Existence of a strange functionA discontinuous constructionChoice function on the countable subsets of the...
Existence of a strange function
A discontinuous constructionChoice function on the countable subsets of the realsMöbius Transform of a Continuous Possibility FunctionProduct of sigma-subadditive functionsKolmogorov doesn't show existence of Dirichlet process for arbitrary measurable spaces. Why?Forcing as a replacement of induction and diagonal argumentsCan the integral of a “generic” bounded measurable function be determined by its values on the rationals?Existence of a Borel measurable function below any positive functionExistence of a specific mad familyDo the Lebesgue-null sets cover “all the sets can naturally be regarded as sort-of-null sets”?A representation of $F_{sigmadelta}$-ideals?
$begingroup$
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
asked 1 hour ago
Dieter KadelkaDieter Kadelka
7116
$endgroup$
add a comment |
$begingroup$
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
asked 1 hour ago
Dieter KadelkaDieter Kadelka
7116
$endgroup$
1
$begingroup$
Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
$endgroup$
– Asaf Karagila
1 hour ago
$begingroup$
Thank you. I shall try it.
$endgroup$
– Dieter Kadelka
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
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
asked 1 hour ago
Dieter KadelkaDieter Kadelka
7116
$endgroup$
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
asked 1 hour ago
Dieter KadelkaDieter Kadelka
7116
asked 1 hour ago
Dieter KadelkaDieter Kadelka
7116
asked 1 hour ago
Dieter KadelkaDieter Kadelka
7116
asked 1 hour ago
Dieter KadelkaDieter Kadelka
7116
asked 1 hour ago
Dieter KadelkaDieter Kadelka
7116
7116
1
$begingroup$
Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
$endgroup$
– Asaf Karagila
1 hour ago
$begingroup$
Thank you. I shall try it.
$endgroup$
– Dieter Kadelka
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
1
$begingroup$
Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
$endgroup$
– Asaf Karagila
1 hour ago
$begingroup$
Thank you. I shall try it.
$endgroup$
– Dieter Kadelka
1 hour ago
$begingroup$
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
1 hour ago
1
1
$begingroup$
Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
$endgroup$
– Asaf Karagila
1 hour ago
$begingroup$
Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
$endgroup$
– Asaf Karagila
1 hour ago
$begingroup$
Thank you. I shall try it.
$endgroup$
– Dieter Kadelka
1 hour ago
$begingroup$
Thank you. I shall try it.
$endgroup$
– Dieter Kadelka
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 54 mins ago
Piotr HajlaszPiotr Hajlasz
8,99343270
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 54 mins ago
Piotr HajlaszPiotr Hajlasz
8,99343270
$endgroup$
add a comment |
$begingroup$
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.
answered 54 mins ago
Piotr HajlaszPiotr Hajlasz
8,99343270
$endgroup$
add a comment |
$begingroup$
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.
answered 54 mins ago
Piotr HajlaszPiotr Hajlasz
8,99343270
$endgroup$
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.
answered 54 mins ago
Piotr HajlaszPiotr Hajlasz
8,99343270
edited 40 mins ago
answered 54 mins ago
Piotr HajlaszPiotr Hajlasz
8,99343270
answered 54 mins ago
Piotr HajlaszPiotr Hajlasz
8,99343270
answered 54 mins ago
Piotr HajlaszPiotr Hajlasz
8,99343270
8,99343270
add a comment |
add a comment |
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1
$begingroup$
Sure, it's not a hard exercise to use a Vitali set to engineer something like this.
$endgroup$
– Asaf Karagila
1 hour ago
$begingroup$
Thank you. I shall try it.
$endgroup$
– Dieter Kadelka
1 hour ago
$begingroup$
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
1 hour ago