Problem: Show that the axiom of choice (AOC) is equivalent to the statement that for any indexed family of non-empty sets, with , the product
Proof: This is pretty immediate when one writes down the actual definition of the product, namely:
So, if one assumes the AOC then one must assume the existence of a choice function
So, then if we consider as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function$
where clearly since we have already indexed out set we have that . So, consider
This is clearly a well-defined mapping and and thus
from where it follows that
Conversely, let be a class of sets and let be an indexing function. We may then index by . Then, by assumption
Thus there exists some
Thus, we have that
is a well-defined mapping with
For each . It follows that we have produced a choice function for and the conclusion follows.
Remark: We have assumed the existence of a bijective indexing function , but this is either A) a matter for descriptive set theory or B) obvious since satisfies the conditions. This depends on your level of rigor.
Problem: Let be a set; let be an indexed family of spaces; and let be an indexed family of functions
a) Prove there is a unique coarsest topology on relative to whish each of the functions is continuous.
and let . Prove that is a subbasis for .
c) Show that the map is continuous relative to if and only if each map is continuous.
d) Let be defined by the equation
Let denote the subspace of of the product space . Prove taht the image under of each element of is an open set in .
a) We first prove a lemma
Lemma: Let be a topology on , then all the mappings are continuous if and only if where is defined in part b).
Proof:Suppose that all the mappings are continuous. Then, given any open set we have that is continuous and so is open and thus from where it follows that .
Conversely, suppose that . It suffices to prove that for a fixed but arbitrary . So, to do this let be open in then and thus by assumption ; but this precisely says that is open in . By prior comment the conclusion follows.
By previous problem is in fact a topology on , and by our lemma we also know that all the mappings are continuous since . To see that it’s the coarsest such topology let be a topology for which all of the are continuous. Then, by the other part of our lemma we know that and thus . So,
And thus is coarser than .
The uniqueness is immediate.
b) It follows from the previous problem that we must merely show that is a subbasis for the topology . The conclusion will follow from the following lemma (which was actually an earlier problem, but we reprove here for referential reasons):
Lemma: Let be a set and be a subbasis for a topology on . Then, the topology generated by equals the intersection of all topologies which contain .
Also, let be the topology generated by the subbasis .
Clearly since we have that .
Conversely, let . Then, by definition to show that it suffices to show that for a fixed but arbitrary . To do this we first note that by definition that
for some . Now, if we know (since ) that and thus
for each . It follows that is the union of sets in and thus . It follows from previous comment that .
The conclusion follows.
The actual problem follows immediately from this.
c) So, let be some mapping and suppose that is continuous for each . Then, given a subbasic open set in we have that
for some and for some open sets in respectively. Thus may be written as
but since each we see that is the finite union of open sets in and thus open in . It follows that is continuous.
Conversely, suppose that is continuous then is continuous since it’s the composition of continuous maps.
d) First note that
from where it follows that the initial topology under the class of maps on is the same as the initial topology given by the single map . So, in general we note that if is given the initial topology determined by then given an open set in we have that which is open in the subspace .
Problem: Suppose that for each the topology on is given by a basis . The collection of all sets of the form
Such that is a basis for with the box topology, denote this collection by . Also, the collection of all sets of the form
Where for finitely many and otherwise is a basis for the product topology on .
Proof: To prove the first part we let be open. Then, by construction of the box topology for each we may find some such that is open in and . So, then for each we may find some such that and thus
Noticing that and every element of is open finishes the argument.
Next, we let be open with respect to the product topology. Once again for each we may find some such that is open in for each and for all but finitely many , call them . So, for each we may find some such that and so
Noting that and is a collection of open subsets of finishes the argument.
Problem: Let be a collection of topological spaces such that is a subspace of for each . Then, is a subspace of if both are given the product or box topology.
Proof: Let denote the topologies inherits as a subspace of and as a product space respectively. Note that are generated by the bases (where is the basis on with the product topology), and
So, let where then
Where is open in , and thus is open in . Also, since for all but finitely many it follows that for all but finitely many . And so . Similarly, if then
Where is open in , but this means that for some open set in and so
From where it follows that and thus are equal.
The case for the box topology is completely analgous.
Problem: Let be a collection of Hausdorff spaces, then is Hausdorff with either the box or product topologies
Proof: It suffices to prove this for the product topology since the box topology is finer.
