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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.