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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.

Sujaan Kunalan
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user222887
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  • 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 $I_n$. – hardmath Mar 11 '15 at 22:34

1 Answers1

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Any finite collection out of your set will have an element with the greatest $n$

Ross Millikan
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    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)? – user222887 Mar 11 '15 at 22:37
  • 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. – Ross Millikan Mar 11 '15 at 22:39
  • 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? – user222887 Mar 11 '15 at 22:45
  • That is correct. You can use any of those. – Ross Millikan Mar 11 '15 at 22:47