Nested homeomorphic sets












4












$begingroup$


Suppose we have a countable collection of sets ${U_n}$ such that $U_nsubset U_{n+1}$ for each $n$ and $U_n$ is homeomorphic to $mathbb{R}$ (or more generally, $X$) for each $n$, then is $bigcup_{n=1}^infty U_n$ homeomorphic to $mathbb{R}$ (or $X$)? I am not sure pushing $n$ to $infty$ works and I couldn't construct an explicit homeomorphism. Thanks!






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try $U_n=[0,n]$.
    $endgroup$
    – Gerry Myerson
    2 hours ago
















4












$begingroup$


Suppose we have a countable collection of sets ${U_n}$ such that $U_nsubset U_{n+1}$ for each $n$ and $U_n$ is homeomorphic to $mathbb{R}$ (or more generally, $X$) for each $n$, then is $bigcup_{n=1}^infty U_n$ homeomorphic to $mathbb{R}$ (or $X$)? I am not sure pushing $n$ to $infty$ works and I couldn't construct an explicit homeomorphism. Thanks!






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Try $U_n=[0,n]$.
    $endgroup$
    – Gerry Myerson
    2 hours ago














4












4








4





$begingroup$


Suppose we have a countable collection of sets ${U_n}$ such that $U_nsubset U_{n+1}$ for each $n$ and $U_n$ is homeomorphic to $mathbb{R}$ (or more generally, $X$) for each $n$, then is $bigcup_{n=1}^infty U_n$ homeomorphic to $mathbb{R}$ (or $X$)? I am not sure pushing $n$ to $infty$ works and I couldn't construct an explicit homeomorphism. Thanks!






general-topology







share|cite|improve this question











$endgroup$




Suppose we have a countable collection of sets ${U_n}$ such that $U_nsubset U_{n+1}$ for each $n$ and $U_n$ is homeomorphic to $mathbb{R}$ (or more generally, $X$) for each $n$, then is $bigcup_{n=1}^infty U_n$ homeomorphic to $mathbb{R}$ (or $X$)? I am not sure pushing $n$ to $infty$ works and I couldn't construct an explicit homeomorphism. Thanks!






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









bof

51.3k457120




51.3k457120










asked 3 hours ago









J.DoeJ.Doe

6814




6814








  • 1




    $begingroup$
    Try $U_n=[0,n]$.
    $endgroup$
    – Gerry Myerson
    2 hours ago














  • 1




    $begingroup$
    Try $U_n=[0,n]$.
    $endgroup$
    – Gerry Myerson
    2 hours ago








1




1




$begingroup$
Try $U_n=[0,n]$.
$endgroup$
– Gerry Myerson
2 hours ago




$begingroup$
Try $U_n=[0,n]$.
$endgroup$
– Gerry Myerson
2 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

Let $$U_i = {e^{2pi i theta} -1 mid theta in [0,1-1/i)} cup {e^{2pi i theta} +1 mid theta in [1/2, 3/2-1/i)}.$$



Then every $U_i$ is a subset of the union of two circles that are glued together at a single point at the origin and is homeomorphic to $mathbb{R}$. The union $bigcup_i geq 1$ is equal to the union of the two circles.




Here's a simpler example for $Xcong [0,1)$ using the same idea. Let $$U_i = {e^{2pi itheta} mid theta in [0,1-1/i)}.$$



Then $U_i cong [0,1)$ for all $i geq 1$, but $bigcup_{i geq 1} U_i = S^1$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think this would be a better example if you simply gave your solution to the $mathbb R$ case (without the "wedge" language, which people may not understand).
    $endgroup$
    – zhw.
    2 hours ago










  • $begingroup$
    Or you could simply take $U_n=[-n,n)$ and $bigcup_nU_n=mathbb R.$ But I'm surprised that you can do the same thing for $Xcongmathbb R$. Could you please give an explanation for dummies?
    $endgroup$
    – bof
    2 hours ago










  • $begingroup$
    @zhw. thanks for the suggestion. Hopefully the edit is a bit easier to read (unfortunately, it's not too easy to write the $U_i$s in a very simple way without talking about gluing/wedges).
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Oh, right. Or $U_n={t:-2lt tle1}cup{e^{it}:0le tltpi-frac1n}$. By the way, you are using the letter $i$ for two different things.
    $endgroup$
    – bof
    1 hour ago



















0












$begingroup$

For $X=Bbb R$, it's homeomorphic with any open interval, and $U_nsubseteq U_{n+1}$ implies here that $U_n$ is a subinterval.

Thus, we can choose decreasing $a_n$ and increasing $b_n$ such that $U_n$ is homeomorphically mapped to $(a_n, b_n)$.

