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Say we have an open set $(a,b)\subset\Bbb{R}$, which has an infinite cover $\mathfrak{C}$. Let us assume $(a,b)$ is not compact. Then we can select an open set $(a_1,b_1)\subset (a,b)$ such that it is not compact. If we cannot select such an $(a_1,b_1)$, this implies $(a,b)$ is compact, which is a contradiction. Similarly, from every $(a_k,b_k)\subset (a_{k-},b_{k-1})$, we can select a non-compact $(a_{k+1},b_{k+1})\subset (a_k,b_k)$. Also, we can ensure $\lim_{n\to\infty}\operatorname{diam}((a_n.b_n))=0$. Taking $(a,b)$ to be $(0,1)$, $(a_k,b_k)=(0,\frac{1}{k+1})$ is one such example.

Now let us take $\bigcap_{i=1}^\infty (a_i,b_i)$. This intersection is a subset of $\bigcap_{i=1}^\infty [a_i,b_i]$, where $[a_i,b_i]$ is the closure of $(a_i,b_i)$. $\bigcap_{i=1}^\infty [a_i,b_i]$ contains one point (as $\Bbb{R}$ is a complete metric space). Hence, $\bigcap_{i=1}^\infty (a_i,b_i)$ is either nonempty, or contains one point. In both instances, $\bigcap_{i=1}^\infty (a_i,b_i)$ can be covered by a finite number of open sets. This is a contradictory to the initial assertion that each of $(a_k,b_k)$ cannot be covered by a finite number of open sets.

I know that $(a,b)$ is not compact. I was just hoping to understand the flaw in my argument. This proof is very similar to that of Heine-Borel theorem.

Thanks in advance!!

2 Answers2

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The $(a_i,b_i)$ are indeed not compact. That does not prevent their intersection from being compact.

Also there is no "initial assertion that each of $(a_k,b_k)$ cannot be covered by a finite number of open sets." anywhere in your argument

Hagen von Eitzen
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  • The infinite intersection is in other words the smallest set that we have selected. And all sets that we select are non-compact, as assumed in the proof. –  Aug 17 '13 at 15:10
  • @AyushKhaitan The intersection is not among the sets we selected. The sets we selected are open intervals, the intersection is not. – Hagen von Eitzen Aug 17 '13 at 15:10
  • In the case of closed sets, where $[a_k.b_k]=[0,\frac{1}{k+1}]$, is $\bigcap_{i=1}^\infty [a_i,b_i]=[0]$ among the closed sets we selected, although it is a cosed set? –  Aug 17 '13 at 17:19
  • I somehow don't understand the difference between this argument and that given in the proof of Heine-Borel theorem. Any help would be greatly appreciated. Thanks! –  Aug 17 '13 at 17:21
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There are two main errors.

  1. The fact that $\bigcap_{k\ge 1}(a_k,b_k)$ is compact in no way contradicts the fact that the sets $(a_k,b_k)$ are not compact: since $\lim_{k\to\infty}(b_k-a_k)=0$, $\bigcap_{k\ge 1}(a_k,b_k)$ is not one of the sets $(a_k,b_k)$, and it’s not true in general that an intersection of non-compact sets must be non-compact.

  2. The statement that a set $A$ is not compact does not mean that $A$ cannot be covered by a finite number of open sets. Every set in every topological space can be covered by a finite number of open sets; in fact, only one open set is needed, because the whole space is an open set.

Brian M. Scott
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