Express Set of Dyadic with union/complementation of closed intervals
1
$begingroup$
Let $S = {{mover 2^n}|minmathbb Z, nin mathbb Ncup{0}}$
- Can $S cap[0,1]$ be expressed as countable union, complement or intersection of closed intervals with non zero, dyadic length and dyadic endpoints ?
- Further, can any subset of $[0,1]$ be expressed with the same rules?
As for 1, as we can do the following:
Let ${a_n}_n$ be an enumeration of dyadics in $[0,1]$
def $A_n = [0,a_n]cap[a_n,1] = {a_n}$
Then take a union of $A_n$'s
Incase $a_n = 0$ or $1$, we can create decreasing/increasing sequence of dyadics and then take complement
Is the solution correct? Please also provide hints/counterexample for 2
general-topology
edited Jan 8 at 2:46
Andrés E. Caicedo
66.1k8160252
asked Jan 8 at 2:46
AnvitAnvit
1,807419
$endgroup$
add a comment |
1
$begingroup$
Let $S = {{mover 2^n}|minmathbb Z, nin mathbb Ncup{0}}$
- Can $S cap[0,1]$ be expressed as countable union, complement or intersection of closed intervals with non zero, dyadic length and dyadic endpoints ?
- Further, can any subset of $[0,1]$ be expressed with the same rules?
As for 1, as we can do the following:
Let ${a_n}_n$ be an enumeration of dyadics in $[0,1]$
def $A_n = [0,a_n]cap[a_n,1] = {a_n}$
Then take a union of $A_n$'s
Incase $a_n = 0$ or $1$, we can create decreasing/increasing sequence of dyadics and then take complement
Is the solution correct? Please also provide hints/counterexample for 2
general-topology
edited Jan 8 at 2:46
Andrés E. Caicedo
66.1k8160252
asked Jan 8 at 2:46
AnvitAnvit
1,807419
$endgroup$
$begingroup$
No and no for real intervals.
$endgroup$
– William Elliot
Jan 8 at 2:55
$begingroup$
@WilliamElliot What is the problem with the solution?
$endgroup$
– Anvit
Jan 8 at 2:59
$begingroup$
The A's have zero length.
$endgroup$
– William Elliot
Jan 8 at 9:22
$begingroup$
I have two questions to clarify yours. Dyadic numbers are exactly numbers belonging to $S$, right? Can we combine operations of countable union, complement or intersection several (maybe countably many) times? Even if so, the answer to Question 2 is negative, because all expessible sets belong to a $sigma$-algebra of Borel subsets, and there are non-Borel subsets (even non-Lebesgue measurable) contained in $[0,1]$.
$endgroup$
– Alex Ravsky
Jan 10 at 11:51
$begingroup$
@AlexRavsky yes and yes, $S$ is exactly all dyadic numbers. Operations have to be countable (can be infinite)
$endgroup$
– Anvit
Jan 11 at 15:30
add a comment |
1
1
1
$begingroup$
Let $S = {{mover 2^n}|minmathbb Z, nin mathbb Ncup{0}}$
- Can $S cap[0,1]$ be expressed as countable union, complement or intersection of closed intervals with non zero, dyadic length and dyadic endpoints ?
- Further, can any subset of $[0,1]$ be expressed with the same rules?
As for 1, as we can do the following:
Let ${a_n}_n$ be an enumeration of dyadics in $[0,1]$
def $A_n = [0,a_n]cap[a_n,1] = {a_n}$
Then take a union of $A_n$'s
Incase $a_n = 0$ or $1$, we can create decreasing/increasing sequence of dyadics and then take complement
Is the solution correct? Please also provide hints/counterexample for 2
general-topology
edited Jan 8 at 2:46
Andrés E. Caicedo
66.1k8160252
asked Jan 8 at 2:46
AnvitAnvit
1,807419
$endgroup$
Let $S = {{mover 2^n}|minmathbb Z, nin mathbb Ncup{0}}$
- Can $S cap[0,1]$ be expressed as countable union, complement or intersection of closed intervals with non zero, dyadic length and dyadic endpoints ?
- Further, can any subset of $[0,1]$ be expressed with the same rules?
As for 1, as we can do the following:
Let ${a_n}_n$ be an enumeration of dyadics in $[0,1]$
def $A_n = [0,a_n]cap[a_n,1] = {a_n}$
Then take a union of $A_n$'s
Incase $a_n = 0$ or $1$, we can create decreasing/increasing sequence of dyadics and then take complement
Is the solution correct? Please also provide hints/counterexample for 2
general-topology
general-topology
edited Jan 8 at 2:46
Andrés E. Caicedo
66.1k8160252
asked Jan 8 at 2:46
AnvitAnvit
1,807419
edited Jan 8 at 2:46
Andrés E. Caicedo
66.1k8160252
asked Jan 8 at 2:46
AnvitAnvit
1,807419
edited Jan 8 at 2:46
Andrés E. Caicedo
66.1k8160252
edited Jan 8 at 2:46
Andrés E. Caicedo
66.1k8160252
edited Jan 8 at 2:46
Andrés E. Caicedo
66.1k8160252
66.1k8160252
asked Jan 8 at 2:46
AnvitAnvit
1,807419
asked Jan 8 at 2:46
AnvitAnvit
1,807419
asked Jan 8 at 2:46
AnvitAnvit
1,807419
1,807419
$begingroup$
No and no for real intervals.
