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Win 7 Ultimate 32 Bit Indir

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Q:

A potential proof that Borel sets are not compact.

I have been thinking about this problem for a while, and I know there are many possible proofs to it.
Consider $X=\left\{0,1\right\}$ and let $O=\left\{\left\{0\right\},\left\{1\right\}\right\}$. Then, $O$ is an open cover of the set $X$. But, since $O$ is a collection of two disjoint sets, it cannot be a finite collection of open sets. Thus, $X$ is not compact.
My question is, how do I state my proof formally. All I know is that I need to find a
(1) sequence $O_n=\left\{\left\{0\right\},\left\{1\right\}\right\}$ such that $\text{diam}(O_n)\rightarrow\infty$,
(2) sequence $x_n\in X$ such that $x_n\in O_n$ for all $n$ and
(3) a sequence $x_n\in X$ such that $x_n\rightarrow x_\infty$ such that
I am not sure how to show that (1) can be done, can someone help me out?

A:

You can just find sequences like
$$
\underbrace{0}_{\text{sequence}}
\;\;,\;\;
\underbrace{1}_{\text{sequence}}
\;\;,\;\;
\underbrace{0,1}_{\text{sequence}}
\;\;,\;\;
\underbrace{0,1,2}_{\text{sequence}}
\;\;,\;\;
\underbrace{0,1,2,0}_{\text{sequence}}
$$

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