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Partitioning the Reals into two Locally Uncountable, Dense Sets

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Partitioning the Reals into two Locally Uncountable, Dense Sets

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2

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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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asked 4 hours ago

Charles HudginsCharles Hudgins

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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

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 $mathbb{R}$ 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

asked 4 hours ago

Charles HudginsCharles Hudgins

3006

$endgroup$

share|cite|improve this question

asked 4 hours ago

Charles HudginsCharles Hudgins

3006

share|cite|improve this question

asked 4 hours ago

Charles HudginsCharles Hudgins

3006

asked 4 hours ago

Charles HudginsCharles Hudgins

3006

asked 4 hours ago

Charles HudginsCharles Hudgins

3006

2 Answers
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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".

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

Eric WofseyEric Wofsey

194k14223354

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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

answered 3 hours ago

Ross MillikanRoss Millikan

302k24201375

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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

edited 3 hours ago

answered 3 hours ago

Eric WofseyEric Wofsey

194k14223354

$endgroup$

add a comment | 

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

edited 3 hours ago

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 $rinmathbb{R}$, 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

edited 3 hours ago

answered 3 hours ago

Eric WofseyEric Wofsey

194k14223354

$endgroup$

share|cite|improve this answer

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

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

answered 3 hours ago

Ross MillikanRoss Millikan

302k24201375

$endgroup$

add a comment | 

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

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.

share|cite|improve this answer

answered 3 hours ago

Ross MillikanRoss Millikan

302k24201375

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Ross MillikanRoss Millikan

302k24201375

answered 3 hours ago

Ross MillikanRoss Millikan

302k24201375

answered 3 hours ago

Ross MillikanRoss Millikan

302k24201375