When I use a word it means exactly what I want it to mean, nothing more and nothing less.
Humpty Dumpty
from Alice in Wonderland
Let
.
Define a set
to be open in
if:
and is open in
.
For
example,
any
and any
.
,
and
is
open in
.
For
example,
any
and any
.
And
,
and
is
open in
.
For
example,
any
and any
.
And
Show that this forms a topology:
is open since
and
is
open in
is open in
since it is open in
That the given subsets are closed under arbitrary unions and finite intersections follows by some simple Boolean algebra . Check that each of the three types of open sets are closed under arbitrary unions and finite intersections. Then check that a union or intersection of one set of type 1. 2. or 3. with another set of type 1. 2. or 3. gives a set of type 1. 2. or 3.
is not Hausdorff. In particular, there do not exist open sets
and
such
that
,
and
Assume
then
is
an open set of
hence
for some
.
Assume
then
for some
But,
Let
be a topological space and let
.
We can define a topology on
which
we write as
be
defining a set
to
be open in
if either
or
is
open in
and
. We
call this the local topology at
.
One checks that this does define a topology and if
has
at least two points then it is not Hausdorff. All non-empty open sets contain
.
Let
be any set. Define
and all finite subsets to be closed. It is immediate that arbitrary
intersections and finite unions of closed sets are closed hence the
complements, the cofinite sets, and
form a topology. If
is infinite this is not Hausdorff. We want to look at the cofinite topology on
Suppose we are given
such that
then
is
continuous.
. More generally, the inverse image of finite (closed) sets are finite
(closed).
Let
Let
be a sequence,
Then
if and only if
is continuous,
having
the cofinite topology.
This is almost a trivial restatement of the various definitions.
By definition,
if for any
there exists an
such that
That is, the inverse image of any open set in
is
cofinite.
Conversely, suppose the inverse image of any open set in
is
cofinite in
Since
is open its inverse image is cofinite that is there exists an
such that
Let
be a subsequence of a convergent sequence
then
Let
be
the functional representation of
.
From the
Theorem,
implies
is continuous.
Letting
generate
the subsequence
,
for all
implies that
Hence
is
continuous as is
, being the composition of continuous functions. Thus, again from the Theorem,