Can 2 manifolds neighbor each other












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I understand the definition of a set being a neighborhood of a point with a disk around it. But can 2 manifolds have elements in common where if m and n are manifolds of the same dimensions, there exists an element x in both m and n making m and n neighborhoods. (They are also open sets and smooth compact manifolds)










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  • 1




    $begingroup$
    “Neighborhood” is always used in the context of some space, here a topological space or a manifold. So “are both $M$ and $N$ neighborhoods of $x$?” is not a well-posed question. If you say “I undersand the definition of […] a neighborhood” – what's the definition you know? The details may be important.
    $endgroup$
    – Luke
    Jan 22 at 20:00


















0












$begingroup$


I understand the definition of a set being a neighborhood of a point with a disk around it. But can 2 manifolds have elements in common where if m and n are manifolds of the same dimensions, there exists an element x in both m and n making m and n neighborhoods. (They are also open sets and smooth compact manifolds)










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    “Neighborhood” is always used in the context of some space, here a topological space or a manifold. So “are both $M$ and $N$ neighborhoods of $x$?” is not a well-posed question. If you say “I undersand the definition of […] a neighborhood” – what's the definition you know? The details may be important.
    $endgroup$
    – Luke
    Jan 22 at 20:00
















0












0








0





$begingroup$


I understand the definition of a set being a neighborhood of a point with a disk around it. But can 2 manifolds have elements in common where if m and n are manifolds of the same dimensions, there exists an element x in both m and n making m and n neighborhoods. (They are also open sets and smooth compact manifolds)










share|cite|improve this question











$endgroup$




I understand the definition of a set being a neighborhood of a point with a disk around it. But can 2 manifolds have elements in common where if m and n are manifolds of the same dimensions, there exists an element x in both m and n making m and n neighborhoods. (They are also open sets and smooth compact manifolds)







general-topology differential-geometry general-relativity






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share|cite|improve this question













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edited Jan 22 at 21:29









Andrés E. Caicedo

65.7k8160250




65.7k8160250










asked Jan 22 at 17:54









bgunerbguner

11




11








  • 1




    $begingroup$
    “Neighborhood” is always used in the context of some space, here a topological space or a manifold. So “are both $M$ and $N$ neighborhoods of $x$?” is not a well-posed question. If you say “I undersand the definition of […] a neighborhood” – what's the definition you know? The details may be important.
    $endgroup$
    – Luke
    Jan 22 at 20:00
















  • 1




    $begingroup$
    “Neighborhood” is always used in the context of some space, here a topological space or a manifold. So “are both $M$ and $N$ neighborhoods of $x$?” is not a well-posed question. If you say “I undersand the definition of […] a neighborhood” – what's the definition you know? The details may be important.
    $endgroup$
    – Luke
    Jan 22 at 20:00










1




1




$begingroup$
“Neighborhood” is always used in the context of some space, here a topological space or a manifold. So “are both $M$ and $N$ neighborhoods of $x$?” is not a well-posed question. If you say “I undersand the definition of […] a neighborhood” – what's the definition you know? The details may be important.
$endgroup$
– Luke
Jan 22 at 20:00






$begingroup$
“Neighborhood” is always used in the context of some space, here a topological space or a manifold. So “are both $M$ and $N$ neighborhoods of $x$?” is not a well-posed question. If you say “I undersand the definition of […] a neighborhood” – what's the definition you know? The details may be important.
$endgroup$
– Luke
Jan 22 at 20:00












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