Not $sigma$-compact set without axiom of choice












0












$begingroup$



Today in measure theory, we introduced the concept of a $sigma$-compact set, which is a set which can be expressed as the countable union of compact set.




Since the set of $sigma$-compact sets seemed to be very large (I couldn't think of a not $sigma$-compact set), I asked for a set which is not $sigma$-compact, but our instructor couldn't come up with a set, which he didn't have to explicitly construct with the axiom of choice, for which we didn't have time.




My question is if there is such an set which can be constructed without the axiom of choice.







measure-theory compactness axiom-of-choice







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    But this is not about "sets" it's about topology.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 21:50






  • 6




    $begingroup$
    The irrational numbers are famously not $sigma$-compact.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 21:51










  • $begingroup$
    $mathbb{R}$ in the discrete topology is not $sigma$-compact, as its only compact subsets are finite.
    $endgroup$
    – Henno Brandsma
    Dec 18 '18 at 22:03






  • 1




    $begingroup$
    @Henno: Since this is about measure theory, I expect the question is really in the context of $Bbb R$ with the standard topology and Lebesgue measure.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:06






  • 1




    $begingroup$
    Viktor, compact is a topological notion. Not a measure theoretic one.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:18
















0












$begingroup$



Today in measure theory, we introduced the concept of a $sigma$-compact set, which is a set which can be expressed as the countable union of compact set.




Since the set of $sigma$-compact sets seemed to be very large (I couldn't think of a not $sigma$-compact set), I asked for a set which is not $sigma$-compact, but our instructor couldn't come up with a set, which he didn't have to explicitly construct with the axiom of choice, for which we didn't have time.




My question is if there is such an set which can be constructed without the axiom of choice.







measure-theory compactness axiom-of-choice







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    But this is not about "sets" it's about topology.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 21:50






  • 6




    $begingroup$
    The irrational numbers are famously not $sigma$-compact.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 21:51










  • $begingroup$
    $mathbb{R}$ in the discrete topology is not $sigma$-compact, as its only compact subsets are finite.
    $endgroup$
    – Henno Brandsma
    Dec 18 '18 at 22:03






  • 1




    $begingroup$
    @Henno: Since this is about measure theory, I expect the question is really in the context of $Bbb R$ with the standard topology and Lebesgue measure.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:06






  • 1




    $begingroup$
    Viktor, compact is a topological notion. Not a measure theoretic one.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:18














0












0








0





$begingroup$



Today in measure theory, we introduced the concept of a $sigma$-compact set, which is a set which can be expressed as the countable union of compact set.




Since the set of $sigma$-compact sets seemed to be very large (I couldn't think of a not $sigma$-compact set), I asked for a set which is not $sigma$-compact, but our instructor couldn't come up with a set, which he didn't have to explicitly construct with the axiom of choice, for which we didn't have time.




My question is if there is such an set which can be constructed without the axiom of choice.







measure-theory compactness axiom-of-choice







share|cite|improve this question









$endgroup$





Today in measure theory, we introduced the concept of a $sigma$-compact set, which is a set which can be expressed as the countable union of compact set.




Since the set of $sigma$-compact sets seemed to be very large (I couldn't think of a not $sigma$-compact set), I asked for a set which is not $sigma$-compact, but our instructor couldn't come up with a set, which he didn't have to explicitly construct with the axiom of choice, for which we didn't have time.




My question is if there is such an set which can be constructed without the axiom of choice.







measure-theory compactness axiom-of-choice




measure-theory compactness axiom-of-choice






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 21:49









Viktor GlombikViktor Glombik

9661527




9661527








  • 1




    $begingroup$
    But this is not about "sets" it's about topology.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 21:50






  • 6




    $begingroup$
    The irrational numbers are famously not $sigma$-compact.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 21:51










  • $begingroup$
    $mathbb{R}$ in the discrete topology is not $sigma$-compact, as its only compact subsets are finite.
    $endgroup$
    – Henno Brandsma
    Dec 18 '18 at 22:03






  • 1




    $begingroup$
    @Henno: Since this is about measure theory, I expect the question is really in the context of $Bbb R$ with the standard topology and Lebesgue measure.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:06






  • 1




    $begingroup$
    Viktor, compact is a topological notion. Not a measure theoretic one.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:18














  • 1




    $begingroup$
    But this is not about "sets" it's about topology.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 21:50






