Partitioning the Reals into two Locally Uncountable, Dense SetsLocally non-enumerable dense subsets of RUncountable dense subset whose complement is also uncountable and denseA question about uncountable, dense sets in RA Question regarding disjoint dense setsThere is no uncountable collection of pairwise disjoint open sets in $mathbb R$Prove that an uncountable space with the countable-closed topology satisfies the countable chain condition.Is $mathbbR$ the disjoint union of finitely many congruent dense sets?Can a countable dense subset be split into two disjoint dense subsets?3 dense uncountable pairwise disjoint subsets of real lineIs the intersection of a dense set and a linear subspace in $mathbb R^n$ dense?
Can SQL Server create collisions in system generated constraint names?
Size of electromagnet needed to replicate Earth's magnetic field
If a planet has 3 moons, is it possible to have triple Full/New Moons at once?
What happened to Captain America in Endgame?
How to denote matrix elements succinctly?
How did Captain America manage to do this?
Should the Death Curse affect an undead PC in the Tomb of Annihilation adventure?
Does tea made with boiling water cool faster than tea made with boiled (but still hot) water?
What are the characteristics of a typeless programming language?
Re-entry to Germany after vacation using blue card
How to fry ground beef so it is well-browned
What is the philosophical significance of speech acts/implicature?
Can we say “you can pay when the order gets ready”?
555 timer FM transmitter
Alignment of various blocks in tikz
Why does nature favour the Laplacian?
How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice?
Why was the Spitfire's elliptical wing almost uncopied by other aircraft of World War 2?
Extension of 2-adic valuation to the real numbers
Why must Chinese maps be obfuscated?
Pre-plastic human skin alternative
Map of water taps to fill bottles
On The Origin of Dissonant Chords
How can I practically buy stocks?
Partitioning the Reals into two Locally Uncountable, Dense Sets
Locally non-enumerable dense subsets of RUncountable dense subset whose complement is also uncountable and denseA question about uncountable, dense sets in RA Question regarding disjoint dense setsThere is no uncountable collection of pairwise disjoint open sets in $mathbb R$Prove that an uncountable space with the countable-closed topology satisfies the countable chain condition.Is $mathbbR$ the disjoint union of finitely many congruent dense sets?Can a countable dense subset be split into two disjoint dense subsets?3 dense uncountable pairwise disjoint subsets of real lineIs the intersection of a dense set and a linear subspace in $mathbb R^n$ dense?
$begingroup$
Is it possible to find two disjoint subsets $X$ and $Y$ of $mathbbR$ such that both are dense in $mathbbR$ and both are locally uncountable?
By a locally uncountable set $X subset mathbbR$, I mean a set which has the property that if I take any nonempty open subset $U$ of $mathbbR$, then $U cap X$ has cardinality strictly larger than the natural numbers.
general-topology measure-theory examples-counterexamples
asked 4 hours ago
Charles HudginsCharles Hudgins
3006
$endgroup$
add a comment |
$begingroup$
Is it possible to find two disjoint subsets $X$ and $Y$ of $mathbbR$ such that both are dense in $mathbbR$ and both are locally uncountable?
By a locally uncountable set $X subset mathbbR$, I mean a set which has the property that if I take any nonempty open subset $U$ of $mathbbR$, then $U cap X$ has cardinality strictly larger than the natural numbers.
general-topology measure-theory examples-counterexamples
asked 4 hours ago
Charles HudginsCharles Hudgins
3006
$endgroup$
add a comment |
$begingroup$
Is it possible to find two disjoint subsets $X$ and $Y$ of $mathbbR$ such that both are dense in $mathbbR$ and both are locally uncountable?
By a locally uncountable set $X subset mathbbR$, I mean a set which has the property that if I take any nonempty open subset $U$ of $mathbbR$, then $U cap X$ has cardinality strictly larger than the natural numbers.
general-topology measure-theory examples-counterexamples
asked 4 hours ago
Charles HudginsCharles Hudgins
3006
$endgroup$
Is it possible to find two disjoint subsets $X$ and $Y$ of $mathbbR$ such that both are dense in $mathbbR$ and both are locally uncountable?
