The set is not compact because the sequence $x_n=frac1n+1+1 $ which lies in $S,forall n in BbbN$, does not have a convergent subsequence in $S$



                    $x_n to 1 $ thus every subsequence of $x_n$ converges to $1$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 22 hours ago









                    Marios Gretsas

                    8,28611337




                    8,28611337




















                        up vote
                        1
                        down vote













                        The logic about the supremum and infimum is correct.



                        This is non-compact, more precisely because compact sets are to be closed, and this set is not closed : $2$ is a limit point of the set(why?), and $2$ is not part of the set. Note that you need not consider the logic of open covers if you are looking at closedness and boundedness.






                        share|cite|improve this answer
























                          up vote
                          1
                          down vote













                          The logic about the supremum and infimum is correct.



                          This is non-compact, more precisely because compact sets are to be closed, and this set is not closed : $2$ is a limit point of the set(why?), and $2$ is not part of the set. Note that you need not consider the logic of open covers if you are looking at closedness and boundedness.






                          share|cite|improve this answer






















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            The logic about the supremum and infimum is correct.



                            This is non-compact, more precisely because compact sets are to be closed, and this set is not closed : $2$ is a limit point of the set(why?), and $2$ is not part of the set. Note that you need not consider the logic of open covers if you are looking at closedness and boundedness.






                            share|cite|improve this answer












                            The logic about the supremum and infimum is correct.



                            This is non-compact, more precisely because compact sets are to be closed, and this set is not closed : $2$ is a limit point of the set(why?), and $2$ is not part of the set. Note that you need not consider the logic of open covers if you are looking at closedness and boundedness.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 22 hours ago









                            астон вілла олоф мэллбэрг

                            33.5k32870




                            33.5k32870



























                                 

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