Is $mathbb{R}^omega$ endowed with the box topology completely normal (or hereditarily normal)?












2












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Just out of curiosity, I'd like to know more properties of box topology. I found Is $mathbb{R}^omega$ a completely normal space, in the box topology? quite interesting, but unfortunately, it hasn't attracted too much attention. I also searched it in MathOverflow, the comment by Ramiro de la Vega in https://mathoverflow.net/questions/314887/is-it-still-an-open-problem-whether-mathbbr-omega-is-normal-in-the-box-top asserted it's been known the answer is negative. However, I can't obtain any further information on the Internet. Could somebody provide a disproof, or at least offer some useful links? Hopefully, this is not a duplicate.






general-topology separation-axioms box-topology







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$endgroup$












  • $begingroup$
    About a year ago on this site I asked whether it was normal and the answer was that it was still unknown.
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 6:04










  • $begingroup$
    Yeah, but completely normality is a strictly stronger notion.
    $endgroup$
    – YuiTo Cheng
    Dec 15 '18 at 6:27








  • 1




    $begingroup$
    @DanielWainfleet not completely normal has been known for quite a long time now. Normal is stil open AFAIK (in ZFC).
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 9:48












  • $begingroup$
    @HennoBrandsma. I haven't checked but it may have been you who answered this to me on my posted Q about this.
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:08












  • $begingroup$
    @DanielWainfleet I’m not sure but I don’t think so.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 18:30
















2












$begingroup$


Just out of curiosity, I'd like to know more properties of box topology. I found Is $mathbb{R}^omega$ a completely normal space, in the box topology? quite interesting, but unfortunately, it hasn't attracted too much attention. I also searched it in MathOverflow, the comment by Ramiro de la Vega in https://mathoverflow.net/questions/314887/is-it-still-an-open-problem-whether-mathbbr-omega-is-normal-in-the-box-top asserted it's been known the answer is negative. However, I can't obtain any further information on the Internet. Could somebody provide a disproof, or at least offer some useful links? Hopefully, this is not a duplicate.






general-topology separation-axioms box-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    About a year ago on this site I asked whether it was normal and the answer was that it was still unknown.
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 6:04










  • $begingroup$
    Yeah, but completely normality is a strictly stronger notion.
    $endgroup$
    – YuiTo Cheng
    Dec 15 '18 at 6:27








  • 1




    $begingroup$
    @DanielWainfleet not completely normal has been known for quite a long time now. Normal is stil open AFAIK (in ZFC).
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 9:48












  • $begingroup$
    @HennoBrandsma. I haven't checked but it may have been you who answered this to me on my posted Q about this.
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:08












  • $begingroup$
    @DanielWainfleet I’m not sure but I don’t think so.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 18:30














2












2








2





$begingroup$


Just out of curiosity, I'd like to know more properties of box topology. I found Is $mathbb{R}^omega$ a completely normal space, in the box topology? quite interesting, but unfortunately, it hasn't attracted too much attention. I also searched it in MathOverflow, the comment by Ramiro de la Vega in https://mathoverflow.net/questions/314887/is-it-still-an-open-problem-whether-mathbbr-omega-is-normal-in-the-box-top asserted it's been known the answer is negative. However, I can't obtain any further information on the Internet. Could somebody provide a disproof, or at least offer some useful links? Hopefully, this is not a duplicate.






general-topology separation-axioms box-topology







share|cite|improve this question









$endgroup$




Just out of curiosity, I'd like to know more properties of box topology. I found Is $mathbb{R}^omega$ a completely normal space, in the box topology? quite interesting, but unfortunately, it hasn't attracted too much attention. I also searched it in MathOverflow, the comment by Ramiro de la Vega in https://mathoverflow.net/questions/314887/is-it-still-an-open-problem-whether-mathbbr-omega-is-normal-in-the-box-top asserted it's been known the answer is negative. However, I can't obtain any further information on the Internet. Could somebody provide a disproof, or at least offer some useful links? Hopefully, this is not a duplicate.






