3.3 Limit Points 3.3.1 Main Deﬁnitions the Motivation Behind the Deﬁnitions of These Section Is the Following

3.3. LIMIT POINTS 109

3.3 Limit Points 3.3.1 Main Definitions The motivation behind the definitions of these section is the following. Consider the problem of finding the limit lim f (x). It implies that x can be made as close x c to c as one needs. However, if the→ domain of f is the set S = (a, b) c where a < b < c, then we cannot make x as close to c as we want since there∪ { } are no points close to c in the domain of f. Intuitively speaking, a limit point of a set S in a space X is a point of X which is both close to S and surrounded by infinitely many points of S. 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. 110 CHAPTER 3. TOPOLOGY OF THE REAL LINE

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. So, (x δ, x + δ) (a, b) x = ∅. − ∩ { } 6 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 x < a is similar and left for the∈ reader to check).∈ Let δ < x b , for x b | − | example δ = | − | Then (x δ, x + δ) (a, b) x = ∅. 2 − ∩ \{ }

Example 3.3.9 Similarly, L ([a, b]) = [a, b].

Example 3.3.10 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 con- − tain points of Z other than x. Let x R. If x is also an integer, say n, then 1 1 ∈ n , n + Z n = ∅ so n / L (Z) for any integer n. If x / Z then − 2 2 ∩ { } ∈ ∈ there exists an integer n such that n 1 < x < n (see homework in the section on the consequences of the completeness− axiom), 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 3.3.11 L (Q) = R. For any real number x 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 3.3. LIMIT POINTS 111

Example 3.3.12 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 3.3.13 If S = : n Z+ then L (S) = 0 (see problems). n ∈ { } 3.3.2 Properties We now understand that for x to be a limit point of a set S, every neighborhood of x has to contain points other than x. The question is how many points. The next result answers this question.

Theorem 3.3.14 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 element 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 3.3.15 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. 112 CHAPTER 3. TOPOLOGY OF THE REAL LINE

Theorem 3.3.16 Let S R ⊆ S = S L (S) ∪ Proof. We show inclusion both ways.

1. S L (S) S From∪ set theory,⊆ we know that if A C and B C then A B C. We already know that S S. We also⊆ know that⊆L (S) S. Therefore,∪ ⊆ S L (S) S. ⊆ ⊆ ∪ ⊆ 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). ∈ ∈ ∪

3.3.3 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. 3.3. LIMIT POINTS 113

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¯. 114 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 known 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 known 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 115

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). ∪ ⊆ 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.

501