Then, $cup_nU_n$ will be homeomorphic with $(inf a_n, sup b_n) $.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This does not appear to answer the question.
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    2 hours ago











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3087802%2fnested-homeomorphic-sets%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Let $$U_i = {e^{2pi i theta} -1 mid theta in [0,1-1/i)} cup {e^{2pi i theta} +1 mid theta in [1/2, 3/2-1/i)}.$$



Then every $U_i$ is a subset of the union of two circles that are glued together at a single point at the origin and is homeomorphic to $mathbb{R}$. The union $bigcup_i geq 1$ is equal to the union of the two circles.




Here's a simpler example for $Xcong [0,1)$ using the same idea. Let $$U_i = {e^{2pi itheta} mid theta in [0,1-1/i)}.$$



Then $U_i cong [0,1)$ for all $i geq 1$, but $bigcup_{i geq 1} U_i = S^1$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think this would be a better example if you simply gave your solution to the $mathbb R$ case (without the "wedge" language, which people may not understand).
    $endgroup$
    – zhw.
    2 hours ago










  • $begingroup$
    Or you could simply take $U_n=[-n,n)$ and $bigcup_nU_n=mathbb R.$ But I'm surprised that you can do the same thing for $Xcongmathbb R$. Could you please give an explanation for dummies?
    $endgroup$
    – bof
    2 hours ago










  • $begingroup$
    @zhw. thanks for the suggestion. Hopefully the edit is a bit easier to read (unfortunately, it's not too easy to write the $U_i$s in a very simple way without talking about gluing/wedges).
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Oh, right. Or $U_n={t:-2lt tle1}cup{e^{it}:0le tltpi-frac1n}$. By the way, you are using the letter $i$ for two different things.
    $endgroup$
    – bof
    1 hour ago
















4












$begingroup$

Let $$U_i = {e^{2pi i theta} -1 mid theta in [0,1-1/i)} cup {e^{2pi i theta} +1 mid theta in [1/2, 3/2-1/i)}.$$



Then every $U_i$ is a subset of the union of two circles that are glued together at a single point at the origin and is homeomorphic to $mathbb{R}$. The union $bigcup_i geq 1$ is equal to the union of the two circles.




Here's a simpler example for $Xcong [0,1)$ using the same idea. Let $$U_i = {e^{2pi itheta} mid theta in [0,1-1/i)}.$$



Then $U_i cong [0,1)$ for all $i geq 1$, but $bigcup_{i geq 1} U_i = S^1$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think this would be a better example if you simply gave your solution to the $mathbb R$ case (without the "wedge" language, which people may not understand).
    $endgroup$
    – zhw.
    2 hours ago










  • $begingroup$
    Or you could simply take $U_n=[-n,n)$ and $bigcup_nU_n=mathbb R.$ But I'm surprised that you can do the same thing for $Xcongmathbb R$. Could you please give an explanation for dummies?
    $endgroup$
    – bof
    2 hours ago










  • $begingroup$
    @zhw. thanks for the suggestion. Hopefully the edit is a bit easier to read (unfortunately, it's not too easy to write the $U_i$s in a very simple way without talking about gluing/wedges).
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Oh, right. Or $U_n={t:-2lt tle1}cup{e^{it}:0le tltpi-frac1n}$. By the way, you are using the letter $i$ for two different things.
    $endgroup$
    – bof
    1 hour ago














4












4








4





$begingroup$

Let $$U_i = {e^{2pi i theta} -1 mid theta in [0,1-1/i)} cup {e^{2pi i theta} +1 mid theta in [1/2, 3/2-1/i)}.$$



Then every $U_i$ is a subset of the union of two circles that are glued together at a single point at the origin and is homeomorphic to $mathbb{R}$. The union $bigcup_i geq 1$ is equal to the union of the two circles.




Here's a simpler example for $Xcong [0,1)$ using the same idea. Let $$U_i = {e^{2pi itheta} mid theta in [0,1-1/i)}.$$



Then $U_i cong [0,1)$ for all $i geq 1$, but $bigcup_{i geq 1} U_i = S^1$.






share|cite|improve this answer











$endgroup$



Let $$U_i = {e^{2pi i theta} -1 mid theta in [0,1-1/i)} cup {e^{2pi i theta} +1 mid theta in [1/2, 3/2-1/i)}.$$



Then every $U_i$ is a subset of the union of two circles that are glued together at a single point at the origin and is homeomorphic to $mathbb{R}$. The union $bigcup_i geq 1$ is equal to the union of the two circles.




