Express Set of Dyadic with union/complementation of closed intervals
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
$begingroup$
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 |
$begingroup$
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
$endgroup$
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
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
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 |
0
active
oldest
votes
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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%2fmath.stackexchange.com%2fquestions%2f3065746%2fexpress-set-of-dyadic-with-union-complementation-of-closed-intervals%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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%2fmath.stackexchange.com%2fquestions%2f3065746%2fexpress-set-of-dyadic-with-union-complementation-of-closed-intervals%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
$begingroup$
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