Mixing
Is there an embedding theorem for non-second-countable manifolds?
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It is well known although I don't know how to prove that any second-countable topological manifold of dimension $n$ can be embedded into $mathbb{R}^{2n}$(we consider Hausdorff manifolds only). I wonder if there is a similar result, whether positive or negative, for non-second-countable manifolds? Of course there are some very strange non-second-countable manifolds like Prüfer surface which I don't want to count in, so I quote a definition from Handbook of Set-theoretic Topology:
Chapter 14, Definition 5.1: A space is $omega$-bounded if every countable subset has compact closure.
Denote the long line by $L$. Many basic examples of non-second-countable manifolds, such as $L$ and $Ltimes L$ are $omega$-bounded. In the same chapter
the structure theorem for $omega$-bounded surfaces(Theorem 5.14, "The Bagpipe Theorem") is proved. So it seems that $omega$-boundedness is a suitable condition.
Question: Is there a "nice" space(a finite dimensional $omega$-bounded manifold, if possible) such that every $omega$-bounded manifold of dimension $n$ can be embedded into it?
I have proved that the space obtained by gluing the diagonal line and $x$-axis of the first octant of $Ltimes L$ cannot be embedded into $L^{n}$ for any $n$, so possibly $L^{n}$ does not work(but the proof is ugly and may be flawed).
general-topology set-theory manifolds
asked Jan 21 at 16:01
1830rbc031830rbc03
41048
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add a comment |
$begingroup$
It is well known although I don't know how to prove that any second-countable topological manifold of dimension $n$ can be embedded into $mathbb{R}^{2n}$(we consider Hausdorff manifolds only). I wonder if there is a similar result, whether positive or negative, for non-second-countable manifolds? Of course there are some very strange non-second-countable manifolds like Prüfer surface which I don't want to count in, so I quote a definition from Handbook of Set-theoretic Topology:
Chapter 14, Definition 5.1: A space is $omega$-bounded if every countable subset has compact closure.
Denote the long line by $L$. Many basic examples of non-second-countable manifolds, such as $L$ and $Ltimes L$ are $omega$-bounded. In the same chapter
the structure theorem for $omega$-bounded surfaces(Theorem 5.14, "The Bagpipe Theorem") is proved. So it seems that $omega$-boundedness is a suitable condition.
Question: Is there a "nice" space(a finite dimensional $omega$-bounded manifold, if possible) such that every $omega$-bounded manifold of dimension $n$ can be embedded into it?
I have proved that the space obtained by gluing the diagonal line and $x$-axis of the first octant of $Ltimes L$ cannot be embedded into $L^{n}$ for any $n$, so possibly $L^{n}$ does not work(but the proof is ugly and may be flawed).
general-topology set-theory manifolds
asked Jan 21 at 16:01
1830rbc031830rbc03
41048
$endgroup$
add a comment |
$begingroup$
It is well known although I don't know how to prove that any second-countable topological manifold of dimension $n$ can be embedded into $mathbb{R}^{2n}$(we consider Hausdorff manifolds only). I wonder if there is a similar result, whether positive or negative, for non-second-countable manifolds? Of course there are some very strange non-second-countable manifolds like Prüfer surface which I don't want to count in, so I quote a definition from Handbook of Set-theoretic Topology:
Chapter 14, Definition 5.1: A space is $omega$-bounded if every countable subset has compact closure.
Denote the long line by $L$. Many basic examples of non-second-countable manifolds, such as $L$ and $Ltimes L$ are $omega$-bounded. In the same chapter
the structure theorem for $omega$-bounded surfaces(Theorem 5.14, "The Bagpipe Theorem") is proved. So it seems that $omega$-boundedness is a suitable condition.
Question: Is there a "nice" space(a finite dimensional $omega$-bounded manifold, if possible) such that every $omega$-bounded manifold of dimension $n$ can be embedded into it?
I have proved that the space obtained by gluing the diagonal line and $x$-axis of the first octant of $Ltimes L$ cannot be embedded into $L^{n}$ for any $n$, so possibly $L^{n}$ does not work(but the proof is ugly and may be flawed).
general-topology set-theory manifolds
asked Jan 21 at 16:01
1830rbc031830rbc03
41048
$endgroup$
It is well known although I don't know how to prove that any second-countable topological manifold of dimension $n$ can be embedded into $mathbb{R}^{2n}$(we consider Hausdorff manifolds only). I wonder if there is a similar result, whether positive or negative, for non-second-countable manifolds? Of course there are some very strange non-second-countable manifolds like Prüfer surface which I don't want to count in, so I quote a definition from Handbook of Set-theoretic Topology:
Chapter 14, Definition 5.1: A space is $omega$-bounded if every countable subset has compact closure.
