# the meaning of “finite” in the finite covering theorem

A textbook I am using to learn analysis states (in reference to just the real line):

Every system of open intervals covering a closed interval contains a finite subsystem that covers the closed interval.

(the textbook is "Mathematical Analysis I" by V.A. Zorich)

Let's say $S$ = {$U_n$} is the system of open intervals $U_n$ in question, which covers the closed interval $I$.

Now, if we take all $U_n$ to be infinitely small in length, but take $S$ to be infinite in cardinality such that all $U_n$ still cover $I$ (does this even make sense?), wouldn't it be impossible to select a finite subsystem of $S$ that covers $I$?

I feel that maybe I am missing an elementary but important distinction in my idea of "infinite"

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It is impossible for an open interval to be "infinitely small in length". The measure of $(a,b)$ is $b-a>0$; an "infinitely small in length" interval would be the singleton set $\{a\}$, which is closed. (Also the closed interval in the quoted statement should be bounded.) –  Tom Cooney Jun 25 '12 at 11:53
@TomCooney: Just a minor quibble: The important point isn't that singletons are closed in the real line, but rather that they are decidedly not open. –  Arthur Fischer Jun 25 '12 at 11:58
@TomCooney: a closed interval in the real line must be bounded, surely? it can't have $\infty$ as an endpoint because it would then contain it, but it's not a real number. –  Ben Millwood Jun 25 '12 at 12:49
@benmachine: You are mistaken, for instance all of $\mathbb R$ is a closed interval. Closed implies that every point of the space that is a limit of points of the subset is in the subset itself, but $\infty$ is not a point of the space. The property that unbounded intervals lack is "compact" (whose definition seems to be considered here), but not closed. –  Marc van Leeuwen Jun 25 '12 at 13:15
@MarcvanLeeuwen: ah, I was interpreting "closed interval" to mean "a thing of the form $[a,b]$" not "a thing that is both closed and an interval". I'd forgotten that the condition of having square brackets at either end was sufficient but not necessary :) –  Ben Millwood Jun 25 '12 at 13:27

Your mistake lies in the assumption that you can choose "$U_n$ infinitely small in length". For each $n$ seperately the set $U_n$ is an open set. An open set contains an open interval and an open interval has a positive measure (the notion of "length" doesn't really make sense for an arbitrary open set, but that is not the problem here). Obviously the open sets may become "smaller" (again, you have to be careful how you measure the size of an open set) if $n$ varies, but that's a different story.

Btw: The property you quote is the definition of compactness. For subsets of the real numbers we have: bounded and closed if and only if compact. Note that the textbook presumably assumes that the interval is bounded.

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First let me tell you that proving compactness of a closed and bounded interval $[a,b]$ is trivial once you know the following two facts:

1. In a metric space sequential compactness and open cover compactness are equivalent

2. Bolzano - Weierstrass is available to us.

By (1) it suffices to prove that for any sequence $a_n \subset [a,b]$ we have that there is a convergent subsequence. But this is trivial because Bolzano - Weierstrass guarantees that we have a convergent subsequence, and the interval being closed guarantees that the subsequence converges to something in $[a,b]$ so that $[a,b]$ is compact.

Now back to your problem. Since the open cover of $[a,b]$ that you are considering consists of just open intervals and not arbitrary open sets then each element $U_n \in S$ will contain a closed and bounded interval $[c_n,d_n]$ of length $d_n - c_n > 0$. I don't understand what you mean by "infinitely small length"; the "length" of each $U_n$ will be at least this $c_n - d_n$ and is a constant. What do you mean by "choose each $U_n$ to be of infinitely small length"?

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I'd think that bringing sequential compactness and Bolzano-Weierstrass into this is possibly above the level of the question. Also, he didn't say "arbitrarily small length", he said something even more impossible... –  Ben Millwood Jun 25 '12 at 12:50
@benmachine I thought the BW property and sequential compactness was standard stuff..... –  user38268 Jun 25 '12 at 13:05
Yeah, but the proof that they're equivalent in a metric space sounds like something you'd learn at the same time as learning what compactness was all about. I'm not sure it's easier than just proving the result directly. But I guess it's a matter of taste which results ought to depend on which. –  Ben Millwood Jun 25 '12 at 13:14
@BenjaLim It is standard, but I'm not there yet. Interestingly, the text I mentioned presents the BW theorem immediately following the finite covering theorem and uses it in its proof, both of them under a subsection "Basic Lemmas Connected with the Completeness of the Real Numbers", alluding to what benmachine mentions. This indirection is why I did not accept your answer over Simon's, but I'm sure it will prove helpful soon, so thank you. –  afsmi Jun 25 '12 at 13:29

Try to write down an open interval that is infinitely small in length. You'll find that you can't (unless you count the empty set, and you can't cover a closed interval in empty sets!).

Note that if you have an interval which is not closed, like $[0,1)$, then you can use a cover like $U_n = [0,\frac{n-1}{n})$, each set covering closer and closer to the right endpoint and hence eventually covering any point to the left of it. This cover has no finite subcover, since any finite collection of such sets includes a biggest one, and that biggest one misses a tiny sliver of the original interval.

What happens when you try to do the same thing with $[0,1]$? Well, you can get closer and closer and closer to $1$, but at some point you've got to hit it, so you've got to have an open interval like $(1-\epsilon,1+\delta)$ in your cover, but then you've just made infinitely many (indeed, all but finitely many) sets of your cover redundant, since you now only need to cover up to $1-\epsilon$, and regardless of how small $\epsilon$ is, there's some $1/n$ less than it, so there's an interval $[1,\frac{n-1}{n})$ that does the job.

Actually, this realisation that if you have infinitely many open sets creeping up on something, the open set actually containing that something is going to let you throw most of them away, is surprisingly close to the formal proof of your result, known as (more-or-less) the Heine-Borel theorem: just for fun, here's a proof in iambic pentameter.

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