So, let be distinct. Then, for some . Now, since is Hausdorff there exists disjoint open neighborhoods of respectively. So, are disjoint open neighborhoods of respectively. The conclusion follows.
Problem: Prove that .
Clearly this is continuous since
Problem: One of the implications holds for theorem 19.6 even in the box topology, which is it?
Proof: If where the latter is given the box topology then we have that each is continuous and thus so is each . #
Problem: Let be a sequence of points in the product space . Prove that converges to if and only if the sequences coverge to for each . Is this fact true if one uses the box topology?
Proof: Suppose that is a neighborhood of such that
is infinite. Notice then that if that from where it follows that is a neighborhood of which does not contain all but finitely many values of contradicting the fact that in .
Conversely, suppose that for each and let be a basic open neighborhood of . Then, letting be the finitely many indices such that . Since each there exists some such that . So, let . Now, note that if then for some . But, since clearly we must have that and thus . It follows that for every we have that . Then, since every neighborhood of contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows.
Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider . Clearly each coordinate converges to zero, but is a neighborhood of in the product topology. But, if one claimed that for every (for some that they’d be wrong. To see this merely note and so and thus .
Problem: Let be the subset of consisting of all eventually zero sequences. What is in the box and product topology?
Proof: We claim that in the product topology . To see this let be a basic non-empty open set in with the product topology. Since we are working with the product topology we know there are finitely many indices such that . So, for each select some and consider where
Clearly then and thus every non-empty open set in intersects and the conclusion follows.
Now, we claim that with the box topology that . To see this let . Then, there exists some subsequence of the sequence which is non-zero. For each form an interval such that . Then, consider
Clearly then is a neighborhood of and since each clearly has an infinite subsequence of non-zero values it is disjoint from . It follows that in with the box topology that is closed and thus as desired.
Problem: Given sequences and of real numbers with define by the equation
Show that if is given the product topology that is a homeomorphism. What happens if is given the box topology?
Proof: Let us first prove that is a bijection. To do this we prove something more general…
Lemma: Let and be two classes of untopologized sets and a collection of bijections . Then, if
we have that is a bijection.
Proof: To prove injectivity we note that if
And by definition of an -tuple this implies that
for each . But, since each is injective it follows that
For each . Thus,
To prove surjectivity we let be arbitrary. We then note that for each fixed we have there is some such that . So, if is the corresponding -tuple of these values we have that
from where surjectivity follows. Combining these two shows that is indeed a bijection.
Lemma: Let and be two classes of non-empty topological spaces and a corresponding class of continuous functions such that . Then, if and are given the product topologies the mapping
Proof: Since the codomain is a product space it suffices to show that
is continuous for each . We claim though that the diagram
commutes where and denote the canonical projections from and to and respectively. To see this we merely note that
which confirms the commutativity of the diagram. But, the conclusion follows since is the composition of two continuous maps (the projection being continuous since is a product space).
The lemma follows by previous comment.
We come to our last lemma before the actual conclusion of the problem.
Lemma: Let and be two classes of non-empty topological spaces and a set of homeomorphisms with . Then,
is a homeomorphism if and are given the product topology.
Proof: Our last two lemmas show that is bijective and continuous. To prove that it’s inverse is continuous we note that
And similarly for the other side. Thus,
Which is continuous since each is continuous and appealing to our last lemma again. Thus, is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows.
Thus, getting back to the actual problem we note that if we denote that each is a homeomorphism. Thus, since it is easy to see that
we may conclude by our last lemma (since we are assuming that we are giving in both the domain and codomain the product topology) that is a homeomorphism.
This is also continuous if we give the box topology. To see this we merely need to note that and thus if all of the are open then so are (since each is continuous) is each and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function.
Problem: Show that the normal formulation of continuity is equivalent to the open set version.
Proof: Suppose that are metric spaces and for every and every there exists some such that . Then, given an open set we have that . To see this let then and since is open by hypothesis there exists some open ball such that and thus by assumption of continuity there is some such that and so and thus is an interior point of .
Conversely, suppose that the preimage of an open set is always open and let and be given. Clearly is open and thus is open. So, since there exists some such that and so
Problem: Suppose that is continuous. If is a limit point of the subset of , is it necessarily true that is a limit point of ?
