Compact Sets in a Metric Space
If
is a metric space and
then a cover or covering of
is a collection of subsets
in
such that:
![{\displaystyle {\begin{aligned}\quad S\subseteq \bigcup _{A\in {\mathcal {F}}}A\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf5a6d042ccca00fc342e94617fabfd74f51dfc0)
Furthermore, we said that an open cover (or open covering) is simply a cover that contains only open sets.
We also said that a subset
is a subcover/subcovering (or open subcover/subcovering if
is an open covering) if
is also a cover of
, that is:
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \quad S \subseteq \bigcup_{A \in \mathcal{S}} A \quad \text{where} \: \mathcal S \subseteq \mathcal F \end{align}}
We can now define the concept of a compact set using the definitions above.
Definition: Let
be a metric space. The subset
is said to be Compact if every open covering
of
has a finite subcovering of
.
In general, it may be more difficult to show that a subset of a metric space is compact than to show a subset of a metric space is not compact. So, let's look at an example of a subset of a metric space that is not compact.
Consider the metric space
where
is the Euclidean metric and consider the set
. We claim that this set is not compact. To show that
is not compact, we need to find an open covering
of
that does not have a finite subcovering. Consider the following open covering:
![{\displaystyle {\begin{aligned}\quad {\mathcal {F}}=\left\{\left((0,1-{\frac {1}{n}}\right):n\in \mathbb {Z} ,n\geq 1\right\}=\left\{\left(0,{\frac {1}{2}},\right),\left(0,{\frac {2}{3}}\right),\left(0,{\frac {3}{4}}\right),...\right\}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b56118d74b6ac7a64a60b21d30ec808e19695d4f)
Clearly
is an infinite subcovering of
and furthermore:
![{\displaystyle {\begin{aligned}\quad (0,1)\subseteq \bigcup _{k=2}^{\infty }\left(0,1-{\frac {1}{k}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cdec6fb27345ed96ab58bb9571b8819ceea458f)
Let
be a finite subset of
containing
elements. Then:
![{\displaystyle {\begin{aligned}\quad {\mathcal {F}}^{*}=\left\{\left(0,1-{\frac {1}{n_{1}}}\right),\left(0,1-{\frac {1}{n_{2}}}\right),...,\left(0,1-{\frac {1}{n_{p}}}\right)\right\}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69f14039eef9e2297af01b8eba0c808dabc602eb)
Let
. Then due to the nesting of the open covering
, we see that:
![{\displaystyle {\begin{aligned}\quad \bigcup _{k=1}^{p}\left(0,1-{\frac {1}{n_{p}}}\right)=\left(0,1-{\frac {1}{n^{*}}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2edf009e4afdb0577b96cadbb420902aa14769b2)
But for
we need
. But
, so
and
, so
. Therefore any finite subset
of
cannot cover
. Hence,
is not compact.