Every closed set in $mathbb R^2$ is boundary of some other subset
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A problem is bugging me many years after I first met it:
Prove that any closed subset of $mathbbR^2$ is the boundary of some set in $mathbbR^2$.
I have toyed with this problem several times over the last 20 years, but I have never managed to either prove it, or conversely prove that the question as posed is in some way wrong.
I can't remember which book I found the question in originally (but I am pretty sure that is exactly the question as it appeared in the book).
Any help, with either the topology or the source would be gratefully received!
general-topology
asked May 29 '12 at 19:10
Old John
17.5k34898
add a comment |Â
up vote
14
down vote
favorite
A problem is bugging me many years after I first met it:
Prove that any closed subset of $mathbbR^2$ is the boundary of some set in $mathbbR^2$.
I have toyed with this problem several times over the last 20 years, but I have never managed to either prove it, or conversely prove that the question as posed is in some way wrong.
I can't remember which book I found the question in originally (but I am pretty sure that is exactly the question as it appeared in the book).
Any help, with either the topology or the source would be gratefully received!
general-topology
asked May 29 '12 at 19:10
Old John
17.5k34898
Does "frontier" mean boundary?
– Chris Eagle
May 29 '12 at 19:14
Yes, I am sure the book used "frontier" to mean "boundary".
– Old John
May 29 '12 at 19:19
add a comment |Â
up vote
14
down vote
favorite
up vote
14
down vote
favorite
A problem is bugging me many years after I first met it:
Prove that any closed subset of $mathbbR^2$ is the boundary of some set in $mathbbR^2$.
I have toyed with this problem several times over the last 20 years, but I have never managed to either prove it, or conversely prove that the question as posed is in some way wrong.
I can't remember which book I found the question in originally (but I am pretty sure that is exactly the question as it appeared in the book).
Any help, with either the topology or the source would be gratefully received!
general-topology
asked May 29 '12 at 19:10
Old John
17.5k34898
A problem is bugging me many years after I first met it:
Prove that any closed subset of $mathbbR^2$ is the boundary of some set in $mathbbR^2$.
I have toyed with this problem several times over the last 20 years, but I have never managed to either prove it, or conversely prove that the question as posed is in some way wrong.
I can't remember which book I found the question in originally (but I am pretty sure that is exactly the question as it appeared in the book).
Any help, with either the topology or the source would be gratefully received!
general-topology
general-topology
asked May 29 '12 at 19:10
Old John
17.5k34898
asked May 29 '12 at 19:10
Old John
17.5k34898
edited Sep 8 at 20:19
user99914
asked May 29 '12 at 19:10
Old John
17.5k34898
asked May 29 '12 at 19:10
Old John
17.5k34898
asked May 29 '12 at 19:10
Old John
17.5k34898
17.5k34898
Does "frontier" mean boundary?
– Chris Eagle
May 29 '12 at 19:14
Yes, I am sure the book used "frontier" to mean "boundary".
– Old John
May 29 '12 at 19:19
add a comment |Â
Does "frontier" mean boundary?
– Chris Eagle
May 29 '12 at 19:14
Yes, I am sure the book used "frontier" to mean "boundary".
– Old John
May 29 '12 at 19:19
Does "frontier" mean boundary?
– Chris Eagle
May 29 '12 at 19:14
Does "frontier" mean boundary?
– Chris Eagle
May 29 '12 at 19:14
Yes, I am sure the book used "frontier" to mean "boundary".
– Old John
May 29 '12 at 19:19
Yes, I am sure the book used "frontier" to mean "boundary".
– Old John
May 29 '12 at 19:19
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
There is an very elementary way to solve this, that is also much more widely applicable.
Let $Y$ be a topological space that can be partitioned into dense subsets $D$ and $E$. If $X subset Y$ is closed, then there is a $V subset X$ such that $operatornameFr V = X$.
Take $V = X setminus (D cap operatornameInt X)$. Then since $V subset X$ we have $operatornameCl V subset operatornameCl X$, and $operatornameFr X subset V$ and $E cap operatornameInt X$ dense in $operatornameInt X$, therefore $operatornameCl V = X$. On the other hand $Y setminus X$ is dense in $Y setminus operatornameInt X$ and $D cap operatornameInt X$ is dense in $operatornameInt X$, therefore $operatornameInt V = emptyset$. It follows that $ operatornameFr V = X$.
Some additional information:
The question probably came from Willard's General topology, problem 3 B.
As Henno Brandsma pointed out in a comment, a space that can be partitioned into
two dense subsets is called resolvable. For some examples of irresolvable spaces, see the answer to "Is a perfect set a boundary?".
