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?

How to limit Drive Letters Windows assigns to new removable USB drives

Minor Revision with suggestion of an alternative proof by reviewer

"Whatever a Russian does, they end up making the Kalashnikov gun"? Are there any similar proverbs in English?

How to fry ground beef so it is well-browned

Is there a way to generate a list of distinct numbers such that no two subsets ever have an equal sum?

What makes accurate emulation of old systems a difficult task?

Philosophical question on logistic regression: why isn't the optimal threshold value trained?

Alignment of various blocks in tikz

As an international instructor, should I openly talk about my accent?

How does Captain America channel this power?

"The cow" OR "a cow" OR "cows" in this context

How to stop co-workers from teasing me because I know Russian?

Size of electromagnet needed to replicate Earth's magnetic field

Partitioning the Reals into two Locally Uncountable, Dense Sets

Checks user level and limit the data before saving it to mongoDB

What's the polite way to say "I need to urinate"?

Can I criticise the more senior developers around me for not writing clean code?

What is the most expensive material in the world that could be used to create Pun-Pun's lute?

Don’t seats that recline flat defeat the purpose of having seatbelts?

Why was the Spitfire's elliptical wing almost uncopied by other aircraft of World War 2?

Map of water taps to fill bottles

Two field separators (colon and space) in awk

Why does nature favour the Laplacian?

Which big number is bigger?



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?













2












$begingroup$


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.










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    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.










    share|cite|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      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.










      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      Charles HudginsCharles Hudgins

      3006




      3006




















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          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".






          share|cite|improve this answer











          $endgroup$




















            2












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













              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
              );



              );













              draft saved

              draft discarded


















              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









              3












              $begingroup$

              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".






              share|cite|improve this answer











              $endgroup$

















                3












                $begingroup$

                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".






                share|cite|improve this answer











                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  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".






                  share|cite|improve this answer











                  $endgroup$



                  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".







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 3 hours ago

























                  answered 3 hours ago









                  Eric WofseyEric Wofsey

                  194k14223354




                  194k14223354





















                      2












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$

















                        2












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$















                          2












                          2








                          2





                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$



                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          Ross MillikanRoss Millikan

                          302k24201375




                          302k24201375



























                              draft saved

                              draft discarded
















































                              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.




                              draft saved


                              draft discarded














                              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





















































                              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







                              Popular posts from this blog

                              Disable / Remove link to Product Items in Cart Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How can I limit products that can be bought / added to cart?Remove item from cartHide “Add to Cart” button if specific products are already in cart“Prettifying” the custom options in cart pageCreate link in cart sidebar to view all added items After limit reachedLink products together in checkout/cartHow to Get product from cart and add it againHide action-edit on cart page if simple productRemoving Cart items - ObserverRemove wishlist items when added to cart

                              Helsingin valtaus Sisällysluettelo Taustaa | Yleistä sotatoimista | Osapuolet | Taistelut Helsingin ympäristössä | Punaisten antautumissuunnitelma | Taistelujen kulku Helsingissä | Valtauksen jälkeen | Tappiot | Muistaminen | Kirjallisuutta | Lähteet | Aiheesta muualla | NavigointivalikkoTeoksen verkkoversioTeoksen verkkoversioGoogle BooksSisällissota Helsingissä päättyi tasan 95 vuotta sittenSaksalaisten ylivoima jyräsi punaisen HelsinginSuomalaiset kuvaavat sotien jälkiä kaupungeissa – katso kuvat ja tarinat tutuilta kulmiltaHelsingin valtaus 90 vuotta sittenSaksalaiset valtasivat HelsinginHyökkäys HelsinkiinHelsingin valtaus 12.–13.4. 1918Saksalaiset käyttivät ihmiskilpiä Helsingin valtauksessa 1918Teoksen verkkoversioTeoksen verkkoversioSaksalaiset hyökkäävät Etelä-SuomeenTaistelut LeppävaarassaSotilaat ja taistelutLeppävaara 1918 huhtikuussa. KapinatarinaHelsingin taistelut 1918Saksalaisten voitonparaati HelsingissäHelsingin valtausta juhlittiinSaksalaisten Helsinki vuonna 1918Helsingin taistelussa kaatuneet valkokaartilaisetHelsinkiin haudatut taisteluissa kaatuneet punaiset12.4.1918 Helsingin valtauksessa saksalaiset apujoukot vapauttavat kaupunginVapaussodan muistomerkkejä Helsingissä ja pääkaupunkiseudullaCrescendo / Vuoden 1918 Kansalaissodan uhrien muistomerkkim

                              Adjektiivitarina Tarinan tekeminen | Esimerkki: ennen | Esimerkki: jälkeen | Navigointivalikko