Prove that the Cantor set is homeomorphic to $(X,mathscr T)$.
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For each $nin mathbb N$, let $X_n={0,2}$ and let $mathscr T_n $ be the discrete topology on $X_n$. Let $X=prod_{n=1}^infty X_n,$ and $mathscr T$ be the product topology on $X$. Prove that the Cantor set is homeomorphic to $(X,mathscr T)$.
Proof. Let $C$ be the Cantor set, Let $(C,mathscr U_C)$ be the subspace topological space of usual topology on $mathbb R$. Let the map $psi:(C,mathscr U_A)to (X,mathscr T)$ be defined as $psi(x)=(x_1,x_2,...,x_n,....),$ where $xin C$ and $x=sum_{i=1}^infty frac{x_i}{3^i},x_i=0$or $2$. $psi$ is well defined if we take the ternary expansion of $1/3,1/9$,...etc with $0$ and $2$ s. We can easily prove that $psi$ is injective.
How do I prove that $psi$ is surjective? What is the guarentee that, If $yin X$, $exists xin C$:$f(x)=y$?
For continuity of $psi$, Let $mathscr B$ be the basis of the product topology on $X$. It is enough to show that $psi^{-1}(B)in mathscr U_C$ for every $Bin mathscr B$. Let $B in mathscr B$, $B=prod_{i=1}^n B_i$, $B_i in mathscr T_i$, $B_i=X_i$ for all but finitely many $i$. I want to prove $exists$ some open interval ($I$(say) in $mathbb R$) such that $psi^{-1}(B)=Ccap I$. How do I prove the existence of $I$?
For completing the proof of homeomorphism, I need to prove that $psi $ is open. Let $Uin mathscr U_C$, then $U=Ccap J$ for some interval $J$ in $mathbb R$. then $psi(U)=psi(Ccap J)=psi(C)cappsi(J)=Xcap psi(J).$ How do I complete the proof. Please help me.
general-topology alternative-proof cantor-set
edited Nov 17 at 23:16
Paul Frost
7,8041527
asked Nov 17 at 17:06
Unknown x
2,44911025
|
show 2 more comments
up vote
3
down vote
favorite
For each $nin mathbb N$, let $X_n={0,2}$ and let $mathscr T_n $ be the discrete topology on $X_n$. Let $X=prod_{n=1}^infty X_n,$ and $mathscr T$ be the product topology on $X$. Prove that the Cantor set is homeomorphic to $(X,mathscr T)$.
Proof. Let $C$ be the Cantor set, Let $(C,mathscr U_C)$ be the subspace topological space of usual topology on $mathbb R$. Let the map $psi:(C,mathscr U_A)to (X,mathscr T)$ be defined as $psi(x)=(x_1,x_2,...,x_n,....),$ where $xin C$ and $x=sum_{i=1}^infty frac{x_i}{3^i},x_i=0$or $2$. $psi$ is well defined if we take the ternary expansion of $1/3,1/9$,...etc with $0$ and $2$ s. We can easily prove that $psi$ is injective.
How do I prove that $psi$ is surjective? What is the guarentee that, If $yin X$, $exists xin C$:$f(x)=y$?
For continuity of $psi$, Let $mathscr B$ be the basis of the product topology on $X$. It is enough to show that $psi^{-1}(B)in mathscr U_C$ for every $Bin mathscr B$. Let $B in mathscr B$, $B=prod_{i=1}^n B_i$, $B_i in mathscr T_i$, $B_i=X_i$ for all but finitely many $i$. I want to prove $exists$ some open interval ($I$(say) in $mathbb R$) such that $psi^{-1}(B)=Ccap I$. How do I prove the existence of $I$?
For completing the proof of homeomorphism, I need to prove that $psi $ is open. Let $Uin mathscr U_C$, then $U=Ccap J$ for some interval $J$ in $mathbb R$. then $psi(U)=psi(Ccap J)=psi(C)cappsi(J)=Xcap psi(J).$ How do I complete the proof. Please help me.
general-topology alternative-proof cantor-set
edited Nov 17 at 23:16
Paul Frost
7,8041527
asked Nov 17 at 17:06
Unknown x
2,44911025
1
For surjectivity, note that both sides of $psi$ are sequences of $0$s and $2$s.
