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Heine-Borel Theorem states that if a set has an open cover and if we can find a finite subcover from that open cover that covers the set, the set would be compact. I got this question while I was trying to prove that Heine-Borel property will imply that the set is closed.

Is there any restriction on open sets that are the contained in the open cover? If not, can't I just make one of the open sets to be a set of real numbers? Then, the set would always have a finite subcover, that is, the set of all real numbers that covers the set. Hence, it can cover $(0,1)$ or any other open intervals but still $(0,1)$ is not closed.

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    $\begingroup$ All sets have some finite open cover. What you want is that all arbitrary open covers have some finite subcover. $\endgroup$ – Crostul May 10 '17 at 17:43
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    $\begingroup$ "Heine-Borel Theorem states that if a set has an open cover and if we can find a finite subcover from that open cover that covers the set, the set would be compact. " Absolutely not! It says that if EVERY open cover (not just one open cover) has a finite subcover the set will be compact (but not by the theorem but by the definition) and that such a set is closed and bounded (and vice versa). $\endgroup$ – fleablood May 10 '17 at 17:48
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    $\begingroup$ "Is there any restriction on open sets that are the contained in the open cover?" Not "the" open cover. ANY open cover. So no restriction. " If not, can't I just make one of the open sets to be a set of real numbers?" Yes, that would be one open cover and it does have a finite subcover. Now show that EVERY other open covers including the ones that do not have all the reals as a set also have a finite subcover. If you can do that then yes (0,1) is closed. (Does {(1/n, 1)} have a finite subcover? It must if (0,1) is closed. And it doesn't. So (0,1) is not closed $\endgroup$ – fleablood May 10 '17 at 17:52
  • $\begingroup$ @fleablood Thank you so much for the clarification, now I understand it better. My book's definition was not very specific about it. I guess the author tried to make open cover arbitrary by saying $an open cover$ but I couldn't catch the intention. $\endgroup$ – user3000482 May 10 '17 at 17:58
  • $\begingroup$ I'm not saying that there aren't bad books out there, but... read it again. I think you are misinterpretting the statement or the context in which it was stated. It would be correct to say. "S is compact if when S has an open cover, the cover has a finite subcover". That means whichever open cover we pick will have a finite subcover, not that "S is compact if it has an open cover that has a finite subcover" Those two statements are very different. $\endgroup$ – fleablood May 10 '17 at 18:06
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The Heine Borel Theorem states that in $\mathbb{R}^n$, a set is compact iff it is closed and bounded. Perhaps in your class you initially defined compactness as being closed and bounded, but in general a set $K$ is compact if every open cover has a finite sub-cover.

An open cover $\mathcal{U}$ for as set $K$ is a collected of sets that are open in $\mathbb{R}^n$ such that $$K\subset\bigcup_{U\in\mathcal{U}}U.$$ So, the sets simply must be open. I think the key point you are missing here is that all open covers have to have a finite subcover, not just some open cover. Otherwise, every subset of $\mathbb{R}^n$ would be compact, because $\mathcal{U}=\{\mathbb{R}^n\}$ is an open cover for every subset of $\mathbb{R}^n$.

To see that every compact set is closed, suppose $K$ is compact and $p\not\in K$. For each $x\in K$, let $U_x$ be an open set containing $x$ and $V_x$ be an open set containing $p$ such that $U_x\cap V_x=\emptyset$. Then $\mathcal{U}=\{U_x:x\in K\}$ is an open cover of $K$. Thus, we can extract a finite subcover $\{U_{x_1},\dots, U_{x_n}\}$. Note that $$\left(\bigcap_{i=1}^n V_{x_i}\right)\cap\left(\bigcup_{i=1}^n U_{x_i}\right)=\emptyset$$ by our construction. Since $K\subset \bigcup_{i=1}^n U_{x_i}$, it follows that $V=\bigcap_{i=1}^n V_{x_i}$ is an open set containing $p$ that contains no element of $K$. Thus, $p$ cannot be a limit point of $K$. It follows that $K$ contains all of its limit points, and is closed.

Note that the point where we used a finite subcover is so that we could take a finite intersection of open sets to ensure $V$ was open.

