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Our lecturer presented to us (at the time we only just had defined $\mathbb{R}$ and some basic terms such as interval, etc.) a sort of pseudo-Heine-Borel theorem for real numbers:

Theorem: Suppose that to each $x\in[0,1]$ we assign a positive $δ_x$.

Then we can find a finite subset $M\subset[0,1]$ so that $[0,1]\subset\bigcup_{x\in M}U_{\delta_x}$ where $U_{\delta_x}=(x-\delta_x,x+\delta_x)$

In other words: The interval $[0,1]$ is a subset of a finite union of intervals (neighbourhoods) $U_{x_1}\cup U_{x_2}\cup\dots\cup U_{x_n}$ for appropriate points $x_1,\dots ,x_n\in[0,1]$

The usual procedure when I can't understand a proof would be going through various sources, eventually understanding the whole proof. This time, such procedure is impossible, as I have found none that would deal with this theorem in such way. I'll go through the proof with bracketed numbers denoting my comments.

Proof: We define

$I=\{x\in[0,1]\mid$ interval $[0,x]$ can be "covered" by finitely many intervals $U_{\delta_x}\}$

Then $I$ is not empty - we only need to consider $x\in (0,\delta_0)$

$I$ is bounded - $I\subset[0,1]$

Let $S=\sup I$. Then $S\in[0,1]$ and $S>0$

To arrive at contradiction, we will assume that $S\in(0,1)$

Then $\exists a\in(S-\delta ,S)$, $a\in I^{(1)}$

We can also find $b\in (S,S+\delta)\cap[0,1]$, interval $[0,b]$ can be covered with finitely many intervals.

$b>\sup I$, $b\in I$, which is a contradiction.

Therefore if $S\in[0,1]$, $S\notin[0,1)$, then $S=1$ & $S\in I^{(2)}$, therefore $I=[0,1]$

My questions:

(1) What is the use of the element $a$? What does it show?

(2) How do we know that $S\in I$?

I realize asking you people to go through a proof for me is a bit beyond what this site is for, but really, I've tried and tried and found no other way! (and it's been months)

Thanks very much for any help!

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$(0,\delta_0)$ does not cover $0$, so $(0,\delta_0)$ is not evidence that $I$ is not empty. –  Thomas Andrews Jan 9 '13 at 13:16
    
Oh yes, that is true. In that case, what would be the alternative? –  Dahn Jahn Jan 9 '13 at 13:18
    
Also, do you mean $U_{\delta_x}\supset(x-\delta_x,x+\delta_x)$ or $U_{\delta_x}=(x-\delta_x,x+\delta_x)$? Because $\supset$ would let you pick $U_{\delta}=[0,1]$... –  Thomas Andrews Jan 9 '13 at 13:19
    
Ah, I am sorry to present this faulty proof, I am really struggling here. I suppose then it'd have to be an equality, I only used the $\supset$ because of the definition of a neighbourhood (has the interval as a subset). I am going to edit this now. –  Dahn Jahn Jan 9 '13 at 13:21
    
I think you want to rewrite the first sentence of the theorem to "Suppose that to each $x$ we assign a positive $\delta_x$". The idea is that if we cover the interval by this "trivial" infinite cover, we will be able to pick finitely many of the $x$, say $x_{i}$, $1 \leq i \leq n$ that will also cover it. –  Pedro Tamaroff Jan 9 '13 at 13:29
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3 Answers 3

up vote 2 down vote accepted

Let's write down some preliminaries first

This first might be taken as a definition, but you might have seen this (not necessarily as a definition) if you know a little bit about metric spaces or topology.

DEF A set $S$ in $\Bbb R$ is called an open set if it is the union of open intervals. In particular, open intervals are open. A set is called closed if it is the complement of an open set. In particular, all closed intervals $[a,b]$ are closed.

DEF Given a set $X$, by a cover of $X$ we mean a family of sets $\mathscr F$ for which $\bigcup \mathscr F\supset X$. If each of the sets $F\in \mathscr F$ is open, we call $\mathscr F$ an open cover of $X$.

THM Let $I=[a,b]$, a closed interval. Suppose that to each $x\in [a,b]$ we assign a $\delta_x>0$, and define $U_x=(x-\delta_x,x+\delta_x)$, $\mathscr U=\{U_x\}_{x\in [a,b]}$. Then there exists finitely many $x_i\in [a,b]$, $i=1,\dots,n$, such that the family $\mathscr U'=\{U_{x_i}\}_{i=1,\dots,n}$ covers $[a,b]$.

Note This is the statement that any closed interval in $\Bbb R$ is compact. It can be proven that compactness is a topological property, that is, if $X$ and $Y$ are homeomorphic and $X$ is compact, so is $Y$. Since $[0,1]$ is homeomorphic to any closed interval $[a,b]$, it suffices to prove that $I=[0,1]$ is compact, which is what we'll do.

