Existence of a strange function












5












$begingroup$


Inspired by A discontinuous construction:
Does there exist a function $a colon [0,1] to (0,infty)$ and a family ${D_x colon x in [0,1]}$ of countable, dense subsets of $[0,1]$ with $bigcup_{x in [0,1]} D_x = [0,1]$ and $sum_{r in D_x} a(r) < infty$ for all $x in [0,1]$,






set-theory measure-theory







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$endgroup$








  • 1




    $begingroup$
    Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
    $endgroup$
    – Asaf Karagila
    Feb 23 at 11:54










  • $begingroup$
    Thank you. I shall try it.
    $endgroup$
    – Dieter Kadelka
    Feb 23 at 12:01










  • $begingroup$
    The Vitali set $V$ allows the definition of $cal{D} := {(v + mathbb{Q}) cap [0,1] colon v in V}$. The $D in cal{D}$ are either identical or disjoint. This is essential for defining $a$ whatever we like. Asaf Karagila thank you again.
    $endgroup$
    – Dieter Kadelka
    Feb 23 at 12:25
















5












$begingroup$


Inspired by A discontinuous construction:
Does there exist a function $a colon [0,1] to (0,infty)$ and a family ${D_x colon x in [0,1]}$ of countable, dense subsets of $[0,1]$ with $bigcup_{x in [0,1]} D_x = [0,1]$ and $sum_{r in D_x} a(r) < infty$ for all $x in [0,1]$,






set-theory measure-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
    $endgroup$
    – Asaf Karagila
    Feb 23 at 11:54










  • $begingroup$
    Thank you. I shall try it.
    $endgroup$
    – Dieter Kadelka
    Feb 23 at 12:01










  • $begingroup$
    The Vitali set $V$ allows the definition of $cal{D} := {(v + mathbb{Q}) cap [0,1] colon v in V}$. The $D in cal{D}$ are either identical or disjoint. This is essential for defining $a$ whatever we like. Asaf Karagila thank you again.
    $endgroup$
    – Dieter Kadelka
    Feb 23 at 12:25














5












5








5





$begingroup$


Inspired by A discontinuous construction:
Does there exist a function $a colon [0,1] to (0,infty)$ and a family ${D_x colon x in [0,1]}$ of countable, dense subsets of $[0,1]$ with $bigcup_{x in [0,1]} D_x = [0,1]$ and $sum_{r in D_x} a(r) < infty$ for all $x in [0,1]$,






set-theory measure-theory







share|cite|improve this question









$endgroup$




Inspired by A discontinuous construction:
Does there exist a function $a colon [0,1] to (0,infty)$ and a family ${D_x colon x in [0,1]}$ of countable, dense subsets of $[0,1]$ with $bigcup_{x in [0,1]} D_x = [0,1]$ and $sum_{r in D_x} a(r) < infty$ for all $x in [0,1]$,






set-theory measure-theory




set-theory measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 23 at 11:47









Dieter KadelkaDieter Kadelka

9116




9116








  • 1




    $begingroup$
    Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
    $endgroup$
    – Asaf Karagila
    Feb 23 at 11:54










  • $begingroup$
    Thank you. I shall try it.
    $endgroup$
    – Dieter Kadelka
    Feb 23 at 12:01










  • $begingroup$
    The Vitali set $V$ allows the definition of $cal{D} := {(v + mathbb{Q}) cap [0,1] colon v in V}$. The $D in cal{D}$ are either identical or disjoint. This is essential for defining $a$ whatever we like. Asaf Karagila thank you again.
    $endgroup$
    – Dieter Kadelka
    Feb 23 at 12:25














  • 1




    $begingroup$
    Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
    $endgroup$
    – Asaf Karagila
    Feb 23 at 11:54










  • $begingroup$
    Thank you. I shall try it.
    $endgroup$
    – Dieter Kadelka
    Feb 23 at 12:01










  • $begingroup$
    The Vitali set $V$ allows the definition of $cal{D} := {(v + mathbb{Q}) cap [0,1] colon v in V}$. The $D in cal{D}$ are either identical or disjoint. This is essential for defining $a$ whatever we like. Asaf Karagila thank you again.
    $endgroup$
    – Dieter Kadelka
    Feb 23 at 12:25








1




1




$begingroup$
Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
$endgroup$
– Asaf Karagila
Feb 23 at 11:54




$begingroup$
Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
$endgroup$
– Asaf Karagila
Feb 23 at 11:54












$begingroup$
Thank you. I shall try it.
$endgroup$
– Dieter Kadelka
Feb 23 at 12:01




$begingroup$
Thank you. I shall try it.
$endgroup$
– Dieter Kadelka
Feb 23 at 12:01












