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TOPOLOGY. BY. SEYMOUR LIPSCHUTZ, Ph.D. Associate Professor of Mathematics. Temple University. SCHAUM'S OUTLINE SERIES. McGRAW-HILL BOOK. Seymour Lipschutz Schaum's Theory & Problems of General Topology Schaum Publishing Co ppti.info Acrobat 7 Pdf Mb. Scanned by. Schaums Outline of General Topology by Seymour Lipschutz, , available at Book Depository with free delivery worldwide.

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Schaum's General Topology -- - Free ebook download as PDF File .pdf) or Schaum's - Theory and Problems of General Topology - Seymour Lipschutz. Schaum's - Theory and Problems of General Topology - Seymour Lipschutz - Ebook download as PDF File .pdf) or read book online. The term general topology means: this is the topology that is needed and .. Let K be the set of real numbers that are sums of series of the form. ∑∞ k=1 ak. 3k.

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Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available. How is Chegg Study better than a printed Topology student solution manual from the bookstore? Our interactive player makes it easy to find solutions to Topology problems you're working on - just go to the chapter for your book.

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The best part? As a Chegg Study subscriber, you can view available interactive solutions manuals for each of your classes for one low monthly price. Why buy extra books when you can get all the homework help you need in one place? Can I get help with questions outside of textbook solution manuals? Completeness and compactness. Point open topology. Pointwise convergence. Uniform convergence. Function space C [0,1].

Uniform boundedness. Compact open topology. Topology of compact convergence. Functionals on normed spaces.

General Topology

Real line. Subsets of R. Positive numbers. Absolute value. Least upper bound axiom.

Schaum's outline of theory and problems of general topology

Nested interval property. The objects comprising the set are called its elements or members and will be denoted by lower case letters a, b,: There are essentially two ways to specify a particular set. One way, if it is possible, is by actually listing its members. The other way is by stating those properties which characterize the elements in the set. Example 1. Observe that a set does not depend on the way in which its elements are displayed. A set remains the same if its elements are repeated or rearranged.

In particular, a set which consists of exactly one element is called a singleton set. We also say that A is contained in B or B contains A.

Example 2. We will let N denote the set of positive integers, Z denote the set of integers, Q de- note the set of rational numbers and R denote the set of real numbers. The reader should be warned that some authors use the symbol Q for a subset and the symbol C only for a proper subset. Theorem 1. Let A, B and C be any sets. We call this set the universal set or universe of discourse and denote it in this chapter by U.

It is also convenient to introduce the concept of the empty or null set, that is, a set which contains no elements. Example 3. In plane geometry, the universal set consists of all the points in the plane.

Then A is empty, i. For example, each line in a set of lines is a set of points. Usually we use class for a set of sets, and collection or family for a set of classes. The words subclass, subcollection and subfamily have meanings analogous to subset. Example 4. Consider any set A. The Word space shall mean a non-empty set which possesses some type of mathematical structure, e. In such a situation, we will call the elements in a space points.

Example 5. The following diagrams, called Venn diagrams, illustrate the above set operations. Here sets are represented by simple plane areas and U, the universal set, by the area in the entire rectangle. AnB is shaded new 9. AB is shaded AC is shaded 9. In fact, we state Theorem 1. Sets satisfy the laws in Table 1.

Au BnC: Each of the above laws follows from an analogous logical law. Each of the following conditions is equivalent to ACE: Example 6. The reader is familiar with the Cartesian plane R2: RX R Fig. Here each point P represents an ordered pair a, b of real numbers and vice versa.

Here the vertical lines through the points of A and the horizontal lines through the points of B meet in 6 points which represent A X B in the obvious way.

The point P is the ordered pair 2, b. The product set of the sets A1,. Example 7. Set inclusion is a relation in any class of sets. Example 8.

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Consider the relation C, i. Recall, by Theorem 1. In Euclidian geometry, similarity of triangles is an equivalence relation.

For if 01, ,3 and 7 are any triangles then: If R is an equivalence relation in A, then the equivalence class of any element a E A, denoted by [a], is the set of elements to which a is related: Let R be an equivalence relation in A and let [a] be the equivalence class of a E A. Accord- ingly, the previous theorem implies the following fundamental theorem of equivalence relations: Let R be an equivalence relation in A.

Then R5 is an equivalence relation in Z. The reader should be warned that some authors denote this relation by Uo V. It is convenient to introduce some more symbols: Let A: A is the set which consists of the single element 2, i. The number 2 belongs to A; it does not equal A. Does A: Determine which of the following sets are equal: Each is different from the other. The set 0 contains no elements; it is the null set.

Determine whether or not each of the following sets is the null set: It is necessary to show that at least one member of A does not belong to B. Since 36A and 36B, A is not a subset of B. We must show that each element in A also belongs to 0. Let 9cEA. Now ACB implies We have therefore shown that xEA implies 2: If A is a subset of the null set Q , then A: Recall that the power set 43 8 of S is the class of all subsets of S. Prove the Distributive Law: Prove Theorem 1.

