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I made a mistake while proving the theorem in the title, however, I don't see where.

Here's the theorem:

Let $C$ be closed and limited subsets of $\Bbb{R}$. Then $\bigcap_{i \in K} (C_i) = \emptyset \Rightarrow \exists F \subseteq K \; \text {finite} | \bigcap_{i \in F} (C_i) = \emptyset$

Proof (proceeds by finding equivalent problems).

$$\Bbb{R} - \bigcap_{i \in K} (C_i) = \Bbb{R} \Rightarrow \exists F \subseteq K \; \text {finite} | \Bbb{R} - \bigcap_{i \in F} (C_i) = \Bbb{R}$$

$$\bigcup_{i \in K} (\Bbb{R} - C_i) = \Bbb{R} \Rightarrow \exists F \subseteq K \; \text {finite} | \bigcup_{i \in F} (\Bbb{R} - C_i) = \Bbb{R}$$

Let $O_i = \Bbb{R} - C_i$. Note that $\forall i \in K. \, O_i$ is open and unlimited subset of $\Bbb{R}$

$$\bigcup_{i \in K} (O_i) = \Bbb{R} \Rightarrow \exists F \subseteq K \; \text {finite} | \bigcup_{i \in F} (O_i) = \Bbb{R}$$

And the last is false since if we take $\forall i \in \Bbb{N}. \, O_i = (-\infty , i), B = \bigcup_{i \in \Bbb{N}} (O_i) = \Bbb{R}$ but there are no finite subsets of N which generates R in that way.

I cannot.find the flaw in my proof. Can you please help me with that?

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  • $\begingroup$ (1) Changing your question after receiving an answer is in very poor taste. Even more so when you're doing so without interacting with the people who bothered to answer your question at first. (2) Changing "compact" to "closed and bounded" is not really a change, since on the real line a set is compact if and only if it is closed and bounded. $\endgroup$
    – Asaf Karagila
    Sep 21, 2018 at 11:54
  • $\begingroup$ I'm sorry, I would not have changed the question if I've seen answers but I hit the edit button without refreshing the page, and on my outdated smartphone browser autorefresh does not work. :-(. I apologize, but when I saw the answer, I had already edited it. (2) What I wanted to clearify is the fact that all the $C_i$ are subsets of R... $\endgroup$
    – LuxGiammi
    Sep 21, 2018 at 12:00

2 Answers 2

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For example you cannot take $ O_i = (-\infty,i) $ as an example because theese $ O_i $ are not the complement of a compact subset of $ \mathbb{R} $.

A proof would be as follows: Fix a $ C_1 $. Since $ \bigcap_{i \in K} C_i = \emptyset $ we have $ \forall x \in C_1 : \exists 1 \neq i \in K : x \notin C_i $. Hence $ \{ C_i^C \}_{i \in K} $ is an open covering of $ C_1 $ and we can choose $ i_1 , ... , i_n \in K$ with $ C_1 \subset \bigcup_{j=1}^{n} {C_{i_j}}^ C.$ Which implies $ C_1 \bigcap_{j=1}^{n} C_{i_j} = \emptyset $.

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The problem is that the complement of $(-\infty,i)$ is $[i,\infty)$ which is closed but it is not compact.

What you have shown is that the requirement of compactness is necessary, and that closure is not enough to conclude that if the intersection is empty, then a finite intersection is already empty.

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  • $\begingroup$ So, as far as I've understood, the counterexample is wrong, because the sets have an infinte complement, but the proof has no problem per se. $\endgroup$
    – LuxGiammi
    Sep 21, 2018 at 11:56
  • $\begingroup$ The word "infinite" is a bit awkward here. After all, $[0,1]$ is infinite too. Unbounded fits better. Non-compact is even better. $\endgroup$
    – Asaf Karagila
    Sep 21, 2018 at 12:07

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