## 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.

1.

**Problem:** Let be a topological space, let . Suppose that for each there is an open set containing such that . Show that is open.

## Munkres Chapter one Section three: Relations

1.

**Problem:** Define two points and of the pane to be equivalent if . Check that this is an equivalence relation and describe the equivalence classes

**Proof:**

**Reflexivity**: Clearly

**Symmetry:** If then

and thus

**Transitivity: **If then and implies that . But, concatenating these inequalities and cutting out the middle gives and so .

2.

**Problem: **Let be a relation on a set . If , define the restriction of to to be the relation . Show that the restriction of an equivalence relation is an equivalence relation

**Proof:**

**Reflexivity:** Note that for each it’s true that and so (by ‘s reflexivity) and so

**Symmetry:** If then which says that and thus

**Transitivity: **If then we first note that . Furthermore, since we see that ‘s transitivity implies that and so by previous comment

3.

**Problem:** Here is a proof that every relation that is both symmetric and transitive is also relfexive

“Since is symmetric, . Since is transitive “

What’s the problem?

**Proof:** He assumed there is a such that

4.

**Problem: **Let be surjective function. Let us define a relation on by setting

if

**a) **Show that this is an equivalence relation

**b) **Let (set of all equivalence classes) Show there is a bijective correspondence between and

**Proof:**

**a)**

**Reflexivity:** Clearly, since is a function, so that

**Symmetry:** If then evidently from where symmetry immediately foll0ws.

**Transitivity:** Equally easy, if then

**b) ** Define

First, it is not clear that this is actually a function (the rule of assignment implicitly stated above). To see this suppose that

then

thus is a well-defined function. To see that this function is bijective we first note that if

and then by the surjectivity of for every there is some such that . Thus, and so is surjective.

*Remark:* All we’ve done is identified the fibers together

5.

**Problem:** LEt 4latex S$ and be the following subsets of

and

**a) **Show that is an equivalence relation on and . Describe the equivalence classes of

**b) **Show that given any collection of equivalence relations on a set , their intersection is an equivalence relation

**Proof:**

**a) **

**Reflexivity:** Clearly for every

**Symmetry:** This follows since

**Transitivity:** Clearly, if then and thus is the difference of two integers and so itself and integer.

To see that we merely note that

**b) **Let be a nonempty class of equivalence relations and let

To see that is an equivalence relation on we first note that since that . Next, we note that if then and so and so . Lastly, if then and so and so

6.

**Problem:** Define a relation on by setting if either , or and . Show that this an order relation on the plane.

**Proof:**

**Comparable**:** **Since the definition of the relation took arbitrary points in and compared them, it’s clearly a linear ordering.

**Non-reflexive:** This is clear since

**Transitive: **Suppose that and . There are several cases all of which are trivial.

7.

**Problem:** Show that the restriction of an order relation is an order relation.

**Proof:** Let be an order relation on and .

**Comparability:** Let then and so or , and so or

**Non-reflexivity:** If then

which is a contradiction.

**Transitivity:** This is identical to the transitivity proof for restrictions of equivalence relations.

8.

**Problem:** Check that the relation given by iff or if then .

**Proof:**

**Comparability:** Let . If then and if then we have by the comparability of the usual ordering on that either or

**Non-reflexivity:** Since and non-reflexivity is immediate.

**Transitivity:** This is just several simple cases again.

9.

**Problem:** Check that if and are two ordered sets, then the dictionary order is an order relation on

**Proof: **This is also relatively menial, but so important that it needs to be done:

**Comparability:** Let be arbitrary. If then the comparability of implies that either or , and so either or . If the comparability of implies that either or and so either or

**Non-reflexivity: **Note that by the non-reflexivity of we see that and so

**Transitivity:** Suppose that and . Now, if then . So, if we see that and so and so . If then and so . Now, if then . If then and so . If then and th same conclusion follows.

10.