general-topology separation-axioms box-topology




general-topology separation-axioms box-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 12:39









YuiTo ChengYuiTo Cheng

2,3144937




2,3144937












  • $begingroup$
    About a year ago on this site I asked whether it was normal and the answer was that it was still unknown.
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 6:04










  • $begingroup$
    Yeah, but completely normality is a strictly stronger notion.
    $endgroup$
    – YuiTo Cheng
    Dec 15 '18 at 6:27








  • 1




    $begingroup$
    @DanielWainfleet not completely normal has been known for quite a long time now. Normal is stil open AFAIK (in ZFC).
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 9:48












  • $begingroup$
    @HennoBrandsma. I haven't checked but it may have been you who answered this to me on my posted Q about this.
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:08












  • $begingroup$
    @DanielWainfleet I’m not sure but I don’t think so.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 18:30


















  • $begingroup$
    About a year ago on this site I asked whether it was normal and the answer was that it was still unknown.
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 6:04










  • $begingroup$
    Yeah, but completely normality is a strictly stronger notion.
    $endgroup$
    – YuiTo Cheng
    Dec 15 '18 at 6:27








  • 1




    $begingroup$
    @DanielWainfleet not completely normal has been known for quite a long time now. Normal is stil open AFAIK (in ZFC).
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 9:48












  • $begingroup$
    @HennoBrandsma. I haven't checked but it may have been you who answered this to me on my posted Q about this.
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 18:08












  • $begingroup$
    @DanielWainfleet I’m not sure but I don’t think so.
    $endgroup$
    – Henno Brandsma
    Dec 15 '18 at 18:30
















$begingroup$
About a year ago on this site I asked whether it was normal and the answer was that it was still unknown.
$endgroup$
– DanielWainfleet
Dec 15 '18 at 6:04




$begingroup$
About a year ago on this site I asked whether it was normal and the answer was that it was still unknown.
$endgroup$
– DanielWainfleet
Dec 15 '18 at 6:04












$begingroup$
Yeah, but completely normality is a strictly stronger notion.
$endgroup$
– YuiTo Cheng
Dec 15 '18 at 6:27






$begingroup$
Yeah, but completely normality is a strictly stronger notion.
$endgroup$
– YuiTo Cheng
Dec 15 '18 at 6:27






1




1




$begingroup$
@DanielWainfleet not completely normal has been known for quite a long time now. Normal is stil open AFAIK (in ZFC).
$endgroup$
– Henno Brandsma
Dec 15 '18 at 9:48






$begingroup$
@DanielWainfleet not completely normal has been known for quite a long time now. Normal is stil open AFAIK (in ZFC).
$endgroup$
– Henno Brandsma
Dec 15 '18 at 9:48














$begingroup$
@HennoBrandsma. I haven't checked but it may have been you who answered this to me on my posted Q about this.
$endgroup$
– DanielWainfleet
Dec 15 '18 at 18:08






$begingroup$
@HennoBrandsma. I haven't checked but it may have been you who answered this to me on my posted Q about this.
$endgroup$
– DanielWainfleet
Dec 15 '18 at 18:08














$begingroup$
@DanielWainfleet I’m not sure but I don’t think so.
$endgroup$
– Henno Brandsma
Dec 15 '18 at 18:30




$begingroup$
@DanielWainfleet I’m not sure but I don’t think so.
$endgroup$
– Henno Brandsma
Dec 15 '18 at 18:30










1 Answer
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oldest

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1












$begingroup$

It is not. Erik van Douwen showed ("The box product of countably many metrizable spaces need not be normal", Fund. Math., link) that if $X_0$ is the irrationals (as a subspace of the reals) and for $n ge 1$, $X_n = omega+1$ (a compact space: a convergent sequence in the reals is homeomorphic to it) then $Box_{n in omega} X_n$ is not normal.



