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 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
    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.
    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
    1 hour 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
    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
    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
    1 hour 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
    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
















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
    1 hour 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
    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














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 1 hour ago

























answered 2 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.
    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
    1 hour 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
    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














  • 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
    1 hour 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
    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








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
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
1 hour 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
1 hour 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
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




$begingroup$
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
    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
    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 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
    1 hour 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
    1 hour 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
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

Knooppunt Holsloot

Image
Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
Read more

Altaar (religie)

Image
Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
Read more

Gregoriusmis

Image
Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
Read more