$endgroup$
– William Elliot
Jan 8 at 2:55
$begingroup$
@WilliamElliot What is the problem with the solution?
$endgroup$
– Anvit
Jan 8 at 2:59
$begingroup$
The A's have zero length.
$endgroup$
– William Elliot
Jan 8 at 9:22
$begingroup$
I have two questions to clarify yours. Dyadic numbers are exactly numbers belonging to $S$, right? Can we combine operations of countable union, complement or intersection several (maybe countably many) times? Even if so, the answer to Question 2 is negative, because all expessible sets belong to a $sigma$-algebra of Borel subsets, and there are non-Borel subsets (even non-Lebesgue measurable) contained in $[0,1]$.
$endgroup$
– Alex Ravsky
Jan 10 at 11:51
$begingroup$
@AlexRavsky yes and yes, $S$ is exactly all dyadic numbers. Operations have to be countable (can be infinite)
$endgroup$
– Anvit
Jan 11 at 15:30
add a comment |
$begingroup$
No and no for real intervals.
$endgroup$
– William Elliot
Jan 8 at 2:55
$begingroup$
@WilliamElliot What is the problem with the solution?
$endgroup$
– Anvit
Jan 8 at 2:59
$begingroup$
The A's have zero length.
$endgroup$
– William Elliot
Jan 8 at 9:22
$begingroup$
I have two questions to clarify yours. Dyadic numbers are exactly numbers belonging to $S$, right? Can we combine operations of countable union, complement or intersection several (maybe countably many) times? Even if so, the answer to Question 2 is negative, because all expessible sets belong to a $sigma$-algebra of Borel subsets, and there are non-Borel subsets (even non-Lebesgue measurable) contained in $[0,1]$.
$endgroup$
– Alex Ravsky
Jan 10 at 11:51
$begingroup$
@AlexRavsky yes and yes, $S$ is exactly all dyadic numbers. Operations have to be countable (can be infinite)
$endgroup$
– Anvit
Jan 11 at 15:30
$begingroup$
No and no for real intervals.
$endgroup$
– William Elliot
Jan 8 at 2:55
$begingroup$
No and no for real intervals.
$endgroup$
– William Elliot
Jan 8 at 2:55
$begingroup$
@WilliamElliot What is the problem with the solution?
$endgroup$
– Anvit
Jan 8 at 2:59
$begingroup$
@WilliamElliot What is the problem with the solution?
$endgroup$
– Anvit
Jan 8 at 2:59
$begingroup$
The A's have zero length.
$endgroup$
– William Elliot
Jan 8 at 9:22
$begingroup$
The A's have zero length.
$endgroup$
– William Elliot
Jan 8 at 9:22
$begingroup$
I have two questions to clarify yours. Dyadic numbers are exactly numbers belonging to $S$, right? Can we combine operations of countable union, complement or intersection several (maybe countably many) times? Even if so, the answer to Question 2 is negative, because all expessible sets belong to a $sigma$-algebra of Borel subsets, and there are non-Borel subsets (even non-Lebesgue measurable) contained in $[0,1]$.
$endgroup$
– Alex Ravsky
Jan 10 at 11:51
$begingroup$
I have two questions to clarify yours. Dyadic numbers are exactly numbers belonging to $S$, right? Can we combine operations of countable union, complement or intersection several (maybe countably many) times? Even if so, the answer to Question 2 is negative, because all expessible sets belong to a $sigma$-algebra of Borel subsets, and there are non-Borel subsets (even non-Lebesgue measurable) contained in $[0,1]$.
$endgroup$
– Alex Ravsky
Jan 10 at 11:51
$begingroup$
@AlexRavsky yes and yes, $S$ is exactly all dyadic numbers. Operations have to be countable (can be infinite)
$endgroup$
– Anvit
Jan 11 at 15:30
$begingroup$
@AlexRavsky yes and yes, $S$ is exactly all dyadic numbers. Operations have to be countable (can be infinite)
$endgroup$
– Anvit
Jan 11 at 15:30
add a comment |
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$begingroup$
No and no for real intervals.
$endgroup$
– William Elliot
Jan 8 at 2:55
$begingroup$
@WilliamElliot What is the problem with the solution?
$endgroup$
– Anvit
Jan 8 at 2:59
$begingroup$
The A's have zero length.
$endgroup$
– William Elliot
Jan 8 at 9:22
$begingroup$
I have two questions to clarify yours. Dyadic numbers are exactly numbers belonging to $S$, right? Can we combine operations of countable union, complement or intersection several (maybe countably many) times? Even if so, the answer to Question 2 is negative, because all expessible sets belong to a $sigma$-algebra of Borel subsets, and there are non-Borel subsets (even non-Lebesgue measurable) contained in $[0,1]$.
$endgroup$
– Alex Ravsky
Jan 10 at 11:51
$begingroup$
@AlexRavsky yes and yes, $S$ is exactly all dyadic numbers. Operations have to be countable (can be infinite)
$endgroup$
– Anvit
Jan 11 at 15:30