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Namely, the sphere is $\{x\in\mathbb{R}^d: \| x\|_2=1\}$. I am going about this by proving that the sphere is bounded and closed. I have proved that it is bounded and I can see that it must be closed but I don't know how to write it out, can this be done without using the continuous map method?

Przemysław Scherwentke
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  • Why would you want to do it without the continuous map method? – Ben Grossmann Oct 31 '14 at 16:35
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    Maybe the easiest way to show it is closed would be to show that its complement is open? If you pick a point in $\mathbb{R}^d$ that isn't on the sphere can you draw a neighborhood around the point that also isn't on the sphere? –  Oct 31 '14 at 16:36
  • Thanks! I can visualise this but I can't formulate a proof. –  Oct 31 '14 at 16:46
  • Given a point $p$ not on the sphere, we know that $||p|| \neq 1$ hence $\delta = | \ ||p|| - 1 \ | > 0$. Thus the open ball $B_{\delta/2}(p)$ does not intersect the sphere. – Simon S Oct 31 '14 at 16:54

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$D=\mathbb{R}^d\setminus(B_1\cup B_2)$, where $B_1=\{x\in\mathbb{R}^d: \| x\|_2<1\}$ and $B_2=\{x\in\mathbb{R}^d: \| x\|_2>1\}$. Note $B_1,B_2$ are open. (You can find a proof about open ball is open here, and proof for $B_2$ is open is similar.) Hence $D$ is closed.

John
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