This space can be seen as a subspace of $mathbb{R}^omega$ in the box topology, so the latter space is not hereditarily normal (so not completely normal). Erik himself showed in the paper (as a "byproduct") that a box product of metrisable spaces cannot be hereditarily normal if infinitely many of them are non-discrete.



A proof of the first result can also be found in Mary Ellen Rudin's "Lectures on Set theoretic topology" (a very nice book that should be read by anyone interested in research in general topology, IMHO), in the chapter on box products; this is where I found it.






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    $begingroup$

    It is not. Erik van Douwen showed ("The box product of countably many metrizable spaces need not be normal", Fund. Math., link) that if $X_0$ is the irrationals (as a subspace of the reals) and for $n ge 1$, $X_n = omega+1$ (a compact space: a convergent sequence in the reals is homeomorphic to it) then $Box_{n in omega} X_n$ is not normal.



    This space can be seen as a subspace of $mathbb{R}^omega$ in the box topology, so the latter space is not hereditarily normal (so not completely normal). Erik himself showed in the paper (as a "byproduct") that a box product of metrisable spaces cannot be hereditarily normal if infinitely many of them are non-discrete.



    A proof of the first result can also be found in Mary Ellen Rudin's "Lectures on Set theoretic topology" (a very nice book that should be read by anyone interested in research in general topology, IMHO), in the chapter on box products; this is where I found it.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      It is not. Erik van Douwen showed ("The box product of countably many metrizable spaces need not be normal", Fund. Math., link) that if $X_0$ is the irrationals (as a subspace of the reals) and for $n ge 1$, $X_n = omega+1$ (a compact space: a convergent sequence in the reals is homeomorphic to it) then $Box_{n in omega} X_n$ is not normal.



      This space can be seen as a subspace of $mathbb{R}^omega$ in the box topology, so the latter space is not hereditarily normal (so not completely normal). Erik himself showed in the paper (as a "byproduct") that a box product of metrisable spaces cannot be hereditarily normal if infinitely many of them are non-discrete.



      A proof of the first result can also be found in Mary Ellen Rudin's "Lectures on Set theoretic topology" (a very nice book that should be read by anyone interested in research in general topology, IMHO), in the chapter on box products; this is where I found it.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        It is not. Erik van Douwen showed ("The box product of countably many metrizable spaces need not be normal", Fund. Math., link) that if $X_0$ is the irrationals (as a subspace of the reals) and for $n ge 1$, $X_n = omega+1$ (a compact space: a convergent sequence in the reals is homeomorphic to it) then $Box_{n in omega} X_n$ is not normal.



        This space can be seen as a subspace of $mathbb{R}^omega$ in the box topology, so the latter space is not hereditarily normal (so not completely normal). Erik himself showed in the paper (as a "byproduct") that a box product of metrisable spaces cannot be hereditarily normal if infinitely many of them are non-discrete.



        A proof of the first result can also be found in Mary Ellen Rudin's "Lectures on Set theoretic topology" (a very nice book that should be read by anyone interested in research in general topology, IMHO), in the chapter on box products; this is where I found it.






        share|cite|improve this answer











        $endgroup$



        It is not. Erik van Douwen showed ("The box product of countably many metrizable spaces need not be normal", Fund. Math., link) that if $X_0$ is the irrationals (as a subspace of the reals) and for $n ge 1$, $X_n = omega+1$ (a compact space: a convergent sequence in the reals is homeomorphic to it) then $Box_{n in omega} X_n$ is not normal.



        This space can be seen as a subspace of $mathbb{R}^omega$ in the box topology, so the latter space is not hereditarily normal (so not completely normal). Erik himself showed in the paper (as a "byproduct") that a box product of metrisable spaces cannot be hereditarily normal if infinitely many of them are non-discrete.



        A proof of the first result can also be found in Mary Ellen Rudin's "Lectures on Set theoretic topology" (a very nice book that should be read by anyone interested in research in general topology, IMHO), in the chapter on box products; this is where I found it.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 15 '18 at 10:00

























        answered Dec 15 '18 at 9:35









        Henno BrandsmaHenno Brandsma

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