Talk:Heine–Borel theorem

Proof
I removed the following proof from the main page:

Here, we consider the version for the real numbers, as stated above.

->
If a subset of the real numbers is compact, then it is bounded.

Let X be a compact subset of the real numbers.

Define a collection of open sets O_n to be the open interval (-n, +n). These sets cover X. Some finite sub-collection must also cover X.  This finite sub-collection is bounded, and so X is bounded.

If a subset of the real numbers is compact, then it is closed.

Suppose (for a contradiction) that it were not closed. Then, there would be a sequence of number x_n in X such that x_n converges to a point x not in the set X.

Define the sequence of open sets O_n where O_n = R - [x - (1/n), x+ (1/n)], i.e. the whole line minus a small closed interval around x. O_n is open and the collection covers X.  Since X is compact, a finite subcover exists. So, every point in X is at least 1/n away from x (n fixed). So, a sequence in X cannot converge to a point outside of X. Thus, X is closed.

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We are given a closed and bounded subset X of R. We are given an open cover {O} of X.  We need to find a finite subcover.

Step 1: Reduce the open cover to a countable subcover. For each rational number r, pick one open set in O_r that contains r. We use the axiom of choice here. Since X is closed, this subcollection also covers X.

Step 2: We now have a countable subcover {O_n}. Suppose that there is no finite subcover. Consider P_n = the union of the first n members of O_n. Each P_n is open, and does not cover X. So, there is some point x_n in X - P_n. Since X is closed and bounded, every sequence in X has some convergent subsequence (A proof of this should also be given; in fact, some books list this as the definition of compactness). This convergent subsequence, call it y_n, converges to a point y. However, since X is closed, y is also in X.  So, y is covered by some O_n, and y is in some P_n. This is a contradiction, so there must be some finite subcover.

We still need to prove that, any sequence in a closed and bounded set has a convergent subsequence. Since the set X is bounded, it is bounded by a closed interval [-n, +n]. Now, one of the two sub-intervales [-n, 0] and [0, +n] must contain an infinite number of points in the sequence. Repeat this process, dividing the interval in half, finding a subsequence contained in that half. Eventually, this yields a convergent subsequence. The axiom of choice is again required.

Important point: the Axiom of Choice was used.

Since this theorem is important in analysis, the Axiom of Choice is also important in analysis.

The first half of the proof is correct, but the second half is not. Step 1, the reduction to a countable subcover, is not correct: the constructed countable subcover need not cover all of X. The Axiom of Choice is in fact not needed for the proof. AxelBoldt 18:52 Oct 4, 2002 (UTC)

What's a correct proof? Overall, I was expecting to see more proofs and examples on the Mathematics pages.

Page Title
Why is this page under a title with an en dash, with the hyphenated page redirecting to here? En dashes are for separating members of ranges (like "3–10 years") whereas hypens are for separting words (as in the title of this article), and having the wrong one both looks irksome and makes a google for the theorem show up incorrectly. I would put in one of those move article vote thingys here but I'm not sure how to initiate that. (By the way I also fiddled with the formatting of the below discussion to separate it from this one.) Quietbritishjim (talk) 09:46, 11 March 2008 (UTC)

Well, looks like I was wrong, the modern interpretation of an en-dash allows this. Still looks wrong to me, but not really important enough to worry about! Quietbritishjim (talk) 18:00, 31 March 2008 (UTC)

Wrong proof?
The current (April 21, 2009) proof of the part under If a set is compact, then it must be closed. is wrong.

Let a set S be compact. A finite collection C of open sets which do not intersect at least one neighborhood of an accumulation point \vec a cannot be an open cover of S, because the intersection of those neighborhoods not met forms an open set, which contains a point in S not covered by C.

1) Every set imaginable will intersect at least one neighborhood of any point; just increase the size of the neighborhood until it intersects your set. So that sentence makes no sense. It is possible that the original author meant to say every neighborhood.

2) The last part is hard to figure out, but I think the author intended to say the union of neighborhoods that are not intersected by C contains a point in S not covered by C. The problem is, if S is a closed set, then it contains every accumulation point.