PROOF 1 Let $I_u=[0,u]$ and let $P(I_u)$ be the statement "given any open cover $\mathscr U$ (as before) there exists a finite subcover $\mathscr U'$ of $I_u$." Define the set

$$A=\{x\in I:P(I_x)\}$$

It follows that $0\in A$, for the "degenerate" interval $[0,0]$ has a finite subcover given any infinite cover we exhibit. Since $A$ is bounded and nonempty, the completeness of $\Bbb R$ guarantees $A$ has a supremum, say $\alpha =\sup A$. Since $1$ is an upper bound of $A$, $\alpha \leq 1$. Suppose for the sake of contradiction that $\alpha <1$, and let $\mathscr U$ be an open cover of $[0,1]$. Then $\alpha$ is in this cover, that is, $\alpha \in U_{x'}$ for some $x'\in I$. Since $\alpha$ is the supremum of $A$, there exists a point $a\in A$ (given this $\delta_{x'}$) such that $\alpha-\delta_{x'}<a< \alpha(<\alpha+\delta_{x'})$. But $\mathscr U$ is a cover of $[0,a]$ so we can cover it with a finite family $\mathscr U'$ by hypothesis. Choose $b\in (a,\alpha+\delta_{x'})$. Then we can cover $[0,b]$ with $\mathscr U'\cup\{U_{x'}\}$, whence $b$ is also in $A$. But this is a contradiction, for $b>\alpha$ yet $\alpha =\sup A$. Thus $\alpha \geq 1$. This in turn means $\alpha =1$, as desired. $\blacktriangle$.

PROOF 2 We use the method of bisection. Suppose for the sake of contradiction, that given any open cover $\mathscr C$ of $I$ there exists no finite subcover $\mathscr C'$. This means there exists a subset $X$ of $I$ which is not covered by any finite subcover. Note that $X$ is necessarily infinite (What would happen if it was finite?) Thus, $X$ is infinite over either half of $I$, or both. (Why?) Let this half be $I_1$. (If $X$ is infinite over both, let's agree to choose the left side, $[0,1/2]$). The same argument yields that $X$ is infinite over one side of $I_1$, call it $I_2$. Follow this construction, to obtain a sequence $\{I_n\}_{n\in \Bbb N}$ of closed intervals of length $\mathscr L(I_n) =1/2^{n}$. Cantor's princple of closed intervals assures there exists $\mu\in \bigcap_{n\in \Bbb N}I_n$. Since $\mu \in I$, it is in one of the open sets (intervals) $(x,y)=O\in \mathscr C$. In particular $(\mu -\delta,\mu+\delta)$ is in $(x,y)=O$ for some $\delta$. Choose now $N$ so that $1/2^N<\delta$. Then $I_n\subset (\mu -\delta,\mu+\delta)\subset O$ whenever $n\geq N$. But then we have covered all these points, that we said couldn't be covered by no finite subcover of $\mathscr C$ using one element of $\mathscr C$, namely $O$! (which is a finite subcover). This contradiction shows that our assumption that no finite subcover exists given an open cover of $I$ must be wrong, thus the assertion follows. $\blacktriangle$.

If anything is unclear, let me know.

ADD The two previous methods are amazingly useful. We can use them to prove Bolzano's Theorem (IVT), Bolzano-Weierstrass (every bounded sequence has a conv. subsequence), Heine-Borel, Heine-Cantor (continuity on compact spaces is uniform), and maybe some more I don't know about.

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Sorry for taking so long to go through this properly. I appreciate this answer very much and in the shortest possible time (that is tomorrow), I'll study it closely. Sadly now I am preoccupied with passing tomorrow's exams in much less interesting subjects such as accounting and insurance law. –  Dahn Jahn Jan 10 '13 at 18:43
    
Here I go: First proof questions. a) How do we know that $\alpha$ is in the cover? b) If we know that $\alpha$ is in the cover, don't we then also know that $[0,\alpha]$ can be covered, thus no need for $a$? c)When we know that $\alpha=1$, then we only need to take the interval $(1-\delta_1,1+\delta_1)$ to cover the whole $[0,1]$ (because we've only covered $[0,1)$ at that point), is that correct? Thanks. –  Dahn Jahn Jan 12 '13 at 9:32
    
Actually: about c) if we know that $\alpha$ is in the cover, then we probably don't need to do that. But that goes against what coffeemath said, or so it seems to me. –  Dahn Jahn Jan 12 '13 at 9:35
    
@DahnJahn I added a tiny bit before the "$\alpha$ is in the cover", hope now it is clear. –  Pedro Tamaroff Jan 12 '13 at 16:08
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It looks like the proof is showing that any open cover of $[0,1]$ contains a finite open subcover. The notation seems vague however. If the open cover is a collection of open sets $U$, then for each specific $x \in [0,1]$ one can choose an open set from the cover which contains $x$, and call that chosen open set $U_x$.

