Chapter 3

Topology of the Real Line

In this chapter, we study the features of R which allow the notions of limits and continuity to be defined precisely. This is what is meant by topology. Though it is done here for the real line, similar notions also apply to more general spaces, called topological spaces. This branch of mathematics, in its modern form, dates back to the mid 1800s and the work of Bolzano (1781-1848), Cantor (1845-1918) and Weierstrass (1815, 1897).

3.1 Interior Points and Neighborhoods

Throughout this chapter, it will be useful to think of intervals (a, b) or [a, b]. Though the various notions studied (open, closed to name a few) apply to more general sets, understanding them in terms of intervals will be helpful.

3.1.1 Main Definitions The notion of interior point tries to capture the idea of how "deep" within a set a point is. Intuitively, the interior of a set is the part of the set which is not on "the edge of the set". Of course these concepts need to be made much more precise mathematically.

Definition 3.1.1 (Interior Point) Let S R and x R. ⊆ ∈ 1. x is said to be an interior point of S if there exists δ > 0 such that (x δ, x + δ) S. − ⊆ 2. If the above is true, S is said to be a neighborhood of x.

3. Given a set S R, the interior of S, denoted Int (S) or S◦, is the set of interior points⊆ of S.

Remark 3.1.2 We will remark the following:

89 90 CHAPTER 3. TOPOLOGY OF THE REAL LINE

1. Intuitively speaking, a neighborhood of a point is a set containing the point, in which you can move the point a little without leaving the set.

2. A neighborhood of a point x R is any set which contains an interval of the form (x δ, x + δ) for some∈ δ > 0. This definition is a little bit more general that− the definition some texts use. 3. (x δ, x + δ) is a neighborhood of x δ > 0. In fact, certain texts use this as− a definition of a neighborhood. They∀ say that a neighborhood of a point x is a set of the form (x δ, x + δ) for some δ > 0. This means that for them, a neighborhood of a− point x has x at its center. Our definition is a little bit more general. 4. It should be clear from the definition that an interior point of a set belongs to the set. We will state it as a result in the proposition below.

Example 3.1.3 Are 1, 1.9, 2, 2.1 interior points of (0, 2)?

Example 3.1.4 Same question for [0, 2].

Example 3.1.5 Conjecture what Int ((0, 2)) and Int ([0, 2]) might be.

Proposition 3.1.6 The following results follow immediately from the defini- tion.

1. If U is a neighborhood of x and U V , then V is also a neighborhood of x. ⊆ 2. Int (S) S. ⊆ Proof. We prove each part separately. Though these proofs are easy and could have been left to the reader, we will do them to illustrate how one works with these new concepts.

Part 1: If U is a neighborhood of x then there exists δ > 0 such that (x δ, x + δ) U. Since U V , we also have (x δ, x + δ) V . Thus,− V is also⊆ a neighborhood⊆ of x. − ⊆ Part 2: If x Int (S) then there exists δ > 0 such that (x δ, x + δ) S and since∈ x (x δ, x + δ), it follows that x S. − ⊆ ∈ − ∈

Example 3.1.7 Every point of (a, b) with a < b is an interior point of (a, b). Also, any point outside of (a, b) is not an interior point. Therefore, Int ((a, b)) = min x a , x b (a, b). This is easy to see. If x (a, b), we let δ = {| − | | − |}. Then, ∈ 2 (x δ, x + δ) (a, b). − ⊆ Example 3.1.8 a and b are not interior points of [a, b] with a b. Every other point of that interval is an interior point. Therefore, Int ([a, b])≤ = (a, b). 3.1. INTERIOR POINTS AND NEIGHBORHOODS 91

Example 3.1.9 Since every interval of positive length contains both a ratio- nal and an irrational, it follows that Q as well as R Q have no interior \ points. If x were an interior point of Q then there would exist δ > 0 such that (x δ, x + δ) Q. But we know this can’t happen since there is at least one irrational− number⊆ between x δ and x + δ. The same argument works for − R Q. Therefore, Int (Q) = ∅ and Int (R Q) = ∅. \ \ Example 3.1.10 Int (R) = R. To see this, we need to prove that every real number is an interior point of R that is we need to show that for every x R, ∈ there is δ > 0 such that (x δ, x + δ) R. Let x R. Then, (x 1, x + 1) R − ⊆ ∈ − ⊆ thus x is an interior point of R.

3.1.2 Properties

Theorem 3.1.11 Let x R, let Ui denote a family of neighborhoods of x. ∈ n

1. Ui is a neighborhood of x. i=1 \ ∞ 2. Ui is not always a neighborhood of x. i=1 \ ∞ 3. Ui is a neighborhood of x. i=1 [ Proof. We prove each part separately.

1. We need to prove that there exists δ > 0 such that (x δ, x + δ) n − ⊆ Ui. Since each Ui is a neighborhood of x, for each i there ex- i=1 \ ists δi such that (x δi, x + δi) Ui. Let δ = min δ1, δ2, ..., δn . − ⊆ { } Then, (x δ, x + δ) Ui for each i = 1, 2..., n. It follows that − n ⊆ n (x δ, x + δ) Ui thus proving that Ui is a neighborhood of − ⊆ i=1 i=1 x. \ \ 1 1 2. It is enough to find a counterexample. Let U = − , . Clearly, i i i   ∞ for each i, Ui is a neighborhood of 0. Yet, Ui = 0 which is { } i=1 \ ∞ not a neighborhood of 0. It remains to prove that Ui = 0 (see { } i=1 problems). \ 3. See problems. 92 CHAPTER 3. TOPOLOGY OF THE REAL LINE

3.1.3 Techniques to Remember To prove x is an interior point of S, one has to find a δ such that • (x δ, x + δ) S. − ⊆ Note the above also proves S is a neighborhood of x. • To prove x is not an interior point of S, one needs to show that no δ works • in other words δ > 0, (x δ, x + δ) contains points not in S. ∀ − 3.1.4 Important Facts to Know and Remember Some of the facts listed below were discussed in the notes, others are addressed in the exercises.