Proof: No. Consider with the suspace topology inherited from with the usual topology. Define
This is clearly continuous since and which are obviously open. But, notice that is a limit point for since given a neighborhood of we must have that there is some which contains it. But, is not a limit point for since that set has no limit points.
Problem: Let and denote a singlet set in the two topologies and respectively. Let be the identity function. Show that
a) is continuous if and only if is finer than
b) is a homeomorphism if and only if
a) Assume that is continuous then given we have that . Conversely, if is finer than we have that given that
b) If is a homeomorphism we see that both it and are continuous and so mimicking the last argument we see that and . Conversely, if then we now that
or equivalently that
which defines the homeomorphic property.
Problem: Given and show that the maps and given by and are topological embeddings.
Proof: Clearly and are continuous since the projection functions are the identity and constant functions. They are clearly injective for if, for example, then by definition of an ordered pair we must have that . Lastly, the inverse function is continuous since is the restriction of the projection to . The same is true for .
Problem: Show that with the usual subspace topology and .
Proof: Define and . These are easily both proven to be homeomorphisms.
Problem: Find a function which is continuous at precisely one point.
Suppose that is continuous at , then choosing sequences of rational and irrationals numbers respectively both converging to . We see by the limit formulation of metric space continuity that
And so if were to be continuous anywhere it would have to be at . To show that it is in fact continuous at we let be given then choosing we see that from where the conclusion follows since this implies that .
a) Suppose that is “continuous from the right”, that is, for each . Show that is continuous when considered as a function from to .
b) Can you conjecture what kind of functions are continuous when considered as maps as . As maps from to ?
a) Note that by the assumption that we know that for every there exists some such that implies that . So, let be open and let . Then, and since is open we see that there is some such that . But, by assumption there exists some such that . But, and thus and thus and so is an interior point for from where it follows that is open and thus is continuous.
b) I’m not too sure, and not too concerned right now. My initial impression is that if is continuous then is open which should be hard to do. Etc.
Problem: Let be an ordered set in the order topology. Let be continuous.
a) Show that the set is closed in
b) Let . Show that is continuous.
a) Let then . Suppose first that there is no and consider
This is clearly open in by the continuity of and is contained in it. Now, to show that let then which with simplification gives the important part that and so but since there is no such that this implies that . Similar analysis shows that and since there is no as was mentioned above this implies that . Thus, and thus .
Now, suppose that there is some such that then letting we once again see that is open and . Furthermore, a quick check shows that if that and so and and so and so so that . The conclusion follows
b) Let and . As was shown in a) both are closed and thus define
Notice that since are both assumed continuous and that we may conclude by the gluing lemma that is in fact continuous. But, it is fairly easy to see that
Problem: Let be a collection of subset of ; let . Let and suppose that is continuous for each
a) Show that if the collection is finite each set is closed, then is continuous.
b) Find an example where the collection is countable and each is closed but is not continuous.
c) An indexed family of sets is said to be locally finite if each point of has a neighborhood that intersects only finitely many elements of . Show that if the family is locally finite and each is closed then is continuous.$
a) This follows since if is closed then it is relatively easy to check that but since each is continuous we see that is closed in . But, since each is closed in it follows that each is closed in . Thus, is the finite union of closed sets in , and thus closed.
b) Give the subspace topology inherited from with the usual topology and consider with
Clearly each i
Lemma: Let be any topological space and be a locally finite collection of subsets of . Then,
Proof: The left hand inclusion is standard, so it suffices to prove the right inclusion. So, let since the collection of sets is locally finite there exists some neighborhood of such that it intersects only finitely many, say , elements of the collection. So, suppose that then is a neighborhood of which does not intersect contradicting the assumption it is in the closure of that set.
Now, once again we let be closed and note that and each is closed in and since is closed in we see that is closed in . So, noting that it is evident from the assumption that is locally finite in that so is and thus (for notational convenience) letting the above lemma implies that
From where it follows that the preimage of a closed set under is closed. The conclusion follows.
Problem: Let and be continuous functions. Let us define a map by the equation . Show that is continuous.
Proof: This follows from noting the two projections of are and . But, both of these are continuous since . To see this we note that if and only if which is true if and only if or in other words . Using this we note that the preimage an open set in will be the product of open sets by the continuity of . It clearly follows both projections, and thus the function itself are continuous.