  • 6




    $begingroup$
    The irrational numbers are famously not $sigma$-compact.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 21:51










  • $begingroup$
    $mathbb{R}$ in the discrete topology is not $sigma$-compact, as its only compact subsets are finite.
    $endgroup$
    – Henno Brandsma
    Dec 18 '18 at 22:03






  • 1




    $begingroup$
    @Henno: Since this is about measure theory, I expect the question is really in the context of $Bbb R$ with the standard topology and Lebesgue measure.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:06






  • 1




    $begingroup$
    Viktor, compact is a topological notion. Not a measure theoretic one.
    $endgroup$
    – Asaf Karagila♦
    Dec 18 '18 at 22:18








1




1




$begingroup$
But this is not about "sets" it's about topology.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 21:50




$begingroup$
But this is not about "sets" it's about topology.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 21:50




6




6




$begingroup$
The irrational numbers are famously not $sigma$-compact.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 21:51




$begingroup$
The irrational numbers are famously not $sigma$-compact.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 21:51












$begingroup$
$mathbb{R}$ in the discrete topology is not $sigma$-compact, as its only compact subsets are finite.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:03




$begingroup$
$mathbb{R}$ in the discrete topology is not $sigma$-compact, as its only compact subsets are finite.
$endgroup$
– Henno Brandsma
Dec 18 '18 at 22:03




1




1




$begingroup$
@Henno: Since this is about measure theory, I expect the question is really in the context of $Bbb R$ with the standard topology and Lebesgue measure.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 22:06




$begingroup$
@Henno: Since this is about measure theory, I expect the question is really in the context of $Bbb R$ with the standard topology and Lebesgue measure.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 22:06




1




1




$begingroup$
Viktor, compact is a topological notion. Not a measure theoretic one.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 22:18




$begingroup$
Viktor, compact is a topological notion. Not a measure theoretic one.
$endgroup$
– Asaf Karagila♦
Dec 18 '18 at 22:18










2 Answers
2






active

oldest

votes


















3












$begingroup$

The axiom of choice is not needed to construct a subset of $Bbb R$ which is not $sigma$-compact (which I guess is ultimately what this question is about).



The irrational numbers form such a set, since any compact subset of $Bbb{Rsetminus Q}$ has an empty interior, and therefore any $sigma$-compact set of irrational numbers is the countable union of closed nowhere dense sets, which means that it is meager. But Baire's theorem shows that the irrational numbers are not meager.



(To the initiated, Baire's theorem is provable in $sf ZF$ when the space is separable, like $Bbb{Rsetminus Q}$ which has $pi+Bbb Q$ as a dense subset.)






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that - in $mathbb{R}$ at least - a set is $sigma$-compact only if (indeed, iff) it is the union of countably many closed sets. Such a set is called $F_sigma$ (or $Pi^0_2$). A fundamental theorem concerning the topology of $mathbb{R}$ is the Baire category theorem, which states that the intersection of countably many dense open sets is dense. An easy corollary is that no countable dense set can be the intersection of countably many open sets (exercise!), and as a consequence the complement of a countable dense set cannot be $F_sigma$.



    So in particular, the set of irrationals is not $sigma$-compact.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Well, BCT is generally equivalent to DC. So just saying "Oh BCT shows ..." is not enough to conclude that choice was not needed. But noting that the irrationals are separable is enough to finish the claim.
      $endgroup$
      – Asaf Karagila♦
      Dec 18 '18 at 21:58










    • $begingroup$
      I assure BCT is the Baire Category Theorem, but what is DC?
      $endgroup$
      – Viktor Glombik
      Dec 18 '18 at 22:15










    • $begingroup$
      @Viktor: Dependent Choice. It is a weak form of choice (or a strong form of countable choice).
      $endgroup$
      – Asaf Karagila♦
      Dec 18 '18 at 22:19











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The axiom of choice is not needed to construct a subset of $Bbb R$ which is not $sigma$-compact (which I guess is ultimately what this question is about).



    The irrational numbers form such a set, since any compact subset of $Bbb{Rsetminus Q}$ has an empty interior, and therefore any $sigma$-compact set of irrational numbers is the countable union of closed nowhere dense sets, which means that it is meager. But Baire's theorem shows that the irrational numbers are not meager.



