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I am looking for an open cover of [1,2) that has no finite subcover.

I'm thinking (1/n, 2-1/n).

Does this work? I think it is certainly an open cover of [1,2), but i'm not sure if it has finite subcover or not.

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  • $\begingroup$ Hint: If you take just finitely many sets of the form $(1/n,2-1/n)$, what is their union? $\endgroup$ – Lee Mosher Dec 7 '15 at 21:43
  • $\begingroup$ would the union be (1,2) $\endgroup$ – samsonite Dec 7 '15 at 21:58
  • $\begingroup$ Well, take just one of them. What is its union? Then two of them, what is their union? $\endgroup$ – Lee Mosher Dec 7 '15 at 22:00
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Suppose that the cover $\{\left(\frac{1}{n},2-\frac{1}{n}\right)\}_{n\in\mathbb N}$ of $[1,2)$ has a finite subcover we can find $k$ natural numbers, $n_1<n_2<\cdots <n_k$ such that $$[1,2)\subseteq\bigcup_{i=1}^{k}\left(\frac{1}{n_i},2-\frac{1}{n_i}\right)$$
Observe that $n_i<n_j \Rightarrow\left(\dfrac{1}{n_i},2-\dfrac{1}{n_i}\right)\subseteq\left(\dfrac{1}{n_j},2-\dfrac{1}{n_j}\right)$

Thus we have $[1,2)\subseteq \left(\dfrac{1}{n_k},2-\dfrac{1}{n_k}\right)$

However, $2-\dfrac{1}{n_k}\in[1,2)$ but $2-\dfrac{1}{n_k}\notin \left(\dfrac{1}{n_k},2-\dfrac{1}{n_k}\right)$ which is a contradiction.

Thus the given cover has no finite subcover.

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  • $\begingroup$ If $0<a<b\Rightarrow \frac{1}{b}<\frac{1}{a}$ and $-\frac{1}{a}<-\frac{1}{b}\Rightarrow 2-\frac{1}{a}<2-\frac{1}{b}$. What is the mistake here? $\endgroup$ – R_D Dec 9 '15 at 2:21

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