1. Definitions and theorems in this section. 2. Let S = a . What is Int (S)? { } 3. Let S be a finite set. What is Int (S)? 4. Let S be an non-empty bounded set. Is sup S Int (S)? ∈ 5. Let a, b and c be three real numbers such that a < b < c. Let S = (a, b) c . What is Int (S)? Same question for S = [a, b] c . ∪ { } ∪ { } 6. What is the interior of known sets such as ∅, N, Z, Q, R, (a, b), [a, b]? 7. A B = Int (A) Int (B). ⊆ ⇒ ⊆ 8. If A and B are two sets, Int (A B) = Int (A) Int (B), but in general Int (A B) = Int (A) Int (B).∩ The most we can∩ say is that Int (A) Int (B)∪ Int6 (A B).∪ ∪ ⊆ ∪ 9. S is open Int (S) = S. ⇐⇒ 3.1.5 Exercises 1. Using the definition of a neighborhood, complete the sentence: "A set S fails to be a neighborhood of x if ...".

∞ 1 1 2. Prove Ui = 0 where Ui = − , . { } i i i=1 \   3. Prove part 3 of theorem 3.1.11. 4. Suppose that x is an interior point of U and U V . Prove that x is also an interior point of V . ⊆

5. Let x R and S R. Prove that the two conditions below are equivalent: ∈ ⊆ (a) S is a neighborhood of x. 3.1. INTERIOR POINTS AND NEIGHBORHOODS 93

(b) There exists two numbers a and b such that x (a, b) S. ∈ ⊆ 6. Let x R and y R. Assume further that x = y. Prove that one can find ∈ ∈ 6 a neighborhood S of x and a neighborhood T of y such that S T = ∅. ∩ 7. Given that S R and x is an upper bound of S, explain why S cannot be a neighborhood⊆ of x.

8. Given that S R, S = ∅, and S is bounded above, explain why neither ⊆ 6 S nor R S can be a neighborhood of sup S. \ 9. Suppose that A and B are subsets of R and that x is an interior point of A B. Is it true that x must be an interior point of A or B? Explain and justify∪ your answer.

10. Suppose that A and B are subsets of R and that x is an interior point of both A and B. Is it true that x must be an interior point of A B? Explain and justify your answer. ∩

11. Let A and B be subsets of R. (a) Prove that if A B then Int (A) Int (B). ⊆ ⊆ (b) Prove that Int (A B) = Int (A) Int (B). ∩ ∩ (c) Is Int (A B) = Int (A) Int (B)? If not, can we prove one of the inclusions?∪ Which one? If∪ you answered yes, then prove the inclusion you think can be proven.

12. Let a R. What is Int ( a )? ∈ { } 13. Let S be a finite set. What is Int (S)?

14. What are Int (∅), Int (N), Int (Z)? 15. Let a, b and c be three real numbers such that a < b < c. Let S = (a, b) c . What is Int (S)? ∪ { } 16. Let S be an non-empty bounded subset of R. Is sup S Int (S)? Explain and justify your answer. ∈

17. Let S R. Prove that Int (Int (S)) = Int (S). ⊆ 94 CHAPTER 3. TOPOLOGY OF THE REAL LINE

3.1.6 Hints for the Exercises For all the problems, remember to first clearly state what you have to prove or do.

1. Using the definition of a neighborhood, complete the sentence: "A set S fails to be a neighborhood of x if ...". Hint: This is just the negation of the definition of a neighborhood.

∞ 1 1 2. Prove Ui = 0 where Ui = − , . { } i i i=1   Hint: First,\ show 0 is an element of this set. Then, show no other real number can be in the set. 3. Prove part 3 of theorem 3.1.11. Hint: straightforward, just use definitions. 4. Suppose that x is an interior point of U and U V . Prove that x is also an interior point of V . ⊆ Hint: straightforward, just use definitions.

5. Let x R and S R. Prove that the two conditions below are equivalent: Hint:∈ Remember⊆ to prove both directions.

(a) S is a neighborhood of x. (b) There exists two numbers a and b such that x (a, b) S. ∈ ⊆ 6. Let x R and y R. Assume further that x = y. Prove that one can find ∈ ∈ 6 a neighborhood S of x and a neighborhood T of y such that S T = ∅. Hint: Make your own examples to see what is going on here. ∩

7. Given that S R and x is an upper bound of S, explain why S cannot be a neighborhood⊆ of x. Hint: See what would happen if S were a neighborhood of x.

8. Given that S R, S = ∅, and S is bounded above, explain why neither ⊆ 6 S nor R S can be a neighborhood of sup S. Hint: similar\ to the previous problem.

9. Suppose that A and B are subsets of R and that x is an interior point of A B. Is it true that x must be an interior point of A or B? Explain and justify∪ your answer. Hint: Make your own examples. What if A and B are intervals and x is an endpoint?

