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02072012, 09:37 PM  #1 
Is Butthurt.  bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework hella studying and shit! 
02072012, 09:46 PM  #2 
Is Butthurt.  Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework here is some notez bro! Math 55a: Honors Advanced Calculus and Linear Algebra Metric topology IV: Sequences and convergence; the spaces B(X; Y ) and C(X; Y ), and uniform convergence Sequences and convergence in metric spaces. [See Rudin, 3.1, 47{51.] The notion of convergence in the metric space R is implicit in such familiar contexts as in nite series and even nonterminating decimals, but was not made explicit until centuries after Euler spent considerable e ort trying to evaluate such series as 1! �� 2! + 3! �� 4! + [sic]. As with many such notions, while our initial interest is in sequences in R or perhaps Rn, the basics are just as easy to formulate in the context of an arbitrary metric space, and we shall have occasion to use this notion in that generality later in the course. Let fpng = fp1; p2; p3; : : :g1 be a sequence in a metric space X (i.e., with each p n 2 X, n = 1; 2; 3; : : :). We say that fpng converges if there is a point p 2 X such that: for every > 0 there is an integer N such that d(pn; p) < for each n > N . Equivalently, for every > 0 we have d(pn; p) < for all but nitely many n. [Why are the two de nitions equivalent?] Such p is called the limit of fpng. The de nite article must be justi ed: we must show that if p; p0 are both limits of fpng then p = p0. This is easily shown as follows [see Rudin, Thm. 3.2b, p.48]. For any > 0, there are integers N;N0 such that d(pn; p) < for each n > N and d(pn; p0) < for each n > N0. Let n be any integer that exceeds both N and N0. Then by the triangle inequality d(p; p0) < 2. Since is an arbitrary positive number, it follows that p = p0. We also use the following notations for \p is the limit of fpng": \fpng converges to p", \pn approaches p", \p = limn!1 pn", and \pn ! p" (or \pn ! p as n ! 1").2 A sequence that does not converge is said to diverge. Note that this notion can depend on X as well as fpng; e.g. f2��ng converges as a sequence in R, but not as a sequence in (0; 1). We can say \fpng converges in X" if the ambient space may otherwise be ambiguous. Like continuity of functions, the limit of a sequence may be de ned topologically: p n ! p if every open set containing p also contains pn for all but nitely many n.3 This lets us de ne lim n!1 pn in an arbitrary topological space, but with the proviso that the limit of a sequence may not be unique without some assumption on the topology. (Challenge: formulate such an assumption that makes limits unique but still allows a nonmetric topology!) It should be clear that pn ! p if and only if the sequence fd(pn; p)g converges 1 We use the braces to distinguish the sequence fpng from its typical element pn. Note that this contains more information than the set fpnjn = 1; 2; 3; : : :g, namely which element of that set occurs as which pn. 2 NB In such contexts 1 is no more than a suggestive symbol; e.g., there is no point \1" in R for fng to approach. 3 NB this is not quite the same as \contains all but nitely many pn"! Cf. the end of footnote 1. 1 to 0 in R. It is easy to see that a sequence f(xn; yn)g in a product metric space X Y converges to (x; y) if and only if xn ! x and yn ! y. Convergence of sequences is also related to several notions already introduced, as follows. Theorem. [Cf. Rudin, 3.2d on p.48] If p is a limit point of E X then there exists a sequence fpng in E with pn ! p. The closure of any E X is the set of limits in X of sequences in E. In particular E is closed if and only if every sequence in E that converges in X also converges in E. Proof : We saw in the second handout that if p is a limit point then there are q n 2 E (n = 1; 2; 3; : : :), with each qn 6= p, such that for each r > 0 we have d (p; qn) < r for all but nitely many n. Then qn ! p. If p 2 E but p is not a limit point then p 2 E, and we obtain fpng by setting each pn = p. Conversely, if pn ! p for some sequence fpng in E, then every neighborhood of p contains a point of E, whence p 2 E. The last statement follows from Rudin 2.27b (also done in the second handout): E = E if and only if E is closed. 2 Theorem. [Cf. Rudin, Thm. 4.2 on p.84] Let f : X ! Y be a function between metric spaces, and p 2 X. Then f is continuous at p if and only if f(pn) ! f(p) for every sequence fpng in X such that pn ! p. Proof : ) is clear from the de nition. For (, suppose on the contrary that f is not continuous at p. Then there exists > 0 such that for each > 0 some p 0 2 X satis es d(p; p0) < but d(f(p); f(p0)) . Let pn be a p0 that works for = 1=n. Then pn ! p but f(pn) does not converge to f(p). 