$bigcup U_{i}$ and $bigcup V_{j}$ are homeomorphic iff there are equally many $U_{i}$ and $V_{j}$ where...
$begingroup$
Let $U$ and $V$ two open sets in $mathbb{R}$.
First, I proved that if $U$ is open, then $U = bigsqcup_{i in mathbb{N}} U_{i}$ where $U_{i}$ is open and $"bigsqcup"$ means disjoint union.
Second, I proved that $U_{i}$ are uniquely determined by $U$, that is, $bigsqcup_{i in mathbb{N}} U_{i}$ is unique, up to order of the factors.
Now, I want to prove that
If $U$ and $V$ are open, $U$ and $V$ are homeomorphic iff there are equally many $U_{i}$ and $V_{j}$.
This makes perfect sense for me, but I dont know how to write correctly.
Can someone help me?
I dont know if it helps, but I wrote $U = bigsqcup I_{x}$ where $I_{x}$ is the largest interval containing $x$ and countained in $U$.
real-analysis general-topology metric-spaces
asked Dec 20 '18 at 20:45
Lucas CorrêaLucas Corrêa
1,6251321
$endgroup$
add a comment |
$begingroup$
Let $U$ and $V$ two open sets in $mathbb{R}$.
First, I proved that if $U$ is open, then $U = bigsqcup_{i in mathbb{N}} U_{i}$ where $U_{i}$ is open and $"bigsqcup"$ means disjoint union.
Second, I proved that $U_{i}$ are uniquely determined by $U$, that is, $bigsqcup_{i in mathbb{N}} U_{i}$ is unique, up to order of the factors.
Now, I want to prove that
If $U$ and $V$ are open, $U$ and $V$ are homeomorphic iff there are equally many $U_{i}$ and $V_{j}$.
This makes perfect sense for me, but I dont know how to write correctly.
Can someone help me?
I dont know if it helps, but I wrote $U = bigsqcup I_{x}$ where $I_{x}$ is the largest interval containing $x$ and countained in $U$.
real-analysis general-topology metric-spaces
asked Dec 20 '18 at 20:45
Lucas CorrêaLucas Corrêa
1,6251321
$endgroup$
add a comment |
$begingroup$
Let $U$ and $V$ two open sets in $mathbb{R}$.
First, I proved that if $U$ is open, then $U = bigsqcup_{i in mathbb{N}} U_{i}$ where $U_{i}$ is open and $"bigsqcup"$ means disjoint union.
Second, I proved that $U_{i}$ are uniquely determined by $U$, that is, $bigsqcup_{i in mathbb{N}} U_{i}$ is unique, up to order of the factors.
Now, I want to prove that
If $U$ and $V$ are open, $U$ and $V$ are homeomorphic iff there are equally many $U_{i}$ and $V_{j}$.
This makes perfect sense for me, but I dont know how to write correctly.
Can someone help me?
I dont know if it helps, but I wrote $U = bigsqcup I_{x}$ where $I_{x}$ is the largest interval containing $x$ and countained in $U$.
real-analysis general-topology metric-spaces
asked Dec 20 '18 at 20:45
Lucas CorrêaLucas Corrêa
1,6251321
$endgroup$
Let $U$ and $V$ two open sets in $mathbb{R}$.
First, I proved that if $U$ is open, then $U = bigsqcup_{i in mathbb{N}} U_{i}$ where $U_{i}$ is open and $"bigsqcup"$ means disjoint union.
Second, I proved that $U_{i}$ are uniquely determined by $U$, that is, $bigsqcup_{i in mathbb{N}} U_{i}$ is unique, up to order of the factors.
Now, I want to prove that
If $U$ and $V$ are open, $U$ and $V$ are homeomorphic iff there are equally many $U_{i}$ and $V_{j}$.
This makes perfect sense for me, but I dont know how to write correctly.
Can someone help me?
I dont know if it helps, but I wrote $U = bigsqcup I_{x}$ where $I_{x}$ is the largest interval containing $x$ and countained in $U$.
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Dec 20 '18 at 20:45
Lucas CorrêaLucas Corrêa
1,6251321
asked Dec 20 '18 at 20:45
Lucas CorrêaLucas Corrêa
1,6251321
asked Dec 20 '18 at 20:45
Lucas CorrêaLucas Corrêa
1,6251321
asked Dec 20 '18 at 20:45
Lucas CorrêaLucas Corrêa
1,6251321
asked Dec 20 '18 at 20:45
Lucas CorrêaLucas Corrêa
1,6251321
1,6251321
add a comment |
add a comment |
1 Answer
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$begingroup$
Hints: your construction $U = bigcup_x I_x$, where $I_x$ is a maximal interval containing $x in U$, is a good way of describing the decomposition of $U$ into its connected components, but you need to pick just one $x$ from each connected component to get a disjoint union. You are right that there will be at most countably many distinct connected components $I_x$, but there need not be infinitely many (e.g., take $U = Bbb{R}$). So your first observation should be that $U = bigsqcup_{j in J}U_j$, where $J subseteq Bbb{N}$ and each $U_j$ is non-empty open interval. Now if you have $V = bigsqcup_{k in K}V_k$ where $K subseteq Bbb{N}$ and each $V_k$ is a non-empty open interval, then, if $J$ and $K$ have the same number of elements you can use a bijection between $J$ and $K$ and your favourite homeomomorphism between any two open intervals $(a, b)$ and $(c, d)$ ($a < b$ and $c < d)$ to construct a homeomorphism between $U$ and $V$. Conversely you can use a homeomorphism between $U$ and $V$ to construct a bijection between $J$ and $K$ (so $J$ and $K$ have "equally many elements").