    (To the initiated, Baire's theorem is provable in $sf ZF$ when the space is separable, like $Bbb{Rsetminus Q}$ which has $pi+Bbb Q$ as a dense subset.)






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The axiom of choice is not needed to construct a subset of $Bbb R$ which is not $sigma$-compact (which I guess is ultimately what this question is about).



      The irrational numbers form such a set, since any compact subset of $Bbb{Rsetminus Q}$ has an empty interior, and therefore any $sigma$-compact set of irrational numbers is the countable union of closed nowhere dense sets, which means that it is meager. But Baire's theorem shows that the irrational numbers are not meager.



      (To the initiated, Baire's theorem is provable in $sf ZF$ when the space is separable, like $Bbb{Rsetminus Q}$ which has $pi+Bbb Q$ as a dense subset.)






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The axiom of choice is not needed to construct a subset of $Bbb R$ which is not $sigma$-compact (which I guess is ultimately what this question is about).



        The irrational numbers form such a set, since any compact subset of $Bbb{Rsetminus Q}$ has an empty interior, and therefore any $sigma$-compact set of irrational numbers is the countable union of closed nowhere dense sets, which means that it is meager. But Baire's theorem shows that the irrational numbers are not meager.



        (To the initiated, Baire's theorem is provable in $sf ZF$ when the space is separable, like $Bbb{Rsetminus Q}$ which has $pi+Bbb Q$ as a dense subset.)






        share|cite|improve this answer









        $endgroup$



        The axiom of choice is not needed to construct a subset of $Bbb R$ which is not $sigma$-compact (which I guess is ultimately what this question is about).



        The irrational numbers form such a set, since any compact subset of $Bbb{Rsetminus Q}$ has an empty interior, and therefore any $sigma$-compact set of irrational numbers is the countable union of closed nowhere dense sets, which means that it is meager. But Baire's theorem shows that the irrational numbers are not meager.



        (To the initiated, Baire's theorem is provable in $sf ZF$ when the space is separable, like $Bbb{Rsetminus Q}$ which has $pi+Bbb Q$ as a dense subset.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 21:56









        Asaf Karagila♦Asaf Karagila

        304k32433764




        304k32433764























            1












            $begingroup$

            Note that - in $mathbb{R}$ at least - a set is $sigma$-compact only if (indeed, iff) it is the union of countably many closed sets. Such a set is called $F_sigma$ (or $Pi^0_2$). A fundamental theorem concerning the topology of $mathbb{R}$ is the Baire category theorem, which states that the intersection of countably many dense open sets is dense. An easy corollary is that no countable dense set can be the intersection of countably many open sets (exercise!), and as a consequence the complement of a countable dense set cannot be $F_sigma$.



            So in particular, the set of irrationals is not $sigma$-compact.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Well, BCT is generally equivalent to DC. So just saying "Oh BCT shows ..." is not enough to conclude that choice was not needed. But noting that the irrationals are separable is enough to finish the claim.
              $endgroup$
              – Asaf Karagila♦
              Dec 18 '18 at 21:58










            • $begingroup$
              I assure BCT is the Baire Category Theorem, but what is DC?
              $endgroup$
              – Viktor Glombik
              Dec 18 '18 at 22:15










            • $begingroup$
              @Viktor: Dependent Choice. It is a weak form of choice (or a strong form of countable choice).
              $endgroup$
              – Asaf Karagila♦
              Dec 18 '18 at 22:19
















            1












            $begingroup$

            Note that - in $mathbb{R}$ at least - a set is $sigma$-compact only if (indeed, iff) it is the union of countably many closed sets. Such a set is called $F_sigma$ (or $Pi^0_2$). A fundamental theorem concerning the topology of $mathbb{R}$ is the Baire category theorem, which states that the intersection of countably many dense open sets is dense. An easy corollary is that no countable dense set can be the intersection of countably many open sets (exercise!), and as a consequence the complement of a countable dense set cannot be $F_sigma$.



