Is the Boundary of an n-Dimensional Space Always n-1 Dimensional?

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SUMMARY

The boundary of an n-dimensional space is definitively n-1 dimensional, as established in the discussion. For instance, a 2-dimensional circle has a 1-dimensional boundary, while a 3-dimensional sphere has a 2-dimensional boundary. However, the concept of "boundary" can become ambiguous in abstract spaces, as illustrated by the graph of the function y = sin(1/x) for x in (0, 1), which raises questions about defining boundaries in pathological cases. Ultimately, the discussion concludes that a line cannot effectively divide a 3-dimensional manifold, as it does not create separate regions.

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

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

I hope that was what you were asking.
 
Thanks
 
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

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

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 x = 0 \wedge y \in [-1, 1])
 
Hurkyl said:
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

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

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 x = 0 \wedge y \in [-1, 1])
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?
 
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 into two separate parts.
 
A sphere is 2-dimensional. It has no boundary. The question is, as pointed out by Halls, meaningless.
 

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