Prove that any bounded open subset of R is the union of disjoint ope.pdf

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MA500-2: Topological Structures 2016 Aisling McCluskey, Daron Anderson [email protected], [email protected] Contents 0 Preliminaries 2 1 Topological Groups 8 2 Morphisms and Isomorphisms 15 3 The Second Isomorphism Theorem 27 4 Topological Vector Spaces 42 5 The Cayley-Hamilton Theorem 43 6 The Arzelà-Ascoli theorem 44 7 Tychonoff ’s Theorem if Time Permits 45 Continuous assessment 30%; final examination 70%. There will be a weekly workshop led by Daron during which there will be an opportunity to boost continuous assessment marks based upon workshop participation as outlined in class. This module is self-contained; the notes provided shall form the module text. Due to the broad range of topics introduced, there is no recommended text. However General Topology by R. Engelking is a graduate-level text which has relevant sections within it. Also Undergraduate Topology: a working textbook by McCluskey and McMaster is a useful revision text. As usual, in-class discussion will supplement the formal notes. 1 0 PRELIMINARIES 0 Preliminaries Reminder 0.1. A topology τ on the set X is a family of subsets of X, called the τ-open sets, satisfying the three axioms. (1) Both sets X and ∅ are τ-open (2) The union of any subfamily is again a τ-open set (3) The intersection of any two τ-open sets is again a τ-open set We refer to (X,τ) as a topological space. Where there is no danger of ambi- guity, we suppress reference to the symbol denoting the topology (in this case, τ) and simply refer to X as a topological space and to the elements of τ as its open sets. By a closed set we mean one whose complement is open. Reminder 0.2. A metric on the set X is a function d: X×X → R satisfying the five axioms. (1) d(x,y) ≥ 0 for all x,y ∈ X (2) d(x,y) = d(y,x) for x,y ∈ X (3) d(x,x) = 0 for every x ∈ X (4) d(x,y) = 0 implies x = y (5) d(x,z) ≤ d(x,y) + d(y,z) for all x,y,z ∈ X Axiom (5) is often called the triangle inequality. Definition 0.3. If d′ : X × X → R satisfies axioms (1), (2), (3) and (5) but maybe not (4) then we call it a pseudo-metric. Reminder 0.4. Every metric on X induces a topology on X, called the metric topology. We define an open ball to be a set of the form B(x,r) = {y ∈ X : d(x,y) < r} for any x ∈ X and r > 0. Then a subset G of X is defined to be open (wrt the metric topology) if for each x ∈ G, there is r > 0 such that B(x,r) ⊂ G. Thus open sets are arbitrary unions of open balls. Topological Structures 2016 2 Version 0.15 0 PRELIMINARIES The definition of the metric topology makes just as much sense when we are working with a pseudo-metric. Open balls are defined in the same manner, and the open sets are exactly the unions of open balls. Pseudo-metric topologies are often neglected because they do not have the nice property of being Hausdorff. Reminder 0.5. Suppose f : X → Y is a function between the topological spaces X and Y . We say f is continuous to mean that whenever U is open in Y ...

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Prove that f is onto. I will rate lifesaver for any help. Thanks! f(x) = 2x/1 - x Solution We need to show that for every y there is an element x such that y= f(x) Let\'s take y = 2x/(1-x) and solve for x first we can rewrite it as 1/y = (1-x)/(2x) 1/y = 1/(2x) - x/(2x)=1/(2x) - 1/2 add 1/2 to both sides 1/y + 1/2 = 1/(2x) get a common denominator on the left side to get (2+y)/(2y) = 1/(2x) now we can once again flip both fractions to get (2y)/(2+y) = 2x divide both sides by 2 to get y/(2+y) = x since y is a real number then x is a real number. Hence for any y we have the number x= y/(2+y) that will give usy=f(x) That\'s it..

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Prove that Dirichlet\'s discontinuous function is not Reimann integrable. Please provide a detailed proof. Hint: The Dirichlet discontinuous function is defined as D(x)={(c when x is rational) and (d when x is irrational)}. Solution hello. today in the calculus class that I AP (along with one other grad student) we introduced Riemann integration and the other said something to the effect that for a function to be integrable, it must not have \"too many holes\" (informally speaking), that is it must only be discontinuous on a finite number of points. I countered from my memory of real analysis (which may be faulty, I do stats, not a mathematician proper :S) a theorem that as long as the function\'s discontinuities are countable set of points, even if they are infinite, the funcion is integrable. But then I thought about the Dirichlet function, which is defined (one of many ways) as f(x)=x for rational x, 0 for irrational x. Since the rationals are countable the function should be integrable if the theorem I recalled is valid but the DIrichlet function is clearly not Riemann integrable (it is Lebesgue integrable, but that is something else altogether). I think I am misstating the theorem- I knew it had something to do with countability, can anyone help me out? THanks. 24.92.85.35 (talk) 01:21, 3 December 2011 (UTC) A function is Riemann integrable if and only if the set of discontinuities has Lebesgue measure zero. Measure zero sets include finite sets, countable sets, and even some uncountable sets (like the cantor set). The set of all irrational numbers in [0,1] has lebesgue measure one. The dirichlet function is discontinuous on all the irrational numbers and therefore NOT Riemann integrable. For the longest time everyone worked hard on this to figure what exactly are the necessary and sufficient conditions on a function to be integrable. Riemann came along and fixed up the integral (meaning defined it properly) and put it upon solid ground and proved many of its properties from this new definition, which everyone had previously taken for granted since Newton. He made the question well posed and hence the integral is named in his honor. And then Henri Lebesgue was the first one who successfully answered this question and also generalized the Riemann integral.71.211.145.44 (talk) 03:02, 3 December 2011 (UTC) To be precise, a function on an interval [a,b] is Riemann integrable if and only if it is bounded and the set of discontinuities has Lebesgue measure zero. 78.13.145.217 (talk) 22:54, 6 December 2011 (UTC).

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