Denote the long line by $L$. Many basic examples of non-second-countable manifolds, such as $L$ and $Ltimes L$ are $omega$-bounded. In the same chapter
the structure theorem for $omega$-bounded surfaces(Theorem 5.14, "The Bagpipe Theorem") is proved. So it seems that $omega$-boundedness is a suitable condition.
Question: Is there a "nice" space(a finite dimensional $omega$-bounded manifold, if possible) such that every $omega$-bounded manifold of dimension $n$ can be embedded into it?
I have proved that the space obtained by gluing the diagonal line and $x$-axis of the first octant of $Ltimes L$ cannot be embedded into $L^{n}$ for any $n$, so possibly $L^{n}$ does not work(but the proof is ugly and may be flawed).
general-topology set-theory manifolds
general-topology set-theory manifolds
asked Jan 21 at 16:01
1830rbc031830rbc03
41048
asked Jan 21 at 16:01
1830rbc031830rbc03
41048
edited Jan 22 at 5:53
1830rbc03
asked Jan 21 at 16:01
1830rbc031830rbc03
41048
asked Jan 21 at 16:01
1830rbc031830rbc03
41048
asked Jan 21 at 16:01
1830rbc031830rbc03
41048
41048
add a comment |
add a comment |
1 Answer
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I think the answer is simple: recall that each Tychonoff space can be embedded in a Tychonoff cube of the same weight. Less simple case concerns closed embeddings into powers of the real line. Spaces admitting such embeddings are exactly realcompact spaces. To them is devoted a small section 3.11 of Engelking’s “General topology” (2nd edn.). I attached its first page below.
answered Jan 22 at 2:08
Alex RavskyAlex Ravsky
42.4k32383
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1
$begingroup$
Thank you very much. I knew this cube and by "nice" I actually mean finite dimensional space...I will edit my question.
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– 1830rbc03
Jan 22 at 5:49
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
I think the answer is simple: recall that each Tychonoff space can be embedded in a Tychonoff cube of the same weight. Less simple case concerns closed embeddings into powers of the real line. Spaces admitting such embeddings are exactly realcompact spaces. To them is devoted a small section 3.11 of Engelking’s “General topology” (2nd edn.). I attached its first page below.
answered Jan 22 at 2:08
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
1
$begingroup$
Thank you very much. I knew this cube and by "nice" I actually mean finite dimensional space...I will edit my question.
$endgroup$
– 1830rbc03
Jan 22 at 5:49
add a comment |
$begingroup$
I think the answer is simple: recall that each Tychonoff space can be embedded in a Tychonoff cube of the same weight. Less simple case concerns closed embeddings into powers of the real line. Spaces admitting such embeddings are exactly realcompact spaces. To them is devoted a small section 3.11 of Engelking’s “General topology” (2nd edn.). I attached its first page below.
answered Jan 22 at 2:08
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
1
$begingroup$
Thank you very much. I knew this cube and by "nice" I actually mean finite dimensional space...I will edit my question.
$endgroup$
– 1830rbc03
Jan 22 at 5:49
add a comment |
$begingroup$
I think the answer is simple: recall that each Tychonoff space can be embedded in a Tychonoff cube of the same weight. Less simple case concerns closed embeddings into powers of the real line. Spaces admitting such embeddings are exactly realcompact spaces. To them is devoted a small section 3.11 of Engelking’s “General topology” (2nd edn.). I attached its first page below.
answered Jan 22 at 2:08
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
I think the answer is simple: recall that each Tychonoff space can be embedded in a Tychonoff cube of the same weight. Less simple case concerns closed embeddings into powers of the real line. Spaces admitting such embeddings are exactly realcompact spaces. To them is devoted a small section 3.11 of Engelking’s “General topology” (2nd edn.). I attached its first page below.
answered Jan 22 at 2:08
Alex RavskyAlex Ravsky
42.4k32383
answered Jan 22 at 2:08
Alex RavskyAlex Ravsky
42.4k32383
answered Jan 22 at 2:08
Alex RavskyAlex Ravsky
42.4k32383
answered Jan 22 at 2:08
Alex RavskyAlex Ravsky
42.4k32383
42.4k32383
1
$begingroup$
Thank you very much. I knew this cube and by "nice" I actually mean finite dimensional space...I will edit my question.
$endgroup$
– 1830rbc03
Jan 22 at 5:49
add a comment |
1
$begingroup$
Thank you very much. I knew this cube and by "nice" I actually mean finite dimensional space...I will edit my question.
$endgroup$
– 1830rbc03
Jan 22 at 5:49
1
1
$begingroup$
Thank you very much. I knew this cube and by "nice" I actually mean finite dimensional space...I will edit my question.
$endgroup$
– 1830rbc03
Jan 22 at 5:49
$begingroup$
Thank you very much. I knew this cube and by "nice" I actually mean finite dimensional space...I will edit my question.
$endgroup$
– 1830rbc03
Jan 22 at 5:49
add a comment |
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