Now for any specific $x \in [0,1],$ since $U_x$ is open and contains $x$, one can choose a positive $\delta_x$ such that $$(x-\delta_x,\ x+\delta_x) \subset U_x.$$ So the subscript $x$ on the $U$ reflects a particular choice of one of the open sets $U$ in the original cover, while the subscript $x$ on the $\delta_x$ reflects a particular choice of a positive $\delta$ for the point $x$.

Now if $S$ is the sup of $I$ and if one supposes $S<1$, then one can apply the above to the specific point $x=S$. This means the specific point $S$ lies in one of the open sets, denoted $U_S$, of the original open cover, and as described above we may choose $\delta_S$ such that $$(S-\delta_S,\ S+\delta_S) \subset U_S.$$

The idea now is that, since $S$ is the sup of $I$, we can choose a point $a \in I$ slightly to the left of $S$ and at the same time to the right of $S-\delta_S.$ By definition of $I$ there are finitely many open sets from the cover which cover the interval $[0,a]$, and at this point we are going to throw in the open set $U_S$ as another open set in that finite open cover of $[0,a]$. This last open set $U_S$ now covers the points between $a$ and $S$, it also covers $S$ itself, and finally it covers points to the right of $S$, up to at least $\min(S+\delta_S,1)$. So if we fix a point $b \in [0,1]$ which is to the right of $S$ but to the left of $S+\delta_S$, we see that the new finite cover, including the old one which covered $[0,a]$ along with the added open set $U_S$, now covers the interval $[0,b]$. But this implies that $b$ is in $I$ by definition of $I$, which since $b>S$ contradicts the definition of $S$ as the sup of $I$.

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That was very easy to read, thanks. Therefore the point $a$ serves to show that $[0,a]$ can be covered, because before that we only knew that $I$ is nonempty? As to the second question, I still do not see how we know that $I=[0,1]$ –  Dahn Jahn Jan 9 '13 at 14:25
    
That $I$ is nonempty follows from the open set $U_0$ which contains all points in $(0-\delta_0,0+\delta_0)$, so that if $0 \le c < \delta_0$ then $c \in I$ because $[0,c]$ is covered by the single open set $U_0$, which thus constitutes a finite open subcover of $[0,c]$ (with in fact only one open set in that cover). For the second question: If you read through it, there was an initial point in the argument where we assumed that $S<1$, and the argument went on to a contradiction. Now we have $S \in [0,1]$ together with the denial of $S<1$, i.e. we know $S \ge 1$. Conclusion $S=1$. –  coffeemath Jan 9 '13 at 14:33
    
Yes, sorry, I haven't made myself clear enough, the part I am confused about is how do we know that $S\in I$ (and thus $I=[0,1]$) –  Dahn Jahn Jan 9 '13 at 14:35
    
We do not know that $S \in I$, and do not need to know it. If you agree the argument shows that $S$ cannot be less than 1, the only possibility left is that $S=1$. Then the one extra open set $U_S=U_1$ definitely covers the point $x=1$ and, along with the other finitely many open sets, covers all points in $[0,1]$. –  coffeemath Jan 9 '13 at 14:53
    
Got it! Thanks. –  Dahn Jahn Jan 9 '13 at 15:04
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There are various things coming together here:

(i) The system ${\mathbb R}$ which is "complete" (i.i) in a metric sense and (i.ii) with respect to its inherent order.

(ii) The idea of "compactness" which is a subtle finiteness property. Intuitively: On a compact space "you can't get lost". Unfortunately the notion of compactness has no simple mathematical description. A space $X$ is compact if, given any neighborhood field $\bigl(U(x)\bigr)_{x\in X}$, you can find a finite set of "watch towers" $x_k\in X$ such that $X=\bigcup_{k=1}^N U(x_k)$.

The Heine-Borel Theorem in your question states that a finite closed interval $[a,b]\subset{\mathbb R}$ is a compact space in this general sense. This theorem is not easy to prove as an immediate consequence of the agreed definitions. In this sense it cannot be regarded as an "exercise".

One possible proof can be sketched as follows: Let $\bigl(U(x)\bigr)_{x\in [0,1]}$ be a neighborhood field on the interval $[0,1]$, and assume that there is no finite set of watch towers $x_k$ $\ (1\leq k\leq N)$ such that $[0,1]=\bigcup_{k=1}^N U(x_k)$. Then there is no finite set of watch towers for at least one of $[0,{1\over2}]$ and $[{1\over2},1]$, and there is no finite set of watch towers for, say, at least one of $[{1\over2},{3\over4}]$, $[{3\over4},1]$, and so on. Using the completeness properties of ${\mathbb R}$ we arrive at a point $\xi\in{\mathbb R}$, such that in the immediate neighborhood of $\xi$ there are arbitrary small intervals admitting no finite set of watch towers. But the given neighborhood $U(\xi)$ would alone cover all these intervals $-$ a contradiction.

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You can see the proof you sketch fully in my answer. –  Pedro Tamaroff Jan 9 '13 at 21:05
    
@Peter Tamaroff: The task was to reach the OP where he stands. A formal proof can be found anyplace. –  Christian Blatter Jan 9 '13 at 21:13
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