10. Suppose that A and B are subsets of R and that x is an interior point of both A and B. Is it true that x must be an interior point of A B? Explain and justify your answer. ∩ Hint: First, look at some examples to understand the problem better. 3.1. INTERIOR POINTS AND NEIGHBORHOODS 95

After you’ve tried some examples, if you think it is true, try to prove the result.

11. Let A and B be subsets of R. (a) Prove that if A B then Int (A) Int (B). Hint: straightforward,⊆ just use definitions.⊆ (b) Prove that Int (A B) = Int (A) Int (B). Hint: straightforward,∩ just use definitions∩ and don’t forget to prove the double inclusion. (c) Is Int (A B) = Int (A) Int (B)? If not, can we prove one of the inclusions?∪ Which one? If∪ you answered yes, then prove the inclusion you think can be proven. Hint: See problem 9.

12. Let a R. What is Int ( a )? Hint:∈ use the definition of{ interior.} Don’tforget to justify whatever answer you find.

13. Let S be a finite set. What is Int (S)? Hint: use the definition of interior. Don’tforget to justify whatever answer you find.

14. What are Int (∅), Int (N), Int (Z)? Hint: use the definition of interior. Don’tforget to justify whatever answer you find. 15. Let a, b and c be three real numbers such that a < b < c. Let S = (a, b) c . What is Int (S)? Hint:∪ use { } the definition of interior. Don’tforget to justify whatever answer you find.

16. Let S be an non-empty bounded subset of R. Is sup S Int (S)? Explain and justify your answer. ∈ Hint: see problem 8.

17. Let S R. Prove that Int (Int (S)) = Int (S). Hint: one⊆ direction is tricky. Write clearly what you have to prove first. 96 CHAPTER 3. TOPOLOGY OF THE REAL LINE

3.2 Open and Closed Sets 3.2.1 Main Definitions Here, we are trying to capture the notion which explains the difference between (a, b) and [a, b] and generalize the notion of closed and open intervals to any sets. The notions of open and closed sets are related. One is defined precisely, the other one is defined in terms of the first one. In these notes, we define precisely what being open means. We then define closed in terms of being open. Some texts do the reverse. They define being closed precisely, then being open in terms of being closed.

Definition 3.2.1 (Open and Closed Sets) Let S R. ⊆ 1. S is said to be open if every point of S is an interior point of S.

2. S is said to be closed if and only if R S is open. \ Example 3.2.2 In the previous section, we saw that Int ((a, b)) = (a, b). Thus (a, b) is open according to our definition. It is why we call it an open interval.

Proposition 3.2.3 The following should be obvious from the definition:

1. S is open if for any x S, there exists δ > 0 such that (x δ, x + δ) S. ∈ − ⊆ 2. S is open if for any x S, there exists a < b such that x (a, b) S. ∈ ∈ ⊆ 3. S is open if for any x S, there exists a neighborhood of x included in S. ∈ 4. Int (S) is open.

Proof. We only prove part 4 and leave the rest as exercises.

Part 4: We want to show that every point of Int (S) is an interior point of Int (S) that is if x Int (S) then there exists δ > 0 such that (x δ, x + δ) Int (S). Since x ∈Int (S), there exists δ > 0 such that (x δ, x−+ δ) S. ⊆ To show (x δ,∈ x + δ) Int (S) we need to show that− every point⊆ of (x δ, x + δ)−is an interior⊆ point of S. Let y (x δ, x + δ). Let  = min− ( y x + δ , x + δ y ) ∈ − | − | | − | then (y , y + ) (x δ, x + δ) S. So, 2 − ⊆ − ⊆ y is an interior point of S. Since y was arbitrary, (x δ, x + δ) Int (S). − ⊆

Remark 3.2.4 It may appear that a set is either open or closed. Nothing could be further from the truth. In fact, the majority of subsets of R are neither open nor closed. 3.2. OPEN AND CLOSED SETS 97

Remark 3.2.5 It should be clear to the reader that S is open if and only if R S is closed. Clearly, if R S is closed, then by definition R (R S) is open. \ \ \ \ But R (R S) = S. Conversely, if S is open, then R S must be closed because \ \ \ if it were not then R (R S) = S would not be open by definition. This would be a contradiction. \ \

Having defined open and closed sets, it is natural to ask what open and closed sets look like. It is also natural to ask what kind of sets are open, closed. We will try to answer these questions in this section. We begin with some examples.

Example 3.2.6 R is open. We noticed above that Int (R) = R.

Example 3.2.7 ∅ is open. For a set not to be open, at least one of its points must fail to be an interior point. Therefore, the set must not be empty.

Example 3.2.8 R is closed. This is true since R R = ∅ is open. \ Example 3.2.9 ∅ is closed since R ∅ = R is open. \ Example 3.2.10 Let a and b be any two real numbers such that a < b. Then, (a, b) is open. This says that every open interval is open. We know this from a previous example in which we noticed that every point of (a, b) was an interior point of (a, b). In the same way, intervals of the form ( , a) or (a, ) are also open. −∞ ∞

Example 3.2.11 Let a and b be any two real numbers such that a < b. Con- sider [a, b). This set is not open because a is not an interior point. Its comple- ment is R [a, b) = ( , a) [b, ). This set is not open either as b is not an interior point..\ Thus−∞[a, b) is∪ not∞ closed either. In conclusion [a, b) is neither open nor closed. The same is true for (a, b].

Example 3.2.12 [a, b] is closed. Its complement is R [a, b] = ( , a) (a, ). This set is open. It can be proven directly. It is also\ a consequence−∞ of∪ theorem∞ 3.2.15.