– John Douma
Nov 17 at 18:06
What if there exists a sequence $(a_1,a_2,...,)$, where $a_i=0$ or $2$. but $sum frac{a_i}{3^i}$ is not in cantor set. Is it possible? If not why?
– Unknown x
Nov 18 at 1:46
If you look at the form of the sum you will see that it corresponds to $.a_1a_2...$ which is a base three fraction. Your sum $is$ a "decimal" expansion of each element of the Cantor set.
– John Douma
Nov 18 at 1:49
@Unknownx: A simple way to go is to observe, from the construction of the Cantor set, that it contains any finite sum $sum_{i=1}^n a_i/3^i$ where $a_i in {0,2}$; indeed, such a value is an endpoint of one of the removed open intervals. Now note that any such infinite sum is by definition a limit of finite sums, and the Cantor set is closed...
– Nate Eldredge
Nov 18 at 2:41
For the last part: (1) the complement of an open set is compact (2) continuous functions take compact sets to compact sets. Indeed, this proves more generally that any continuous bijection between compact Hausdorff spaces is automatically a homeomorphism.
– Nate Eldredge
Nov 18 at 2:43
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For each $nin mathbb N$, let $X_n={0,2}$ and let $mathscr T_n $ be the discrete topology on $X_n$. Let $X=prod_{n=1}^infty X_n,$ and $mathscr T$ be the product topology on $X$. Prove that the Cantor set is homeomorphic to $(X,mathscr T)$.
Proof. Let $C$ be the Cantor set, Let $(C,mathscr U_C)$ be the subspace topological space of usual topology on $mathbb R$. Let the map $psi:(C,mathscr U_A)to (X,mathscr T)$ be defined as $psi(x)=(x_1,x_2,...,x_n,....),$ where $xin C$ and $x=sum_{i=1}^infty frac{x_i}{3^i},x_i=0$or $2$. $psi$ is well defined if we take the ternary expansion of $1/3,1/9$,...etc with $0$ and $2$ s. We can easily prove that $psi$ is injective.
How do I prove that $psi$ is surjective? What is the guarentee that, If $yin X$, $exists xin C$:$f(x)=y$?
For continuity of $psi$, Let $mathscr B$ be the basis of the product topology on $X$. It is enough to show that $psi^{-1}(B)in mathscr U_C$ for every $Bin mathscr B$. Let $B in mathscr B$, $B=prod_{i=1}^n B_i$, $B_i in mathscr T_i$, $B_i=X_i$ for all but finitely many $i$. I want to prove $exists$ some open interval ($I$(say) in $mathbb R$) such that $psi^{-1}(B)=Ccap I$. How do I prove the existence of $I$?
For completing the proof of homeomorphism, I need to prove that $psi $ is open. Let $Uin mathscr U_C$, then $U=Ccap J$ for some interval $J$ in $mathbb R$. then $psi(U)=psi(Ccap J)=psi(C)cappsi(J)=Xcap psi(J).$ How do I complete the proof. Please help me.
general-topology alternative-proof cantor-set
edited Nov 17 at 23:16
Paul Frost
7,8041527
asked Nov 17 at 17:06
Unknown x
2,44911025
For each $nin mathbb N$, let $X_n={0,2}$ and let $mathscr T_n $ be the discrete topology on $X_n$. Let $X=prod_{n=1}^infty X_n,$ and $mathscr T$ be the product topology on $X$. Prove that the Cantor set is homeomorphic to $(X,mathscr T)$.
Proof. Let $C$ be the Cantor set, Let $(C,mathscr U_C)$ be the subspace topological space of usual topology on $mathbb R$. Let the map $psi:(C,mathscr U_A)to (X,mathscr T)$ be defined as $psi(x)=(x_1,x_2,...,x_n,....),$ where $xin C$ and $x=sum_{i=1}^infty frac{x_i}{3^i},x_i=0$or $2$. $psi$ is well defined if we take the ternary expansion of $1/3,1/9$,...etc with $0$ and $2$ s. We can easily prove that $psi$ is injective.