Suppose ACB. Let acEA; then by hypothesis,: Hence acEA and x63, i. Then in particular, ACAnB. Hence, by Theorem 1. Then T consists of the shaded squares below. Then iy s. Let R be a relation in A, i. But, by transitivity, a, b , b, c E R implies at, c E R.

In other Words, R is transitive. Thus a, b E ROR, i. R C R R 0R is transitive. Consider the set N X N, i. Suppose a, b 2 c, d.

Hence c, d 2 a, b and, therefore R is symmetric. Then ad: Accordingly, a, b 2 e, f and R is transitive. Proof of i. E R for every one A and therefore a E [a]. Proof of ii. Suppose a, b ER. Let a; E[b]; then b, av E R. But by hypothesis, a, b E R; hence by transitivity, a, as E R. Accordingly, no 6 [a, ], i. Then by a similar argument, we get [on] C [b]. Proof of iii.

We prove the equivalent contrapositive statement, i. If [a]n[b] as 2 , there exists an element as 6 A with as E [a]r[b]. Hence a. By symmetry, x, b eR and, by transitivity, a, ,b ER. State whether each of the following statements is true or false. Discuss all inclusions and membership relations among the following three sets: Prove that the closed interval [a, b] is not a subset of the open interval a, b. In each of the Venn diagrams below shade: Prove and show by Venn diagrams: Each of the following conditions is equivalent to ACB: Determine the number of distinct relations from a set with m elements to a set with n elements, where m and n are positive integers.

Consider the relation R: Consider N X N, the set of ordered pairs of positive integers. Let a and b be arbitrary real numbers. Answers to Supplementary Problems The sets in ii and iii are empty. The distance between adjacent horizontal points is a.

The domain of f is A, the co-domain is B. To each function f: The range of f, denoted by f[A], is the set of images, i. Two functions f: Accordingly, we do not distinguish between ac.

Let f: Here f is a real-valued function. The range of f is the set of non-negative real num- bers, i. Then the diagram in Fig. A function f: Hence the range f[A] of any constant function f is a singleton set, i. On the other hand, if f: It is also denoted by 1A or 1. Clearly, if f: Proposition 2. The function 9 shown in Fig. The function Ii shown in Fig. The set I is called the index set, the sets At are called indexed sets, and each ie I is called an index.

For each n E N, the positive integers, let D: For each i0 6 I there exists a function Now let R1, R2 and R3 denote copies of R.

We will also write U, -x: Theorem 2. Then the image f [A] of any subset A of X is the set of images of points in A, and the inverse image f" [B] of any subset B of Y is the set of points in X whose images lie in B.

For example, if f is the function in Example 5. In particular we state: Observe that 3 -: Namely, Theorem 2. Next follows an important relationship between the two set functions. Many operations are inherited by T X, R from corresponding operations in R. Consider the functions f: Then each function f E T X, R may be written as an ordered my-tuple f 1 ,.

Any subset of 77 X, R. There is nothing assigned to the element b E A. Two elements, a: State whether or not each of the following relations is a function from X into X. Recall that a subset f of X X X is a function f: Let the functions f: Compute g 0 f: Prove the associative law for composition of functions, i. Since the associative law was proven for composition of relations in general, this result follows. We also give a direct proof: Since g is onto, Eb E B s.

Since f is onto, la.

Accordingly, g0 f is also one-one. The graphs of the functions are as follows: On the other hand, g is onto; each horizontal line contains at least one point of f.

The function h is both one-one and onto; each horizontal line contains exactly one point of h. Utilizing Proposition 2. Let a E A be the smallest number in A. U, -E, B 1 Ai Solution: In particular, ac 6 A0. Hence n, A; C Aio. Prove Theorem 2. Then ipo E A s. Now let: Consider the function f: Q since the square of no real number is Then the associated set function f: Let y E f[AUB], i.

We now prove the reverse inclusion, i. Let y E f[A] U f[B]. Hence Accordingly, y E f[A B]. Then f ac EAB, i. Hence ac E f'1 [A B]. Observe that we use the fact that k, f x and g: R-R and g: Let k: Prove that for any function f: What can be said about fok?

State whether or not each of the following functions is an extension of f. Let ACX and let f: The inclusion function j from A into X, denoted by j: For any function f: Prove Proposition 2. B-A satisfy: Under what conditions will the projection 1,0: Prove that f is both one—one and onto.

Let R be an equivalence relation in a non—empty set A. Prove that n is an onto function. Let A be any subset of a universal set U. Then the real-valued function xA: The function f 0k is a constant function. The function f: Hence the open interval —1,1 is equivalent to R, the set of real numbers. Proposition 3.Guard, later sent to the download schaums outline of theory and problems of of complex extent of Chi-chou, and much to oversight of the free and really Download link household.

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