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Here is an extract from https://math.uc.edu/~halpern/calc.1/Ho/Heineborelthm.pdf

Some words and sentences were cut and modified to avoid wordiness

Theorem: Let $\mathcal F$ be a family of open intervals (set) that covers the closed interval $[a.b]$. Then $F$ has a finite subcover.

Proof. Let $B = \{x : a\leq x \leq b, \text{and} \; [a,x] \; \text{has a finite subcover} \}$. Then $B$ is nonempty since it contains $a$ and has an upper bound $b$, so the set $B$ has a least upper bound $x_0$. First we see that $x_0$ is in $B$. If not, then let $(c,d)$ be the open interval (set) in $\mathcal F$ that contains $x_0$, then there has to be some element $x \in B$ in the open set $(c,x_0)$. Since $x \in B$, we have $F_1,\dots, F_n$ in $\mathcal F$ that covers $[a,x]$. Now the intervals $F_1, \dots, F_n$, $(c,d)$ in $\mathcal F$ cover $[a,x_0]$. So $x_0 \in B$. Second we show $x_0 = b$. Suppose $x_0 < b$, then $[a,x_0]$ the finite family $F_1, \dots, F_n, (c,d) = F_{n+1}$ in $\mathcal F$ would cover $[a,d]$ and $x_0$ would not be an upper bound of $B$. So we have found finitely many intervals (sets) that cover $[a,b]$

For the proof of the Hein-Borel Theorem, how does $F_{n+1}$ cover $[a,d]$? The open set $(c,d)$ doesn't touch $d$ and I think we are to assume $d < b$. In fact the other cover, $F_1, \dots, F_n$ doesn't touch anywhere near $d$ either.

EDIT

I also have a counterexample to the proof

enter image description here

I have three open sets: blue, red, and purple. Blue set only covers up until $x$, purple set covers a small portion around $x_0$, leaving a gap in between blue and purple. The red set is what covers the rest of the interval.

Remark I looked at Spivak's Manifold book and I have noticed he uses the words "interval" and $U$ (open set) separately. Is there a reason for this?

enter image description here

EDIT#2 okay I figured out the yellow text. Basically he is saying the values less than $\alpha$ (he calls them "$x$") must live in $A$. If I am wrong, correct me. Also, why exactly is my picture wrong then?

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As I've reminded you before: to avoid future link rot, please copy down as much of the relevant information as possible into your question. At minimum, take a screenshot and include it as an image in your question. –  Zev Chonoles Jun 28 '13 at 4:39
    
I didn't think the link would be broken since it came from a university source. –  sidht Jun 28 '13 at 4:41
    
(And still, copy-typing the text would be preferred) –  Hagen von Eitzen Jun 28 '13 at 5:29
    
One moment please –  sidht Jun 28 '13 at 5:35
    
@sidht Thank you –  Hagen von Eitzen Jun 28 '13 at 5:50
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3 Answers 3

The text should better read "... if $(c,d)$ is an open interval in $\mathcal F$ that contains $x_0$ ..." in the first part. Also, you are right: we (immediately) only get that $ F_{n+1}$ covers $[a,d)$, but that still shows that $x_0$ is not an upper boud for $B$, as witnessed by any element of the non-empty set $(x_0,b]\cap [a,d)$, for example by $x_1=\min\{\frac{x_0+d}2,b\}$.

Also note that we cannot a priori assume $d\le b$ as the covering property of $\mathcal F$ allows it to actually cover more than just $[a,b]$ (and that is necesarily the case).

