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The sphere $S^n$ with the requirement that $x_{n+1} \geq 0$ is homeomorphic(topological equivalent) to the ball $B^n$. I mean the euclidean sphere with the euclidean metric, even if that makes a difference.

So I just wondering, is it the case that if you cut a small piece out of sphere i.e. take 1 cut point in the sphere is that the same ball.

As I imagine if you do take a cut point. The sphere would have a hole in it? why can't you then stretch that hole such that it looks like the top hemisphere of a sphere.

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    Your use of the term "cut point" is unfortunate, because that already has a different meaning. The correct term is "puncture".2011-09-30

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I'm not entirely sure I'm understanding the question correctly, but I think you're asking what happens if you remove a point from the $n$-sphere $S^n$. The answer is that -- e.g. via stereographic projection -- the resulting space is homeomorphic to $\mathbb{R}^n$. It follows from this that $S^n$ minus a single point is not compact (which was clear anyway, since removing a non-isolated point from a compact (Hausdorff!) space never yields a compact space), so it cannot be homeomorphic to the closed ball $B^n$. Of course it is homeomorphic to the open ball, and both the open and closed ball have the same homotopy type: they are both contractible spaces.

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    @thinker: See https://en.wikipedia.org/wiki/Stereographic_projection#Other_formulations_and_generalizations2016-05-24