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I'm reading Spivak's calculus on manifolds, and it defines an open cover $\mathcal{O}$ of $A$ as "A collection of open sets such that every point $x \in A$ is in some open set in the collection $\mathcal{O}$"

It says that $\mathbb{R}$ cannot be covered by a finite number of sets, but since $\mathbb{R}$ itself is open, doesn't $\mathbb{R}$ cover itself?

Does the open sets in the open cover have to be bounded or something? I can't find any other conditions though.

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    $\begingroup$ What page? what does it say exactly? $\endgroup$ – Jonas Meyer Oct 22 '17 at 17:51
  • $\begingroup$ LMAO you are right, thank you!!! There is a part where it says "no finite number of the open sets in $\mathcal{O}$ will cover $\mathbb{R}$" but I thought it was referring to the general $\mathcal{O}$ it was talking about early instead of a specific one. $\endgroup$ – CRX111011 Oct 22 '17 at 18:02
  • $\begingroup$ You can use MathJax in the titles of your posts $\endgroup$ – Chase Ryan Taylor Oct 22 '17 at 18:33
  • $\begingroup$ No, R is not a cover of R because it not a collection of open sets. {R} is a cover of R. $\endgroup$ – William Elliot Oct 22 '17 at 19:00
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Almost certainly what is meant here is that $\mathbb{R}$ is not compact, meaning for any open cover of $\mathbb{R}$, it is not guaranteed we can find a finite subcover.

$\{\mathbb{R}\}$ is a perfectly valid open cover of the reals.

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    $\begingroup$ No, it is not quite what it is saying, but you wouldn't know that because they didn't tell you in the question. It is giving one particular example of an open cover of $\mathbb R$ by open intervals of length $1$ and saying that no finite subcollection will cover $\mathbb R$, which yes implies that $\mathbb R$ is not compact. $\endgroup$ – Jonas Meyer Oct 22 '17 at 17:52
  • $\begingroup$ Comment edited to add context. $\endgroup$ – Jonas Meyer Oct 22 '17 at 17:53
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Looking up "open cover" in the index to see what is actually said, I see as expected that essential context is left out. It says that $\mathbb R$ cannot be covered by finitely many intervals of the form $(a,a+1)$.

You may have a question about why that is the case. Or, you may have misinterpreted it as saying what you wrote in your question. Either way, including in your question what is actually said in the book would be helpful.

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    $\begingroup$ You are correct, sorry for the confusion :) $\endgroup$ – CRX111011 Oct 22 '17 at 18:04
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You're correct. Are you sure no other conditions are imposed on the cover? To add another example,

$$ \mathcal{O}_{\lambda} = \{(-\infty,\lambda),(-\lambda,\infty)\} $$

covers $\mathbb{R}$ for each $\lambda > 0$.

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  • $\begingroup$ They wrongly paraphrased the book, there's much other context. $\endgroup$ – Jonas Meyer Oct 22 '17 at 17:54
  • $\begingroup$ Yes I was wrong oops, it was talking about a specific cover. $\endgroup$ – CRX111011 Oct 22 '17 at 18:03

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