Problem: Let . We say that is continuous in eahc variable separately if for each , the map is continuous and for each the map is continuous. Show that if is continuous then is continuous in each variable separately.
Proof: If is continuous then clearly it is continuous in each variable since if we denote by the mapping we see that where but the RHS is the composition of continuous maps and thus continuous. A similar analysis holds for the other variable.
Problem: Let be given by
a) Show that is continuous in each variable separately.
b) Compute .
c) Show that is not continuous
a) Clearly both and are continuous for since they are the quotient of polynomials and the denominator is non-zero. Lastly, they are both continuous at since it is trivial to check that $
c) This clearly proves that is not continuous with is not continuous since if is the diangonal we have that
and so in particular
Problem: Let ; let be continuous and let be Hausdorff. Prove that if may be extended to a continuous function , then is uniquely determined by .
Proof: I am not too sure what this question is asking, but assuming it’s asking that if this extension existed it’s unique we can do this two ways
Way 1(fun way!):
Lemma: Let be any topological space and a Hausdorff space. Suppose that are continuous and define . Then, is closed in
Proof: Note that is clearly continuous since and . It is trivial then to check that and since is Hausdorff we have that is closed and the conclusion follows.
From this we note that if agree on such that we have that
From where it follows that and so . So, thinking of as a subspace of we see that and thus clearly is dense in . So, the conclusion readily follows by noting that if are two continuous extensions then by definition .
Way 2(unfun way): Let be two extensions of and suppose there is some . Clearly and thus is a limit point of . So, by assumption and so using the Hausdorffness of we may find disjoint neighborhoods of them respectively. Thus, are neighborhoods of in . Thus, is a neighborhood of . But, clearly there can be no otherwise . It follows that is a neighborhood of disjoint from which contradicts the density of in . The conclusion follow.
Problem: Prove that every simply ordered set with the order topology is Hausdorff. The product of two Hausdorff spaces is Hausdorff. A subspace of a Hausdorff space is Hausdorff.
Proof: Let have the order topology and let be distinct and assume WLOG that . If there does not exists a such that then are disjoint neighborhoods of respectively. They are disjoint for to suppose that would imply that contradictory to our assumption. If there is some then are disjoint neighborhoods of respectively.
Let be Hausdorff and be distinct. Since are distinct it follows we may assume WLOG that . But, since is Hausdorff there exists disjoint neighborhoods of . So, consider these are clearly neighborhoods of in and to assume would imply .
Lastly, suppose that is Hausdorff and let be a subspace of . If are distinct there exists disjoint neighborhoods of them in . Thus, are disjoint neighborhoods of them in .
Problem: Let be a collection of subsets of the set . Suppose that , and that the finite union and arbitrary intersection of elements of are in . Prove that the collection is a topology on .
Proof: Clearly and are in . Now, suppose that then and since and is closed under arbitrary intersection it follows that and thus . Lastly, suppose that then and since and is closed under finite union it follows that and thus .
Problem: Show that if is a closed in and is closed in that is closed in .
Proof: Since is closed in and is a subspace of we have that for some closed set but since is closed it follows that is the intersection of two closed sets in and thus closed in .
Problem: Show that if is closed in and is closed in then is closed in
Proof: This follows immediately from question 9.
Problem: Show that if is open in and is closed in , then is open in and is closed in .
Proof: This follows immediately from the fact that and .
Problem: Let be an ordered set in the order topology. Show that . Under what conditions does equality hold?
Proof: Let then is a neighborhood of disjoint from and thus . Equality will hold when are limit points of or said otherwise whenever we have that there are some $lated d,e$ such that .
Problem: Let and denote subsets of . Prove that if then , , and . Give an example where this last inclusion is strict.
Proof: We choose to prove the second part first. Let then there is a neighborhood of such that and thus and so . Conversely, suppose that then and and so there are neighborhoods of such that clearly then is a neighborhood of such that and thus .
Using this, if then we have that from where it follows that .
Let then there is a neighborhood of such that and thus for every and thus for every and so finally we may conclude that .
To see when inclusion can be strict consider with the usual topology. Then,
Problem: Criticize proof (see book).
Proof: There is no guarantee that the for which intersected will b e the same that will intersect if you pick another .
Problem: Let and denote subsets of . Determine whether the following equations hold, if any equality fails determine which inclusion holds.
a) Equation does not hold. Consider that but . The inclusion always holds.
b) This follows from a) that equality needn’t hold. Once again the inclusion is true.
c) This need be true either . The inclusion holds.