A somewhat stronger result can be found in "Any two disjoint open sets are the interior and exterior of some set".
edited Apr 13 '17 at 12:21
Community♦
1
answered May 31 '12 at 1:22
Niels J. Diepeveen
5,7541830
Another way to phrase your result is that in a topological space $Y$, "every closed set is a boundary" is equivalent to "$Y$ is a boundary".
– Chris Eagle
May 31 '12 at 4:17
@Chris Eagle: That is a very nice way to put it. In fact it can be proved along the same lines that a closed subset of any boundary is also a boundary.
– Niels J. Diepeveen
Jun 1 '12 at 1:30
1
Such a space $Y$ is called resolvable. Just FYI.
– Henno Brandsma
May 4 '14 at 11:43
add a comment |Â
up vote
15
down vote
Any subspace of $mathbbR^2$ is second countable, and hence separable. So if $X$ is your closed subspace, then let $A$ be a countable subset of $X$ that is dense in $X$. Then the closure of $A$ (in $mathbbR^2$) is $X$, while the interior of $A$ is empty (since any nontrivial open set in $mathbbR^2$ is uncountable). Thus $X$ is the boundary of $A$.
answered May 29 '12 at 19:21
Chris Eagle
28.6k26694
1
That would seem to do it - many thanks. I think I was somewhat blinded by the fact that the problem appeared in one of the first chapters of the book, well before concepts such as second countable had been introduced.
– Old John
May 29 '12 at 19:26
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
There is an very elementary way to solve this, that is also much more widely applicable.
Let $Y$ be a topological space that can be partitioned into dense subsets $D$ and $E$. If $X subset Y$ is closed, then there is a $V subset X$ such that $operatornameFr V = X$.
Take $V = X setminus (D cap operatornameInt X)$. Then since $V subset X$ we have $operatornameCl V subset operatornameCl X$, and $operatornameFr X subset V$ and $E cap operatornameInt X$ dense in $operatornameInt X$, therefore $operatornameCl V = X$. On the other hand $Y setminus X$ is dense in $Y setminus operatornameInt X$ and $D cap operatornameInt X$ is dense in $operatornameInt X$, therefore $operatornameInt V = emptyset$. It follows that $ operatornameFr V = X$.
Some additional information:
The question probably came from Willard's General topology, problem 3 B.
As Henno Brandsma pointed out in a comment, a space that can be partitioned into
two dense subsets is called resolvable. For some examples of irresolvable spaces, see the answer to "Is a perfect set a boundary?".
A somewhat stronger result can be found in "Any two disjoint open sets are the interior and exterior of some set".
edited Apr 13 '17 at 12:21
Community♦
1
answered May 31 '12 at 1:22
Niels J. Diepeveen
5,7541830
Another way to phrase your result is that in a topological space $Y$, "every closed set is a boundary" is equivalent to "$Y$ is a boundary".
– Chris Eagle
May 31 '12 at 4:17
@Chris Eagle: That is a very nice way to put it. In fact it can be proved along the same lines that a closed subset of any boundary is also a boundary.
– Niels J. Diepeveen
Jun 1 '12 at 1:30
1
Such a space $Y$ is called resolvable. Just FYI.
– Henno Brandsma
May 4 '14 at 11:43
add a comment |Â
up vote
9
down vote
accepted
There is an very elementary way to solve this, that is also much more widely applicable.
Let $Y$ be a topological space that can be partitioned into dense subsets $D$ and $E$. If $X subset Y$ is closed, then there is a $V subset X$ such that $operatornameFr V = X$.
Take $V = X setminus (D cap operatornameInt X)$. Then since $V subset X$ we have $operatornameCl V subset operatornameCl X$, and $operatornameFr X subset V$ and $E cap operatornameInt X$ dense in $operatornameInt X$, therefore $operatornameCl V = X$. On the other hand $Y setminus X$ is dense in $Y setminus operatornameInt X$ and $D cap operatornameInt X$ is dense in $operatornameInt X$, therefore $operatornameInt V = emptyset$. It follows that $ operatornameFr V = X$.
Some additional information:
The question probably came from Willard's General topology, problem 3 B.
As Henno Brandsma pointed out in a comment, a space that can be partitioned into
two dense subsets is called resolvable. For some examples of irresolvable spaces, see the answer to "Is a perfect set a boundary?".
A somewhat stronger result can be found in "Any two disjoint open sets are the interior and exterior of some set".
edited Apr 13 '17 at 12:21
Community♦
1
answered May 31 '12 at 1:22
Niels J. Diepeveen
5,7541830
Another way to phrase your result is that in a topological space $Y$, "every closed set is a boundary" is equivalent to "$Y$ is a boundary".
