Express Set of Dyadic with union/complementation of closed intervals












1












$begingroup$



Let $S = {{mover 2^n}|minmathbb Z, nin mathbb Ncup{0}}$




  1. Can $S cap[0,1]$ be expressed as countable union, complement or intersection of closed intervals with non zero, dyadic length and dyadic endpoints ?

  2. 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










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$endgroup$












  • $begingroup$
    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
















1












$begingroup$



Let $S = {{mover 2^n}|minmathbb Z, nin mathbb Ncup{0}}$




  1. Can $S cap[0,1]$ be expressed as countable union, complement or intersection of closed intervals with non zero, dyadic length and dyadic endpoints ?

  2. 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










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$



Let $S = {{mover 2^n}|minmathbb Z, nin mathbb Ncup{0}}$




  1. Can $S cap[0,1]$ be expressed as countable union, complement or intersection of closed intervals with non zero, dyadic length and dyadic endpoints ?

  2. 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










share|cite|improve this question











$endgroup$





Let $S = {{mover 2^n}|minmathbb Z, nin mathbb Ncup{0}}$




  1. Can $S cap[0,1]$ be expressed as countable union, complement or intersection of closed intervals with non zero, dyadic length and dyadic endpoints ?

  2. 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






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edited Jan 8 at 2:46









Andrés E. Caicedo

66.1k8160252




66.1k8160252










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


















  • $begingroup$
    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










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