Problem: Prove that if then in the product topology on .
Proof: Let . Let be a basic open set in which contains . Since we can choose some point . Then, . It follows that
Conversely, let . Let be any set such that . Since is open in it contains some point . Then, . It follows that . A similar technique works for .
10., 11,. 12 Covered in theorem stated and proved at the beginning of the post.
Problem: Prove that is Hausdorff if and only if the diagonal is closed in with the product topology.
Proof: Suppose that is Hausdorff then given we may find disjoint neighborhoods of them. So, and since . It follows that is closed.
Conversely, suppose is closed in and are distinct. Then, and so there exists a basic open neighborhood of such that and so are neighborhoods of in which are disjoint. For, to suppose they were not disjoint would to assume that contradicting the assumption that .
Problem: In the cofinite topology on to what point or points does the sequence converge to?
Proof: It converges to every point of . To see this let be arbitrary and let be any neighborhood of it. Then, is finite and in particular is finite. If it’s empty we’re done, so assume not and let then clearly for all such that we have that and thus is in . The conclusion follows.
Problem: Prove that the axiom is equivalent to the condition that for each pair of points there are neighborhoods of each which doesn’t contain the other.
Proof: Suppose that is then given distinct the sets are obviously neighborhoods of respectively which don’t contain the other.
Conversely, suppose the opposite is true and let then there is a neighborhood of it such that and thus is open and is therefore closed.
Problem: Consider the five topologies on given in exercise 7 of section 13 (my section 2).
a) Deterime the closure of under each of these topologies.
b) Which of these topologies are Hausdorff? Which are
a) As a reminder the topologies are
For the first one we easily see that .
For the second one we can see that . To see this note that is open by definition and thus being the complement of it is closed. Thus,
For the third one . To see this note that we in a sense already proved this in 14, but for any and any neighborhood of it we have that and thus if (assuming it’s non-empty) we see that . Thus, given any point of and any neighborhood of it we have that is infinite, and thus clearly . The conclusion follows from that.
For the fourth topology we note that which is open in with the upper limit topology. Remember that
For the last one . Clearly and since is closed and is the intersection of all closed supersets of it follows that . Now, suppose that then which is a neighborhood of which doesn’t intersect and thus . So, . The conclusion follows.
Lemma: If is Hausdorff, then it is
Proof: Let and by assumption there exists disjoint neighborhoods of respectively and so clearly and thus is open and so is closed.
: Clearly the first topology is Hausdorff since it is an order topology and we proved in the initial post that all order topologies are Hausdorff.
Lemma: Let be a set and two topologies on such that is finer than . Then, if is Hausdorff with the topology it is Hausdorff with the topology.
Proof: Clearly if are distinct we may find disjoint neighborhoods of them in the topology given by and thus the same neighborhoods work in consideration of the topology given by .
: From this lemma it follows that having a finer topology than with the usual topology is Hausdorff
: The cofinite topology is but not Hausdorff. To see that it’s it suffices to prove that is closed for any . But, this is trivial since being the complement of a finite set is open, thus closed. But, it is not Hausdorff since it cannot support two disjoint, non-empty open sets. To see this suppose that is open in with the cofinite topology then is finite and since a disjoint set would have to be a subset of it follows that is finite and thus it’s complement not finite. Thus, is not open.
: Once again this topology finer than that of with the usual topology since
: This isn’t even . To see this we must merely note that if is any set containing we must have that there is some basic open set such that , but this means that . So, there does not exist a neighborhood of which does not contain .
Problem: Consider the lower limit topology on and the topology given by the basis . Determine the closure of the intervals and in these two topologies.
Proof: We first prove more generally that if then . To see this we first note that is open since which we claim is open. To see this we note that . So, since is the intersection of all closed supersets of and is closed we see that . So, we finish the argument by showing that . To see this we show that is a limit point for from where the conclusion will follow. So, let be any neighborhood of , then we may find some basic open neighborhood such that , but clearly contains infinitely many points from where it follows that is a limit point of . From this we may conclude for that and
We now claim that in the topology generated by that and . More generall, let us prove that in this topology
Clearly if then choosing some rational number and some rational number then and so that . Also, if then choosing some such that we see that and . Thus, the only possibilities for are . But, just as before if is any neighborhood we may find some open basic neighborhood $latex [p,q)$ such that but clearly and thus since was arbitrary it follows that .