Example 3.2.13 Q is neither open nor closed. Earlier, we saw that none of the points in Q were interior points thus Q is not open. We also saw that its complement R Q shared the same property, so it is not open. Hence, Q is not closed. \

Example 3.2.14 R Q is neither open nor closed for the same reason. \ In the examples, we noted that R and ∅ were both closed and open. You may wonder how many sets share this property. It can be proven that these are the only sets with this property. 98 CHAPTER 3. TOPOLOGY OF THE REAL LINE

3.2.2 Properties Theorem 3.2.15 (Unions and intersections of open and closed sets) Let Ui denote an infinite family of open sets and Hi denote an infinite family {of closed} sets. { }

∞ 1. Ui is open. i=1 [ n

2. Ui is open. i=1 \ n 3. Hi is closed. i=1 [ ∞ 4. Hi is closed. i=1 \ Proof. We prove some of the parts to illustrate how to work with these concepts. The other parts are left as exercises.

∞ ∞ 1. We need to prove that if x Ui then x is an interior point of Ui ∈ i=1 i=1 [ [ ∞ ∞ that is there exists δ > 0 such that (x δ, x + δ) Ui. If x Ui − ⊆ ∈ i=1 i=1 then x Ui for some i. Since Ui is open, x is an interior[ point of [Ui so ∈ there exists δi > 0 such that (x δi, x + δi) Ui. Then, (x δi, x + δi) − ⊆ − ⊆ ∞ Ui. i=1 [ 2. See problems. As you do the proof, make sure you understand why the ∞ proof you do would not work for Ui i=1 \ n n 3. To prove Hi is closed, we prove R Hi is open. By De Morgan’s \ i=1 i=1 [n n [ laws, R Hi = (R Hi). Since Hi is closed for every i, R Hi is \ \ \ i=1 i=1 [ n \ open. Thus, (R Hi) is open by part 2 of this theorem. \ i=1 \ 4. See problems. 3.2. OPEN AND CLOSED SETS 99

You will recall that the completeness axiom stated that every non-empty subset of R which was bounded above had a supremum in R. If we add the restriction that the set also be closed, we get a much stronger result as the theorem below shows.

Theorem 3.2.16 (Existence of a largest member) Every non-empty closed set bounded above must have a largest member. Similarly, every non-empty closed set bounded below must have a smallest member. Proof. We prove the largest member part. Let H be a non-empty closed set bounded above. To prove that H has a largest member, we prove that sup H H. From the hypothesis, we know that sup H exists (why?). Let α = sup H.∈ We do a proof by contradiction. Suppose α / H. Then, α R H which is open ∈ ∈ \ since H is closed. Thus, we can choose δ > 0 such that (α δ, α + δ) R H. Since no element of H can be greater than α and (α δ, α−+ δ) / H it⊆ follows\ that α δ is an upper bound of H. This is a contradiction− since ∈α δ < α and α = sup−H. −

3.2.3 Closure of a Set

Definition 3.2.17 (Closure) Let S R and x R. ⊆ ∈ 1. x is said to be close to S if δ > 0, (x δ, x + δ) S = ∅. ∀ − ∩ 6 2. The closure of S, denoted S is defined by

S = x R : x is close to S { ∈ } Remark 3.2.18 Obviously, condition 1 of the definition is satisfied for every point of a set. Thus implying that every point of a set is close to the set and therefore S S ⊆ for every set S.

Example 3.2.19 The closure of (a, b) is [a, b] in other words (a, b) = [a, b]. From the remark above, (a, b) (a, b). Also, δ > 0, (a δ, a + δ) (a, b) = ∅. Thus, a (a, b). Similarly,⊆ one can show∀ that b (−a, b). This∩ shows6 that [a, b] (a,∈ b). Can (a, b) contain anything else? We prove∈ that if x > b or x < a then ⊆x / (a, b) thus proving that (a, b) = [a, b]. We prove it in the case x > b. ∈ x b The case x < a is similar. Let δ = | − |. Then (x δ, x + δ) (a, b) = ∅. 2 − ∩ Hence, x is not close to (a, b), so x is not in the closure of (a, b).

Example 3.2.20 Q = R. Let x R. For any δ > 0, (x δ, x + δ) must ∈ − contain some rational number thus (x δ, x + δ) Q = ∅. − ∩ 6 Example 3.2.21 R Q = R. The same argument as above works here. \ Example 3.2.22 ∅ = ∅. 100 CHAPTER 3. TOPOLOGY OF THE REAL LINE

Theorem 3.2.23 (Important facts about closure) Let A, B and S be sets of real numbers.

1. S S ⊆ 2. A B = A B ⊆ ⇒ ⊆ 3. S is always closed.

Proof. We prove each part separately.

