Section 31: 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
9. Let
be the set of all points of
of the form
, for
rational; let
be the set of all points of this form for
irrational. If
is an open set of
containing
, show there exists no open set
containing
that is disjoint from
, as follows:
(a) Let
consist of all irrational numbers
in
such that
is contained in
. Show
is the union of the sets
and countably many one-point sets.
(b) Use Exercise 5 of §27 to show that some set
contains an open interval
of
.
(c) Show that
contains the open parallelogram consisting of all points of the form
for which
and
.
(d) Conclude that if
is a rational number with
, then the point
of
is a limit point of
.
(a) For every irrational point
there is a basis neighborhood
contained in
, therefore, if
then
. The rest of
, i.e. the rational points, is a countable set of points.
(b) According to the result of the Exercise 5 of §27, since
has non-empty interior in
being compact and Hausdorff, one of the closed sets in the union must have a non-empty interior in
as well. Therefore, there is
such that
has non-empty interior, and contains a non-empty interval
.
(c) I show a little different result: for
found in (b) there is a non-empty interval
and
such that for every
,
. Take
where
and
are such that
(found in (b)). Let
.
is not empty. For each
and
there is a point
. Therefore,
.
(d) Given (c), for every point
and every neighborhood
of
there is
such that
.