So, now we split into the two cases. First assume that then given any neighborhood of we may find some basic neighborhood such that , but since we see that from where it follows that and thus since was arbitrary it follows that and thus . Now suppose that , then is clearly a neighborhood of that doesn’t intersect and thus so that
The conclusion follows.
Problem: If , we define the boundary of by .
a) Prove that and are disjoint and
b) Prove that i and only if is both open and closed
c) Prove that is open if and only if
d) If is open, is it true that ?
a) Clearly if then there exists a neighborhood of whose intersection with the complement of is empty, thus . Now, let now if there exists a neighborhood of such that then and if not then every neighborhood of contains points of and since it follows that it also contains points of . Thus, either or , thus . Conversely, if then either there exists a neighborhood of such that and thus . Conversely, if then for every neighborhood of we have that and in particular and so .
b) Suppose first that . If we’re done, so assume not and let . Since there is a neighborhood of such that but since it follows that and thus is open. Conversely, letting we see that and so by the same logic there exists a neighborhood of such that and thus is open and so closed.
Conversely, suppose that is both open and closed and suppose that . If then for every neighborhood of we must have that and thus , which is a contradiction. Conversely, if then for every neighborhood of we must have that and thus which is a contradiction.
c) Suppose that is open and let , then every neighborhood of contains points of by definition, but since (it can’t be an interior point, thus ) they must be points of and thus . So, . Conversely, if then must be a limit point of which is not in and thus every neighborhood contains points of and so
d) No, it is not true. With the usual topology on the set is open, but
We choose to omit 18, 19, 21 on the grounds that they are monotonous and easy, monotonous and easy, and way, way to long.
This begins a substantial effort to complete all of (except the first chapter) the problems in James Munkre’s Topology I have chosen to start at chapter 2 considering the first chapter is nothing but prerequisites. Thus, without a minute to spare let us being.
Problem: Let be a topological space, let be a subset of . Suppose that for each there is an open set containing such that . Prove that is open.
Proof: We claim that but this is obvious since for each we have that . Conversely, since each we have that the union of all of them is contained in , namely . Thus, is the union of open sets and thus open.
Problem: Compare the nine topologies on given in example 1.
Solution: This is simple.
Problem: Show that given a set if we denote to be cocountable topology (a set is open if it’s complement is countable or the full space) that is a topological space. Is it still a topological space if we let ?
Proof: Clearly for the first part . Now, if is a collection of open sets then we note that is finite and thus for any . Thus, is finite and thus in . Now, if we have that and thus is the finite union of finite sets and thus finite, so .
If we redefine the topology as described it is not necessarily a topology. For example, give that topology and note that is a collection of elements of but whose complement is neither infinite, empty, or the full space. Thus, this isn’t a topology.
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 unique topology on containing all the collections , and a unique largest topology contained in all of the .
c) If , let and . Find the smallest topology containing and the larges topology contained in .
a) Let and let be arbitrary. Then, by assumption we have that for every but since this was assumed to be a topology we have that for every that and thus . Now, if we must have that for every and thus for every and thus . Thus, noting that for every we must have that the conclusion follows.
No, the union of two topologies needn’t be a topology. Let be defined as in part c and note that but
b) This follows immediately from part a. For the first part let (where is the set of all topologies) and let , this clearly satisfies the conditions. For the second one merely take
c) We (as pointed out very poignantly in the previous problem) just take the union of the two topologies. So, we claim that the desired smallest topology is in fact . But this is just computation and we leave it to the reader. For the second part just intersect the two topologies.
Problem: Show that if is a base for a topology on , then the topology generated by equals the intersection of all topologies on which contain . Prove the same if is a subbase
Proof: Let be the intersection of all topologies on which contain and the topology generated by . Clearly since is itself a topology on containing . Conversely, let we show that where is any topology on containing . But, this is obvious since for some and thus is the union of open sets in and thus in . The conclusion follows.
Next, let be above except now is a subbase. For the same reasons as above we have that . Conversely, for any topology containing we have that if then where each is the finite union of elements of . But, by construction it follows that each is open (it is the finite intersection of open sets in ) and thus is the union of open sets in and thus .