**Problem:**

**a) **Show that the map :x\mapsto\frac{x}{1-x^2}$ is order preserving.

**b) **Show that the equation defines an inverse for

**Proof:**

**a) **Suppose that then and so and so and so finally . Next, suppose that then and so the above shows that and so latex x>0,y<0$ this obvious.

**b) **This is just function composition and is simple.

11.

**Problem: **Show that an element of an ordered set has at most one immediate successor and at most one immediate predecessor. Show that a subset of an ordered set has at most one smallest element and at most one largest element.

**Proof:** Let and suppose that and are both immediate predecessor to . Suppose that , then contradicting ‘s status as an immediate predecessor. The proof for immediate successor is similar.

Let both be least elements of a set . Then, since we have that . But, since we have that . Since and is impossible, it follows that . The proof for largest elements is similar.

12. I’ve done this, but I don’t feel like writing it up. I will at request though.

13.

**Problem: **Prove the following theorem:

” If an ordered set has the l.u.b. property, then it has the g.l.b. property.”

**Proof:** Let have the l.u.b. property and let be bounded below. Consider the set . Then, (since is bounded below) and it’s bounded above (by any element of ). Thus, by assumption exists. We claim that . To see this, first note that , since otherwise there is some such that and so clearly is an upper bound for which is smaller than , contradicting it’s definition as . Now, to see that for every we suppose not, then there is some such that , but this contradicts that is an upper bound for . It follows that and that . Thus, and so .

*Remark:* That may have been poorly written (I’m in a rush to finish these), if you have any questions just ask.

14.

**Problem:** If is a relation on a set , define a new relation on by letting if

**a) **Show that is symmetric iff

**b) **Show that if is an order relation, then is an order relation.

**c) **Prove the converse of the theorem in exercise thirteen.

**Proof:**

**a) **This is absolutely clear. If then .

**b) **

**Comparability:** Let be arbitrary. We may assume WLOG that and so

**Non-reflexivity: **Suppose that then , which is a contradiction.

**Transitivity: **Let be arbitrary, then and so and therefore

**c) **It’s pretty much the same procedure. For a non-empty bounded above set we have that is non-empty and bounded below. Thus, exists and using the same logic as before one can show that .

15.

**Problem:** Assume that the real line has the l.u.b. property.

**a) **Show that the sets and have the l.u.b.

**b) **Does in the dictionary order have the l.u.b? What about ? What about

**Proof:**

**a) **Let be bounded above by . By the l.u.b property of we have that $ exists. Clearly, we have that and , thus

Let be nonempty and bounded above by . We have that and certainly .

**b)** Ask me for these answers if there’s a strong desire too.

## Munkres Chapter one Section two: Functions

1.

**Problem:** Let . Let and

**a) **Show that and that equality holds iff

**b) **Show that and that equality holds iff is surjective

**Proof:**

**a) **Let , then$ latex f(x)\in f\left(A_0\right)$ and so . Now, suppose that is injective and let clearly then . It follows by injectivity that (if this isn’t apparent, note that by definition means that for some . Now, by injectivity we see that and so the result becomes clear). Conversely, suppose that for every . Then, in particular we see that and so

from where injectivity follows.

**b) **Let , then for some . It follows then that .

Now, suppose that is surjective and let . Then, and so . Conversely, suppose that for every . Then, in particular we see that and surjectivity follows.

2.

**Problem:** Let and let and for . Show that preserves inclusions, unions, intersections, and differences of sets:

**a) **

**b) **

**c) **

**d) **

Show that preserves inclusions and unions only:

**e) **

**f) **

**g) **; show that equality holds iff is injective

**h) **; show that equality holds iff is injective

**Proof:**

**a) **Let , then and so and thus

**b) **See number three

**c) **See number three

**d) **Let then , or it follows that and (otherwise ). This is equivalent to saying that . Conversely, let , then and . Thus, and . It follows that or

**e) **Let then for some and thus (since ) we see that for some which is equivalent to saying that .

**f) **See number three

**g) **See number three for the first part. Clearly, it suffices to prove the second part for two sets. So, suppose that is injective and let then and so and and so and , or . The proposed equality follows.