answered Dec 20 '18 at 21:27
Rob ArthanRob Arthan
29.2k42966
$endgroup$
$begingroup$
I got the first hint. For converse, I try to take a homeomorphism between $U$ and $V$ and take the restrictions $f|_{U_{i}}: U_{i} to A$. Since $f$ is open, $A$ must be an open set. So, how can I show that $J$ and $K$ are "equal"?
$endgroup$
– Lucas Corrêa
Dec 21 '18 at 1:43
$begingroup$
The homeomorphism must give you a bijection between the connected components of $U$ and $V$ and thus a bijection between the sets $J$ and $K$ that we are using to index the connected components.
$endgroup$
– Rob Arthan
Dec 21 '18 at 1:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hints: your construction $U = bigcup_x I_x$, where $I_x$ is a maximal interval containing $x in U$, is a good way of describing the decomposition of $U$ into its connected components, but you need to pick just one $x$ from each connected component to get a disjoint union. You are right that there will be at most countably many distinct connected components $I_x$, but there need not be infinitely many (e.g., take $U = Bbb{R}$). So your first observation should be that $U = bigsqcup_{j in J}U_j$, where $J subseteq Bbb{N}$ and each $U_j$ is non-empty open interval. Now if you have $V = bigsqcup_{k in K}V_k$ where $K subseteq Bbb{N}$ and each $V_k$ is a non-empty open interval, then, if $J$ and $K$ have the same number of elements you can use a bijection between $J$ and $K$ and your favourite homeomomorphism between any two open intervals $(a, b)$ and $(c, d)$ ($a < b$ and $c < d)$ to construct a homeomorphism between $U$ and $V$. Conversely you can use a homeomorphism between $U$ and $V$ to construct a bijection between $J$ and $K$ (so $J$ and $K$ have "equally many elements").
answered Dec 20 '18 at 21:27
Rob ArthanRob Arthan
29.2k42966
$endgroup$
$begingroup$
I got the first hint. For converse, I try to take a homeomorphism between $U$ and $V$ and take the restrictions $f|_{U_{i}}: U_{i} to A$. Since $f$ is open, $A$ must be an open set. So, how can I show that $J$ and $K$ are "equal"?
$endgroup$
– Lucas Corrêa
Dec 21 '18 at 1:43
$begingroup$
The homeomorphism must give you a bijection between the connected components of $U$ and $V$ and thus a bijection between the sets $J$ and $K$ that we are using to index the connected components.
$endgroup$
– Rob Arthan
Dec 21 '18 at 1:56
add a comment |
$begingroup$
Hints: your construction $U = bigcup_x I_x$, where $I_x$ is a maximal interval containing $x in U$, is a good way of describing the decomposition of $U$ into its connected components, but you need to pick just one $x$ from each connected component to get a disjoint union. You are right that there will be at most countably many distinct connected components $I_x$, but there need not be infinitely many (e.g., take $U = Bbb{R}$). So your first observation should be that $U = bigsqcup_{j in J}U_j$, where $J subseteq Bbb{N}$ and each $U_j$ is non-empty open interval. Now if you have $V = bigsqcup_{k in K}V_k$ where $K subseteq Bbb{N}$ and each $V_k$ is a non-empty open interval, then, if $J$ and $K$ have the same number of elements you can use a bijection between $J$ and $K$ and your favourite homeomomorphism between any two open intervals $(a, b)$ and $(c, d)$ ($a < b$ and $c < d)$ to construct a homeomorphism between $U$ and $V$. Conversely you can use a homeomorphism between $U$ and $V$ to construct a bijection between $J$ and $K$ (so $J$ and $K$ have "equally many elements").
answered Dec 20 '18 at 21:27
Rob ArthanRob Arthan
29.2k42966
$endgroup$
$begingroup$
I got the first hint. For converse, I try to take a homeomorphism between $U$ and $V$ and take the restrictions $f|_{U_{i}}: U_{i} to A$. Since $f$ is open, $A$ must be an open set. So, how can I show that $J$ and $K$ are "equal"?
$endgroup$
– Lucas Corrêa
Dec 21 '18 at 1:43
$begingroup$
The homeomorphism must give you a bijection between the connected components of $U$ and $V$ and thus a bijection between the sets $J$ and $K$ that we are using to index the connected components.