1. If x S then clearly for any δ > 0, (x δ, x + δ) S contains at least x and∈ is therefore not empty. So, x S.− ∩ { } ∈ 2. We must prove that assuming A B, if x A¯ then x B¯ that is x is ⊆ ∈ ∈ close to B in other words, δ > 0, (x δ, x + δ) B = ∅. Let x A¯. Then, (x δ, x + δ) B contains∀ (x −δ, x + δ) ∩A since6 A B.∈ But − ∩ − ∩ ⊆ since x A¯, (x δ, x + δ) A = ∅. Therefore, (x δ, x + δ) B = ∅. ∈ − ∩ 6 − ∩ 6 3. We prove that S is closed by proving that R S is open. Before we proceed with the proof, it is important to understand\ what these sets represent. S is the set of elements close to S, thus R S is the set of elements not \ close to S. We show R S is open by showing that every element of \ R S is an interior point of R S. In other words, if x R S, then \ \ ∈ \ δ > 0 : (x δ, x + δ) R S. If x R S then x is not close to S, ∃ − ⊆ \ ∈ \ therefore we can pick δ > 0 such that (x δ, x + δ) S = ∅. We wish to − ∩ show that (x δ, x + δ) R S. If y (x δ, x + δ) then y cannot be − ⊆ \ ∈ − close to S (why?) thus y R S. Since y was an arbitrary element of ∈ \ (x δ, x + δ), it follows that (x δ, x + δ) R S. − − ⊆ \

Theorem 3.2.24 Let S R. The following are equivalent: ⊆ 1. S is closed. 2. S = S

Proof. See problems.

3.2.4 Techniques to Remember To prove that a set is open, one can use one of the following: • — Use the definition, that is prove that every point in the set is an interior point. — Prove that its complement is closed. — Prove that it can be written as the intersection of a finite family of open sets or as the union of a family of open sets. 3.2. OPEN AND CLOSED SETS 101

To prove that a set is not open, one can use one of the following: • — Find one of its point which is not an interior point. — Prove that its complement is not closed.

To prove that a set is closed, one can use one of the following: • — Prove that its complement is open. — Prove that it can be written as the union of a finite family of closed sets or as the intersection of a family of closed sets. — Prove that it is equal to its closure.

To prove that a set is not closed, one can use one of the following: • — Prove that its complement is not open. — Prove that it is not equal to its closure.

3.2.5 Important Facts to Know and Remember 1. Definitions and theorems in this section.

2. Being closed is not the opposite of being open. In other words, a set is not either open or closed. It can be neither.

3. Only R and ∅ are both open and closed. 4. Any set containing a finite number of elements is closed.

5. A B = A¯ B¯ ∪ ∪ 6. A B A¯ B¯ ∩ ⊆ ∩ 7. Let S be a non-empty bounded set. Is sup S S? ∈ 8. Given a set S, Int (S) is open, S is closed and Int (S) S S. ⊆ ⊆ 3.2.6 Exercises 1. Prove the statements in proposition 3.2.3.

2. Prove that S is open Int (S) = S. ⇐⇒ 3. Prove that ( , a] and [a, ) are closed subsets of R. −∞ ∞ 4. Finish proving theorem 3.2.15.

5. Prove the second half of theorem 3.2.16.

6. Prove theorem 3.2.24. 102 CHAPTER 3. TOPOLOGY OF THE REAL LINE

7. Let U be an open set and H be a closed set.

(a) Prove that U H is open. \ (b) Prove that H U is closed. \ 8. Give an example of an infinite family of open sets whose intersection fails to be open. 9. Give an example of an infinite family of closed sets whose union fails to be closed. 10. Give an example of two sets A and B such that neither set is open but their union is.

11. Let x R. ∈ (a) Prove that x is closed. { } (b) Using part a, prove that every finite set must be closed.

12. Let A and B be non-empty open subsets of R such that A B = ∅. Can A B be finite? Explain. ∩ 6 ∩ 13. Let S = [0, 1) (1, 2). ∪ (a) Let U = Int (S). What is U? (b) What is U? (c) Give an example of a set S of real numbers such that if U = Int (S), then U = S? 6 1 14. Let S = : n Z+ . Evaluate S. n ∈   15. Let A and B be subsets of R. (a) If A B and A is open, prove that A Int (B). ⊆ ⊆ (b) If A B and B is closed, prove that A B. ⊆ ⊆ 16. Let A and B be subsets of R. (a) Prove that A B = A¯ B¯. ∪ ∪ (b) Prove that A B A¯ B¯. ∩ ⊆ ∩ (c) Give an example which shows why we do not have equality in part b.

17. Prove that if S R then R S = Int (R S). ⊆ \ \ 18. Make a conjecture regarding whether sup S S and prove your conjecture. ∈ 3.2. OPEN AND CLOSED SETS 103

3.2.7 Hints for the Exercises For all the problems, remember to first clearly state what you have to prove or do.

1. Prove the statements in proposition 3.2.3. Hint: Simply use the definitions.

2. Prove that S is open Int (S) = S. Hint: Don’t forget to⇐⇒ prove both directions. Each direction is straight- forward, use definitions and results from this section and the previous one.

3. Prove that ( , a] and [a, ) are closed subsets of R. Hint: Prove−∞ their complement∞ is open.

4. Finish proving theorem 3.2.15. Hint: Straightforward, use definitions.

5. Prove the second half of theorem 3.2.16. Hint: The proof of the second half is similar to that of the first half which is in your notes.

6. Prove theorem 3.2.24. Hint: Don’tforget to prove both directions. In one direction, you have to prove S = S. Since we already know that S S. it is enough to prove that S S. For this, try a proof by contradiction.⊆ The other direction is easy and⊆ quick.

7. Let U be an open set and H be a closed set.

(a) Prove that U H is open. Hint: Look at\ examples and draw pictures to really understand what is happening. Then, try to write U H as an intersection of two open sets. \ (b) Prove that H U is closed. Hint: Look at\ examples and draw pictures to really understand what is happening. Then, try to write H U as an intersection of two open sets. \

8. Give an example of an infinite family of open sets whose intersection fails to be open. Hint: The homework on sequences of sets has such examples.

