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Why is any unbounded subset of the reals not compact?

I have that a subset of the reals is compact if every open cover of this subset has a finite sub-cover.

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  • $\begingroup$ because the real line is not compact $\endgroup$ – qbert Oct 12 '16 at 21:34
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Take the cover $\{(-n,n)\}_{n\in\Bbb N}$.

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  • $\begingroup$ This was actually suggested to be used but I'm not sure how to apply it. And isn't this just one particular subset of the reals? $\endgroup$ – Remy Oct 12 '16 at 21:20
  • $\begingroup$ @JohnH No, this is an open cover that is very unlikely to have a finite subcover, whatever your unbounded set of reals may be... $\endgroup$ – Jean-Claude Arbaut Oct 12 '16 at 21:21
  • $\begingroup$ @JohnH, suppose that exists a finite subcover of the set. How will be the set? $\endgroup$ – Martín-Blas Pérez Pinilla Oct 12 '16 at 21:22
  • $\begingroup$ negative to positive infinity? That is one set (finite), but it has infinitely many elements? $\endgroup$ – Remy Oct 12 '16 at 21:27
  • $\begingroup$ Sorry, I am having trouble grasping the concepts of covers and sub covers $\endgroup$ – Remy Oct 12 '16 at 21:28
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For your unbounded subset of the reals, consider the open cover that consists of the sets $\{O_1, O_2, \dotsc\}$ where $O_i$ is the open ball centered at the origin with radius $i$. This cover admits no finite subcover.

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An alternative definition for compactness, which is logically equivalent to the other one says that a subset of $\mathbb R$ is compact if and only if it is closed and bounded.

Some analysis textbooks take this as THE definition, other use it as an if and only if theorem.

Now you know. It can't be compact because it's not bounded.

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