Conversely, suppose that for every . We see then in particular the following line of reasoning

and thus

and clearly singletons are disjoint iff they’re single elements are not equal. Thus, condensing

and injectivity follows.

**h) **Let , then and . This, in particular says that for some and (otherwise . Thus, for some , or .

Now, suppose that is injective and let , then by injectivity and so and . Thus, and . So, . Conversely, if then we see that

but iff . It follows that and so .

3.

**Problem:** Show that b), c), f), and g) of the last exercise hold for arbitrary unions and intersections

**Proof:**

**a) **Let then . Thus, there exists some such that and so . Therefore, .

Conversely, if let , then for some . Therefore, so that . So finally, we may conclude that

**c) **We merely need note that

(where the first equality is gotten noticing that and using property d) from the previous problem). Then, recalling the last problem we can see that this is equal to

and so comparing the LHS of the first equality with the RHS of the last leads to the desired result.

**f) **Let . Then, for some . Thus, for some for some . Thus, and so .

Conversely, let . Then, for some . It follows that for some and so for some . Therefore,

**g)** Let . Then, for some . It follows that for some for every . Thus, for every and so

4.

**Problem: **Let and

**a) **If , show that

**b) **If and are injective, show that is injective.

**c) **If is injective, what can you say about the injectivity of and ?

**d) **If and are surjective, prove that is surjective

**e) **If is surjective, what can you say about the surjectivity of and ?

**Proof:**

**a) **Let , then but this clearly implies that . Conversely, if we see that and so . Therefore,

**b) **If then by the injectivity of we see that and so by the injectivity of it follows that .

**c) **If is injective then is injective. To see this, suppose not; then there exists such that but which contradicts ‘s injectivity.

**d) **We note that from where the conclusion follows.

**e) **It must be true that is surjective. To see this suppose not. Then,

which contradicts the surjectivity of

5.

**Problem: **In general, let us denote the identity function on a set by . That is, define

Given we say that a function is a left inverse for if ; and we see that is a right inverse for if .

**a) **Show that if has a left inverse, is injective; and if has a right inverse, is surjective

**b) **Give an example of a function that has a left inverse but no right inverse

**c) **Give an example of a function which has a right inverse but no left inverse.

**d) **Can a function have more than one left inverse? More than one right inverse?

**e) **Show that if has both a left inverse and a right inverse , then is bijective and

**Proof:**

**a) **If we see that and so is injective. Conversely, let be arbitrary, we know that and so is the required element which maps to under

**b) **The inclusion map . Clearly it has no right inverse or it’d be surjective. But, the mapping

surely satisfies the left inverse requirement since

.

**c) **Consider the function . Clearly this possesses no left inverse, otherwise it’d be injective. But, has the quality that and so is a suitable right inverse

**d) **Yes, in the first example we could have taken and the second example we could have had , yet both are left and right inverses respectively.

**e) **Clearly by a) and b) we see that is bijective. Furthermore, let be arbitrary (we can say since it’s surjective), then

and

where is furnished by ‘s injectivity. Thus, putting the two together gives .

6.

**Problem: **Let . By restricting the domain and range of obtain from a bijective function .

**Proof:** Merely take and and so then the function is bijective and a restriction of (in fact, you could take the vaccuous function )

## Munkres Chapter one Section one

1.

**Problem: **Check the distributive laws for union and intersection. Also, check DeMorgan’s laws.

**Proof:** Our first goal is to check that and . To verify the fist we notice that

which by the distributive laws of ‘and’ and ‘or’ (easily checked by a truth table) we see that this is equivalent to

from where the conclusion follows. The other case is completely analogous.

