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Let $I_n=\left(\frac{1}{n},1\right)$. Show that $(0,1)$ is not compact: show that any finite collection of $\{I_n\}$ will not cover $(0,1)$.

Give me a hint.

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  • $\begingroup$ This is a straightforward proof. Take any finite subset of $\{I_n\}$ and exhibit an element of $(0,1)$ not contained in any of the selected $I_n$. $\endgroup$
    – hardmath
    Mar 11 '15 at 22:34
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Any finite collection out of your set will have an element with the greatest $n$

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    $\begingroup$ If some element has the greatest n, then I need to choose the interval (1/(n+1),1) so that it is not contained in any I_n but it is in (0,1)? $\endgroup$
    – user222887
    Mar 11 '15 at 22:37
  • $\begingroup$ You just have to find one point that is in $(0,1)$ but not contained in one of the intervals in the finite collection. That proves this finite collection is not a cover of $(0,1)$. Then you argue you can do that no matter what finite collection you are given. $\endgroup$ Mar 11 '15 at 22:39
  • $\begingroup$ I can say that 1/(n+1) qualifies as a point in (0,1)? My argument is that no matter what natural number n we have, there will always be a 1/(n+1) outside of the finite collection but still in (0,1). I can even say 1/(n+2) or 1/(n+3) or 1/(n+60000) if I wanted to, right? $\endgroup$
    – user222887
    Mar 11 '15 at 22:45
  • $\begingroup$ That is correct. You can use any of those. $\endgroup$ Mar 11 '15 at 22:47

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