3) I'm not really sure about this, but do we need to assume that we're working in $$\mathbb R^n$$ in order to prove that compactness => closure? According to the statement of the theorem in the introduction, in the general case we need to prove completeness, not closure. —Preceding unsigned comment added by 75.18.231.173 (talk) 05:30, 22 April 2009 (UTC)


 * The part of the proof that you mentioned is not wrong, it is just hard to follow. I'll try to make it clearer if I can. --Bdmy (talk) 09:28, 22 April 2009 (UTC)


 * I think unsigned is misinterpreting the text of the proof, because it is harder to be specific in English (and other natural languages) than in symbolic logic. The quote is not saying that the open sets in C don't intersect any neighborhood of a at all. It is saying that, for each set in C, there is some neighborhood of a that it does not intersect. I think that perhaps the current wording ("such that each open set U is disjoint from some neighborhood ...") makes this clearer. --Hccrle (talk) 22:56, 19 March 2012 (UTC)

Provability of uniform space (or metric space) version from ultrafilter principle
I don't think this can actually be done. The proofs I have seen, including the one given in the citation, use a filter- or net-based definition of completeness rather than the sequential one given here. The following is of course not conclusive, but on p253 of Mícheál Ó Searcóid's Elements of Abstract Analysis, the author states that "it cannot be shown in ZF that a sequentially closed subset of an arbitrary metric space is necessarily closed, simply because it cannot be proven that an arbitrary infinite set has a countable subset."

I'm talking about the 2002 edition of Searcóid's book and the 1996 edition of Schechter's.

At the risk of being an idiot, just think about it: How am I supposed to use the completeness of a space if I can't even construct a sequence to my specifications? —Preceding unsigned comment added by 67.175.99.81 (talk) 04:12, 2 September 2010 (UTC)

ZF is set theory without the Axiom of Choice (AC). They're just saying that without AC, you can't prove that an infinite set has a countable subset, or that the countable union of countable sets is countable, and all kinds of other weirdness. Without AC, all kinds of things can happen, like every set of reals can be Lebesgue measurable (see Solovay model). That's why AC is used in mathematics like real analysis. --Hccrle (talk) 23:38, 19 March 2012 (UTC)

Removed link
I removed the following link from the references section First of all because it wasn't a reference for any material on the page, secondly because it isn't notable enough to warrent inclusion as an external link, and finally because at the critical part of the proof (where the completeness of the reals is actually used) they, in their own words, "omit crucial but obvious details".
 * A simple and humorous proof of the Heine-Borel theorem using model theory, by M. Macauley et al

To expand on the third point: I haven't thoroughly read the paper, but on skimming it it doesn't make sufficiently clear where the completeness of the reals is used. Suspiciously, near the beginning the authors say that when proving Heine-Borel one must "summon ... the completeness of the reals, one of those blatantly obvious analysis facts that you had no idea how to prove". Well, normally you don't prove it, you assume it as an axiom. You can prove it if you construct the reals, that's really just to check that a set satisfying the axioms exists, not to prove things about the reals; in any case, if that's the view they take, they should state what construction of the reals that they're using, and how their proof uses that. Quietbritishjim (talk) 16:40, 20 October 2010 (UTC)
 * The paper has not been published yet apparently, so removing the link is not inappropriate. On the other hand, the idea is to use the compactness of the space of binary trees.   You show that the interval [0,1] is compact by showing that it is the image of a compact set.  Where do they use the completeness of the reals?  If they did, their argument would be circular.  Having said that, it is possible that that's the reason it did not get published :)  Tkuvho (talk) 14:07, 21 October 2010 (UTC)

Circularity?
The proof of the Heine–Borel theorem uses Cantor's intersection theorem, and the proof of Cantor's intersection theorem uses the Heine–Borel theorem. The two theorems each use the other to prove themselves! Nick Levine (talk) 08:13, 6 August 2013 (UTC)

You're absolutely right, and I've just fixed this. I based my changes on the proof given in Rudin's Functional Analysis, in the appendix on compactness. Knightofairplanes (talk) 08:02, 25 October 2014 (UTC)

This is not a sentence.
"Indeed, the intersection of the finite family of sets V_U is a neighborhood W of a in R^n." 212.149.241.8 (talk) 14:51, 7 February 2014 (UTC)

It is. Perhaps you're confused by the phrase "is a neighbourhood W of a", which is being used as shorthand for "is a neighbourhood (which we'll call W) of a". This is very common in mathematical writing, but perhaps it's not very good English. It could be expanded out as:


 * Let W be the intersection of the finite family of sets VU. Then W is a neighborhood of a in Rn.