How do I prove that $psi$ is surjective? What is the guarentee that, If $yin X$, $exists xin C$:$f(x)=y$?
For continuity of $psi$, Let $mathscr B$ be the basis of the product topology on $X$. It is enough to show that $psi^{-1}(B)in mathscr U_C$ for every $Bin mathscr B$. Let $B in mathscr B$, $B=prod_{i=1}^n B_i$, $B_i in mathscr T_i$, $B_i=X_i$ for all but finitely many $i$. I want to prove $exists$ some open interval ($I$(say) in $mathbb R$) such that $psi^{-1}(B)=Ccap I$. How do I prove the existence of $I$?
For completing the proof of homeomorphism, I need to prove that $psi $ is open. Let $Uin mathscr U_C$, then $U=Ccap J$ for some interval $J$ in $mathbb R$. then $psi(U)=psi(Ccap J)=psi(C)cappsi(J)=Xcap psi(J).$ How do I complete the proof. Please help me.
general-topology alternative-proof cantor-set
general-topology alternative-proof cantor-set
edited Nov 17 at 23:16
Paul Frost
7,8041527
asked Nov 17 at 17:06
Unknown x
2,44911025
edited Nov 17 at 23:16
Paul Frost
7,8041527
asked Nov 17 at 17:06
Unknown x
2,44911025
edited Nov 17 at 23:16
Paul Frost
7,8041527
edited Nov 17 at 23:16
Paul Frost
7,8041527
edited Nov 17 at 23:16
Paul Frost
7,8041527
7,8041527
asked Nov 17 at 17:06
Unknown x
2,44911025
asked Nov 17 at 17:06
Unknown x
2,44911025
asked Nov 17 at 17:06
Unknown x
2,44911025
2,44911025
1
For surjectivity, note that both sides of $psi$ are sequences of $0$s and $2$s.
– John Douma
Nov 17 at 18:06
What if there exists a sequence $(a_1,a_2,...,)$, where $a_i=0$ or $2$. but $sum frac{a_i}{3^i}$ is not in cantor set. Is it possible? If not why?
– Unknown x
Nov 18 at 1:46
If you look at the form of the sum you will see that it corresponds to $.a_1a_2...$ which is a base three fraction. Your sum $is$ a "decimal" expansion of each element of the Cantor set.
– John Douma
Nov 18 at 1:49
@Unknownx: A simple way to go is to observe, from the construction of the Cantor set, that it contains any finite sum $sum_{i=1}^n a_i/3^i$ where $a_i in {0,2}$; indeed, such a value is an endpoint of one of the removed open intervals. Now note that any such infinite sum is by definition a limit of finite sums, and the Cantor set is closed...
– Nate Eldredge
Nov 18 at 2:41
For the last part: (1) the complement of an open set is compact (2) continuous functions take compact sets to compact sets. Indeed, this proves more generally that any continuous bijection between compact Hausdorff spaces is automatically a homeomorphism.
– Nate Eldredge
Nov 18 at 2:43
|
show 2 more comments
1
For surjectivity, note that both sides of $psi$ are sequences of $0$s and $2$s.
– John Douma
Nov 17 at 18:06
What if there exists a sequence $(a_1,a_2,...,)$, where $a_i=0$ or $2$. but $sum frac{a_i}{3^i}$ is not in cantor set. Is it possible? If not why?
– Unknown x
Nov 18 at 1:46
If you look at the form of the sum you will see that it corresponds to $.a_1a_2...$ which is a base three fraction. Your sum $is$ a "decimal" expansion of each element of the Cantor set.
– John Douma
Nov 18 at 1:49
@Unknownx: A simple way to go is to observe, from the construction of the Cantor set, that it contains any finite sum $sum_{i=1}^n a_i/3^i$ where $a_i in {0,2}$; indeed, such a value is an endpoint of one of the removed open intervals. Now note that any such infinite sum is by definition a limit of finite sums, and the Cantor set is closed...
