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
    1 hour 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
    1 hour 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 1 hour ago









bof

51.3k457120




51.3k457120










asked 1 hour ago









J.DoeJ.Doe

6814




6814








  • 1




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














  • 1




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








1




1




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




$begingroup$
Try $U_n=[0,n]$.
$endgroup$
– Gerry Myerson
1 hour 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.
    46 mins 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
    39 mins 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
    38 mins 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
    13 mins 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
    1 hour ago










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    1 hour 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.
    46 mins 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
    39 mins 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
    38 mins 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
    13 mins 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.
    46 mins 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
    39 mins 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
    38 mins 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
    13 mins 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 39 mins ago

























answered 1 hour 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.
    46 mins 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
    39 mins 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
    38 mins 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
    13 mins 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.
    46 mins 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
    39 mins 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
    38 mins 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
    13 mins 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.
46 mins 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.
46 mins 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
39 mins 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
39 mins 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
38 mins 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
38 mins 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
13 mins 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
13 mins 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
    1 hour ago










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    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
    1 hour ago










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    1 hour 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 1 hour ago

























answered 1 hour ago









BerciBerci

60.2k23672




60.2k23672












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










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    1 hour ago


















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










  • $begingroup$
    Your edit has not changed much.
    $endgroup$
    – Dan Rust
    1 hour ago
















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




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












$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
1 hour ago




$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
1 hour 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

CARDNET

Image
CARDNET は日本カードネットワークが運営するクレジット決済ネットワーク。日本カードネットワークの略称としてもCARDNETが使用されている。 事件・事故 2017年4月15日午前11時8分からクレジットカードの決済がしづらくなる障害が発生した。障害は同日午後5時18分に復旧した。原因は、複数あるCARDNETセンターの拠点のうち1つで、2重化してあるL3スイッチの片方が故障し、1系統にトラフィックが集中したことによる輻輳が発生したため。対処としては、故障したL3スイッチ交換とシステム再起動が行われた [1] 。 出典 ^ 金子寛人 ( 2017年4月17日 ). “CARDNETのクレジット決済に6時間強障害、原因はL3スイッチ故障”. 日経コンピュータ. 2018年10月18日 閲覧。 この項目は、コンピュータに関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（PJ:コンピュータ/P:コンピュータ）。 この項目は、経済に関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（ポータル 経済学、プロジェクト 経済）。 This page is only for reference, If you need detailed information, please check here
Read more

Boot-repair Failure: Unable to locate package grub-common:i386

Image
0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h...
Read more

濃尾地震

Image
濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸...
Read more