Given a cover of a compact metric space, all small subsets are subset of some cover set
In topology, the Lebesgue covering lemma is a useful tool in the study of compact metric spaces.
Given an open cover of a compact metric space, a Lebesgue's number of the cover is a number
such that every subset of
having diameter less than
is contained in some member of the cover.
The existence of Lebesgue's numbers for compact metric spaces is given by the Lebesgue's covering lemma:
- If the metric space
is compact and an open cover of
is given, then the cover admits some Lebesgue's number
.
The notion of Lebesgue's numbers itself is useful in other applications as well.
Let
be an open cover of
. Since
is compact we can extract a finite subcover
.
If any one of the
's equals
then any
will serve as a Lebesgue's number.
Otherwise for each
, let
, note that
is not empty, and define a function
by
![{\displaystyle f(x):={\frac {1}{n}}\sum _{i=1}^{n}d(x,C_{i}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/650de14149f820ac39df9cf89d8c194a54ac5b3d)
Since
is continuous on a compact set, it attains a minimum
.
The key observation is that, since every
is contained in some
, the extreme value theorem shows
. Now we can verify that this
is the desired Lebesgue's number.
If
is a subset of
of diameter less than
, choose
as any point in
, then by definition of diameter,
, where
denotes the ball of radius
centered at
. Since
there must exist at least one
such that
. But this means that
and so, in particular,
.
Proof by Contradiction
[edit]
Suppose for contradiction that that
is sequentially compact,
is an open cover of
, and the Lebesgue number
does not exist. That is: for all
, there exists
with
such that there does not exist
with
.
This enables us to perform the following construction:
![{\displaystyle \delta _{1}=1,\quad \exists A_{1}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{1})<\delta _{1}\quad {\text{and}}\quad \neg \exists \beta (A_{1}\subset U_{\beta })}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48d995aa8597848e403bd9f50ef6a23791b7048b)
![{\displaystyle \delta _{2}={\frac {1}{2}},\quad \exists A_{2}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{2})<\delta _{2}\quad {\text{and}}\quad \neg \exists \beta (A_{2}\subset U_{\beta })}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fdd8c643dc3b79ea765ba547efde0790a7a2b56)
![{\displaystyle \vdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8039d9feb6596ae092e5305108722975060c083)
![{\displaystyle \delta _{k}={\frac {1}{k}},\quad \exists A_{k}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{k})<\delta _{k}\quad {\text{and}}\quad \neg \exists \beta (A_{k}\subset U_{\beta })}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a435d8e3fafc0c89b98b9af77b460c937e705239)
![{\displaystyle \vdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8039d9feb6596ae092e5305108722975060c083)
Note that
for all
, since
. It is therefore possible by the axiom of choice to construct a sequence
in which
for each
. Since
is sequentially compact, there exists a subsequence
(with
) that converges to
.
Because
is an open cover, there exists some
such that
. As
is open, there exists
with
. Now we invoke the convergence of the subsequence
: there exists
such that
implies
.
Furthermore, there exists
such that
. Hence for all
, we have
implies
.
Finally, define
such that
and
. For all
, notice that:
, because
.
, because
entails
.
Hence
by the triangle inequality, which implies that
. This yields the desired contradiction.