Nested homeomorphic sets
$begingroup$
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
edited 1 hour ago
bof
51.3k457120
asked 1 hour ago
J.DoeJ.Doe
6814
$endgroup$
add a comment |
$begingroup$
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
edited 1 hour ago
bof
51.3k457120
asked 1 hour ago
J.DoeJ.Doe
6814
$endgroup$
1
$begingroup$
Try $U_n=[0,n]$.
$endgroup$
– Gerry Myerson
1 hour ago
add a comment |
$begingroup$
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
edited 1 hour ago
bof
51.3k457120
asked 1 hour ago
J.DoeJ.Doe
6814
$endgroup$
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
edited 1 hour ago
bof
51.3k457120
asked 1 hour ago
J.DoeJ.Doe
6814
edited 1 hour ago
bof
51.3k457120
asked 1 hour ago
J.DoeJ.Doe
6814
edited 1 hour ago
bof
51.3k457120
edited 1 hour ago
bof
51.3k457120
edited 1 hour ago
bof
51.3k457120
51.3k457120
asked 1 hour ago
J.DoeJ.Doe
6814
asked 1 hour ago
J.DoeJ.Doe
6814
asked 1 hour ago
J.DoeJ.Doe
6814
6814
1
$begingroup$
Try $U_n=[0,n]$.
$endgroup$
– Gerry Myerson
1 hour ago
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $bigcup_{i geq 1} U_i = S^1$.
answered 1 hour ago
Dan RustDan Rust
22.7k114884
$endgroup$
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.
36 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
30 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
28 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
4 mins ago
add a comment |
$begingroup$
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) $.
answered 1 hour ago
BerciBerci
60.2k23672
$endgroup$
$begingroup$
This does not appear to answer the question.
$endgroup$
– Dan Rust
1 hour ago
$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
52 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $bigcup_{i geq 1} U_i = S^1$.
answered 1 hour ago
Dan RustDan Rust
22.7k114884
$endgroup$
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.
36 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
30 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
28 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
4 mins ago
add a comment |
$begingroup$
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 $bigcup_{i geq 1} U_i = S^1$.
answered 1 hour ago
Dan RustDan Rust
22.7k114884
$endgroup$
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.
36 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
30 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
28 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
4 mins ago
add a comment |
$begingroup$
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 $bigcup_{i geq 1} U_i = S^1$.
answered 1 hour ago
Dan RustDan Rust
22.7k114884
$endgroup$
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 $bigcup_{i geq 1} U_i = S^1$.
answered 1 hour ago
Dan RustDan Rust
22.7k114884
edited 29 mins ago
answered 1 hour ago
Dan RustDan Rust
22.7k114884
answered 1 hour ago
Dan RustDan Rust
22.7k114884
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.
36 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
30 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
28 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
4 mins ago
add a comment |
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.
36 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
30 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
28 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
4 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.
36 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.
36 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
30 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
30 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
28 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
28 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
4 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
4 mins ago
add a comment |
$begingroup$
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) $.
answered 1 hour ago
BerciBerci
60.2k23672
$endgroup$
$begingroup$
This does not appear to answer the question.
$endgroup$
– Dan Rust
1 hour ago
$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
52 mins ago
add a comment |
$begingroup$
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) $.
answered 1 hour ago
BerciBerci
60.2k23672
$endgroup$
$begingroup$
This does not appear to answer the question.
$endgroup$
– Dan Rust
1 hour ago
$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
52 mins ago
add a comment |
$begingroup$
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) $.
answered 1 hour ago
BerciBerci
60.2k23672
$endgroup$
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) $.
answered 1 hour ago
BerciBerci
60.2k23672
edited 1 hour ago
answered 1 hour ago
BerciBerci
60.2k23672
answered 1 hour ago
BerciBerci
60.2k23672
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
52 mins ago
add a comment |
$begingroup$
This does not appear to answer the question.
$endgroup$
– Dan Rust
1 hour ago
$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
52 mins 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
52 mins ago
$begingroup$
Your edit has not changed much.
$endgroup$
– Dan Rust
52 mins ago
add a comment |
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$begingroup$
Try $U_n=[0,n]$.
$endgroup$
– Gerry Myerson
1 hour ago