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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". –  Niels Diepeveen Sep 30 '11 at 17:47

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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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I was trying to think of an easy way to prove that n-sphere is path-connected. So if I can show that it's homeophorphic to an open ball or closed ball, then it's easy to prove. I can see that it's the same as open ball $B^n$ now. However, can't see how it the whole space. Suppose if you just make the ball bigger. Hmmm, is $R^n$, homeomorphic to open ball $B^n$? I didn't know that. It would make sense. Will try and prove it. –  simplicity Sep 30 '11 at 17:07
    
@simplicity: you can prove the sphere is path-connected just by writing down the path. For instance on the $2$-sphere you can get from one point to any other point by following an arc of one latitudinal circle and then switching to an arc of a longitudinal circle. Similar constructions work in $n$-dimensions. –  Pete L. Clark Sep 30 '11 at 17:14
    
Wouldn't you need some complex calculations using spherical coordinates? or can you just put $x_1=e^{i\theta_{1}}$ e.t.c. to $x_n$, I suppose it's easy generalization. –  simplicity Sep 30 '11 at 17:18

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