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  • $\begingroup$ So $V$ only contains $p$ as its element? Does this also contradict that $V$ has to be an open set because singleton cannot be an open set? $\endgroup$ – user3000482 May 10 '17 at 18:16
  • $\begingroup$ No, each $V_{x_i}$ was on open set containing $p$ which was disjoint from $U_{x_i}$. As a finite intersection of open sets, $V$ is open, so contains infinitely many points. Its just that none of these points are in $K$, because we showed $V$ and $K$ are disjoint. $\endgroup$ – helloworld112358 May 10 '17 at 18:19
  • $\begingroup$ Is this the only way of proving the compact set is closed? I don't think I would have come up with this proof even if I spent a month on trying. $\endgroup$ – user3000482 May 10 '17 at 18:45
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    $\begingroup$ This is in sense some the most general way of doing it, since it works in a large class of topological spaces (Hausdorff spaces, which are spaces where for any $x\neq y$ you can find disjoint open sets $U$ and $V$ containing $x$ and $y$ respectively). In $\mathbb{R}^n$ (and in fact any metric space), sequential compactness is equivalent to compactness. A set $K$ is sequentially compact if every sequence in $K$ has a convergent subsequence with a limit in $K$. It is perhaps easier to show $K$ is closed using this characterization. $\endgroup$ – helloworld112358 May 10 '17 at 18:47
  • $\begingroup$ As for coming up with this proof yourself, this is perhaps the type of proof that you may need to see other proofs that use this idea of disjoint neighborhoods to come up with this idea. It's okay if you couldn't come up with it on first try. What's important is that you understand the idea of the proof and could reproduce on your own now (or maybe in a few days to make sure you actually understand). And perhaps most important is that you could apply the same idea to other proofs now. $\endgroup$ – helloworld112358 May 10 '17 at 18:51
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That is not what the Heine-Borel Theorem states. The Heine-Borel Theorem states that if $S\subseteq \mathbb{R}^n$ (with the usual topology/metric), then $S$ is compact if and only if $S$ is closed and bounded.

What you stated is more like the definition of compact: $S$ is compact provided that for every open cover of $S$, there is a finite subcover. That "every" is extremely important. Sure, $\{(0,1)\}$ is an open cover of $(0,1)$ and it has a finite subcover, but there exist open covers of $(1,2)$ with no finite subcover. Thus $(0,1)$ is not compact.

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  • $\begingroup$ This is repeated in several different answers here, so picking yours out is a little unfair, but I don't want to repeat the comment for everyone. In the OP's case, I would guess that the book - which is clearly an intro real analysis book - is only concerned with $\Bbb R$ and defines compact as "closed and bounded". Under that definition, the content of the H-B theorem is indeed "every cover has a finite subcover if and only if the set is compact". $\endgroup$ – Paul Sinclair May 10 '17 at 20:24
  • $\begingroup$ Thanks, that is a good point which I had not considered. $\endgroup$ – kccu May 10 '17 at 22:02
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It has to work for any open cover rather than just a particular open cover.

For exampl, consider the open cover $\{\left( \frac1{N}, 1 \right):N \geq 1\}$, finitely many of them will not cover $(0,1)$.

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"Heine-Borel Theorem states that if a set has an open cover and if we can find a finite subcover from that open cover that covers the set, the set would be compact."

1) that's not the theorem that states that but the definition of compact and

2) VERY IMPORTANT It's not that if you can fine one open cover with a finite subcover. It's that EVERY open cover must have a finite subcover.

So, yes, $\{\mathbb R\}$ is an open cover of $(0,1)$ and $\{\mathbb R\}$ is itself finite. That is true$*$.

But $\{\mathbb R\}$ is only one open cover. There are others. And all the others must also have finite subcovers for $(0,1)$ to be compact.

$\{(\frac 1n, 1)| n \in \mathbb N\}$ is a different open cover. It does not have a finite subcover. So it is a counter example that proves that not every open cover of $(0,1)$ has a finite subcover.

So $(0,1)$ is not compact. And by the Heine-Borel Theorem is not closed and bounded.

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$*$[ $\{\mathbb R\}$ is an open cover for every subset of $\mathbb R$ which means every set is compact which means every set is closed and bounded which ... not only is obviously not true, would make the word "compact" utterly useless. Also "closed" and "bounded" would be meaningless.]

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  • $\begingroup$ Regarding 1, it can, depending on your definition, very well be a theorem. It most likely is one in first semester mathematics. $\endgroup$ – Carsten S May 10 '17 at 21:18

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