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I am trying to show that $(a,b)$ is not compact while $[a,b]$ is compact, purely based on definition of compactness. Here are examples, which I write it better: $[a,b]$ is covered by the intervals $(a-\dfrac{1}{n} , b+\dfrac{1}{n})$ for $n$ in $N$. However, any single one of those intervals will suffice as a finite subcover, so $[a,b]$ is compact.

Now about $(a,b)$: Suppose $O[n] = (a+\dfrac{1}{n} , b )$. The union of $O[n]$ sets is $(a,b)$, but there is no finite union of of those to be superset of $(a,b)$.

Those explanations can't be true; because I can say $(a,b)$ is also covered by the intervals $(a-\dfrac{1}{n} , b+\dfrac{1}{n})$ for $n$ in $N$. And, any single one of those intervals will suffice as a finite subcover, so $(a,b)$ is compact.

Please help me to prove that $(a,b)$ is not compact, $[a,b]$ is compact, and $[a,b)$ is not compact; I mean a simple proof and just based on the definition of compactness. All answers in this link, requires other knowledge or they are not clear and easy.

Thank you.

Community
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    Your first paragraph is incorrect, in the sense that it is not a proof that $[a,b]$ is compact. It only shows that for the specific covering you are considering it is possible to (trivially) find a finite subcovering. This does not say anything about whether the same holds for the infinitely many subcoverings you did not consider. Similarly, your third paragraph is incorrect, because it is irrelevant whether some coverings admit finite subcoverings (this always happens, for every set, compact or not), your second paragraph already showed a covering that does not. – Andrés E. Caicedo Jan 31 '15 at 07:40
  • To be compact, any given open cover must have a finite subcover, so giving an example of a cover for $[0,1]$ isn't sufficient. To show that $(0,1)$ is not compact is easier, since you only need to give one open cover that doesn't have a finite subcover. To make it easier, use the real line instead since it is homeomorphic to $(0,1)$. – LDL Jan 31 '15 at 07:41

2 Answers2

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Your proof that $[a,b]$ is compact is incorrect, as is your proof that $(a,b)$ is compact. It is not enough to take a single open cover and show it has a finite subcover; all open covers must have a finite subcover. Your example with $O[n]$ is enough to show $(a,b)$ is not compact. We have found an open cover with no finite subcover, thus it cannot be true that every open cover has a finite subcover.

To show $[a,b]$ is compact is quite tricky, due to the aforementioned need to show every open cover has a finite subcover. I would recommend looking at Wikipedia's proof of the Heine-Borel Theorem.

Jason
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For the first part of the question, see Jason's answer.

(i) Define $I_n = \left(a, \, b - \dfrac{1}{n} \right)$ to show that $(a, \, b)$ is not compact.

(ii) Same $I_n$ as part (i) will suffice.

(iii) This one is involved, it's best to use Heine-Borel. $[a, b]$ is closed and bounded and therefore it is compact.

MathMajor
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