            So in particular, the set of irrationals is not $sigma$-compact.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Well, BCT is generally equivalent to DC. So just saying "Oh BCT shows ..." is not enough to conclude that choice was not needed. But noting that the irrationals are separable is enough to finish the claim.
              $endgroup$
              – Asaf Karagila♦
              Dec 18 '18 at 21:58










            • $begingroup$
              I assure BCT is the Baire Category Theorem, but what is DC?
              $endgroup$
              – Viktor Glombik
              Dec 18 '18 at 22:15










            • $begingroup$
              @Viktor: Dependent Choice. It is a weak form of choice (or a strong form of countable choice).
              $endgroup$
              – Asaf Karagila♦
              Dec 18 '18 at 22:19














            1












            1








            1





            $begingroup$

            Note that - in $mathbb{R}$ at least - a set is $sigma$-compact only if (indeed, iff) it is the union of countably many closed sets. Such a set is called $F_sigma$ (or $Pi^0_2$). A fundamental theorem concerning the topology of $mathbb{R}$ is the Baire category theorem, which states that the intersection of countably many dense open sets is dense. An easy corollary is that no countable dense set can be the intersection of countably many open sets (exercise!), and as a consequence the complement of a countable dense set cannot be $F_sigma$.



            So in particular, the set of irrationals is not $sigma$-compact.






            share|cite|improve this answer









            $endgroup$



            Note that - in $mathbb{R}$ at least - a set is $sigma$-compact only if (indeed, iff) it is the union of countably many closed sets. Such a set is called $F_sigma$ (or $Pi^0_2$). A fundamental theorem concerning the topology of $mathbb{R}$ is the Baire category theorem, which states that the intersection of countably many dense open sets is dense. An easy corollary is that no countable dense set can be the intersection of countably many open sets (exercise!), and as a consequence the complement of a countable dense set cannot be $F_sigma$.



            So in particular, the set of irrationals is not $sigma$-compact.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 21:56









            Noah SchweberNoah Schweber

            125k10150287




            125k10150287












            • $begingroup$
              Well, BCT is generally equivalent to DC. So just saying "Oh BCT shows ..." is not enough to conclude that choice was not needed. But noting that the irrationals are separable is enough to finish the claim.
              $endgroup$
              – Asaf Karagila♦
              Dec 18 '18 at 21:58










            • $begingroup$
              I assure BCT is the Baire Category Theorem, but what is DC?
              $endgroup$
              – Viktor Glombik
              Dec 18 '18 at 22:15










            • $begingroup$
              @Viktor: Dependent Choice. It is a weak form of choice (or a strong form of countable choice).
              $endgroup$
              – Asaf Karagila♦
              Dec 18 '18 at 22:19


















            • $begingroup$
              Well, BCT is generally equivalent to DC. So just saying "Oh BCT shows ..." is not enough to conclude that choice was not needed. But noting that the irrationals are separable is enough to finish the claim.
              $endgroup$
              – Asaf Karagila♦
              Dec 18 '18 at 21:58










            • $begingroup$
              I assure BCT is the Baire Category Theorem, but what is DC?
              $endgroup$
              – Viktor Glombik
              Dec 18 '18 at 22:15










            • $begingroup$
              @Viktor: Dependent Choice. It is a weak form of choice (or a strong form of countable choice).
              $endgroup$
              – Asaf Karagila♦
              Dec 18 '18 at 22:19
















            $begingroup$
            Well, BCT is generally equivalent to DC. So just saying "Oh BCT shows ..." is not enough to conclude that choice was not needed. But noting that the irrationals are separable is enough to finish the claim.
            $endgroup$
            – Asaf Karagila♦
            Dec 18 '18 at 21:58




            $begingroup$
            Well, BCT is generally equivalent to DC. So just saying "Oh BCT shows ..." is not enough to conclude that choice was not needed. But noting that the irrationals are separable is enough to finish the claim.
            $endgroup$
            – Asaf Karagila♦
            Dec 18 '18 at 21:58












            $begingroup$
            I assure BCT is the Baire Category Theorem, but what is DC?
            $endgroup$
            – Viktor Glombik
            Dec 18 '18 at 22:15




            $begingroup$
            I assure BCT is the Baire Category Theorem, but what is DC?
            $endgroup$
            – Viktor Glombik
            Dec 18 '18 at 22:15












            $begingroup$
            @Viktor: Dependent Choice. It is a weak form of choice (or a strong form of countable choice).
            $endgroup$
            – Asaf Karagila♦
            Dec 18 '18 at 22:19




            $begingroup$
            @Viktor: Dependent Choice. It is a weak form of choice (or a strong form of countable choice).
            $endgroup$
            – Asaf Karagila♦
            Dec 18 '18 at 22:19


















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