Section 24: Problem 12 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
(a) If
is a bijection preserving the order, then
is a bijection preserving the order, similar for
. Vice versa, if there are two order preserving bijections from
and
to
and
then the joint function is the order preserving bijection needed.(b)
and
has the order type of
implies, by applying (a) two times, that
has the order type of
. Vice versa, we construct bijective order preserving functions
and the resulting function is an order preserving surjective function onto
.(c) Transfinite induction: we show that if it holds for any
then it holds for
. If
has an immediate predecessor
then
is either empty or has the order type of
, and
has it as well. By (a), the property holds for
. If
has no immediate predecessor, then the set of all elements lower than
has it as the supremum (for any
there is
) and we can construct a sequence of points converging to
. This is not a trivial fact, I think (and I don’t remember it was covered anywhere before). Indeed, if we consider
, for example, then
is a limit point of
but there is no sequence converging to it (any sequence is a countable subset and has an upper bound in
). However, in the case of
we prove it as follows:
is countable; let us enumerate all its elements with positive integer indexes:
; then let
be such that
, let
be the minimum index greater than
such that
, let
be the minimum index such that
and
, etc. (we can always do this as there is no largest element in
, therefore, there is infinite number of elements greater than any
); note that for every
,
and
for all
; if
then
for some
and
for all
. Now, once we have the sequence converging to
, by using (b),
has the order type of
. By the principle of transfinite induction, this holds for any
.(d) We show that any point is path connected to
where
is the immediate successor of
. Indeed,
for
is path connected to
(
is homeomorphic to
) and for
any point
,
, is path connected to
which is path connected to
(by (a) and (c),
has the same order type as
and the order topology on a convex subset is the same as the subspace topology). Therefore,
is path connected as the union of path connected subsets having a common point.(e) An open interval
in
has the order type of an open interval in
. Indeed, by (c),
has the order type of
. Since for convex subsets the order topology is the same as the subspace topology,
has the order type of
. Therefore, for any
there is an open neighborhood
with the order type of
. Moreover, there is a bijective order preserving function from one set onto the other, and by 7(a), it is a homeomorphism.(f) If
could be embedded into
then
would be homeomorphic to a subspace of
which has a countable basis for the topology. Therefore, there must be a countable basis of the topology of
. We show that there is no. For any
:
is open and contains
but does not contains
for
. Therefore, there is a basis element such that it contains
but does not contain
for
. Call it
. Note that for
:
. Since
is uncountable, there must be uncountable number of distinct sets in the basis.