Simplifying Section 13 By Joseph Pang

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1 1 Simplifying Section 13 By Joseph Pang Motivation: This is a study week, but actually it is a March Break taken place in February. Everyone always goes out, and have no time to do their BIG B43 Assignment 5, and at the end, people realize that the due date is coming soon and try to attempt the homework. However, they find that it is mission impossible to understand one single abstract concepts in Section 13. This handout is designed for this situation. If you are feeling panic rite now, relax, and read the handout before you proceed your work again. If you finish the homework during the study week, you can just take a look at the examples in the handout for preparing section 14. Let s Start! First and most important ideas we have to understand - neighborhood. First comes with formal definition and then simplified version: 1. Definition: Neighborhood of x, N (x;ε ) (Version 1) Let x and let ε > 0. A neighborhood of x is a set of the form N(x;ε ) = { y : x - y < ε }. The number ε is referred to as the radius of N(x;ε ). Basically, a neighborhood of x of radius ε is the open interval (x-ε, x +ε ) of length 2ε centered at x. Another way to look at it is that ε is the distance from the point x, so it also implies that ε > 0 and ε 0. The above saying leads to simplified version of neighborhood of x:

2 2 2. Definition: Neighborhood of x, N (x;ε ) Neighborhood of x is the interval ( x-ε, x +ε ) where ε > 0 Here is the picture about the definition: x-ε x x+ε NOTE: ε must not be equal to zero or it is just point x itself! The Neighborhood of x is usually talking about the shaded region above and this region MUST include x. But what is Deleted Neighborhood? 3. Definition: Deleted Neighborhood of x, N* (x;ε ) N* (x;ε ) = N (x;ε ) \ {x} In another word, a deleted neighborhood of x is the neighborhood that does not include x. Here is the picture of the definition: x x-ε x +ε Notice the shaded region does not cover the point x.

3 3 eady to understand the eal stuff! In this section 13, there are two basic definitions: Interior Point and Boundary Point. For Interior, first, I will offer a formal one and then an easy one. (This is your personal preference of which one you use.) Let S be an arbitrary set in the real line. 4. Definition: Interior point (Version 1) A point x is said to be interior to a subset A of if there exists an ε >0 such that (x-ε, x +ε ) A, that is, such that y x < ε ----> y A If x is interior to A, you can say that A is a neighborhood of x. Another way of saying it is that x is interior to A.The set of all interior points of A is called the interior of A. This results the simplify version of Interior point: 5. Definition: Interior point (Version 2) A point x A is called interior point of A if there exist a neighborhood of A completely contained in A. The set of all interior points of A is called the interior, denoted by int(a). Here is the picture: Suppose x 1 and x 2 both belongs to A and x 1 x 2 and ε ε 1 2

4 4 x 1 + ε 1 x x 1 + ε1 1 x 2 + ε 1 x x 2 + ε1 2 Note that the neighborhood is entirely lies in A. Example 1: egion of Set A Q has no interior points (i.e., it has empty interior), because every open interval contains an irrational number. Assuming a < b, the point a belongs to [a,b) but not to its interior, the interior of [a,b] is (a,b) (that will be explained later). After Interior, we have boundary point: 6. Definition: Boundary point A point b is called boundary point of A if every non-empty neighborhood of b intersects A and the complement of A. The set of all boundary points of S is called the boundary of A, denoted by bd(a). Here is the picture: Suppose x 1 and x 2 both belongs to A and x 1 x 2 and ε ε 1 2

5 5 egion of A c x ε x 1 x 1 + ε 1 ε x x ε1 x egion of Set A Notice that part of the region of the neighborhood intersects with set A and part of it intersects with A c. In short, that is N(x;ε ) A φ and also, N(x;ε ) A c φ Example 2: What is the boundary and the interior point of (0,4)? Prove each assertion. Here is the picture of the boundary points: Interval for (0,4) c ε ε 1 ε ε 1 egion for Set (0,4)

6 6 FIND THE BOUNDAY POINTS: The boundary of (0,4) is the set consisting of the two elements {0,4}. Every neighborhood of these two points contains points both from the interval (0,4) and from the complement of that interval. Therefore, both form the boundary. FIND THE INTEIO POINTS: Here is the picture for finding the interior points: x 4-x 0 x 4 Here I give an informal proof for the interior points. To find the interior point of (0,4), we have to apply the definition #1. Let S be the open interval of (0,4) and let x S. If ε = min {x, 4-x}, then we claim that N(x;ε ) S. Indeed for all y N(x;ε ) we have y - x < ε, so that -x - ε < y x < ε 4-x Thus, 0 < y < 5 and y belongs to S. It follows that every point in S is an interior point of S. So by definition of Interior point of version 2, we can conclude that S = int(s) You may ask why we have ε = min {x, 4-x}, the reason is that the we have smallest epsilon to ensure the neighborhood of x covers all the points in the set. If you look back Example 1, why the interior of [a,b] is (a,b)? Since a and b are boundary points of [a,b]! They are not part of the interior point of [a,b]. Now, we have to introduce another two main concepts: Open and Closed. In here I will give the definition of these two terms and follow by a corollary. 7. Definition: Open Set Let A. We say that A is an open set if, given any x A, we can always find an open interval (x-ε, x +ε ) which contains x, and which is a subset of A. In other words, x (x-ε, x +ε ) A From this, we can also say a set S is open if and only if A=int(A).