9. Give an example of an infinite family of closed sets whose union fails to be closed. Hint: The homework on sequences of sets has such examples. 104 CHAPTER 3. TOPOLOGY OF THE REAL LINE

10. Give an example of two sets A and B such that neither set is open but their union is. Hint: Think of intervals.

11. Let x R. ∈ (a) Prove that x is closed. Hint: What{ is} its complement? (b) Using part a, prove that every finite set must be closed. Hint: Use part a!

12. Let A and B be non-empty open subsets of R such that A B = ∅. Can A B be finite? Explain. ∩ 6 Hint:∩ Use the previous question.

13. Let S = [0, 1) (1, 2). ∪ (a) Let U = Int (S). What is U? Hint: Make sure you justify your answer using definitions and results from this section and the previous one. (b) What is U? Hint: Make sure you justify your answer using definitions and results from this section and the previous one. (c) Give an example of a set S of real numbers such that if U = Int (S), then U = S? Hint: Not6 the integers and not the reals!

1 14. Let S = : n Z+ . Evaluate S. n ∈   Hint: Remember that S S. So you only need to think about the points you might be adding to S⊆by taking the closure.

15. Let A and B be subsets of R.

(a) If A B and A is open, prove that A Int (B). Hint:⊆ Combination of two results, one⊆ about the interior of two sets, the other one about open sets and their interior. (b) If A B and B is closed, prove that A B. Hint:⊆ Combination of two results, one about⊆ the closure of two sets, the other one about closed sets and their closure.

16. Let A and B be subsets of R.

(a) Prove that A B = A¯ B¯. Hint: You can∪ either do∪ a direct proof or use results about closure of a set and results about the union of sets. 3.2. OPEN AND CLOSED SETS 105

(b) Prove that A B A¯ B¯. Hint: You can∩ either⊆ do∩ a direct proof or use results about closure of a set and results about the intersection of sets. (c) Give an example which shows why we do not have equality in part b. Hint: Think of two open intervals.

17. Prove that if S R then R S = Int (R S). Hint: Don’t forget⊆ to prove\ inclusion both\ ways. For each inclusion, first write precisely what you need to show, then use the definitions of this section to prove it. 18. Make a conjecture regarding whether sup S S and prove your conjecture. Hint: To prove your conjecture, you will use∈ a "famous" theorem studied in class that we have used many times. It is the theorem which tells us when an upper bound of a set is the supremum of a set. 106 CHAPTER 3. TOPOLOGY OF THE REAL LINE

3.3 Limit Points 3.3.1 Main Definitions Intuitively speaking, a limit point of a set S in a space X is a point of X which can be approximated by points of S other than x as well as one pleases. The notion of limit point is an extension of the notion of being "close" to a set in the sense that it tries to distinguish between being close to a set but isolated and being close to a set and surrounded by many points of the set. For example, c is close to the set S = (a, b) c but if a < b < c then there are ∪ { } a + b no other points of S near c. On the other hand, is close to (a, b) and 2 surrounded by infinitely many points of (a, b). We will see that to be a limit point of a set, a point must be surrounded by an infinite number of points of the set. We now give a precise mathematical definition. In what follows, R is the reference space, that is all the sets are subsets of R.

Definition 3.3.1 (Limit point) Let S R, and let x R. ⊆ ∈ 1. x is a limit point or an accumulation point or a cluster point of S if δ > 0, (x δ, x + δ) S x = ∅. ∀ − ∩ \{ } 6 2. The set of limit points of a set S is denoted L (S)

Remark 3.3.2 Let us remark the following:

1. In the above definition, we can replace (x δ, x + δ) by a neighborhood of x. Therefore, x is a limit point of S if any− neighborhood of x contains points of S other than x.

2. Being a limit point of a set S is a stronger condition than being close to a set S. It requires any neighborhood of the limit point x to contain points of S other than x.

3. The above two remarks should make it clear that L (S) S. We can also ⊆ prove it rigorously. If x L (S) then δ > 0 : (x δ, x + δ) S x = ∅. Let y (x δ, x + δ) S∈ x . Then,∃ y (x δ,− x + δ) and∩ y\{S} 6 x . ∈ − ∩ \{ } ∈ − ∈ \{ } So, y S thus (x δ, x + δ) S = ∅. Hence x S. ∈ − ∩ 6 ∈ 4. However, S is not always a subset of L (S).

Let us first look at easy examples to understand what a limit point is and what the set of limit points of a given set might look like.

Example 3.3.3 Let S = (a, b) and x (a, b). Then x is a limit point of (a, b). Let δ > 0 and consider (x δ, x + δ).∈ This interval will contain points of (a, b) other than x, infinitely many− points in fact. 3.3. LIMIT POINTS 107

Example 3.3.4 a and b are limit points of (a, b). Given δ > 0, the interval (a δ, a + δ) contains infinitely many points of (a, b) a thus showing a is a limit− point of (a, b). The same is true for b. \{ }

Example 3.3.5 Let S = [a, b] and x [a, b]. Then x is a limit point of [a, b]. This is true for the same reasons as above.∈

At this point you may think that there is no difference between a limit point and a point close to a set. Consider the next example.

Example 3.3.6 Let S = (0, 1) 2 . 2 is close to S. For any δ > 0, 2 ∪ { } { } ⊆ (2 δ, 2 + δ) S so that (2 δ, 2 + δ) S = ∅. But 2 is not a limit point of S. − ∩ − ∩ 6 (2 .1, 2 + .1) S 2 = ∅. − ∩ \{ } Remark 3.3.7 You can think of a limit point as a point close to a set but also surrounded by many (infinitely many) points of the set.