Quietbritishjim (talk) 15:02, 7 February 2014 (UTC)

Two versions of the same paper
The article cites two versions of A Pedagogical History of Compactness by Raman–Sundström, one an arXiv preprint and one the final published version from American Mathematical Monthly. I did not want to update the arXiv citation to point to the published version, because I don't know what revisions the arXiv version might have undergone before publication. But it looks strange to cite two versions of the same paper, and this ought to be cleaned up at some point. —Mark Dominus (talk) 18:27, 7 December 2015 (UTC)

Error/ambiguity in proof?
In the proof, when showing that a closed subset of a compact space is compact, there's an error...sort of. The article never clarifies what it means by an open cover; since we're only working in $$\mathbb{R}^n$$, an open cover of a set $$X$$ could mean a collection of open subsets of $$\mathbb{R}^n$$ whose union includes $$X$$, in which case the given proof is fine. If, on the other hand, an open cover of $$X$$ refers to a collection of open subsets of $$X$$ whose union is $$X$$, then the proof given is false. In particular, it claims that, given an open cover $$C_K$$ of $$K$$, then $$C_K\cup \{U\}$$ is an open cover of $$T$$ (where $$U$$ is the complement of $$K$$). But an open subset of $$K$$ is only relatively open in $$T$$, it need not actually be open. The correct proof then would be to union $$U$$ onto every set in $$C_K$$.

Given that $$U$$ is defined as $$\mathbb{R}^n\backslash K$$, the author presumably intended an open cover to mean open subsets of $$R^n$$ - but either this should be made explicit, or the proof should be changed. I'm not sure which is the better option, so I've just put this comment, rather than making any actual changes. — Preceding unsigned comment added by 130.88.123.219 (talk) 13:24, 11 January 2017 (UTC)

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Property ?
This page is in the category «Properties of topological spaces». Can anyone explain why it is the case ? It seems not to be a property, but a theorem with conditions and conclusion. Should we remove this category ? — Preceding unsigned comment added by Arthur MILCHIOR (talk • contribs) 04:21, 4 March 2018 (UTC)

About "generalizations"
Hello! The definition of the Heine–Borel property that you give in the section "Generalizations" "A metric space (or, for instance, a topological vector space) is said to have the Heine–Borel property if each of its open covers has a finite subcover" -- is in fact the definition of a compact space in topology. This is a misunderstanding. No topological vector spaces satisfy this condition. The correct definition is the following: "A topological vector space $X$ is said to have the Heine–Borel property if each bounded set in $X$ is relatively compact (people say also totally bounded)." This is from the book:

(Theorem 28.) I think, we should change the text. (If you allow me, I'll do this.) Eozhik (talk) 11:27, 31 March 2018 (UTC)


 * Dear friends, another comment. This is not called a generalization of the Heine-Borel property:
 * A subset of a metric space is compact if and only if it is complete and totally bounded.
 * In the books on topology, for instance in R.Engelking's "General topology" this is just a characterization of compact uniform spaces (Theorem 8.3.16). I think we should delete the subsection which is now called "In the theory of uniform spaces". Eozhik (talk) 14:23, 14 April 2018 (UTC)


 * Friends, I think, this is also a misunderstanding:"S is compact, that is, every open cover of S has a finite subcover. This is the defining property of compact sets, called the Heine–Borel property" I did not see this term being used in this sense. We should either find a sourse or delete the mentioning of the Heine–Borel property in this phrase. Eozhik (talk) 14:00, 15 April 2018 (UTC)


 * Gentlemen, I'll delete this section and this definition of the Heine–Borel property. This proposition is not called the Heine–Borel property, and it's proof is contained in many textbooks, for example in R. Engelking's "General topology" or in A.N.Kolmogorov, S.V.Fomin's "Elements of the Theory of Functions and Functional Analysis". Eozhik (talk) 05:10, 17 April 2018 (UTC)