7 7 Well, you will also find there is a difference between our textbook on page 117 and in here. If you really want to use that, here is my interpretation of this definition. 8. Definition: Open Set (book + my interpretation) If bd(a) \ A, then A is said to be open. (book s definition) If bd(a) does not belong to A then A is said to be open. (my interpretation) Enough! How come there are so many different versions of Open set? EMEMBE, they are all equivalent, they just define it in different ways! The reason to offer so many versions is to give you more choice to understand one thing. You can choose whichever you feel comfortable with. Now, we have open and we should have closed too! First we start with book definition: 9. Definition: Closed (Version 1) A set A is said to be Closed it bd(a) A. (textbook s definition) What the heck does that mean? Let s visualize first (Using version 2 s definition): Suppose there is a subset S and a, b and a < b then, a Notice that a and b DOES belong to the set S. b Here is another definition of Closed that is useful for assignment purposes :

8 8 10. Definition: Closed (Version 2) A set A is called Closed if the complement of A, \A, is open. Here is the picture of the Closed (version 1) a b Now notice that a and b does not belong to the complement of A, so where does a and b belongs to? The only choice is a and b belongs to the Set A! This is what version 1 of Closed talk about! Therefore it tells you that both versions of definition are equivalent. You can find or feel there is obvious difference. For open, a and b must not belong to the set while for closed, a and b must belong to the set. Then your following question you will ask me, Do open and closed ha s relationship? The answer is the following.. ; -) 11. Theorem. For a subset A of i) A is open A c is closed. ii) A is closed A c is open. In terms of Solving assignment question, # 11 can do the following: 1. If you try to prove something open then show A c is closed. Here is the picture: egion of Set A egion of A c, show they are closed!

9 9 2. If you try to prove something closed then show A c is open. That is: egion of Set A egion of A c, show they are open! Let s look at two examples: Example 3 Let s look at two sets ( -3,3) and [4,7] To show (-3,3) is open, you have two choices, one is using definition #9 or you can use definition #7. In here, I will use definition #9 here. Notice that bd(-3,3) (-3,3), so we can conclude that (-3,3) is open. If you want to use theorem # 12 to show (-3,3) is open, then notice that 3, 3 (-3,3), and its complement is [-,-3] [3, ]. Hence (-3,3) is open. To show [4,7] is closed, you also have two choices, one is using definition # 10 and the other is using theorem # 12 (actually they are saying the same thing). Using definition # 10, first notice that 4,7 [4,7] and then its complement consists of the two open sets (-,4) and (7, ). So we can conclude that [4,7] is closed. The above definition and theorem on open and closed set is not good enough to help us to solve more sophisticated question, so we need to know some properties about them. 12. Open Set Properties i) The union of an arbitrary collection of open subsets in is open. ii) The intersection of any finite collection of open sets in is open.

10 10 Here is the Closed Set Properties 13. Closed Set Properties i) The intersection of an arbitrary collection of closed sets in is closed. ii) The union of any finite collection of closed sets in is closed. Example 4 For any set S. Let F denotes as the union of finite number of the closed sets contained in S. Show that F is a closed set. Proof: Given that F 1, F 2,., F n are closed in and let F:= F 1 F 2. F n. By the De Morgan identity the complement of F is given by F C C C = F F F C since each set F is open, it follows from (9) that i it F is closed. C n C F is open. Hence, by Theorem #12, C You will ask why F is open? Well, the answer is that from the example 3, the complement of any closed set is open, so it leads the above result. Getting Deep!! The most difficult concepts to understand in this section are accumulation point, isolated point and closure. First, I will give you definitions of accumulation point and isolated point, then offer two examples about the concept. 14. Definition: Accumulation Point (Version 1) Let A be a subset in. A point x is called an accumulation point of A if every N(x; ε ) contains at least one point of A distinct from x.

