# a question about lebesque numbers

• May 5th 2008, 08:24 PM
squarerootof2
so we have that {U_alpha} where alpha is in A (indexing set) is a finite open cover of a compact metric space X. i did the first part of problem which was to show that there exists ε>0 such that for each x in X, the open ball B(x;ε) is contained in one of the U_alpha's.

so here such an ε is called the lebesque number of the cover (definition)

now i need to show that if at least one of the U_alpha's is a proper subset of X, then there is a largest lebesque number for the cover.

the hint given is that if one of the open sets in the cover is proper, the lebesque numbers are bounded. and since cover is finite, the least upper bound of lebesque number is again a lebesque number.

can someone explain the problem? thanks.
• May 6th 2008, 06:52 AM
Opalg
Quote:

Originally Posted by squarerootof2
so we have that {U_alpha} where alpha is in A (indexing set) is a finite open cover of a compact metric space X. i did the first part of problem which was to show that there exists ε>0 such that for each x in X, the open ball B(x;ε) is contained in one of the U_alpha's.

so here such an ε is called the lebesque number of the cover (definition)

now i need to show that if at least one of the U_alpha's is a proper subset of X, then there is a largest lebesque number for the cover.

the hint given is that if one of the open sets in the cover is proper, the lebesque numbers are bounded. and since cover is finite, the least upper bound of lebesque number is again a lebesque number.

can someone explain the problem? thanks.

There's something wrong with the statement of this problem. For the result to be true, it is necessary that every $U_\alpha$ (not just one of them) should be a proper subset of X. Otherwise, if $U_{\alpha} = X$ for some α, then clearly every ball, whatever its centre and radius, will lie in that $U_{\alpha}$.

So suppose that all the sets $U_{\alpha}$ are proper subsets of X. Choose x_0 in X. Since a compact metric space is bounded, some ball centred at x_0, with sufficiently large radius R, will be the whole of X (and thus will not be a subset of any of the $U_{\alpha}$s). Therefore a Lebesgue number of the cover must be less that R. Hence the set of all Lebesgue numbers of the cover is bounded above. You have already proved that it is nonempty. Therefore it has a supremum, call it δ. We want to show that δ is the largest Lebesgue number of the cover.

The problem is that the supremum of a set need not always belong to the set. So we need to prove that δ is a Lebesgue number of the cover. To do this, we have to show that every ball of radius less than δ is contained in some member of the cover. So let B(x,r) be a ball centred at some point x, with radius r<δ. By the definition of supremum, there exists a number ε with r<ε<δ, such that ε is a Lebesgue number of the cover. But since r<ε it follows from the definition of Lebesgue number that any ball of radius r (in particular B(x,r)) must lie in some $U_{\alpha}$, which is what we wanted to prove.

[I'm sure that if Henri Lebesgue were still alive, he would want to point out that his name is spelt with a g, not a q.]
• May 6th 2008, 07:07 AM
ThePerfectHacker
Quote:

Originally Posted by Opalg
[I'm sure that if Henri Lebesgue were still alive, he would want to point out that his name is spelt with a g, not a q.]

Is this a common mistake? I been spelling it with a q for a long time. Never realized it.
• May 6th 2008, 07:09 PM
squarerootof2
i checked the book and it turns out the book actually spelled his name with a g, somehow i saw it as a q... sorry about that mistake.