$begingroup$
The Vitali set $V$ allows the definition of $cal{D} := {(v + mathbb{Q}) cap [0,1] colon v in V}$. The $D in cal{D}$ are either identical or disjoint. This is essential for defining $a$ whatever we like. Asaf Karagila thank you again.
$endgroup$
– Dieter Kadelka
Feb 23 at 12:25




$begingroup$
The Vitali set $V$ allows the definition of $cal{D} := {(v + mathbb{Q}) cap [0,1] colon v in V}$. The $D in cal{D}$ are either identical or disjoint. This is essential for defining $a$ whatever we like. Asaf Karagila thank you again.
$endgroup$
– Dieter Kadelka
Feb 23 at 12:25










1 Answer
1






active

oldest

votes


















7












$begingroup$

The relation
$$
xsim y quad iff quad x-yinmathbb{Q}
$$
is an equivalence relation that partitions $[0,1]$ into countable sets of the form $[t]=(t+mathbb{Q})cap [0,1]$.

The set ${[t]:,tin [0,1]}$ (the set of all equivalence classes of the relation) has the same cardinality as $[0,1]$.
Let
$$
psi:[0,1]to{[t]:, tin [0,1]}
$$
be a bijection. For $xin [0,1]$ we define $D_x=psi(x)$. Since the sets $D_x$ are precisely equivalence classes of $sim$, we have that they are countable, dense and $bigcup_{xin [0,1]} D_x=[0,1]$.
Each of the sets $[t]=(t+mathbb{Q})cap [0,1]$ is countable infinite. Let
$$
phi_{[t]}:[t]to{n^{-2}:, ninmathbb{N}}
$$ be a bijection defined for each of the equivalence classes.
Finally we define
$$
a:[0,1]to (0,infty)
quadtext{by}quad
a(x)=phi_{[x]}(x).
$$
It is easy to see that the function $a$ has the desired property since
$D_x=[t]$ for some $t$ and hence
$$
sum_{rin D_x} a(r)=
sum_{rin [t]}phi_{[r]}(r)=
sum_{rin [t]}phi_{[t]}(r)=
sum_{ninmathbb{N}} n^{-2}=frac{pi^2}{6}.
$$
In the second equality we used the fact that $[r]=[t]$ for $rin [t]$ which is a property of any equivalence relation.






share|cite|improve this answer











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    1 Answer
    1






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    active

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    7












    $begingroup$

    The relation
    $$
    xsim y quad iff quad x-yinmathbb{Q}
    $$
    is an equivalence relation that partitions $[0,1]$ into countable sets of the form $[t]=(t+mathbb{Q})cap [0,1]$.

    The set ${[t]:,tin [0,1]}$ (the set of all equivalence classes of the relation) has the same cardinality as $[0,1]$.
    Let
    $$
    psi:[0,1]to{[t]:, tin [0,1]}
    $$
    be a bijection. For $xin [0,1]$ we define $D_x=psi(x)$. Since the sets $D_x$ are precisely equivalence classes of $sim$, we have that they are countable, dense and $bigcup_{xin [0,1]} D_x=[0,1]$.
    Each of the sets $[t]=(t+mathbb{Q})cap [0,1]$ is countable infinite. Let
    $$
    phi_{[t]}:[t]to{n^{-2}:, ninmathbb{N}}
    $$     be a bijection defined for each of the equivalence classes.
    Finally we define
    $$
    a:[0,1]to (0,infty)
    quadtext{by}quad
    a(x)=phi_{[x]}(x).
    $$
    It is easy to see that the function $a$ has the desired property since
    $D_x=[t]$ for some $t$ and hence
    $$
    sum_{rin D_x} a(r)=
    sum_{rin [t]}phi_{[r]}(r)=
    sum_{rin [t]}phi_{[t]}(r)=
    sum_{ninmathbb{N}} n^{-2}=frac{pi^2}{6}.
    $$
    In the second equality we used the fact that $[r]=[t]$ for $rin [t]$ which is a property of any equivalence relation.






    share|cite|improve this answer











    $endgroup$


















      7












      $begingroup$

      The relation
      $$
      xsim y quad iff quad x-yinmathbb{Q}
      $$
      is an equivalence relation that partitions $[0,1]$ into countable sets of the form $[t]=(t+mathbb{Q})cap [0,1]$.