Now, we aim to prove that and . To do this we let . Now, if we’d have that and so and so and so . Conversely, if we see that . If or we’d have that or , and either way . It follows that is in neither, i.e. . The conclusion follows.

To get the other DeMorgan’s law we merely note that (changing to )

2.

**Problem: **Determine which of the following statements are true for all sets and . If a double implication fails, determine whether one or the other of the possible implications holds. IF an equality fails, determine whether the statement becomes true if the equals sign is replaced by an inclusion.

a) and iff

b) or iff

c) and iff

d) or iff

e)

f)

g)

h)

i)

j) and implies

k) The converse of j)

l) The converse of j) assuming that and are non-empty

m)

n)

o)

p)

q)

**Proof:**

a)

: Let then and thus and so

: This isn’t necessarily true. Consider .

b)

: Let then or and so . Thus, .

: This is also wrong, the same counterexample being applicable.

c)

: Let then and and so . Thus, .

: Clearly since

d)

: Not true. Take and

: This is true, considering it’s a weaker formulation of the second direction in the last problem.

e) Clearly not since the LHS is a subset of . For example, .

f) No. If then the LHS is non-empty and the RHS is.

g) This is true. The RHS may be manipulated into the LHS as follows

(the last part by DeMorgan’s) which upon distribution of gives

h) Not true. Note that if this says that

i) This makes sense, if you read it it would say “Everything that’s in both and plus everything that is in but not in gives everything in .” Formally

and so by distribution this equals

j) This is true. Let , then and , thus .

k) This is not always true. Let and then the LHS and RHS of the inclusion are empty yet .

l) Then, this is true. Let , since there exists some , and thus . Thus, and so . A similar procedure shows

*Remark:* This shows why sometimes it’s perilous to just go “let …”

m) This is false. Let . Evidently but

n) This is true. Let then

or

.

or equivalently

or equivalently

o) This is true. Let , then:

or (and this looks weird)

or

or

or

And so . Conversely, let . Then,

(now, if , , and . But, since the first and the last are impossible since the first part of the statement says we may conclude that the second is true) thus

or

or

p) This is true. Note that iff which is true iff . But, this is true iff . (this ilast part was trick since we know that .) which is true iff .

q) This isn’t true in general. Try and

*Remark:* I did these in more detail earlier, if you search for them.

3.

**Problem: **

**a) **Write the contrapositive and converse of the statement “If , then ” and determine which of the statements are true

**b) **Do the same for the statement “If , then

**Proof:**

**a) **The contrapositive is ” If then which is false . The converse is “If , then ” which is false “

**b) **The contrapositive is “If , then which is true. The converse is ” if , then ” which is not true

4.

**Problem: **Let and be sets of real numbers. Write the negation of each of the following statements

a) For every , it is true that

b) For at least one , it is true that

c) For every , it is true that

d) For at least one , it is true that

**Proof:**

I’m going to write these in set notation

**a) **

**b) **

**c) **

**d) **

5.

**Problem: **Let be a nonempty class of sets. Determine the truth of each of the following statements and their converses

**a) **

** b) **

**c) **

**d) **

**Proof:**

**a) **This implication is true, as well as it’s converse (this is the definition of union)

**b) **The implication is false. Consider , then clearly but

**c) **The implication is true. The converse is false though. In the above, class of sets we see that but

**d) **The implication, and it’s converse are true. This is the definition of intersection.

6.

**Problem: **Write the contrapositive of each statements in exercise 5

**Proof: **

**a) **If for every then

**b) **If there exists such that , then

**c) **If for every it is true that then

**d) **If there exists some such that , then

7.

**Problem: **Given sets and , express each of the following sets in terms of and , using the smbols and .

**Proof:**

**a) **

**b) **

**c) **

8.

**Problem: **If a set has two elements, show that has four elements. How many elements does have if has one element? Three elements? No elements? Why is called the power set of ?

**Proof:**

**Lemma: **Let . Then,

**Proof: **Let where

To see that is injective suppose that , then WLOG there is some and so , thus . Next, to prove surjectivity let and let , evidently . It follows that is a bijection, and the lemma is proved.

