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From Planetmath:

Lebesgue number lemma: For every open cover $\mathcal{U}$ of a compact metric space $X$, there exists a real number $\delta > 0$ such that every open ball in $X$ of radius $\delta$ is contained in some element of $\mathcal{U}$.

Any number $\delta$ satisfying the property above is called a Lebesgue number for the covering $\mathcal{U}$ in $X$.

I feel hard to picture and understand the significance of this result. I was wondering if there are some explanation for this lemma? Intuitively,

  1. a number bigger or smaller than a Lebesgue number may not be a Lebesgue number. So is a Lebesgue number simultaneously measuring how separated open subsets in an open cover are between each other, and how big each of them is?
  2. how is a metric space being compact make the existence of a Lebesgue number possible?
  3. Added: Is the lemma equivalent to say that for any open cover, there exist a positive number $\delta$, s.t. any open cover consisting of open balls with radius $\delta$ is always a refinement of the original open cover?

Thanks and regards!

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    $\begingroup$ planetmath.org/encyclopedia/ProofOfLebesgueNumberLemma.html $\endgroup$
    – yoyo
    Feb 3, 2012 at 15:41
  • $\begingroup$ +1 because I also found it hard to picture and understand the significance. I don't remember much about it now, except that it was used in the proof of some other result not long after. $\endgroup$
    – GeoffDS
    Feb 3, 2012 at 15:54
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    $\begingroup$ Any number smaller than the Lebesgue number for a particular cover is also a Lebesgue number for that cover. $\endgroup$ Feb 3, 2012 at 16:46
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    $\begingroup$ As a sidenote, it is used in one of the proofs of Seifert-van Kampen Theorem. For example, the one in Lee's Topological Manifolds. $\endgroup$ Feb 3, 2012 at 20:34
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    $\begingroup$ And it is equivalent to the fact that the cover of all balls of radius $\delta$ is a refinement fo the original open cover. Lemma 2.27 here: arxiv.org/abs/1511.02057v2 $\endgroup$ Jun 28, 2017 at 15:29

6 Answers 6

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The intuition is the following : for each $x$ of $X$ is included in at least one $U$ of $\mathcal U$. Since $U$ is open, it contains a ball centered at $x$ and with positive radius, right ? Let $r(x, U)$ denote the supremum of all such radius. And let $r(x)$ the supremum of all the $r(x,U)$, with $U\in\mathcal U$.

So $r(x)$ is a continuous positive function on $X$. But if $X$ is not compact, you may find a sequence $(x_n)$ such that $r(x_n)$ tends to zero.

However, if your metric space $X$ is compact, then $r$ shall have a minimum. This minimum is of course positive since $r$ is. This minimum is a Lebesgue number, the greater one.

And in fact, the intuition is a proof !

A covering with positive Lebesgue number is called a uniform covering.

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    $\begingroup$ +1. Thanks! I was wondering what you mean by uniform covering? The definition I learned from Wiki is that a family of uniform covering is a filter on the collection of covers of $X$ wrt the star refinement order. It is not unique. $\endgroup$
    – Tim
    Feb 3, 2012 at 20:00
  • $\begingroup$ Is the lemma equivalent to say that for any open cover, there exist a positive number $\delta$, s.t. any open cover consisting of open balls with radius $\delta$ is always a refinement of the original open cover? $\endgroup$
    – Tim
    Feb 3, 2012 at 20:23
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    $\begingroup$ Great explanation $\endgroup$ Oct 5, 2015 at 4:47
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    $\begingroup$ This is excellent. $\endgroup$ Nov 8, 2016 at 1:43
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    $\begingroup$ Possible nitpick: I think $r$ is only lower semi-continuous. $\endgroup$ Oct 3, 2021 at 17:39
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The characteristic property of a compact space is that every open cover has a finite subcover, which is used in proving the Lebesgue number lemma. I'm sure you've read a proof of the Lebesgue number lemma somewhere; otherwise I would recommend reading this one, which I found to be quite intuitive. Quickly, the idea of the proof is as follows: Start off with a given cover $\mathcal{U}$. For every point in $X$, there is an $\epsilon$ neighborhood contained in an element of $\mathcal{U}$. These neighborhoods form an open cover, which must have a finite subcover since $X$ is compact. Now if we pick $\delta$ to be $1/2$ times the least $\epsilon$, we can show that it satisfies the requirements of being a Lebesgue number.

I also rather like fer the proof by contradiction: suppose there was no such $\delta$. In that case, the negation of the Lebesbue number lemma says: there exists a cover $\mathcal{U}$ such that for every $\delta>0$, there exists an open ball of radius $\delta$ not contained in any element of $\mathcal{U}$. So, for each $n$ , we can choose an $x\in X$ such that no $U$ contains $B_{1/n}(x_n)$ . Now, $X$ is compact so there exists a subsequence $(x_{n_k})$ of the sequence of points $(x_n)$ that converges to some $y\in X$. But since $\mathcal{U}$ is an open cover, there is some $\gamma$ such that $B_\gamma(y) \subseteq U$ for some $U\in \mathcal{U}$. Pick $k$ large enough so that $|x_{n_k}-y|\leq \gamma/2$ and $1/n_k < \gamma/2$. Use the triangle inequality to show that $B_{1/n_k}\subseteq U$.

In general, my (not entirely accurate) intuition for this is that for compact spaces, finite covers are sufficient. If a Lebesgue number did not exist, we would get a sequence of sets which are not included in the elements of the cover, which would contradict the finiteness of the cover.

