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 2 hours ago









bof

51.3k457120




51.3k457120










asked 2 hours 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.
    1 hour 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
    55 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
    53 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
    29 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.
    1 hour 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
    55 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
    53 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
    29 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.
    1 hour 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
    55 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
    53 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
    29 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 55 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.
    1 hour 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
    55 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
    53 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
    29 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.
    1 hour 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
    55 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
    53 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
    29 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.
1 hour 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.
1 hour 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
55 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
55 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
53 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
53 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
29 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
29 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

What other Star Trek series did the main TNG cast show up in?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75
Read more

Berlina muro

Image
La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l
Read more

Berlina aerponto

Image
Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa
Read more