2 We deduce the following for sequences of real or complex numbers: Theorem. [Rudin, Thm. 3.3 on p.49] Let fsng; ftng be sequences of real or complex numbers. Assume that sn ! s, tn ! t. Then: the sequence fsn + tng converges to s + t; the sequence fsn �� tng converges to s �� t; the sequence f sntng converges to st; and, if s 6= 0 and sn 6= 0 for each n, the sequence f1=sng converges to 1=s. Proof : This follows immediately from the previous theorem together with the continuity of the functions (s; t) 7! s t, (s; t) 7! st, s 7! 1=s shown in the previous (third) handout. 2 [Note that this looks nothing like the proof in Rudin, which nevertheless should be familiar; what's going on here?] 2 Sequences in function spaces; uniform convergence. Let Y be a metric space, and X an arbitrary set. We noted that the set Y X of functions from X to Y becomes a metric space under the sup metric if Y is bounded, or if X is nite, but that there is a problem if Y is unbounded and X is in nite: the purported distance between two functions might be in nite! The usual way around this problem is to restrict attention to bounded functions from X to Y , i.e. functions f : X ! Y such that f(X) is a bounded set in Y . The set of such functions is denoted B(X; Y ), and it becomes a metric space under the sup metric d(f; g) := supx2X d(f(x); g(x)). Note that B(X; Y ) is simply Y X in the known cases of bounded Y or nite X. Now let ffng be a sequence in B(X; Y ). What does it mean for ffng to converge to f? Unwinding the de nition, we nd that the condition is: For each > 0 there exists an integer N such that d(fn(x); f(x)) < for all x 2 X and n > N. If fn; f are any functions (not necesssarily bounded) from X to Y , we say that f n converges uniformly to f if the above condition is satis ed. [Cf. Rudin, Def. 7.7 on p.147, which is the special case Y = R or C.] Note that if each fn is bounded then so is f (if fn(X) is contained in a neighborhood of radius M, and d (f; fn) < , then f(X) is contained in a neighborhood of radius M + with the same center). If fn ! f uniformly, then certainly fn(x) ! f(x) for each x 2 X, i.e. fn also converges pointwise to f. Note that pointwise convergence of fn to f, de ned by For each x 2 X and each > 0 there exists an integer N such that d (fn(x); f(x)) < for all n > N, is a strictly weaker notion than uniform convergence, since N is allowed to depend on x as well as . (Cf. our earlier distinction between continuity and uniform continuity.) For instance, let X = Y = [0; 1], and fn(x) = xn. Then fn converges pointwise, but not uniformly, to the function f (x) := 1 ; if x = 1; 0; otherwise. If X is itself a metric space4 then we are often interested in the subspace C(X; Y ) of B(X; Y ) consisting of continuous bounded functions. The key fact here is: Theorem. C(X; Y ) is closed in B(X; Y ) for any metric spaces X; Y . That is, the uniform limit of bounded continuous functions is again bounded and continuous. We have already shown that the uniform limit of bounded functions 4 More generally X, but not Y, can be a general topological space. 3 is bounded. We next show that the uniform limit of continuous functions is again continuous: Theorem. [Cf. Rudin, Thms. 7.11 and 7.12 on p.149{150] Suppose X; Y are metric spaces, and ffng a sequence of continuous functions from X to Y . If f fng converges uniformly to a function f : X ! Y , then f is continuous. Proof : Fix x 2 X. We show more generally that if fn is continuous at x and f n ! f uniformly then f is continuous at x. Indeed, let be any positive real number. Since fn ! f uniformly, there exists N such that d(fn; f) < =3 for all n > N. Fix one such n. Since fn is continuous at x, there exists > 0 such that d(fn(x); fn(x0)) < =3 provided d(x; x0) < . Then for all such x we have d (f(x); f(x0)) d(f(x); fn(x)) + d(fn(x); fn(x0)) + d(fn(x0); f(x0)) < 3 + 3 + 3 = : Thus f is continuous at x as claimed. 2 This also completes the proof that C(X; Y ) is closed in B(X; Y ). Note that the example of fn(x) = xn shows that the pointwise limit of continuous functions may fail to be continuous. Must the pointwise limit of bounded functions be bounded? 4 
02072012, 09:48 PM  #3  
in need of a μse TheBagman is Offline Join Date: Apr 2005 Location: gated community, ivory tower, depends on who you ask
Posts: 40,695
MIIDAJ? Scrill: 3,664,724,214,985  Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework I don't attend Harvard.  
02072012, 09:53 PM  #4 
Is Butthurt.  Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework pwnt! are you claiming that natalie portman is teh hella be smartz0rz than you? 
02072012, 10:03 PM  #5 
Is Butthurt.  Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework 
02072012, 10:04 PM  #6  
in need of a μse TheBagman is Offline Join Date: Apr 2005 Location: gated community, ivory tower, depends on who you ask
Posts: 40,695
MIIDAJ? Scrill: 3,664,724,214,985  Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework Probably. She was a psychology major at Harvard.  
02072012, 10:05 PM  #7 
Is Butthurt.  Re: bagman is so hella pwnt and mad that he quit surfing the interwebz and returned to his math55A homework bagman gets psycho analyzed by pilgrims behind plymoth roc. 
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