      The set ${[t]:,tin [0,1]}$ (the set of all equivalence classes of the relation) has the same cardinality as $[0,1]$.
      Let
      $$
      psi:[0,1]to{[t]:, tin [0,1]}
      $$
      be a bijection. For $xin [0,1]$ we define $D_x=psi(x)$. Since the sets $D_x$ are precisely equivalence classes of $sim$, we have that they are countable, dense and $bigcup_{xin [0,1]} D_x=[0,1]$.
      Each of the sets $[t]=(t+mathbb{Q})cap [0,1]$ is countable infinite. Let
      $$
      phi_{[t]}:[t]to{n^{-2}:, ninmathbb{N}}
      $$       be a bijection defined for each of the equivalence classes.
      Finally we define
      $$
      a:[0,1]to (0,infty)
      quadtext{by}quad
      a(x)=phi_{[x]}(x).
      $$
      It is easy to see that the function $a$ has the desired property since
      $D_x=[t]$ for some $t$ and hence
      $$
      sum_{rin D_x} a(r)=
      sum_{rin [t]}phi_{[r]}(r)=
      sum_{rin [t]}phi_{[t]}(r)=
      sum_{ninmathbb{N}} n^{-2}=frac{pi^2}{6}.
      $$
      In the second equality we used the fact that $[r]=[t]$ for $rin [t]$ which is a property of any equivalence relation.






      share|cite|improve this answer











      $endgroup$
















        7












        7








        7





        $begingroup$

        The relation
        $$
        xsim y quad iff quad x-yinmathbb{Q}
        $$
        is an equivalence relation that partitions $[0,1]$ into countable sets of the form $[t]=(t+mathbb{Q})cap [0,1]$.

        The set ${[t]:,tin [0,1]}$ (the set of all equivalence classes of the relation) has the same cardinality as $[0,1]$.
        Let
        $$
        psi:[0,1]to{[t]:, tin [0,1]}
        $$
        be a bijection. For $xin [0,1]$ we define $D_x=psi(x)$. Since the sets $D_x$ are precisely equivalence classes of $sim$, we have that they are countable, dense and $bigcup_{xin [0,1]} D_x=[0,1]$.
        Each of the sets $[t]=(t+mathbb{Q})cap [0,1]$ is countable infinite. Let
        $$
        phi_{[t]}:[t]to{n^{-2}:, ninmathbb{N}}
        $$         be a bijection defined for each of the equivalence classes.
        Finally we define
        $$
        a:[0,1]to (0,infty)
        quadtext{by}quad
        a(x)=phi_{[x]}(x).
        $$
        It is easy to see that the function $a$ has the desired property since
        $D_x=[t]$ for some $t$ and hence
        $$
        sum_{rin D_x} a(r)=
        sum_{rin [t]}phi_{[r]}(r)=
        sum_{rin [t]}phi_{[t]}(r)=
        sum_{ninmathbb{N}} n^{-2}=frac{pi^2}{6}.
        $$
        In the second equality we used the fact that $[r]=[t]$ for $rin [t]$ which is a property of any equivalence relation.






        share|cite|improve this answer











        $endgroup$



        The relation
        $$
        xsim y quad iff quad x-yinmathbb{Q}
        $$
        is an equivalence relation that partitions $[0,1]$ into countable sets of the form $[t]=(t+mathbb{Q})cap [0,1]$.

        The set ${[t]:,tin [0,1]}$ (the set of all equivalence classes of the relation) has the same cardinality as $[0,1]$.
        Let
        $$
        psi:[0,1]to{[t]:, tin [0,1]}
        $$
        be a bijection. For $xin [0,1]$ we define $D_x=psi(x)$. Since the sets $D_x$ are precisely equivalence classes of $sim$, we have that they are countable, dense and $bigcup_{xin [0,1]} D_x=[0,1]$.
        Each of the sets $[t]=(t+mathbb{Q})cap [0,1]$ is countable infinite. Let
        $$
        phi_{[t]}:[t]to{n^{-2}:, ninmathbb{N}}
        $$         be a bijection defined for each of the equivalence classes.
        Finally we define
        $$
        a:[0,1]to (0,infty)
        quadtext{by}quad
        a(x)=phi_{[x]}(x).
        $$
        It is easy to see that the function $a$ has the desired property since
        $D_x=[t]$ for some $t$ and hence
        $$
        sum_{rin D_x} a(r)=
        sum_{rin [t]}phi_{[r]}(r)=
        sum_{rin [t]}phi_{[t]}(r)=
        sum_{ninmathbb{N}} n^{-2}=frac{pi^2}{6}.
        $$
        In the second equality we used the fact that $[r]=[t]$ for $rin [t]$ which is a property of any equivalence relation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 23 at 12:48

























        answered Feb 23 at 12:33









        Piotr HajlaszPiotr Hajlasz

        9,14843472




        9,14843472






























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