Is $mathbb{R}^omega$ endowed with the box topology completely normal (or hereditarily normal)?
$begingroup$
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
asked Dec 14 '18 at 12:39
YuiTo ChengYuiTo Cheng
2,3144937
$endgroup$
add a comment |
$begingroup$
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
asked Dec 14 '18 at 12:39
YuiTo ChengYuiTo Cheng
2,3144937
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 14 '18 at 12:39
YuiTo ChengYuiTo Cheng
2,3144937
$endgroup$
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
asked Dec 14 '18 at 12:39
YuiTo ChengYuiTo Cheng
2,3144937
asked Dec 14 '18 at 12:39
YuiTo ChengYuiTo Cheng
2,3144937
asked Dec 14 '18 at 12:39
YuiTo ChengYuiTo Cheng
2,3144937
asked Dec 14 '18 at 12:39
YuiTo ChengYuiTo Cheng
2,3144937
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 15 '18 at 9:35
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039316%2fis-mathbbr-omega-endowed-with-the-box-topology-completely-normal-or-hered%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 15 '18 at 9:35
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 15 '18 at 9:35
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 15 '18 at 9:35
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
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.
answered Dec 15 '18 at 9:35
Henno BrandsmaHenno Brandsma
116k349127
edited Dec 15 '18 at 10:00
answered Dec 15 '18 at 9:35
Henno BrandsmaHenno Brandsma
116k349127
answered Dec 15 '18 at 9:35
Henno BrandsmaHenno Brandsma
116k349127
answered Dec 15 '18 at 9:35
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039316%2fis-mathbbr-omega-endowed-with-the-box-topology-completely-normal-or-hered%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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