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Sorry, isn't $(x_0, b] \cap [a, d) = (x_0, d)$? –  sidht Jun 28 '13 at 5:51
    
Do you mind reading my edit? Thanks –  sidht Jun 29 '13 at 2:38
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As worded and stated, the proof is indeed incorrect, since the cover given by $F_1, \dots, F_n, (c, d) = F_{n+1}$ covers only the interval $[a, d)$. It is possible that $d > b$, for example, the collection $\{ (-2, \frac{3}{4}), (\frac{1}{2}, 2) \}$ is an open cover of the interval $[0, 1]$. If $d>b$ we're done, since we already have a finite subcover of $[a, b]$ and can disregard the details regarding upper bounds of $B$ and so forth. However, since you can't assume $d>b$, take for the sake of argument that $d \le b$. Then $d \in [a, b]$ (since $(c, d)$ contains an element, namely $x_0$, of $[a, b]$), so there is some element of $\mathcal{F}$, call it $F_{n+2}$ such that $d \in F_{n+2}$. Then $F_1, F_2, \dots , F_{n+2}$ forms an open cover of $[a, d]$, so that $d \in B$ contradicting that $x_0$ is an upper bound of $B$. This completes and corrects the proof from the UC link.

Regarding the remark about Spivak's use of "interval" and "open set", recall that $[a, b], [a, b), (a, b], (a, b)$ are all considered intervals in the real line by a lot of authors (for example Rudin, Ross, Stein, Stewart), and I believe Spivak does as well, but only $(a, b)$ is open in the usual sense here. He's saying that all points in some interval $(c, \alpha]$ lie in this set $A$. I think this is what you were saying in edit #2 as well.

Finally, regarding your picture, my understanding is that you're arguing the gap between the blue set and the purple set means our finite collection omits some points in $[a, b]$ and cannot be an open cover, where the blue set is $\bigcup\limits_{k=1}^n F_k$. However, your picture has $x$ outside the interval $(c, d)$, which contradicts the very definition of $x$ in the proof, so that the picture is not a counterexample of the proof.

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This is from your excerpt from the UC link. "First we see that x0 is in B. If not, then let (c,d) be the open interval (set) in  that contains x0, then there has to be some element x∈B in the open set (c,x0)." This clearly defines where $x$ is, contradicting your picture. –  grantfgates Jul 5 '13 at 20:48
    
The $\alpha$ in Spivak's proof is the $x_0$ in the UC proof. The fact that $\alpha$ is the least upper bound implies that $x \in B$, so that there is a finite covering $F_1, \dots, F_n$ of $[a, x]$. The location of $x$, so long as we maintain $x < \alpha$, is independent of the fact that $\alpha$ is a least upper bound. –  grantfgates Jul 5 '13 at 22:49
    
$\alpha$ being the least upper bound of $B$ means that for any $\epsilon > 0$, there is some $x \in B$ such that $|s-\alpha|<\epsilon$. Since $(c, d)$ is an open interval containing $x_0 = \alpha$, we can pick some $\epsilon > 0$ such that $(\alpha - \epsilon, \alpha+\epsilon) \subset (c, d)$ (this is the definition of openness). Using both of these we can pick and $x \in B$ such that $x \in (c, d)$ as well. –  grantfgates Jul 6 '13 at 2:14
    
OK, the first thing to keep clear is that $B$ is not a set of finite subcovers, but rather the set of all points $x$ in $[a, b]$ such that $[a, x]$ has a finite subcover. So now you're asking why $(c, d)$ overlaps with the finite subcover of $[a, x]$ for some $x \in B$. The logic is exactly the same logic used in my last comment, word for word. (By the way, in the first line of my last comment I meant $|x-\alpha| < \epsilon$, not $|s-\alpha|$.) –  grantfgates Jul 6 '13 at 7:04
    
Look back at the definition of $B$. $B$ is the set of all such $x$. That's how you read set builder notation. However, if Spivak defined $A$ differently than the UC link defined $B$, I wouldn't know, because you don't have the definition of $A$ posted in your original question. –  grantfgates Jul 7 '13 at 0:09
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I think I know what was wrong with my initial problem. I kept thinking that a particular open set gave rise to a contradiction. But the whole proof hinges on the fact that it is saying there is some interval that does this job.

Sure you can find one that doesn't work, but it's saying that even if one interval works, then it's all good.

In my picture $x$ still lives in the red interval, making it a member of $A$.

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