Partitioning the Reals into two Locally Uncountable, Dense Sets












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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.










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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.










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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.










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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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      Charles HudginsCharles Hudgins

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          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".






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            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.






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              2 Answers
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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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                $begingroup$

                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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                  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














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                  edited 3 hours ago

























                  answered 3 hours ago









                  Eric WofseyEric Wofsey

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                  194k14223354























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                      $begingroup$

                      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



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                          answered 3 hours ago









                          Ross MillikanRoss Millikan

                          302k24201375




                          302k24201375






























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