Existence of a strange function
$begingroup$
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 Feb 23 at 11:47
Dieter KadelkaDieter Kadelka
9116
$endgroup$
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 Feb 23 at 11:47
Dieter KadelkaDieter Kadelka
9116
$endgroup$
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
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 Feb 23 at 11:47
Dieter KadelkaDieter Kadelka
9116
$endgroup$
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 Feb 23 at 11:47
Dieter KadelkaDieter Kadelka
9116
asked Feb 23 at 11:47
Dieter KadelkaDieter Kadelka
9116
asked Feb 23 at 11:47
Dieter KadelkaDieter Kadelka
9116
asked Feb 23 at 11:47
Dieter KadelkaDieter Kadelka
9116
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
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
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
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 Feb 23 at 12:33
Piotr HajlaszPiotr Hajlasz
9,14843472
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f323914%2fexistence-of-a-strange-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 Feb 23 at 12:33
Piotr HajlaszPiotr Hajlasz
9,14843472
$endgroup$
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 Feb 23 at 12:33
Piotr HajlaszPiotr Hajlasz
9,14843472
$endgroup$
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 Feb 23 at 12:33
Piotr HajlaszPiotr Hajlasz
9,14843472
$endgroup$
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 Feb 23 at 12:33
Piotr HajlaszPiotr Hajlasz
9,14843472
edited Feb 23 at 12:48
answered Feb 23 at 12:33
Piotr HajlaszPiotr Hajlasz
9,14843472
answered Feb 23 at 12:33
Piotr HajlaszPiotr Hajlasz
9,14843472
answered Feb 23 at 12:33
Piotr HajlaszPiotr Hajlasz
9,14843472
9,14843472
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f323914%2fexistence-of-a-strange-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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