By a locally uncountable set $X subset mathbbR$, I mean a set which has the property that if I take any nonempty open subset $U$ of $mathbbR$, then $U cap X$ has cardinality strictly larger than the natural numbers.
general-topology measure-theory examples-counterexamples
general-topology measure-theory examples-counterexamples
asked 4 hours ago
Charles HudginsCharles Hudgins
3006
asked 4 hours ago
Charles HudginsCharles Hudgins
3006
asked 4 hours ago
Charles HudginsCharles Hudgins
3006
asked 4 hours ago
Charles HudginsCharles Hudgins
3006
asked 4 hours ago
Charles HudginsCharles Hudgins
3006
3006
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes. For instance, for each interval with rational endpoints, pick a Cantor set in the interval, and let $X$ be the union of all these Cantor sets. Then $X$ has uncountable intersection with every open interval, but it is meager (or measure zero, if the Cantor sets you chose all have measure zero) so its complement $Y$ has uncountable intersection with every open interval as well.
Here's another construction which gives uncountably many such subsets at once. For each $rinmathbbR$, let $X_r$ be the set of real numbers whose decimal expansion agrees with $r$'s decimal expansion on all but finitely many of the even digits. Then each $X_r$ has uncountably many points in any open interval, but $X_r$ and $X_s$ are disjoint unless all but finitely many of the even digits of $r$ agree with those of $s$ (in which case $X_r=X_s$).
With heavy use of the axiom of choice, you can get even crazier examples. For instance, you can construct a family of uncountably many disjoint subsets of $mathbbR$, each of which has uncountable intersection with every uncountable closed subset of $mathbbR$. As a sketch of the proof, note that there are $2^aleph_0$ such uncountable closed sets and each has $2^aleph_0$ elements, so you can one by one pick elements to put in each of the sets you're building in a transfinite recursion of length $2^aleph_0$. For more details of this and related constructions, look up "Bernstein sets".
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
add a comment |
$begingroup$
Here is a choice free version. Let $A$ be the set of reals whose decimal expansions eventually consist of only $3$s and $4$s and $B$ be the set of reals whose decimal expansions eventually consist of only $5$s and $6$s. These are both locally uncountable dense subsets of $Bbb R$ and there are lots of reals I haven't used.
answered 3 hours ago
Ross MillikanRoss Millikan
302k24201375
$endgroup$
add a comment |
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%2f3203999%2fpartitioning-the-reals-into-two-locally-uncountable-dense-sets%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. For instance, for each interval with rational endpoints, pick a Cantor set in the interval, and let $X$ be the union of all these Cantor sets. Then $X$ has uncountable intersection with every open interval, but it is meager (or measure zero, if the Cantor sets you chose all have measure zero) so its complement $Y$ has uncountable intersection with every open interval as well.
Here's another construction which gives uncountably many such subsets at once. For each $rinmathbbR$, let $X_r$ be the set of real numbers whose decimal expansion agrees with $r$'s decimal expansion on all but finitely many of the even digits. Then each $X_r$ has uncountably many points in any open interval, but $X_r$ and $X_s$ are disjoint unless all but finitely many of the even digits of $r$ agree with those of $s$ (in which case $X_r=X_s$).
With heavy use of the axiom of choice, you can get even crazier examples. For instance, you can construct a family of uncountably many disjoint subsets of $mathbbR$, each of which has uncountable intersection with every uncountable closed subset of $mathbbR$. As a sketch of the proof, note that there are $2^aleph_0$ such uncountable closed sets and each has $2^aleph_0$ elements, so you can one by one pick elements to put in each of the sets you're building in a transfinite recursion of length $2^aleph_0$. For more details of this and related constructions, look up "Bernstein sets".
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
add a comment |
$begingroup$
Yes. For instance, for each interval with rational endpoints, pick a Cantor set in the interval, and let $X$ be the union of all these Cantor sets. Then $X$ has uncountable intersection with every open interval, but it is meager (or measure zero, if the Cantor sets you chose all have measure zero) so its complement $Y$ has uncountable intersection with every open interval as well.
Here's another construction which gives uncountably many such subsets at once. For each $rinmathbbR$, let $X_r$ be the set of real numbers whose decimal expansion agrees with $r$'s decimal expansion on all but finitely many of the even digits. Then each $X_r$ has uncountably many points in any open interval, but $X_r$ and $X_s$ are disjoint unless all but finitely many of the even digits of $r$ agree with those of $s$ (in which case $X_r=X_s$).
With heavy use of the axiom of choice, you can get even crazier examples. For instance, you can construct a family of uncountably many disjoint subsets of $mathbbR$, each of which has uncountable intersection with every uncountable closed subset of $mathbbR$. As a sketch of the proof, note that there are $2^aleph_0$ such uncountable closed sets and each has $2^aleph_0$ elements, so you can one by one pick elements to put in each of the sets you're building in a transfinite recursion of length $2^aleph_0$. For more details of this and related constructions, look up "Bernstein sets".
