Partitioning the Reals into two Locally Uncountable, Dense SetsLocally non-enumerable dense subsets of...

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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 $mathbb{R}$ 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?













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Is it possible to find two disjoint subsets $X$ and $Y$ of $mathbb{R}$ such that both are dense in $mathbb{R}$ and both are locally uncountable?



By a locally uncountable set $X subset mathbb{R}$, I mean a set which has the property that if I take any nonempty open subset $U$ of $mathbb{R}$, then $U cap X$ has cardinality strictly larger than the natural numbers.






general-topology measure-theory examples-counterexamples







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    $begingroup$


    Is it possible to find two disjoint subsets $X$ and $Y$ of $mathbb{R}$ such that both are dense in $mathbb{R}$ and both are locally uncountable?



    By a locally uncountable set $X subset mathbb{R}$, I mean a set which has the property that if I take any nonempty open subset $U$ of $mathbb{R}$, then $U cap X$ has cardinality strictly larger than the natural numbers.






    general-topology measure-theory examples-counterexamples







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Is it possible to find two disjoint subsets $X$ and $Y$ of $mathbb{R}$ such that both are dense in $mathbb{R}$ and both are locally uncountable?



      By a locally uncountable set $X subset mathbb{R}$, I mean a set which has the property that if I take any nonempty open subset $U$ of $mathbb{R}$, then $U cap X$ has cardinality strictly larger than the natural numbers.






      general-topology measure-theory examples-counterexamples







      share|cite|improve this question









      $endgroup$




      Is it possible to find two disjoint subsets $X$ and $Y$ of $mathbb{R}$ such that both are dense in $mathbb{R}$ and both are locally uncountable?



      By a locally uncountable set $X subset mathbb{R}$, I mean a set which has the property that if I take any nonempty open subset $U$ of $mathbb{R}$, then $U cap X$ has cardinality strictly larger than the natural numbers.






      general-topology measure-theory examples-counterexamples




      general-topology measure-theory examples-counterexamples






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked 3 hours ago









      Charles HudginsCharles Hudgins

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          $begingroup$

          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 $rinmathbb{R}$, 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 $mathbb{R}$, each of which has uncountable intersection with every uncountable closed subset of $mathbb{R}$. 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











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            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









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              2 Answers
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              2 Answers
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              active

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              active

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              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 $rinmathbb{R}$, 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 $mathbb{R}$, each of which has uncountable intersection with every uncountable closed subset of $mathbb{R}$. 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 $rinmathbb{R}$, 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 $mathbb{R}$, each of which has uncountable intersection with every uncountable closed subset of $mathbb{R}$. 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 $rinmathbb{R}$, 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 $mathbb{R}$, each of which has uncountable intersection with every uncountable closed subset of $mathbb{R}$. 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 $rinmathbb{R}$, 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 $mathbb{R}$, each of which has uncountable intersection with every uncountable closed subset of $mathbb{R}$. 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






























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