In a topological space
,
a collection
of open subsets of
is said to form a basis for the topology on
if every open subset of
can
be written as a union of sets in
.
An open covering of a topological space
is a collection of open subsets of
such that every point in
lies in at least one of these open subsets.
A topological space
is said to be compact if every open covering of
admits
a finite refinement.
That is, given any open covering of
,
it is possible select a finite subset of the open covering, that still forms
an open covering of
.
Any basis is an open covering since
itself is an open set and hence must be the union of basis elements
An open covering may not be a basis . For example,
is always an open covering of
is
not compact.
is an open covering without a finite refinement.
Every closed subspace
of a compact space
is
also compact.
Let
be an open covering of
.
Let
be
a collection of open sets in
such
that
.
Since
is
closed
is
a covering of
.
Since
is
compact, let
and
possibly
is a finite subcovering of
.
Hence
is
a finite subcovering of
.
A subset
is
bounded if it is contained in some closed
-cube.
That is
A subset
is
compact if and only if is closed and bounded.
We will show that
is compact. Thus, since
is a closed subset of a compact space it is compact.
Now suppose
is
compact, we need to show it is closed and bounded.
To show it is closed, let
for each point
we
can find
and
such
that
.
Since
is
an open cover of
we can find a finite subcover
.
And since
and
,
is
a open set containing
in
.
Finally, since is
is arbitrary,
is open so is
closed.
To show
is
bounded, fix
and consider the cover
.
Choose a finite subcover
.
Let
Homework due March 16:
Show that if
and
hence
is
bounded.
Assume
and
.
Then