As for your followup questions:

  1. Any number smaller than a Lebesgue number is also a Lebesgue number.
  2. As you saw above, compactness shows up prominently in the proof for the lemma.
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    $\begingroup$ For the first proof using the half radius trick, it is crucial that we set $\delta$ to be half the minimum of $\epsilon_i$, $i = 1, \cdots, N$ where the half radius balls $B(x_i, \epsilon_i / 2)$ form a finite cover of $X$ such that each of the whole radius balls $B(x_i, \epsilon_i)$ is contained in an element of $\mathcal U$. This is different from setting $\delta$ to be half the minimum of $\epsilon_i$, $i = 1, \cdots, N$, where the total radius balls $B(x_i, \epsilon_i)$ are constructed only to form a finite cover of $X$ such that each of them is contained in an element of $\mathcal U$. $\endgroup$
    – Jisang Yoo
    Dec 30, 2013 at 16:02
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    $\begingroup$ Difference is that only the former choice of $\delta$ can complete the proof. $\endgroup$
    – Jisang Yoo
    Dec 30, 2013 at 16:03
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    $\begingroup$ Link is currently broken. $\endgroup$ Apr 23, 2014 at 8:53
  • $\begingroup$ The first argument is wrong / makes no sense as given. (Also noted above.) $\endgroup$
    – Louis
    Mar 13, 2018 at 15:21
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A non-compact counterexample: consider the open cover of $\mathbb{R}$ $$ \{(n,n+1) : n\in\mathbb{Z}\}\cup\{(n-\epsilon_n,n+\epsilon_n) : n\in\mathbb{Z}\} $$ where $\epsilon_n\to0$ as $|n|\to\infty$. Balls centered at integer points need to get smaller and smaller the further you are from the origin, so a Lebesgue number doesn't exist.

Also note that a number smaller than a Lebesgue number is a Lebesgue number.

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For 2.:

Let $\cal U$ be an open cover of the compact space $X$. Suppose that for each positive integer $n$, the open ball $B_{1/n}(x_n)$ is not contained in any element of $\cal U$. Since $X$ is compact, it is sequentially compact, and hence the sequence $\{x_n\}$ has a limit point $x$. Let $U$ be an element of $\cal U$ containing $x$. Since $U$ is an open set, there is an open ball $O\subset U$ with $x\in O$. For sufficiently large $n$, we have $$ B_{1/n}(x_n)\subseteq O\subseteq U, $$ which contradicts the selection of the $x_n$.

I'm not sure how to address 1.); but this may help to understand what motivates the Lebesgue number:

One way to prove that a sequentially compact space is compact (note in the above, one need only assumed $X$ is sequentially compact) is to use the concept of Lebesgue number and total boundedness of $X$. Both of these properties are enjoyed by sequentially compact metric spaces.

Given an open cover of the sequentially compact space $X$, one can cover it with a finite number of balls of radius less than half of the Lebesgue number. These balls are contained in elements of the open cover; so the open cover has a finite subcover.


One way to think about the Lebesgue number is that if $\cal U$ has Lebesgue number $\delta$, then there are elements of $\cal U$ that cover "big chunks" of $X$ (namely open balls with radius less than $\delta$). Contrast this with the open cover of $(0,1)$ $$ {\cal U} =\{ (\textstyle{1\over n+1},{1\over n-1} ) :n=2,3,\ldots \} $$ This open cover is "small" in the sense that the open ball $(0,\epsilon)$ is contained in no element of it.

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  • $\begingroup$ +1. Thanks! Is the lemma equivalent to say that for any open cover, there exist a positive number $\delta$, s.t. any open cover consisting of open balls with radius $\delta$ is always a refinement of the original open cover? $\endgroup$
    – Tim
    Feb 3, 2012 at 20:26
  • $\begingroup$ @Tim Yes the lemma is equivalent to your statement. In fact, any open cover of open balls of radius $\leq \delta$ is also always a refinement of the original cover. $\endgroup$
    – Aru Ray
    Feb 3, 2012 at 20:53
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[This is from Dieudonne's Analysis book]

Here is an intuitive proof : For any point $ x $ , there is an $ r_x (> 0) $ such that $ B(x, r_x) $ is contained in some element of the cover. Note $ \{ B(x, r_x / 2 ) : x \in X \} $ is an open cover, so by compactness it has a finite subcover $ \{ B(x_1, r_{x_1} / 2) , …. , B(x_n, r_{x_n} / 2 ) \} $. Now setting $ \delta := min\{ r_{x_1} / 2 , …. , r_{x_n} / 2 \} > 0 $ will do [Because let $ x $ be an arbitrary point. There is a $ B(x_i, r_{x_i} / 2 ) $ containing it. Now note $ B(x, \delta) $ is contained in $ B(x_i, r_{x_i}), $ which in turn is contained in some element of the cover]

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  1. Here is yet another explanation of the Lebesgue Number Lemma: It says that for compact metric spaces we have a very nice extension of the notion of an open set to an open cover. Indeed, Lebesgue Number Lemma guarantees that for a compact metric space $X$ we have that $\exists \delta>0, \forall x\in X, \exists U_x\in \mathcal{U}: B_\delta(x)\subseteq U_x$. This means that the openness condition of open sets holds for the cover (this is trivial), and it holds "set-by-set". So it does not happen that we can find an open ball separated by the sets in the cover (and there is no other set in the cover that contains it).

As a side note, we can also use this lemma to define an equivalence relation on the open balls of $X$, once we make the cover disjoint. Though I am guessing this might turn out to be a chore to pursue.

  1. As pointed in the above answers, compactness prevents taking limits so that $\delta$ turns out to be $0$, in which case the balls are only points, and the result is trivial.

  2. Yes.

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