– Chris Eagle
May 31 '12 at 4:17
@Chris Eagle: That is a very nice way to put it. In fact it can be proved along the same lines that a closed subset of any boundary is also a boundary.
– Niels J. Diepeveen
Jun 1 '12 at 1:30
1
Such a space $Y$ is called resolvable. Just FYI.
– Henno Brandsma
May 4 '14 at 11:43
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
There is an very elementary way to solve this, that is also much more widely applicable.
Let $Y$ be a topological space that can be partitioned into dense subsets $D$ and $E$. If $X subset Y$ is closed, then there is a $V subset X$ such that $operatornameFr V = X$.
Take $V = X setminus (D cap operatornameInt X)$. Then since $V subset X$ we have $operatornameCl V subset operatornameCl X$, and $operatornameFr X subset V$ and $E cap operatornameInt X$ dense in $operatornameInt X$, therefore $operatornameCl V = X$. On the other hand $Y setminus X$ is dense in $Y setminus operatornameInt X$ and $D cap operatornameInt X$ is dense in $operatornameInt X$, therefore $operatornameInt V = emptyset$. It follows that $ operatornameFr V = X$.
Some additional information:
The question probably came from Willard's General topology, problem 3 B.
As Henno Brandsma pointed out in a comment, a space that can be partitioned into
two dense subsets is called resolvable. For some examples of irresolvable spaces, see the answer to "Is a perfect set a boundary?".
A somewhat stronger result can be found in "Any two disjoint open sets are the interior and exterior of some set".
edited Apr 13 '17 at 12:21
Community♦
1
answered May 31 '12 at 1:22
Niels J. Diepeveen
5,7541830
There is an very elementary way to solve this, that is also much more widely applicable.
Let $Y$ be a topological space that can be partitioned into dense subsets $D$ and $E$. If $X subset Y$ is closed, then there is a $V subset X$ such that $operatornameFr V = X$.
Take $V = X setminus (D cap operatornameInt X)$. Then since $V subset X$ we have $operatornameCl V subset operatornameCl X$, and $operatornameFr X subset V$ and $E cap operatornameInt X$ dense in $operatornameInt X$, therefore $operatornameCl V = X$. On the other hand $Y setminus X$ is dense in $Y setminus operatornameInt X$ and $D cap operatornameInt X$ is dense in $operatornameInt X$, therefore $operatornameInt V = emptyset$. It follows that $ operatornameFr V = X$.
Some additional information:
The question probably came from Willard's General topology, problem 3 B.
As Henno Brandsma pointed out in a comment, a space that can be partitioned into
two dense subsets is called resolvable. For some examples of irresolvable spaces, see the answer to "Is a perfect set a boundary?".
A somewhat stronger result can be found in "Any two disjoint open sets are the interior and exterior of some set".
edited Apr 13 '17 at 12:21
Community♦
1
answered May 31 '12 at 1:22
Niels J. Diepeveen
5,7541830
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered May 31 '12 at 1:22
Niels J. Diepeveen
5,7541830
answered May 31 '12 at 1:22
Niels J. Diepeveen
5,7541830
answered May 31 '12 at 1:22
Niels J. Diepeveen
5,7541830
5,7541830
Another way to phrase your result is that in a topological space $Y$, "every closed set is a boundary" is equivalent to "$Y$ is a boundary".
– Chris Eagle
May 31 '12 at 4:17
@Chris Eagle: That is a very nice way to put it. In fact it can be proved along the same lines that a closed subset of any boundary is also a boundary.
– Niels J. Diepeveen
Jun 1 '12 at 1:30
1
Such a space $Y$ is called resolvable. Just FYI.
– Henno Brandsma
May 4 '14 at 11:43
add a comment |Â
Another way to phrase your result is that in a topological space $Y$, "every closed set is a boundary" is equivalent to "$Y$ is a boundary".
– Chris Eagle
May 31 '12 at 4:17
@Chris Eagle: That is a very nice way to put it. In fact it can be proved along the same lines that a closed subset of any boundary is also a boundary.
– Niels J. Diepeveen
Jun 1 '12 at 1:30
1
Such a space $Y$ is called resolvable. Just FYI.
– Henno Brandsma
May 4 '14 at 11:43
Another way to phrase your result is that in a topological space $Y$, "every closed set is a boundary" is equivalent to "$Y$ is a boundary".
