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The boundary of a space

  1. Oct 22, 2007 #1
    If a space is of n dimension, then the boundary of this space is n-1 dimension or not?
  2. jcsd
  3. Oct 22, 2007 #2
    If by the edge, then that is correct. If you have a 2 dimensional circle, the outer rim or boundry is a curved one dimensional line. If you have a sphere, the outer edge is a cirved two dimentional hollow sphere.

    Basically, the boundary of a space of dimension n is n-1, however, the curving of the boundry is in n space.

    I hope that was what you were asking.
  4. Oct 23, 2007 #3
  5. Oct 24, 2007 #4


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    Alas, his question was incredibly vague; as stated it doesn't make any sense, because the concept of "boundary" doesn't really make sense for an abstract space, and there are lots of pathologies even for "usual" spaces.

    For example, consider the graph of the function

    [tex]y = \sin \left( \frac{1}{x} \right) \quad \quad x \in (0, 1).[/tex]

    How are you going to define the boundary of this curve? Once you've chosen a definition, is it zero-dimensional? (Note that the closure of the graph of this curve consists of the entire line segment [itex]x = 0 \wedge y \in [-1, 1][/itex])
  6. Oct 24, 2007 #5
    Say it more clearly, why we use a line or curve to divide the 2 dimension manifold, why we use a 2 dimension surface to divide the 3 dimension manifold?
    Why we can't use a line to divide the 3 dimension manifold?
  7. Oct 24, 2007 #6


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    Because it doesn't divide it! If you draw a line in 3 dimensions, you can draw a smooth curve from any point, not on the circle, to any other point, not on the circle, without crossing the line. A line does NOT divide 3 dimensional space in to two separate parts.
  8. Oct 24, 2007 #7

    matt grime

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    A sphere is 2-dimensional. It has no boundary. The question is, as pointed out by Halls, meaningless.
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