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
add a comment |
$begingroup$
Yes. For instance, for each interval with rational endpoints, pick a Cantor set in the interval, and let $X$ be the union of all these Cantor sets. Then $X$ has uncountable intersection with every open interval, but it is meager (or measure zero, if the Cantor sets you chose all have measure zero) so its complement $Y$ has uncountable intersection with every open interval as well.
Here's another construction which gives uncountably many such subsets at once. For each $rinmathbbR$, let $X_r$ be the set of real numbers whose decimal expansion agrees with $r$'s decimal expansion on all but finitely many of the even digits. Then each $X_r$ has uncountably many points in any open interval, but $X_r$ and $X_s$ are disjoint unless all but finitely many of the even digits of $r$ agree with those of $s$ (in which case $X_r=X_s$).
With heavy use of the axiom of choice, you can get even crazier examples. For instance, you can construct a family of uncountably many disjoint subsets of $mathbbR$, each of which has uncountable intersection with every uncountable closed subset of $mathbbR$. As a sketch of the proof, note that there are $2^aleph_0$ such uncountable closed sets and each has $2^aleph_0$ elements, so you can one by one pick elements to put in each of the sets you're building in a transfinite recursion of length $2^aleph_0$. For more details of this and related constructions, look up "Bernstein sets".
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
$endgroup$
Yes. For instance, for each interval with rational endpoints, pick a Cantor set in the interval, and let $X$ be the union of all these Cantor sets. Then $X$ has uncountable intersection with every open interval, but it is meager (or measure zero, if the Cantor sets you chose all have measure zero) so its complement $Y$ has uncountable intersection with every open interval as well.
Here's another construction which gives uncountably many such subsets at once. For each $rinmathbbR$, let $X_r$ be the set of real numbers whose decimal expansion agrees with $r$'s decimal expansion on all but finitely many of the even digits. Then each $X_r$ has uncountably many points in any open interval, but $X_r$ and $X_s$ are disjoint unless all but finitely many of the even digits of $r$ agree with those of $s$ (in which case $X_r=X_s$).
With heavy use of the axiom of choice, you can get even crazier examples. For instance, you can construct a family of uncountably many disjoint subsets of $mathbbR$, each of which has uncountable intersection with every uncountable closed subset of $mathbbR$. As a sketch of the proof, note that there are $2^aleph_0$ such uncountable closed sets and each has $2^aleph_0$ elements, so you can one by one pick elements to put in each of the sets you're building in a transfinite recursion of length $2^aleph_0$. For more details of this and related constructions, look up "Bernstein sets".
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
edited 3 hours ago
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
answered 3 hours ago
Eric WofseyEric Wofsey
194k14223354
194k14223354
add a comment |
add a comment |
$begingroup$
Here is a choice free version. Let $A$ be the set of reals whose decimal expansions eventually consist of only $3$s and $4$s and $B$ be the set of reals whose decimal expansions eventually consist of only $5$s and $6$s. These are both locally uncountable dense subsets of $Bbb R$ and there are lots of reals I haven't used.
answered 3 hours ago
Ross MillikanRoss Millikan
302k24201375
$endgroup$
add a comment |
$begingroup$
Here is a choice free version. Let $A$ be the set of reals whose decimal expansions eventually consist of only $3$s and $4$s and $B$ be the set of reals whose decimal expansions eventually consist of only $5$s and $6$s. These are both locally uncountable dense subsets of $Bbb R$ and there are lots of reals I haven't used.
answered 3 hours ago
Ross MillikanRoss Millikan
302k24201375
$endgroup$
add a comment |
$begingroup$
Here is a choice free version. Let $A$ be the set of reals whose decimal expansions eventually consist of only $3$s and $4$s and $B$ be the set of reals whose decimal expansions eventually consist of only $5$s and $6$s. These are both locally uncountable dense subsets of $Bbb R$ and there are lots of reals I haven't used.
answered 3 hours ago
Ross MillikanRoss Millikan
302k24201375
$endgroup$
Here is a choice free version. Let $A$ be the set of reals whose decimal expansions eventually consist of only $3$s and $4$s and $B$ be the set of reals whose decimal expansions eventually consist of only $5$s and $6$s. These are both locally uncountable dense subsets of $Bbb R$ and there are lots of reals I haven't used.
answered 3 hours ago
Ross MillikanRoss Millikan
302k24201375
answered 3 hours ago
Ross MillikanRoss Millikan
302k24201375
answered 3 hours ago
Ross MillikanRoss Millikan
302k24201375
answered 3 hours ago
Ross MillikanRoss Millikan
302k24201375
302k24201375
add a comment |
add a comment |
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%2f3203999%2fpartitioning-the-reals-into-two-locally-uncountable-dense-sets%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
