23. The Lebesgue Number of a Covering

(1875-1941)

Review:

23.1 Definition:

Given a Metric Space $\QTR{Large}{(M,d)}$ , and a subset we say $\QTR{Large}{x}$ is a limit point of $\QTR{Large}{A\ }$ if

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That is $\QTR{Large}{x\ }$ is in the closure of

Notes:

The definition does not claim that
It is not necessarily the case that the set of limit points of $\QTR{Large}{A\ }$ is the closure of $\QTR{Large}{A}$ . For example, a singleton set $\QTR{Large}{\{x\}}$ has no limit points but is its own closure.
Remember, $\QTR{Large}{x\ }$ is in the closure of means that for any 0

23.2 Definitions:

A sequence in $\QTR{Large}{M}$ is just a map Notationally
Given a one to one order preserving map , the map

is called a subsequence. We will usually just write " is a subsequence."
A covering of a Set M is a collection of subsets such that .

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23.3 Definition:

A Metric Space, $\QTR{Large}{(M,d)}$ is called

limit compact if every infinite subset has a limit point.
sequentially compact if Every sequence in $\QTR{Large}{M}$ has a convergent subsequence. That is given we can find a subsequence
countably compact if every covering by a countable number of open sets , , contains a finite subcover. That is there is some finite subset such that
"totally"(my term) compact if every covering by a open sets contains a finite subcover.

23.4 Theorem:

For Metric Spaces these four forms of compactness are equivalent.

In General Topology these four forms of compactness are not equivalent.

Proof:

1. 2. This is 21.4.

We will not present counter-examples to the general theorem.

The remainder of this Page is devoted to completing the proofs of the various equivalences.

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Review (Continued)

21.4.1 Corollary:

Every sequentially/limit compact Metric Space is complete:

21.5 Theorem:

Given a sequentially/limit compact Metric Spaces $\QTR{Large}{(M,d)}$ :

For any 0 and some n there exists points

such that .
In particular, M is bounded in the sense that there exists some b0 such that b for any .

21.6 Theorem(Restated):

Every sequentially/limit compact metric space has a countable dense subset.

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23.5 Theorem: (Lebesgue Number)

Given a sequentially/limit compact Metric Space $\QTR{Large}{(M,d)}$ and a covering by open sets

( ), there exists a real number such that every open ball of radius is contained in some element of .

The number is called a Lebesgue number for the covering.

Proof:

Suppose that no Lebesgue number existed. Then there exists an open cover such that for all there exists an such that no contains . In particular for each n we can choose a sequence such that

for any .

The proof continues in the same way as several of the proofs on Page 21:

Since $\QTR{Large}{M}$ is compact choose a convergent subsequence with for some . Since is an open cover, we know there is some and some with . Again, as before, choose n such that and for i $\QTR{Large}{>\ }$ n. Check that

23.6 Corollary:

Every open covering, of a compact Metric Space $\QTR{Large}{(M,d)}$ has a finite subcovering. In particular 1. 2.4.3.

Proof:

Let be a Lebesgue number for . Applying 21.5, we know there is some

n and points such that MATH . But since is a Lebesgue number for each i we can find such that and hence

23.4 Theorem:

Let $\QTR{Large}{(M,d)}$ be countabily compact, then every infinite set contains a limit point. In particular, 3.1. Hence, 1. 2. 3. 4.

Proof:

Let $\QTR{Large}{A}$ be an infinite subset of $\QTR{Large}{M}$ without limit points, then . Without loss of generality we can assume that $\QTR{Large}{A}$ is countable. Writing , and $\QTR{Large}{M-A}$ is open since $\QTR{Large}{A}$ has no limit points. More generally is open. One notes that,

and
There is no finite subset that covers $\QTR{Large}{M}$ . In particular,