It follows that if then .

*Remark:* Alternatively, one may find a recurrence relation solution to this problem. Namely, let . Then, if one paritions into those sets which contain and those that don’t one can see that each block of the partition has elements, and thus

and thus noting that one may conclude that

9.

**Problem: **Formulate and prove DeMorgan’s laws for arbitrary unions and intersections.

**a) **

**Claim: **Let be a class of subsets of some universal set , then

**Proof:** If then for some . Thus, and so . Conversely, suppose that then there exists some such that . It follows that and so finally . The conclusion follows.

**Corollary:**

**Proof:** We merely note that

which by our last proven claim is

10.

**Problem: **Let denote the set of real numbers. For each of the following subsets of , determine whether it is equal to the cartesian product of two subsets of

**a) **

**b) **

**c) **

**d) **

**e) **

**Proof:**

**a) **This is, namely

**b) **This is, namely

**c) **This is not. To see this suppose that . We see that and thus . Also, and so . But, and thus , which is a contradiction.

**d) **This is, namely

**e) **This is not, using the same argument as in c), first noting that

## Munkres Chapter 1 Section 2

**1.**

**Problem: **Let . Let and .

**a) **Show that and that equality holds if is injective

**b) **Show that and equality holds if and only if is surjective

**Proof:**

**a) **The first part is obvious enough. If then and so from where it follows that .

Now, suppose that is injective but . Then, there is some such that . But, by definition since there exists some such that but since and it follows that and this contradicts injectivity.

Conversely, suppose that . Then, if then and so which means that as desired.

**b) **The first part is clear again. If then and so from where it follows that .

Now, suppose that is surjective. By the previous part it suffices to show the reverse inclusion. So, let (we know that any element of is of the form by surjectivity), then and so and so from where the conclusion follows.

Conversely, suppose that and let . Then, in particular from where it follows that and so surjectivity is guaranteed.

**2. **

**Problem: **Let and let and for . Show that preserves inclusions, unions, intersections, and differences of sets:

**a) **

**b) **

**c) **

**d) **

Show that preserves inclusions and unions only:

**e) **

**f) **

**g) ** and show that equality holds precisely when is injective

**h) ** and show that equality holds precisely when is surjective.

**Proof:**

We assume in all cases that the sets are non-empty and in the case of intersections and set differences, intersecting. For, if not this is trivial.

**a) **Let then and so and so , from where it follows that

**b) **Let then and so and so and so . Noting that all the “and so”s above were actually “if and only if”s shows the reverse inclusion.

**c) **We note that if and only if which is true if and only if which is true if and only if which is true if and only if

**d) **We note that if and only if which is true if and only if . Now, clearly the first part is true if and only if and (noticing that )the second part is true if and only if and so putting these together we get the statement in the previous sentence is true if and only if which is true if and only if

**e) **Let then for some . But, since it follows that for some and so from where it follows that .

**f) **Let then for some and thus for some and so either for some or for some and so which is true if and only if

**g) **Let then for some and so for some and for some and so and so from where it follows that .

Now, suppose that is injective. By the last part to show equality it suffices to show the reverse inclusion. To do this we let then and thus (by injectivity) and so and so . The conclusion follows by previous comment.

Conversely, suppose that then and so

from where it follows that

*Remark:* I technically took to be any subset of

**h) **Let then and and so for some but this and so for some and so

Now, suppose that is injective. From the last part to prove equality it suffices to show the reverse inclusion. So, let then (by injectivity) and so and so (second part by injectivity) and thus

Conversely, if then

from where it follows that

Injectivity follows.

*Remark:* The same remark applies.

**3.**

**Problem:** Show that b), c), f) and g) of the last exercise hold for arbitrary unions and intersections.

**Proof:**

We once again assume that all the following are non-empty since the proof otherwise is trivial.