We look at more examples.

Example 3.3.8 L ((a, b)) = [a, b]. From the first two examples of this section, we know that L ((a, b)) [a, b]. What is left to show is that if x / [a, b] then x ⊆ ∈ is not a limit point of (a, b). Let x R such that x / (a, b). Assume x > b (the case a < a is similar and left for the∈ reader to check).∈ Let δ < x b . Then | − | (x δ, x + δ) (a, b) x = ∅. − ∩ \{ } Example 3.3.9 L (Z) = ∅. Let x R. For x to be a limit point, we must have that for any δ > 0 the interval (∈x δ, x + δ) would have to contain points − of Z other than x. Since for every real number x there exists an integer n such that n 1 x < n, if δ < min ( x n + 1 , x n ), the interval (x δ, x + δ) − ≤ | − | | − | − contains no integer therefore (x δ, x + δ) Z x = ∅. So, no real number − ∩ \{ } can be a limit point of Z.

Example 3.3.10 L (Q) = R. For any real number and for any δ > 0, the interval (x δ, x + δ) contains infinitely many points of Q other than x. Thus, − (x δ, x + δ) Q x = ∅. So every real number is a limit point of Q. − ∩ \{ } 6 Example 3.3.11 If S = (0, 1) 2 then S = [0, 1] 2 but L (S) = [0, 1]. Using arguments similar to the∪ arguments { } used in the∪ previous { } examples, one can prove that every element in [a, b] is a limit point of S and every element outside of [a, b] is not. You will notice that 2 S but 2 / L (S). Also, 0 and 1 are in L (S) but not in S. So, in the most general∈ case, one∈ cannot say anything regarding whether S L (S) or L (S) S. ⊆ ⊆ 1 Example 3.3.12 If S = : n Z+ then L (S) = 0 (see problems). n ∈ { }   108 CHAPTER 3. TOPOLOGY OF THE REAL LINE

3.3.2 Properties

Theorem 3.3.13 Let x R and S R. ∈ ⊆ 1. If x has a neighborhood which only contains finitely many members of S then x cannot be a limit point of S. 2. If x is a limit point of S then any neighborhood of x contains infinitely many members of S. Proof. We prove each part separately. Part 2: This part follow immediately from part 1. Part 1: Let U be a neighborhood of x which contains only a finite number of points of S that is U S is finite. Then, U S x is also finite. ∩ ∩ \{ } Suppose U S x = y1, y2, ..., yn . We show there exists δ > 0 such that (x∩ δ,\{ x +}δ) {U does not} contain any member of S x . Since both (x− δ, x + δ)∩and U are neighborhood of x, so is their\{ inter-} section. This− will prove there is a neighborhood of x containing no el- ement of S x hence proving x is not a limit point of S. Let δ = \{ } min x y1 , x y2 , ..., x yn . Since x is not equal to yi, δ > 0. Then{|(x− δ,| x|+−δ) |U contains| − no|} points of S other than x thus proving our claim.− ∩

Corollary 3.3.14 No finite set can have a limit point. Proof. Follows immediately from the theorem. The next theorem is very important. It helps understand the relationship between the set of limit points of a set and the closure of a set. Theorem 3.3.15 Let S R ⊆ S = S L (S) ∪ Proof. We show inclusion both ways. 1. S L (S) S We∪ already⊆ know that S S. We now show that L (S) S. This will imply the result. Suppose⊆ that x L (S). Then δ > 0, ⊆(x δ, x + δ) ∈ ∀ − ∩ S x = ∅. It follows that (x δ, x + δ) S = ∅, since S x S. Therefore,\{ } 6 x is in the closure of −S. ∩ 6 \{ } ⊆ 2. S S L (S) Let⊆x ∪S. We need to prove that x S L (S). Either x S or x / S. If x ∈S then x S L (S). If x /∈S then∪ x is close to ∈S. Therefore,∈ ∈ ∈ ∪ ∈ δ > 0, (x δ, x + δ) S = ∅. Since x / S, S = S x , therefore ∀ − ∩ 6 ∈ \{ } (x δ, x + δ) S x = ∅. It follows that x L (S) and therefore x −S L (S).∩ So\{ we see} 6 that in all cases, if we assume∈ that x S then we∈ must∪ have x S L (S). ∈ ∈ ∪ 3.3. LIMIT POINTS 109

3.3.3 Important Facts to Remember Definitions properties and theorems in this section. • L (A) L (B) = L (A B) (see problems) • ∪ ∪ S = S L (S) • ∪ L (S) S • ⊆ L (S) is closed (see problems). • If U is open then L (U) = U¯ (see problems). • 3.3.4 Exercises 1 1. Prove that if S = : n Z+ then L (S) = 0 n ∈ { }   2. In each situation below, give an example of a set which satisfies the given condition.

(a) An infinite set with no limit point. (b) A bounded set with no limit point. (c) An unbounded set with no limit point. (d) An unbounded set with exactly one limit point. (e) An unbounded set with exactly two limit points.

3. Prove that if A and B are subsets of R and A B then L (A) L (B). ⊆ ⊆ 4. Prove that if A and B are subsets of R then L (A) L (B) = L (A B). ∪ ∪ 5. Is it true that if A and B are subsets of R then L (A) L (B) = L (A B)? Give an answer in the following cases: ∩ ∩

(a) A and B are closed. (b) A and B are open. (c) A and B are intervals. (d) General case.

