Supplementary Exercises*: Nets: Problem 9 Solution
Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
If there is a subnet
converging to
then for any neighborhood
of
there is
such that
implies
. Let
.
is cofinal in
. Indeed, for any
there is
such that
and
such that
is greater than both
and
. Then
and
. Now, the set of all points from
such that
contains
, therefore, it is cofinal in
.The other direction. For any neighborhood
of
let
be the cofinal subset of those
such that
. We need to define a directed set
and
such that
implies
,
is cofinal in
and for any neighborhood
of
there is
such that
implies
. Consider two different neighborhoods
and
of
and their intersection
. Note that
. However, we want to distinguish all indexes in
from the same indexes in
and
. This suggests to take
as the union of all pairs
where
and define
. Moreover, for a given
we want both:
and
. Therefore, we define the partial order on
as follows:
iff
and
. This is a partial order on
(this can be easily checked even in general for the partial order defined this way on the product of two partially ordered sets). Now, for any pair of elements of
,
and
: find
, then
such that
(here we use the fact that all
are cofinal), then we have
. Therefore,
is a directed set. Moreover,
implies
and
is cofinal in
. For any neighborhood
of
the set
is not empty, and
implies
.