# A Course in Mathematical Analysis

Javascript is not enabled in your browser. Enabling JavaScript in your browser will allow you to experience all the features of our site. Learn how to enable JavaScript on your browser. NOOK Book. See All Customer Reviews. Shop Books. Add to Wishlist. USD Sign in to Purchase Instantly. Overview The three volumes of A Course in Mathematical Analysis provide a full and detailed account of all those elements of real and complex analysis that an undergraduate mathematics student can expect to encounter in their first two or three years of study.

Containing hundreds of exercises, examples and applications, these books will become an invaluable resource for both students and instructors. This first volume focuses on the analysis of real-valued functions of a real variable. Besides developing the basic theory it describes many applications, including a chapter on Fourier series.

## A Course in Real Analysis

It also includes a Prologue in which the author introduces the axioms of set theory and uses them to construct the real number system. Volume II goes on to consider metric and topological spaces and functions of several variables. Volume III covers complex analysis and the theory of measure and integration. About the Author D.

## A First Course in Real Analysis | M.H. Protter | Springer

He has 50 years' experience of teaching undergraduate students in most areas of pure mathematics, but particularly in analysis. Table of Contents Introduction; Part I. Prologue: The Foundations of Analysis: 1. The axioms of set theory; 2. Number systems; Part II. The reader should verify that in each instance the conditions for being a topology are satised.

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This is the indiscrete topology. This is the discrete topology.

## A First Course in Mathematical Analysis

Suppose that Y is a subset of a topological space X,. Y, Y is then a topological subspace of X,. Topological subspaces inherit many, but not all, of the properties of the larger space. Suppose that q is a mapping of a topological space X, onto a set S. In many cases, it is very badly behaved, and quotient topologies are a rich source of idiosyncracies and counterexamples.

If X is an innite set, let f be the collection of subsets of X with a nite complement, together with the empty set. This is the conite topology. If X is an uncountable set, let be the collection of subsets of X with a countable complement, together with the empty set. This is the cocountable topology. Let P denote the vector space of complex polynomials in two variables. This denition can be extended to other settings in algebraic geometry and in ring theory, where it is an important tool. It does not have any clear use in analysis. We now establish some elementary results about topological spaces.

In many cases, the arguments are similar to those for the real line, or for metric spaces, and the details are left to the reader. Then there is a unique topology on X for which B is a base. Proof Let be the collection of unions of sets in B. Then the empty set is in the union of the empty set of subsets of B and X , by ii. Clearly the union of sets in is in , and so it remains to show that nite intersections of sets in are in.

It follows from the construction that is unique. For if is a topology on X for which B is a base, then , by the denition of a base, and , since the union of open sets is open. Thus b is not in the closure of A.

### Your Answer

Proof This follows by making obvious modications to the proof of Proposition A subset O of X is open if and only if it is a neighbourhood of each of its points. Proof i Since x N , N is not empty.

If O is open, then it follows from the denition of neighbourhood that O is a neighbourhood of each of its points. The analogues of Theorem Composition also works well; the proof is easy, but the result is of fundamental importance.

If f is continuous at a X and g is continuous at f a , then g f is continuous at a. The next result corresponds to Theorem We give some details of the proof, though the proof is essentially the same. The following are equivalent. Proof Suppose that f is continuous, that U is open in Y and that x f 1 U.

### A course in mathematical analysis

Since f is continuous at x, f 1 U Nx , and so x is an interior point of f 1 U. Then there exists an open set U in Y such that f a U N. Then a f 1 U f 1 N , and f 1 U is open, by hypothesis, so that f 1 N is a neighbourhood of a. The proof of the equivalence of iii and iv is exactly the same as the proof in Theorem Then f A is dense in the topological subspace f X of Y,. The following are equivalent: i f is a homeomorphism. What about sequences? We shall see that there are some positive results, but that, in general, sequences are inadequate for the denition of topological properties.

The converses of i and ii are false.

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Proof i If M b is a punctured neighbourhood of b then there exists j0 N such that aj M b for j j0. The proof of ii is exactly similar. We shall use the same example to show that the converses do not hold. Suppose that X is an uncountable set, with the cocountable topology described above.

Suppose that A is any uncountable proper subset of X. Suppose that aj A and that aj b as j.

If f is continuous at a then f aj f a as j. The converse is false. There exists j0 N such that 1 aj f N for j j0. Let X be an uncountable set, let be the cocountable topology on X and let be the discrete topology. The identity mapping i from X, into X, has no points of continuity, but a sequence converges in X, if and only if it is eventually constant, in which case it converges in the discrete topology.

Show that f is continuous on X if and only if its restriction to A and its restriction to B are continuous. Is the same true if closed is replaced by open? What if A is open and B is closed? What are the convergent sequences?