– Nate Eldredge
Nov 18 at 2:41
For the last part: (1) the complement of an open set is compact (2) continuous functions take compact sets to compact sets. Indeed, this proves more generally that any continuous bijection between compact Hausdorff spaces is automatically a homeomorphism.
– Nate Eldredge
Nov 18 at 2:43
1
1
For surjectivity, note that both sides of $psi$ are sequences of $0$s and $2$s.
– John Douma
Nov 17 at 18:06
For surjectivity, note that both sides of $psi$ are sequences of $0$s and $2$s.
– John Douma
Nov 17 at 18:06
What if there exists a sequence $(a_1,a_2,...,)$, where $a_i=0$ or $2$. but $sum frac{a_i}{3^i}$ is not in cantor set. Is it possible? If not why?
– Unknown x
Nov 18 at 1:46
What if there exists a sequence $(a_1,a_2,...,)$, where $a_i=0$ or $2$. but $sum frac{a_i}{3^i}$ is not in cantor set. Is it possible? If not why?
– Unknown x
Nov 18 at 1:46
If you look at the form of the sum you will see that it corresponds to $.a_1a_2...$ which is a base three fraction. Your sum $is$ a "decimal" expansion of each element of the Cantor set.
– John Douma
Nov 18 at 1:49
If you look at the form of the sum you will see that it corresponds to $.a_1a_2...$ which is a base three fraction. Your sum $is$ a "decimal" expansion of each element of the Cantor set.
– John Douma
Nov 18 at 1:49
@Unknownx: A simple way to go is to observe, from the construction of the Cantor set, that it contains any finite sum $sum_{i=1}^n a_i/3^i$ where $a_i in {0,2}$; indeed, such a value is an endpoint of one of the removed open intervals. Now note that any such infinite sum is by definition a limit of finite sums, and the Cantor set is closed...
– Nate Eldredge
Nov 18 at 2:41
@Unknownx: A simple way to go is to observe, from the construction of the Cantor set, that it contains any finite sum $sum_{i=1}^n a_i/3^i$ where $a_i in {0,2}$; indeed, such a value is an endpoint of one of the removed open intervals. Now note that any such infinite sum is by definition a limit of finite sums, and the Cantor set is closed...
– Nate Eldredge
Nov 18 at 2:41
For the last part: (1) the complement of an open set is compact (2) continuous functions take compact sets to compact sets. Indeed, this proves more generally that any continuous bijection between compact Hausdorff spaces is automatically a homeomorphism.
– Nate Eldredge
Nov 18 at 2:43
For the last part: (1) the complement of an open set is compact (2) continuous functions take compact sets to compact sets. Indeed, this proves more generally that any continuous bijection between compact Hausdorff spaces is automatically a homeomorphism.
– Nate Eldredge
Nov 18 at 2:43
|
show 2 more comments
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1
For surjectivity, note that both sides of $psi$ are sequences of $0$s and $2$s.
– John Douma
Nov 17 at 18:06
What if there exists a sequence $(a_1,a_2,...,)$, where $a_i=0$ or $2$. but $sum frac{a_i}{3^i}$ is not in cantor set. Is it possible? If not why?
– Unknown x
Nov 18 at 1:46
If you look at the form of the sum you will see that it corresponds to $.a_1a_2...$ which is a base three fraction. Your sum $is$ a "decimal" expansion of each element of the Cantor set.
– John Douma
Nov 18 at 1:49
@Unknownx: A simple way to go is to observe, from the construction of the Cantor set, that it contains any finite sum $sum_{i=1}^n a_i/3^i$ where $a_i in {0,2}$; indeed, such a value is an endpoint of one of the removed open intervals. Now note that any such infinite sum is by definition a limit of finite sums, and the Cantor set is closed...
– Nate Eldredge
Nov 18 at 2:41
For the last part: (1) the complement of an open set is compact (2) continuous functions take compact sets to compact sets. Indeed, this proves more generally that any continuous bijection between compact Hausdorff spaces is automatically a homeomorphism.
– Nate Eldredge
Nov 18 at 2:43