Problem: Show that the topologies on and the Sorgenfrey line aren’t comparable
Proof: See the last part of the next problem
Problem: Consider the following topologies on:
For each determine which of the others contain it.
:Clearly we have that since the defining open base for is contained entirely in .
: But, since but and thus not in . Now, to prove the inclusion indicated we know that for each open set in the cofinite topology we have that and so if we assume WLOG that then
And thus is open in the usual topology.
: But, . To see this it suffices to show that is open in since this is a base for . But, to see this we must merely note that .
: Lastly, . To see this we must merely note that .
For the result is obvious except possibly how it relates to . But, in fact they aren’t comparable. To see this we first show that is not open in . To see this we show it can’t be written as the union of sets of the form and but it clearly suffices to do this for the latter sets. Now, to see that can’t be written as the union of sets of the form we recall from basic real number topology that is not open ( is not an interior point) and thus it can’t be written as the union of intervals or it would be open in the usual topology on . Also, consider .
Problem: Show that the countable collection is a base for the usual topology on
Proof: This follows from the density of . It suffices to show that given any and any that there is some element such that . But, from basic analysis we know there is some rational number such that and similarly there is some such that . Thus, . The conclusion follows. .
Problem: Show that the collection generates a different topology from the one on (the lower limit topology)$.
Proof: Clearly is open in but we show that it can’t be written as the union of elements of . So, suppose that where . Then, there exists some such that . Now, since we must have that but and thus and thus . Contradiction.
Problem: Prove that ever closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is a set which is countable.
Proof: It is easy to see that the above proof (proof 27.) holds in any space which is second countable, in particular for separable metric spaces. Thus, if is closed we surely have that . Thus, but since and we have then that which is the union of a perfect and countable set respectively.
Problem: Prove that every open set may be written as the disjoint union of countably many open intervals.
Proof: We need to prove a quick lemma
Lemma: Let be a topological space and let be a class of connected subspace of such that then is a connected subspace of .
Proof: Suppose that is a separation of . We may assume WLOG that for some . So, now we see that otherwise would be non-empty disjoint subsets of whose union is contradicting that is connected. Thus, it easily follows that for any we have that so that and thus by a similar reasoning we see that . Thus, since was arbitrary it follows that contradicting that .
So, now for each define
And let and finally we prove that
is a countable class of disjoint open intervals whose union is . The fact that each is an open interval follows since it is the union of open connected spaces with non-empty intersection (each element of contains ) it is also an open connected subspace of (note that each element of is open in but since is open it is also open in . But, it was proven in the book the only connected subspace of are intervals and thus is an interval for each .
Now, to see that they are disjoint we show that if then from where the conclusion will follow. So, to see this we first note that if is non-empty then is an open connected subspace of containing both and and thus and but this implies that and respectively from where the conclusion follows.
Now, to see that is countable we notice that for each we have that and so if we let be a fixed but arbitrary then
is an injection since the elements of are pairwise disjoint. The fact that is countable follows immediately.
Thus, is a countable collection of open intervals and
Thus, the conclusion follows.
Problem: Prove that if where each is a closed subset of then at least one has non-empty interior.
Lemma: Let be a complete metric space and a descending sequence of non-empty closed subsets of such that . Then, contains one point.
Proof: Clearly does not contain more than one point since . So, now for each choose some and let . Now, if were somehow finite a similar argument to that used in the first case of problem 26. is applicable, so assume that is infinite. But, since it is evident that is a Cauchy sequence and thus by assumption it converges to some point . We claim that . To see this we note similarly to problem 26. that since is infinite it is easy to see that is a limit point of and so if for some then is a neighborhood of containing only finitely many points of which clearly contradicts that it is a limit point. The conclusion follows.
So, now suppose each had empty interior (i.e. nowhere dense) . Then, since is open and is nowhere dense there exists an open ball of radius less than one such that . Let be the concentric closed ball of whose radius is half that of . Since is nowhere dense contains an open ball of radius less than one-half which is disjoint from . Let be the concentric closed ball of whose radius is one-half that of . Since is nowhere dense we have that contains an open ball of radius less than one-fourth which is disjoint from . Let be the concentric closed ball of whose radius is half that of . Continuing in this we get a descending sequence of non-empty closed subsets for which . Thus, by our lemma we have that there is some . This point is clearly not in any of the ‘s from where the conclusion follows.