**b) **Let then and so in particular for some . Thus, and so . It follows that

Conversely, let then for some . Thus, and so so that . It follows that

The problem follows.

**c) ** Let then and so for every . Thus, for every and thus

Conversely, if we have that for every and so for every . Thus, and so . It follows that

from where the problem follows.

**f) **Let then where . So, where for some . It follows that and thus from where it follows that

Conversely, let then for some . Thus, for some thus for some and so from where it follows that

and so the problem follows.

**g) **Let , then for some . It follows that for some for every . Thus, from where it follows that

Now, assume that is injective and let . Then, for every . Thus, by injectivity we have that for every and so . Thus, and thus

from where the problem follows.

**4.**

**Problem: **Let and .

**a) **If show that

**b) **If and are injective show that is injective.

**c) **If is injective what can you say about the injectivity or ?

**d) **If and are surjective prove that is surjective.

**e) **If is surjective what can you say about and ‘s surjectivity?

**f) **Summarize your answers to b-e in the form of a theorem.

**Proof:**

**a) **Let , then and so . So, and so finally .

Conversely, if then and so so that .

The conclusion follows.

**b) **If then (by ‘s injectivity) and so (by ‘s injectivity). The conclusion follows.

**c) **We claim that is injective implies that is injective. To do this we prove the contrapositive. Suppose that was not injective, then there is such that and so clearly then and so is not injective. The conclusion follows.

**d) **This follows since as desired.

**e) **We claim that is surjective implies that is surjective. One again, we do this by proving the contrapositive. We note that from where the conclusion follows.

**f) **I’m not sure what’s desired but I guess it would be that b) and d) imply that if are bijections then is a bijection. Also, c) and e) prove that if is a bijection then is injective and surjective.

**5.**

**Problem:** In genereal, let us denote the identity function for a set by . That is define .

Given we say the function is a *left inverse* for if . Also, we say is a *right inverse* if .

**a) **Prove that if has a left inverse, is injective, and if has a right inverse, is surjective.

**b) **Give an example of a function has a left inverse but no right inverse

**c) **Give an example of a function that has a right inverse but no left inverse.

**d) **Can a function have more than left inverse? More than one right inverse?

**e) **Show that if has both a left and a right inverse then is bijective and

**Proof:**

**a) **Suppose that has a left inverse and let . We know then that which would be impossible (since is a function$ if .

Conversely, let have right inverse and let be arbitrary. We know then that and since it is the desired value whose image is .

**b) **Consider the function . Clearly this has a left inverse but no right inverse.

**c) **How about ?

**d) **A left inverse need not be unique by this definition. For example, consider the mapping . Then, we can define by

Clearly then for every . But, we may take to be anything so that is not unique.

Right inverses are not unique either. For example, consider

We then consider two function

We then note that

and

So that

But, a similar analysis shows that if is defined by

also has the property that

And since it follows that right inverses need not be unique.

**e) **Clearly by our previous problems we have that having a left and right inverse respectively implies that $laetx f$ is injective and surjective respectively and thus bijective. Now, Let be arbitrary and suppose that

then we see by ‘s injectivity that

which is clearly absurd. Also, let be arbitrary, then by surjectivity we have that for some . Then, if this implies that

which is also a contradiction. It follows that and and thus as desired.

**6.**

**Problem:** Let . By restricting the domain and range of appropriately obtain from a bijective function .

**Proof:** We note that and so is increasing on . Thus,

but since is continuous it follows from the intermediate value theorem that and thus noting that is injective on (since it’s continuous and monotone) we may conclude that is bijective.

## Munkres Chapter 1 Section 1

**1.**

**Problem:** Prove the distributive laws for union and intersection, and prove DeMorgan’s laws.

**Proof:**

**a) **We claim that . To see this we denote the right side by and the left side by , then if and only if but by the distributive laws of the “and” and “or” operators (just draw a simple truth table) this is equivalent to

but this is just the logical statement that . The exact same, (but reversing the arrows…haha not category theory) gives that . It follows that as desired.

