Section 32: Problem 2 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
Take a point
and a point
or a closed set
contained in
(if
contains one point only, we are done). For every
fix
. For every
let
be the product of
with
for all
. For a subset
let
. If the product is Hausdorff,
and
can be separated by basis neighborhoods
and
. Then
and
are open (the projection is an open map) and disjoint (if there is a common point
then
belongs to both
and
) neighborhoods of
and
. Now, if the product is regular, then there are disjoint basis neighborhoods of
and
, and their projections into
are open and disjoint (for the same reason) neighborhoods of
and
. If the product is normal and
is a closed subset of
that does not intersect
, then
and
can be separated by disjoint basis neighborhoods whose projections into
separate
and
.
As Yuan says in a comment below, Munkres defines both regular and normal spaces with the additional requirement that singletons are closed. This is actually a common definition now, which guarantees that every normal space is regular etc. So, I missed that point in the original solution. In other words, we need also to show that the space is a
space. I leave this as a very easy exercise to the reader.
Of course, I am not leaving you alone. Moreover, here is an alternative way to prove almost all statements in this exercise. We can actually use the hereditary properties of spaces. Indeed, each
is homeomorphic to a subspace of the product space (which one?), and so if the product is
, Hausdorff or regular, so is
. Done. Almost done. Because being a normal space is not a hereditary property, so we still need to show that each
is normal if so is the product.