– Chris Eagle
May 31 '12 at 4:17
Another way to phrase your result is that in a topological space $Y$, "every closed set is a boundary" is equivalent to "$Y$ is a boundary".
– Chris Eagle
May 31 '12 at 4:17
@Chris Eagle: That is a very nice way to put it. In fact it can be proved along the same lines that a closed subset of any boundary is also a boundary.
– Niels J. Diepeveen
Jun 1 '12 at 1:30
@Chris Eagle: That is a very nice way to put it. In fact it can be proved along the same lines that a closed subset of any boundary is also a boundary.
– Niels J. Diepeveen
Jun 1 '12 at 1:30
1
1
Such a space $Y$ is called resolvable. Just FYI.
– Henno Brandsma
May 4 '14 at 11:43
Such a space $Y$ is called resolvable. Just FYI.
– Henno Brandsma
May 4 '14 at 11:43
add a comment |Â
up vote
15
down vote
Any subspace of $mathbbR^2$ is second countable, and hence separable. So if $X$ is your closed subspace, then let $A$ be a countable subset of $X$ that is dense in $X$. Then the closure of $A$ (in $mathbbR^2$) is $X$, while the interior of $A$ is empty (since any nontrivial open set in $mathbbR^2$ is uncountable). Thus $X$ is the boundary of $A$.
answered May 29 '12 at 19:21
Chris Eagle
28.6k26694
1
That would seem to do it - many thanks. I think I was somewhat blinded by the fact that the problem appeared in one of the first chapters of the book, well before concepts such as second countable had been introduced.
– Old John
May 29 '12 at 19:26
add a comment |Â
up vote
15
down vote
Any subspace of $mathbbR^2$ is second countable, and hence separable. So if $X$ is your closed subspace, then let $A$ be a countable subset of $X$ that is dense in $X$. Then the closure of $A$ (in $mathbbR^2$) is $X$, while the interior of $A$ is empty (since any nontrivial open set in $mathbbR^2$ is uncountable). Thus $X$ is the boundary of $A$.
answered May 29 '12 at 19:21
Chris Eagle
28.6k26694
1
That would seem to do it - many thanks. I think I was somewhat blinded by the fact that the problem appeared in one of the first chapters of the book, well before concepts such as second countable had been introduced.
– Old John
May 29 '12 at 19:26
add a comment |Â
up vote
15
down vote
up vote
15
down vote
Any subspace of $mathbbR^2$ is second countable, and hence separable. So if $X$ is your closed subspace, then let $A$ be a countable subset of $X$ that is dense in $X$. Then the closure of $A$ (in $mathbbR^2$) is $X$, while the interior of $A$ is empty (since any nontrivial open set in $mathbbR^2$ is uncountable). Thus $X$ is the boundary of $A$.
answered May 29 '12 at 19:21
Chris Eagle
28.6k26694
Any subspace of $mathbbR^2$ is second countable, and hence separable. So if $X$ is your closed subspace, then let $A$ be a countable subset of $X$ that is dense in $X$. Then the closure of $A$ (in $mathbbR^2$) is $X$, while the interior of $A$ is empty (since any nontrivial open set in $mathbbR^2$ is uncountable). Thus $X$ is the boundary of $A$.
answered May 29 '12 at 19:21
Chris Eagle
28.6k26694
answered May 29 '12 at 19:21
Chris Eagle
28.6k26694
answered May 29 '12 at 19:21
Chris Eagle
28.6k26694
answered May 29 '12 at 19:21
Chris Eagle
28.6k26694
28.6k26694
1
That would seem to do it - many thanks. I think I was somewhat blinded by the fact that the problem appeared in one of the first chapters of the book, well before concepts such as second countable had been introduced.
– Old John
May 29 '12 at 19:26
add a comment |Â
1
That would seem to do it - many thanks. I think I was somewhat blinded by the fact that the problem appeared in one of the first chapters of the book, well before concepts such as second countable had been introduced.
– Old John
May 29 '12 at 19:26
1
1
That would seem to do it - many thanks. I think I was somewhat blinded by the fact that the problem appeared in one of the first chapters of the book, well before concepts such as second countable had been introduced.
– Old John
May 29 '12 at 19:26
That would seem to do it - many thanks. I think I was somewhat blinded by the fact that the problem appeared in one of the first chapters of the book, well before concepts such as second countable had been introduced.
– Old John
May 29 '12 at 19:26
add a comment |Â
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Does "frontier" mean boundary?
– Chris Eagle
May 29 '12 at 19:14
Yes, I am sure the book used "frontier" to mean "boundary".
– Old John
May 29 '12 at 19:19