Section 13: Problem 8 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) Apply Lemma 13.2 to show that the countable collection
is a basis that generates the standard topology on
.
(b) Show that the collection
is a basis that generates a topology different from the lower limit topology on
.
(a) The standard topology is clearly finer than the topology generated by
. To see that they are equivalent consider any set
open in the standard topology. Take any point
. Since
is open, and the set of all open intervals is a basis for the standard topology, there is an interval
that contains
and lies in
. There are two rational points
and
such that
. The interval
contains
, lies in
and is a basis element in
.
(b) The lower limit topology is finer than the topology generated by
. Now, for point
having an open neighborhood
in the lower limit topology, there is no basis element in
that would contain
being a subset of
.
The topology is strictly finer than the standard topology, strictly coarser than the lower limit topology, and not comparable to either the
-topology or the countable complement topology or the upper limit topology.
P.S. Based on the comment below. Note that I do not show that these two collections are bases, as I interpret this as given. But if it is not, the proof still remains where each collection is now considered as a subbasis, and you can simply add that they are also closed under finite intersections. In few words, to show what we are mainly interested in, we do not need the fact that the two collections are bases, and if you feel like this is something not given but something we are also asked to prove, just add that both collections are closed under finite intersections.