**b) **We now claim that . Once again we denote the LHS by and the RHS by . Then, implies that , which implies that . Once again, constructing a truth table if necessary, we use the distributivity laws of the logical operators “and” and “or” to get the above implies

which says that . Once again, using the exact same logic in reverse we get that from where it follows that .

**c) **We claim in general that if are subsets of some universal set , then

To see this we once again denote the LHS by and the RHS by . Then, if then and so for some . It follows that and thus . Conversely, if then f0r some and thus and so and so . The conclusion follows.

We now claim that

To do this we note that

And thus if we keep to our LHS-RHS notation we see that from where it follows from elementary set theory that .

**2.**

**Problem: **Determine which of he following statements are true for all sets (contained in some universal set ). If a double implications fails, determine whether one or the other statement of the possible implication holds, If an equality fails, determine whether the statement becomes true if the “equals” is related by one or the other of the inclusion symbols or .

**a) **

**b) **

**c) **

**d) **

**e) **

**f) **

**g) **

**h) **

**i) **

**j) **

**k) **The converse of j)

**l)** The converse of j), assuming that and are non-empty.

**m) **

**n)**

**o) **

**p) **

**q) **** **

**Proof:**

**a) **The right hand implication is clearly true since . That said, the left hand implication is false. Clearly given any non-empty set we have that but it is not true that .

**b) **If the LHS statement is true then for each we have that and so , i.e. . The other direction is not correct though. Consider the sets then but it is not true that .

**c) **This two-sided implication is true. To see the right implications we note that if then and so that and so it follows that . Conversely, if then for each we have that and so .

**d) **The right hand implications is not false. Take for example and . Then it is true that but it is not true that . The converse is true since it is weaker than the second part of the last statement.

**e) **The equality is not always true. Consider for example as the universal set. Then, if and then which is equal to . The inclusion is true thought since means that or that but since it follows that . Thus, .

**f) **Let’s just see what the LHS is equal to. but this is equal to (DeMorgan’s law) but this is equal but this is equal to since the left set is a subset of . It clearly follows that but the inclusion may be strict by considering

**g) **The LHS is equal to and the RHS is which is equal to . Thus, they are, in fact, equal.

**h) **The RHS is (prime denotes complement in the universal set) which is equal to by distributivity laws . The LHS is . Clearly the neither inclusion holds (in general) since the LHS is always contains elements of and the RHS is contains only elements of (if it contains any).

**i) **The LHS is which distributing gives which is equal to which is equal . So, yes, they are equal.

**j) **This implication is true. If then and thus and so

**k) **No, the converse is not true. Consider then . But, clearly is not true.

**l) **In this case it is true. Let then for some (we know there exists some, since ) we have that and so . The other direction works the same.

**m) **This equality is false. Consider the intervals . Note then that (once can picture this as taking the square of side length two whose bottom left hand corner is on the origin. Then, if we carve this square into four smaller squares in the obvious way the LHS is just the upper right and lower left ones whereas the RHS is the full square). In general . To see this we note that if then we may assume WLOG that and thus and so and so

**n)** This is true. To see this we note that if then and so and so in particular or . Conversely, if then and which says that and and so and so that

**o) **This is true. Note that if and only if which is true if and only if which is true if and only if or and so

**p) **This is true. Note that if and only if which is true if and only if . But, this is true if and only if (because by the first part we *must *have and so the last two are just “tricks”) which is true if and only if which is just a fancy way of saying

**q) **This equality does not hold in general. If then but . The, inclusion does hold though. To see this let then and so in particular but and so

*Remark: *God that was horrendous…

**3.**

**Problem: **

**a)**Write the contrapostive and converse of the following statement: “If latex x^2-x>0$” and determine which (if any) of the three statements are true.