Here's a simpler example for $Xcong [0,1)$ using the same idea. Let $$U_i = {e^{2pi itheta} mid theta in [0,1-1/i)}.$$



Then $U_i cong [0,1)$ for all $i geq 1$, but $bigcup_{i geq 1} U_i = S^1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 3 hours ago









Dan RustDan Rust

22.7k114884




22.7k114884








  • 1




    $begingroup$
    I think this would be a better example if you simply gave your solution to the $mathbb R$ case (without the "wedge" language, which people may not understand).
    $endgroup$
    – zhw.
    2 hours ago










  • $begingroup$
    Or you could simply take $U_n=[-n,n)$ and $bigcup_nU_n=mathbb R.$ But I'm surprised that you can do the same thing for $Xcongmathbb R$. Could you please give an explanation for dummies?
    $endgroup$
    – bof
    2 hours ago










  • $begingroup$
    @zhw. thanks for the suggestion. Hopefully the edit is a bit easier to read (unfortunately, it's not too easy to write the $U_i$s in a very simple way without talking about gluing/wedges).
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Oh, right. Or $U_n={t:-2lt tle1}cup{e^{it}:0le tltpi-frac1n}$. By the way, you are using the letter $i$ for two different things.
    $endgroup$
    – bof
    1 hour ago














  • 1




    $begingroup$
    I think this would be a better example if you simply gave your solution to the $mathbb R$ case (without the "wedge" language, which people may not understand).
    $endgroup$
    – zhw.
    2 hours ago










  • $begingroup$
    Or you could simply take $U_n=[-n,n)$ and $bigcup_nU_n=mathbb R.$ But I'm surprised that you can do the same thing for $Xcongmathbb R$. Could you please give an explanation for dummies?
    $endgroup$
    – bof
    2 hours ago










  • $begingroup$
    @zhw. thanks for the suggestion. Hopefully the edit is a bit easier to read (unfortunately, it's not too easy to write the $U_i$s in a very simple way without talking about gluing/wedges).
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Oh, right. Or $U_n={t:-2lt tle1}cup{e^{it}:0le tltpi-frac1n}$. By the way, you are using the letter $i$ for two different things.
    $endgroup$
    – bof
    1 hour ago








1




1




$begingroup$
I think this would be a better example if you simply gave your solution to the $mathbb R$ case (without the "wedge" language, which people may not understand).
$endgroup$
– zhw.
2 hours ago




$begingroup$
I think this would be a better example if you simply gave your solution to the $mathbb R$ case (without the "wedge" language, which people may not understand).
$endgroup$
– zhw.
2 hours ago












$begingroup$
Or you could simply take $U_n=[-n,n)$ and $bigcup_nU_n=mathbb R.$ But I'm surprised that you can do the same thing for $Xcongmathbb R$. Could you please give an explanation for dummies?
$endgroup$
– bof
2 hours ago




$begingroup$
Or you could simply take $U_n=[-n,n)$ and $bigcup_nU_n=mathbb R.$ But I'm surprised that you can do the same thing for $Xcongmathbb R$. Could you please give an explanation for dummies?
$endgroup$
– bof
2 hours ago












$begingroup$
@zhw. thanks for the suggestion. Hopefully the edit is a bit easier to read (unfortunately, it's not too easy to write the $U_i$s in a very simple way without talking about gluing/wedges).
$endgroup$
– Dan Rust
2 hours ago




$begingroup$
@zhw. thanks for the suggestion. Hopefully the edit is a bit easier to read (unfortunately, it's not too easy to write the $U_i$s in a very simple way without talking about gluing/wedges).
$endgroup$
– Dan Rust
2 hours ago












$begingroup$
Oh, right. Or $U_n={t:-2lt tle1}cup{e^{it}:0le tltpi-frac1n}$. By the way, you are using the letter $i$ for two different things.
$endgroup$
– bof
1 hour ago




$begingroup$
Oh, right. Or $U_n={t:-2lt tle1}cup{e^{it}:0le tltpi-frac1n}$. By the way, you are using the letter $i$ for two different things.
$endgroup$
– bof
1 hour ago











0












$begingroup$

For $X=Bbb R$, it's homeomorphic with any open interval, and $U_nsubseteq U_{n+1}$ implies here that $U_n$ is a subinterval.

Thus, we can choose decreasing $a_n$ and increasing $b_n$ such that $U_n$ is homeomorphically mapped to $(a_n, b_n)$.

Then, $cup_nU_n$ will be homeomorphic with $(inf a_n, sup b_n) $.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This does not appear to answer the question.
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    2 hours ago
















0












$begingroup$

For $X=Bbb R$, it's homeomorphic with any open interval, and $U_nsubseteq U_{n+1}$ implies here that $U_n$ is a subinterval.

Thus, we can choose decreasing $a_n$ and increasing $b_n$ such that $U_n$ is homeomorphically mapped to $(a_n, b_n)$.

