Munkres Chapter two Section 12 & 13: Topological Spaces and Bases
Point of post: This is the solutions to Munkres Chapter two Section 12 as the heading indicates.
Problem: Let be a topological space, let . Suppose that for each there is an open set containing such that . Show that is open.
Proof: So, we know that for each there exists some open such that . Clearly
and so is the union of open sets, and thus open.
Problem: Consider the nine topologies on the set indicated in Example 1 of section 12. Compare them.
Proof: This is incredibly laborious to type out, and really not that difficult, so we omit it.
Problem: Show that the collection given in example 4 of section 12 is a topology on the set . Is the collection
a topology on ?
Proof: This is just the cocountable topology. We can see that since it’s complement is all of . Also, since which is finite. Also, we notice that if is a nonempty subset of that
and so the complement of is a subset of a countable set and thus countable. Similarly, if then
and so the complement of is the finite union of countable sets, and thus countable, so . It follows that really is a topology on
Now, is not a topology on any infinite set. To see this let be infinite and let be any surjection. Then, we can see that and are in but their intersection is which is not in
a) If is a family of topologies on , show that is a topology on . Is ?
b) Let be a family of topologies on . Show that there is a unique smallest topology on $laetx X$ containing all the collections and a unique largest topology on contained in all the
c) If let and . Find the smallest topology containing and the largest topology contained in
a) Clearly since for every we have that . Now, if then for every . But, since each is a topology it follows that for every and so . Lastly, if we have that for every . But, since these are topologies this implies that for every and so
b) To show that there is a unique topology which contains all the we define
We first notice that is not empty since . So, let
Clearly then is a topology which contains all the , and any other topology which contains them must contain . More formally, is a topology on such that . Furthermore, given a topology which contains every we see that so that . Also, if were another smallest topology containing all the ‘s that both the inclusions and and so , from where it follows that the smallest such topology is unique.
Using the exact same logic, the smallest topology contained in all of the ‘s is .
c) One can verify that is the smallest topology containing both and is a topology, and thus the smallest contained in both.
Problem: Show that if is a basis for a topology on , then the topology generated by equals the intersection of all topologies on that contain . Show the same is true if is a subbasis.
Proof:Let be the set of all topologies on and let
and the topology generated by . Clearly since is a topology on which contains we have that . Conversely, let , then where . But, for any topology containing we must have that (since it’s closed under arbitrary unions). Thus, is contained in all topologies which are supersets of , namely it is in their intersection . It follows that as desired.
Now, let be a subbasis and
and the topology generated by .
Clearly being a topology on containing guarantees that . Conversely, if , then for some topology on which contains . But, this then means that cannot be an arbitrary union of finite intersections of elements of , since any topology containing must contain them. Thus, . It follows that as required.
Problem: Show that the topologies and aren’t comparable.
Proof: We claim that but . The fact that follows from the definition of the basis. Now, suppose there was a basis element such that . Clearly otherwise . But, by the Archimedean principle there exists some such that , thus but this contradicts that . It follows that no such exists.
We now claim that but . To see this suppose that is a basic element for such that . Now, clearly otherwise . But, by the Archimedean principle we may choose such that and so and thus , but this contradicts that
Problem: Consider the following topologies on
Determine, for each of these topology, which of the others contain it.
Proof:We know from the book that .
Next, we claim that . To see this let , then where we may assume WLOG that . So then, which is the finite union of open sets in , thus
We claim that . To see this we must merely note that
so that every basic set, and thus every set, in may be written as the union of basic open sets in from where the conclusion follows.
It’s also clear that since
and so every element of is the union of open sets in and so
Now, clearly and and so it suffices to check how it relates to . We claim that they aren’t comparable. To see this we note that is open in the upper limit topology, but any set of the form must have and so it will entrap points not in that aren’t removed by . Also, using a very similar argument one can show that is not open in the upper limit topology.
The last inclusion we must check is how relates to and we claim they aren’t comparable. Clearly sine it’s complement is infinite but it’s in . Also, but it’s not in since any basic set containing , must contain
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 generates a topology different than
a) Clearly since is contained is the usual basis for the topology generated by it is coarser than the usual one. Conversely, let be any basic open set in the usual topology on and arbitrary. We know from the structure of the real numbers that there is some such that and some such that . Thus, and from where it follows that the topology generated by is finer than the usual topology. Thus, it follows that the usual topology and the one generated by are equal.
b) We claim that is not in the topology generated by . To see this suppose that could be written as the union of sets of the form , say . We see that since that and since it’s rational we may conclude that . But, this means that for every we have that and thus contradictory to our assumption.
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