Given a cover of a compact metric space, all small subsets are subset of some cover set
In topology, the Delta number, is a useful tool in the study of compact metric spaces. It states:
- If the metric space
is compact and an open cover of
is given, then there exists a number
such that every subset of
having diameter less than
is contained in some member of the cover.
Such a number
is called a Delta number of this cover. The notion of a Delta number itself is useful in other applications as well.
Direct Proof[edit]
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 Delta 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 Delta 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.
References[edit]