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Finally, a chapter relates antidifferentiation to Lebesgue theory, Cauchy integrals, and convergence of parametrized integrals. Nearly exercises allow students to develop their skills in the area. Audience: This text is appropriate for any second course in real analysis or mathematical analysis, whether at the undergraduate or graduate level. This text covers similar material to the books we give US list prices for current editions at Amazon on December 18, : Real and Complex Analysis by W.

## A Course in Mathematical Analysis (Volume 2)

Currrent version released May 11, We now make another denition, similar enough to be confusing. Thus b is a closure point of A if there are points of A arbitrarily close to b. Let i A be the set of isolated points of A. A set is perfect if and only if it is closed, and has no isolated points. We can characterize limit points and closure points of A in terms of convergent sequences. Then aj b as j. The closure of a bounded set is bounded.

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Proof Certainly diam A diam A. For example, the rationals are dense in R. Proof i follows trivially from the denition of closure. Suppose that Y is a metric subspace of a metric space X, d. How are the closed subsets of Y related to the closed subsets of X? Thus y A. Here are some fundamental properties of the collection of closed subsets of a metric space X, d. Proof i The empty set is closed, since there is nothing to go wrong, and X is trivially closed. But then a A. Here is an important example. It then follows from Theorem In other words, all the points suciently close to a are in A; we can move a little way from a without leaving A. The interior A of A is the set of interior points of A. The collection of open subsets of X, d is called the topology of X, d. Although a normed space has plenty of closed linear subspaces, it has only one open linear subspace. Proof Since F is a linear subspace, 0 F. Proof i follows directly from the denition. A metric space is separable if it has a countable dense subset.

Thus R, with its usual metric, is a separable metric space. There are interesting metric spaces which are not separable: Proposition Then fA is bounded. Suppose that G is a dense subset of BX S. Then c is a mapping of H into P S. Thus 2 BX S is not separable. Show that the following are equivalent i a is an isolated point of X, d. Show that the collection of open intervals of R with rational endpoints is a basis for the usual topology of R.

Show that c0 is a separable closed linear subspace of l.

## A Course in Mathematical Analysis: Volume 1, Foundations and Elementary Real Analysis

Show that U is countable. Show that if X, d is separable then the set of strict local maxima is countable. Consider a countable basis B for the topology, and consider the sets of B on which f is bounded above, and attains its supremum at a unique point. Show that if X, d is separable then f has only countably many removable discontinuities.

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Verify that the p-adic metric on Q is an ultrametric. Many, but by no means all, of the properties of a metric space X, d and of mappings from X, d into a metric space Y, , can be dened in terms of the topologies of X, d and Y,. These are called topological properties.

## Mathematical Analysis I | Vladimir A. Zorich | Springer

A collection N of subsets of X is called a base of neighbourhoods of x if each N N is a neighbourhoood of x, and if each neighbourhood of x contains an element. Continuity is also a topological property. Let us make this explicit. Proof a Suppose that f is continuous at a and that N is a neighbourhood of f a. Conversely, suppose the condition is satised. Then f x U. Since this holds for all x f 1 U , f 1 U is open: i implies ii. Conversely, suppose that ii holds.

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Suppose that a X and that N is a neighbourhood of f a. Thus f is continuous at a. Since this is true for all a X, ii implies i. Suppose that iii holds, and that A X. Thus iii implies iv.

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Suppose that iv holds, and that B is closed in Y. By hypothesis, f f 1 B f f 1 B.

Thus iv implies iii. Then f A is dense in the metric subspace f X of Y,. In particular, if X is separable, then so is f X.

The following are equivalent: 1. U is open in X, d if and only if f U is open in Y,. B is closed in X, d if and only if f B is closed in Y,. There are two points to notice about this theorem and its proof. The rst is that one needs facility at handling images and inverse images of sets. The second and more important point is that the conditions, in terms of open sets and closed sets, that we have given for a function to be continuous involve the inverse images of sets in Y , and not the images of sets in X.

Show that a notion or property is topological if it can be dened in terms of each of the following. Show that g x is a continuous strictly positive function on X. This is easy. Its extension to certain topological spaces is Urysohns lemma, which we shall prove later Theorem We now generalize this, by introducing the notion of a topological space.