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







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$endgroup$








  • 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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

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active

oldest

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active

oldest

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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


















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濃尾地震

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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
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