6. Given that S R, S = ∅ and S bounded above but max S does not exist, prove that sup⊆S must6 be a limit point of S. State and prove a similar result for inf S.

7. Let S R. Prove that L (S) must be closed. ⊆ 8. Prove that if U is open then L (U) = U¯. 110 CHAPTER 3. TOPOLOGY OF THE REAL LINE

3.3.5 Hints for the Exercises For all the problems, remember to first clearly state what you have to prove or do. 1 1. Prove that if S = : n Z+ then L (S) = 0 n ∈ { } Hint: Show that 0is a limit point and that no number other than 0 can be a limit point. Consider the three cases: x = 0, x > 0 or x < 0. 2. In each situation below, give an example of a set which satisfies the given condition.

(a) An infinite set with no limit point. Hint: Think of know sets we often use. (b) A bounded set with no limit point. Hint: Review the notes to see what kind of sets do not have limit points. (c) An unbounded set with no limit point. Hint: Think of know sets we often use. (d) An unbounded set with exactly one limit point. Hint: Remembering that L (A B) = L (A) L (B) it is enough to find an unbounded set with no∪ limit point and∪ a set with exactly one limit point (why?) (e) An unbounded set with exactly two limit points. Hint: Similar to previous problem.

3. Prove that if A and B are subsets of R and A B then L (A) L (B). Hint: straightforward, use the definition. ⊆ ⊆

4. Prove that if A and B are subsets of R then L (A) L (B) = L (A B). Hint: straightforward, use the definitions and results∪ of this section.∪

5. Is it true that if A and B are subsets of R then L (A) L (B) = L (A B)? Give an answer in the following cases: ∩ ∩ Hint: think of intervals.

(a) A and B are closed. (b) A and B are open. (c) A and B are intervals. (d) General case.

6. Given that S R, S = ∅ and S bounded above but max S does not exist, prove that sup⊆S must6 be a limit point of S. State and prove a similar result for inf S. Hint: Use the definitions of this section as well as a famous theorem studied in class which gives conditions for an upper bound to be a supremum. 3.3. LIMIT POINTS 111

7. Let S R. Prove that L (S) must be closed. ⊆ Hint: You can either prove that L (S) = L (S) or that R L (S) is open. Either way, you will use the definitions and theorems from\ this section. 8. Prove that if U is open then L (U) = U¯. Hint: we already know that L (U) U¯ since we learned in class that U¯ = U L (U). We need to show that⊆ U¯ L (U). ∪ ⊆ 3.5. TOPOLOGY OF THE REAL LINE: REVIEW QUESTIONS 113

3.5 Topology of the Real Line: Review Ques- tions

To help you understand the difference and the importance between the various notions studied in this chapter, you should know the answers to the questions below:

1. Let x be a real number and S R. Answer True or False. If the result is true, explain why. If it is false,⊆ give a counter example.

(a) If x is an interior point of S then x S. ∈ (b) If x is an interior point of S then x could be in S. (c) If x S then x must be an interior point of S. ∈ (d) If x S then x S. ∈ ∈ (e) If x S then x could be in S. ∈ (f) If x S then x S. ∈ ∈ (g) If x L (S) then x S. ∈ ∈ (h) If x L (S) then x could be in S. ∈ (i) If x S then x L (S). ∈ ∈ (j) If x L (S) then x S. ∈ ∈ (k) If x S then x L (S). ∈ ∈ 2. Find the following sets:

(a) The set of interior points of Z (b) Z¯ (c) L (Z).

3. Same question for Q.

4. Same question for R. 1 5. Same question for S = : n Z+ n ∈   1 6. Same question for S = 1 + : n Z+ n ∈   7. Same question for S = (0, 1) 2 ∪ { } 8. What can be said about intersections and unions of closed sets? of open sets?

9. What can be said about: 114 CHAPTER 3. TOPOLOGY OF THE REAL LINE

(a) the interior of the intersection of two sets? (b) the interior of the union of two sets? (c) the closure of the intersection of two sets? (d) the closure of the union of two sets? (e) the set of limit points of the intersection of two sets? (f) the set of limit points of the union of two sets?

10. Fill out the table below where a, b, c are real numbers such that a < b < c. You should be able to justify each answer. Set S Complete? Open? Closed? Int (S) S L (S) N Z Q R Q \ R ∅ (a, b) [a, b] (a, b) c [a, b] ∪ {c } a ∪ { } finite{ } set S 1 : n Z+ n ∈  1  c + : n Z+ n ∈   11. What do open sets look like? What do closed sets look like? Bibliography

[B] Bartle, G. Robert, The elements of Real Analysis, Second Edition, John Wiley & Sons, 1976.

[C1] Courant, R., Differential and Integral Calculus, Second Edition, Volume 1, Wiley, 1937. [C2] Courant, R., Differential and Integral Calculus, Second Edition, Volume 2, Wiley, 1937.

[DS] Dangello, Frank & Seyfried Michael, Introductory Real Analysis, Houghton Miffl in, 2000. [F] Fulks, Watson, Advanced Calculus, an Introduction to Analysis, Third Edi- tion, John Wiley & Sons, 1978.

[GN] Gaskill, F. Herbert & Narayanaswami P. P., Elements of Real Analysis, Prentice Hall, 1998. [LL] Lewin, J. & Lewin, M., An Introduction to Mathematical Analysis, Second Edition, McGraw-Hill, 1993. [MS] Stoll, Manfred, Introduction to Real Analysis, Second Edition, Addison- Wesley Higher Mathematics, 2001.

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