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Let $S$ be a unit sphere in $\mathbb{R}^3$ - I am told that $\partial S = \emptyset $, but why does the unit sphere have no boundary?

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  • $\begingroup$ For some point on the sphere, it's not hard to find a sequence of points not in $S$ that converge to that point. It's, however, hard to ind a sequence of points in the interior of $S$ that does the same. You need both for a boundary point $\endgroup$
    – Coolwater
    Apr 14, 2016 at 17:55
  • $\begingroup$ Maybe you are thinking to a ball instead of a sphere (hollow ball) ? $\endgroup$
    – Jean Marie
    Apr 14, 2016 at 18:01
  • $\begingroup$ @JeanMarie the quetion says unit sphere in $R^3$ $\endgroup$
    – user331520
    Apr 14, 2016 at 18:39

1 Answer 1

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The term boundary is ambiguous when talking about manifolds. In a sense the term boundary usually refers to either topological boundary or manifold boundary. Topological boundary is the one in which I think you are used to. If we view $S^2$ as a subset of $\mathbb{R}^3$ then for any point $x \in S^2$ there exists a neighborhood (in $\mathbb{R}^3$) which intersect $S^2$ and its complement, non-trivially i.e $\partial(S^2) = S^2$. In view of this, it should become clear that topological boundary has much to do with the topology placed on the ambient space. In the above I took $\mathbb{R}^3$ with its usual topology i.e if $A \subset \mathbb{R}^n$ then $\partial A = A$.

However, manifold boundary is a more intrinsic property. It is only concerned with the object itself. We say $x \in \partial M^n$ if $\phi(x) = 0$ for some chart $(U,\phi)$ on the manifold $M^n$. In other words,

$$\partial M^n = \{p \in M: \phi(p) = (x^1(p),...,x^n(p)); x_n(p) = 0\}$$

In the case of $S^2$, we can map it by charts to the upper-half plane $\mathbb{H}^2 = \{(x^1,x^2):x^2>0\}$ i.e $\partial S^2 = \emptyset$. To make sure this is clear, think about why $\partial(S^2 \setminus B^2) = S^1$ where $B^2$ is an open $2$-disk on the boundary of the unit sphere.

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