Then, $cup_nU_n$ will be homeomorphic with $(inf a_n, sup b_n) $.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This does not appear to answer the question.
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    2 hours ago














0












0








0





$begingroup$

For $X=Bbb R$, it's homeomorphic with any open interval, and $U_nsubseteq U_{n+1}$ implies here that $U_n$ is a subinterval.

Thus, we can choose decreasing $a_n$ and increasing $b_n$ such that $U_n$ is homeomorphically mapped to $(a_n, b_n)$.

Then, $cup_nU_n$ will be homeomorphic with $(inf a_n, sup b_n) $.






share|cite|improve this answer











$endgroup$



For $X=Bbb R$, it's homeomorphic with any open interval, and $U_nsubseteq U_{n+1}$ implies here that $U_n$ is a subinterval.

Thus, we can choose decreasing $a_n$ and increasing $b_n$ such that $U_n$ is homeomorphically mapped to $(a_n, b_n)$.

Then, $cup_nU_n$ will be homeomorphic with $(inf a_n, sup b_n) $.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 2 hours ago









BerciBerci

60.2k23672




60.2k23672












  • $begingroup$
    This does not appear to answer the question.
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    2 hours ago


















  • $begingroup$
    This does not appear to answer the question.
    $endgroup$
    – Dan Rust
    2 hours ago










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    2 hours ago
















$begingroup$
This does not appear to answer the question.
$endgroup$
– Dan Rust
2 hours ago




$begingroup$
This does not appear to answer the question.
$endgroup$
– Dan Rust
2 hours ago












$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
2 hours ago




$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
2 hours ago


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3087802%2fnested-homeomorphic-sets%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

الفوسفات في المغرب

Image
قطار ينقل فوسفات من منطقة خريبكة متجه نحو الدار البيضاء. ديار عمال بمصانع الفوسفات بمدينة آسفي. المحتوى هنا ينقصه الاستشهاد بمصادر. يرجى إيراد مصادر موثوق بها. أي معلومات غير موثقة يمكن التشكيك بها وإزالتها. (أغسطس 2015) الفوسفات في المغرب هو ثالث مصدر عالمي للفوسفات، ويتوفر المغرب على %75 من الاحتياطي العالمي  [بحاجة لمصدر] ويتوزع بنسب متفاوتة بين عدة مناطق أساسية وهي: وادي زم بن جرير بوكراع خريبكة اليوسفية أولاد عبدون كنتور واد الذهب مستقلة. هذه الأخيرة لم تفتح بعد للاستغلال. الصادرات الحامض الفوسفوري P205 تمتص الهند %50 من ا لصادرات الإجمالية آسيا (دون روسيا)%59 أوروبا %30 أمريكا اللاتينية %11 أسمدة صلبة أوروبا 38 % أمريكا اللاتينية %30 آسيا (دون روسيا) 22 % أستراليا 9 % بوابة المغرب هذه بذرة مقالة عن علم الاقتصاد أو موضوع متعلق به بحاجة للتوسيع. شارك في تحريرها. This page is only for reference, If you need detailed information, please check here
Read more

Four equal circles intersect: What is the area of the small shaded portion and its height

Image
3 $begingroup$ In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this. geometry circle share | cite | improve this question edited 2 hours ago Abdulhameed asked 2 hours ago ...
Read more

جامعة ليفربول

Image
جامعة ليفربول University of Liverpool الأسماء السابقة كلية ليفربول الجامعية (1882-1884) جامعة فيكتوريا في ليفربول (1884-1903) معلومات التأسيس 1882 (منذ 137 سنة) الإنتماءات رابطة الجامعات الأوروبية مجموعة راسل جامعات القرميد الأحمر المنحة المالية 143.5 مليون £ النوع جامعة عامة الكليات 6 الموقع الجغرافي إحداثيات 53°24′22″N 2°58′01″W  /  53.406°N 2.967°W  / 53.406; -2.967   المدينة ليفربول المكان   إنجلترا البلد   المملكة المتحدة رقم الهاتف 7942000 151 (44+) إحصاءات الطلاب 17,405 (سنة 2008) عضوية مجموعة رسل [1] أورسيد [2]   طلاب الدراسات العليا 3,860 الموقع www.liverpool.ac.uk تعديل   جامعة ليفربول مبنى فكتوريا جامعة ليفربول (بالإنجليزية: The University of Liverpool )) إحدى الجامعات البريطانية العريقة، تقع في مدينة ليفربول على الساحل الشمالي الغربي من أنجلترا . وهي عضو في مجموعة راسل للجامعات وتعتبر من ضمن جامعات الطوب الأحمر الستة. محتويات ...
Read more