11 11 Here is the picture of this definition b-ε b b+ε Point in neighborhood egion of Set A The whole idea of accumulation is that there exist at least one point in (b-ε, b+ε ). Now, here is another version of accumulation point. 15. Definition: Accumulation Point (Version 2) Let A be a subset in. A point x is called an accumulation point of A if every N(x;ε ) contains infinitely many distinct points of A. The accumulation point of A denoted as A. These two definitions are equivalent. In fact, version 2 also tells you one important issue: a set cannot have an accumulation point unless it contains infinitely many points to begin with. But the converse is not true, for example the set of integer {1,2,3,.} is an infinite set with no accumulation points. Now here is the definition for isolated point. 16. Definition: Isolated Point (Our book s definition) If x A but x is not an accumulation point of A, then x is called an isolated point of A. What does that mean? Here is the interpretation:

12 12 Isolated point Accumulation point of S The following is the relationship between closed set and accumulation point. 17. elationship between Closed set and accumulation point Let A be a subset in. A set A is closed if and only if A contains all of its accumulation points. Example 5 Find the isolated and accumulation points, if any, for the set {1, 1/4, 1/6,1/12,.} {0} Here is the picture: 0 1/12 1/6 1/4 Infinitely many points approaching to 0 Analysis: emember 0 is not part of the set in the form x = 1 2n, and notice that the next element after 1/4 is 1/6, so there is no point between two points. If you take the neighborhood of 1/4, you find that they are empty interval. Also, if you take a neighborhood for 0 you will find many distinct point lying inside the neighborhood. Proof: Notice that every point except 0 is an isolated point. First it is easy to find a small enough 1 neighborhood for any point of the form x = that does not contain any point from the 2n

13 13 set but x = 1 2n. Therefore every point x = 1 2n is isolated. (Think about it, 1/4 is after 0 from the sequence, is there any point between these two points? Also, please look back definition 16 and its explanation below, we can t have accumulation unless its neighborhood contains infinitely many points.) On the other hand, if (-b, b) is any small neighborhood of 0, then if n is large enough, 1 x = 0. Notice that the neighborhood of 0 has many distinct points in its 2 n neighborhood, so 0 is not isolated and it is accumulation point of the set. Example 6: Denote that S is all the accumulation point of set S. Show (S ) S. That is show the accumulation of the accumulation point of S is the subset of the accumulation point of S. Proof For every x (S ), then x is an accumulation point of S. Since S is closed (by #13) then x must belong to S as every closed set contains all of its accumulation points. Hence, we can conclude that (S ) S. Now we come to the last point of the section Closure! We are almost done..! Here is the definition and associated properties of Closure. 18. Definition: Closure (Version 1) cl(a) = A A where A is the set of all accumulation points of A. If you like to think in terms of neighborhood, here is one for you: 19. Definition: Closure (Version 2) A point x is in cl(a) if and only if every neighborhood of x intersects A. Note x is not necessarily in the set A.

14 14 Here is the picture of the definition of Closure of Version 2 (for x does not belong to A but belongs to A ) Suppose we have set A which has a and b such that a < b then, a b isolated points c The holes are the points in interior that is missing. The black dots are isolated points. The vertical lines indicated the sequence approaching some arbitrary points c (note: C is not necessarily an element of A). Here is the picture of the definition of A (accumulation point of set A) a Here is the picture of the closure of A (cl(a) = A A b c a b isolated points c The effect of closure is very simple. It just adds in the boundary points and terminates the sequences and fills in the holes. That s all!

15 Properties of Closure i) cl(s) is a closed set. ii) S is closed iff S = cl S iii) cls = S bd S The proof # 16 can be found in the course handout Some Worked Proofs. Example 7 1 Find the closure of S = { : n Ν} 2n Solution First expand the sequence to see the clear picture. {1, 1/4, 1/6,1/12,.} {0}. and from the previous example, 0 is the accumulation point. So by the definition of closure of version 1, then cl(s) = S S and given that S is {0} so the answer is cl(s) = S {0}. eference 1. Apostol, T.M., Mathematical Analysis., Second Edition, Addison-Wesley, Philippines, Bartle,.G., Introduction to eal Analysis, Third Edition, John Wiley & Sons, New York, Berberian, Sterling K., A First Course in eal Analysis, First Edition, Springer Verlag, New York, Lay, Steven., Analysis With An Introduction to Proof, Third Edition, Prentice Hall, New Jersey, 2001.

16 16 5. udin, Walter., Principles of Mathematical Analysis, McGraw-Hill, New York, Interactive eal Analysis - Topology, 1/3/ Analysis Web Notes, 2/15/2003 That s it Folks! Hope this handout can help your assignment! Joseph Pang Thanks for Dr. Adrian Butscher s valuable opinion - The End of the handout -