Munkres Chapter two Section 12 & 13:

Topological Spaces and Bases

13 Votes

Point of post: This is the solutions to Munkres Chapter two Section 12 as the heading indicates.

1.

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.

2.

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.

3.

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

4.

Problem:

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

Proof:

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.

5.

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.

6.

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

7.

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

8.

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

Proof:

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

Munkres Chapter 2 Section 1

4 Votes

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.

1.

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.

2.

Problem: Compare the nine topologies on given in example 1.

Solution: This is simple.

3.

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.

4.

Problem:

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 .

Proof:

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.
5.

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 .

6.

Problem: Show that the topologies on and the Sorgenfrey line aren’t comparable

Proof: See the last part of the next problem

7.

Problem: Consider the following topologies on :

For each determine which of the others contain it.

Solution:

: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 .

8.

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. .

9.

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.

Munkres Chapter 2 Section 18

11 Votes

1.

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

2.

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.

3.

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

Proof:

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.

4.

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 .

5.
Problem: Show that with the usual subspace topology and .

Proof: Define

and . These are easily both proven to be

homeomorphisms.

6.

Problem: Find a function which is continuous at precisely one point.

Proof: Define

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 .

7.

Problem:

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 ?

Proof:

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.

8.

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.

Proof:

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

9.
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.$

Proof:

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.

10.

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.

11.

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.

12.

Problem: Let be given by

a) Show that is continuous in each variable separately.

b) Compute .

c) Show that is not continuous

Proof:

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 $

b) Evidently

c) This clearly proves that is not continuous with is not continuous since if is

the diangonal we have that

and so in particular

13.

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

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