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 Post subject: Metric space
PostPosted: Tue, 20 Sep 2011 02:00:36 UTC 
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I found this cute, I got it from a friend on IRC:

Let X be a compact metric space, and f:X\to X be distance preserving. Show f is surjective.

Spoiler:
The way I know is by contradiction, but can you come up with another proof?

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 Post subject: Re: Metric space
PostPosted: Tue, 20 Sep 2011 05:17:23 UTC 
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Spoiler:
There are at least two proofs by contradiction that I know of --- considering maximal cardinality of \epsilon-separated set, or that f^k(x^*), for fixed x^*\in X-f(X), having no convergent subsequence.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Metric space
PostPosted: Tue, 20 Sep 2011 05:20:21 UTC 
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Spoiler:
From what I understand you can produce the sequence by filtering via the complements of your \epsilon cover.

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