$endgroup$
– Rob Arthan
Dec 21 '18 at 1:56
add a comment |
$begingroup$
Hints: your construction $U = bigcup_x I_x$, where $I_x$ is a maximal interval containing $x in U$, is a good way of describing the decomposition of $U$ into its connected components, but you need to pick just one $x$ from each connected component to get a disjoint union. You are right that there will be at most countably many distinct connected components $I_x$, but there need not be infinitely many (e.g., take $U = Bbb{R}$). So your first observation should be that $U = bigsqcup_{j in J}U_j$, where $J subseteq Bbb{N}$ and each $U_j$ is non-empty open interval. Now if you have $V = bigsqcup_{k in K}V_k$ where $K subseteq Bbb{N}$ and each $V_k$ is a non-empty open interval, then, if $J$ and $K$ have the same number of elements you can use a bijection between $J$ and $K$ and your favourite homeomomorphism between any two open intervals $(a, b)$ and $(c, d)$ ($a < b$ and $c < d)$ to construct a homeomorphism between $U$ and $V$. Conversely you can use a homeomorphism between $U$ and $V$ to construct a bijection between $J$ and $K$ (so $J$ and $K$ have "equally many elements").
answered Dec 20 '18 at 21:27
Rob ArthanRob Arthan
29.2k42966
$endgroup$
Hints: your construction $U = bigcup_x I_x$, where $I_x$ is a maximal interval containing $x in U$, is a good way of describing the decomposition of $U$ into its connected components, but you need to pick just one $x$ from each connected component to get a disjoint union. You are right that there will be at most countably many distinct connected components $I_x$, but there need not be infinitely many (e.g., take $U = Bbb{R}$). So your first observation should be that $U = bigsqcup_{j in J}U_j$, where $J subseteq Bbb{N}$ and each $U_j$ is non-empty open interval. Now if you have $V = bigsqcup_{k in K}V_k$ where $K subseteq Bbb{N}$ and each $V_k$ is a non-empty open interval, then, if $J$ and $K$ have the same number of elements you can use a bijection between $J$ and $K$ and your favourite homeomomorphism between any two open intervals $(a, b)$ and $(c, d)$ ($a < b$ and $c < d)$ to construct a homeomorphism between $U$ and $V$. Conversely you can use a homeomorphism between $U$ and $V$ to construct a bijection between $J$ and $K$ (so $J$ and $K$ have "equally many elements").
answered Dec 20 '18 at 21:27
Rob ArthanRob Arthan
29.2k42966
answered Dec 20 '18 at 21:27
Rob ArthanRob Arthan
29.2k42966
answered Dec 20 '18 at 21:27
Rob ArthanRob Arthan
29.2k42966
answered Dec 20 '18 at 21:27
Rob ArthanRob Arthan
29.2k42966
29.2k42966
$begingroup$
I got the first hint. For converse, I try to take a homeomorphism between $U$ and $V$ and take the restrictions $f|_{U_{i}}: U_{i} to A$. Since $f$ is open, $A$ must be an open set. So, how can I show that $J$ and $K$ are "equal"?
$endgroup$
– Lucas Corrêa
Dec 21 '18 at 1:43
$begingroup$
The homeomorphism must give you a bijection between the connected components of $U$ and $V$ and thus a bijection between the sets $J$ and $K$ that we are using to index the connected components.
$endgroup$
– Rob Arthan
Dec 21 '18 at 1:56
add a comment |
$begingroup$
I got the first hint. For converse, I try to take a homeomorphism between $U$ and $V$ and take the restrictions $f|_{U_{i}}: U_{i} to A$. Since $f$ is open, $A$ must be an open set. So, how can I show that $J$ and $K$ are "equal"?
$endgroup$
– Lucas Corrêa
Dec 21 '18 at 1:43
$begingroup$
The homeomorphism must give you a bijection between the connected components of $U$ and $V$ and thus a bijection between the sets $J$ and $K$ that we are using to index the connected components.
$endgroup$
– Rob Arthan
Dec 21 '18 at 1:56
$begingroup$
I got the first hint. For converse, I try to take a homeomorphism between $U$ and $V$ and take the restrictions $f|_{U_{i}}: U_{i} to A$. Since $f$ is open, $A$ must be an open set. So, how can I show that $J$ and $K$ are "equal"?
$endgroup$
– Lucas Corrêa
Dec 21 '18 at 1:43
$begingroup$
I got the first hint. For converse, I try to take a homeomorphism between $U$ and $V$ and take the restrictions $f|_{U_{i}}: U_{i} to A$. Since $f$ is open, $A$ must be an open set. So, how can I show that $J$ and $K$ are "equal"?
$endgroup$
– Lucas Corrêa
Dec 21 '18 at 1:43
$begingroup$
The homeomorphism must give you a bijection between the connected components of $U$ and $V$ and thus a bijection between the sets $J$ and $K$ that we are using to index the connected components.
$endgroup$
– Rob Arthan
Dec 21 '18 at 1:56
$begingroup$
The homeomorphism must give you a bijection between the connected components of $U$ and $V$ and thus a bijection between the sets $J$ and $K$ that we are using to index the connected components.
$endgroup$
– Rob Arthan
Dec 21 '18 at 1:56
add a comment |
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