**b) **Do the same for the statement “if then .

**Proof:**

**a) **Contrapositive says then . The negation says that if then

We note that then . Thus, the contrapositive is true. The negation is not true. The actual statement, being logically equivalent to the contrapositive, is true.

**b) **This is exactly the same idea.

**4.
**

**Problem: **Let and $laetx B$ be sets of real numbers. Write the negation of each of the following statements:

**a) **For every it is true that

**b) **For at least one it is true that

**c) **For every it is true that

**d) **For at least one , it is true that

**Proof:**

**a) **There exists some such that

**b) **For every it is true that

**c)** There exists some such that

**d) **For every it is true that

**5.**

**Problem: **Let be a non-empty collection of sets. Determine the truth of each of the following statements about their converses:

**a) **

**b) **

**c) **

**d) **

**Proof:**

**a) **Of course this is true, it’s the definition.

**b) **Clearly not. Take the elements of to be disjoint.

**c) **Of course! By definition if it’s in the intersections it’s in all of them

**d) **What I said in the last one.

**6.**

**Problem:** Write the contrapositive of each of the statements of Exercise 5.

**a) **If for every it is true that then

**b) **If for some it is true that then

**c) **If for some then

**d) **IF for every then

**7.**

**Problem: **Given sets and , express each of the following sets in terms of and using the symbols and .

**Proof:**

**a) **The set is

**b) **The set is

**c) **The set is

**8.**

**Problem:** If a set has two elements, show that has four elements. How may elements does have if has one element? Three elements? No elements? Why is called the power set of ?

**Proof: **We prove that . To do this we let .

Partition into two blocks, namely

and

.

Note, that clearly . Now, given we have two choices, either or . Thus, excluding for a second each element of is paired with an element of is a unique way. It follows that . But, noting that in fact for each we have that it follows that . Thus,

Thus, noting that we can easily prove by induction that

**9.**

**Problem: **Formulate and prove a generalized version of DeMorgan’s theorem.

**Proof:** Already did, cf. problem one.

**10.**

**Problem: **Let denote the set of real numbers. For each of the following subsets of , determine whether it is equal to the Cartesian product of two subsets of

**a) **

**b) **

**c) **

**d) **

**e) **

**Proof:**

**a) **Yes,

**b) **Yes,

**c) **No, notice that as well as but .

**d) **Yes,

**e) **No, notice that but .

## Munkres Chapter 2 Section 19 (Part II)

**9.**

**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

Such that

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.

**10.**

**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.

**b) **Let

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 .

**Proof:**

**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.

So, let

and let

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 .

**Proof: **Let

and

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

where each

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 .

## Munkres Chapter 2 Section 19 (Part I)

**1.**

**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

Where

Noting that and is a collection of open subsets of finishes the argument.

**2.**

**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.

**3.**

**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.

**4.**

**Problem:** Prove that .

**Proof:** Define

Clearly this is continuous since

**5.**

**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 . #

**6.**

**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 .

**7.**

**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.

**8.**

**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

Then,

And by definition of an -tuple this implies that

for each . But, since each is injective it follows that

For each . Thus,

as desired.

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

is continuous.

**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

and

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.

## Munkres Chapter 2 Section 18

**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.

## Munkres Chapter 2 Section 17

**Theorem 17.11**

**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 .

**1.**

**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 .

**2.**

**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 .

**3.**

**Problem: **Show that if is closed in and is closed in then is closed in

**Proof: **This follows immediately from question 9.

**4.**

**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 .

**5.**

**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 .

**6.**

**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,

**7.**

**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 .

**8.**

**Problem: **Let and denote subsets of . Determine whether the following equations hold, if any equality fails determine which inclusion holds.

**a) **

**b) **

**c) **

**Proof:**

**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.

**9.**

**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.**

**13.**

**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 .

**14.**

**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.

**15.**

**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.

**16.**

**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

**Proof:** …

**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.

**b)**

**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 .

**17.**

**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 , 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 ?

**Proof:**

**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.**