Math 426 Homework 4 Due 3 November 2017

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1 Math 46 Homework 4 Due 3 November Given a metric space X,d) and two subsets A,B, we define the distance between them, dista,b), as the infimum inf a A, b B da,b). a) Prove that if A is compact and B is closed, then dista,b) = 0 if and only if A B. b) Give an example to show that this may fail if A is merely assumed to be closed, not compact. a) The condition dista,b) means that for any ɛ we can find a A and b B such that da,b) < ɛ. Therefore, we can find sequences a n ) in A and b n ) in B such that da n,b n ) < 1/n. Since A is compact, a n ) has a convergent subsequence, a ni ) i a A. We will show a A B. By the triangle inequality da,b ni ) da, a ni ) + da ni,b ni ) and both terms on the right converge to 0 as i, so the sequence b ni ) i converges to a. Since B is closed, a B. b) Consider two continuous functions R R. The first is π, projection onto the y coordinate. The second is f x, y) = x y 1. Now consider π 1 0), the x-axis, and f 1 0), the hyperbola given by y = 1/x. These are both closed, by construction, and they are obviously disjoint by consideration of y-coordinates. Moreover, for any n N, the point n,1/n) lies on the hyperbola while the point n,0) lies on the x-axis. The distance between these two points is 1/n, and so the distance between the two sets is 0.. a) We defined a metric space X to be totally bounded if, for all ɛ > 0, the space X can be written as a union of finitely many balls of the form Bx,ɛ) X. Suppose X is a totally bounded metric space and A X is a subspace. Show that A is totally bounded warning: the centres of the balls in X may not lie in A. b) Let p [1, ], let ɛ > 0 and let B p 0,ɛ) denote an ɛ ball in l p. The metric space l p is defined in the supplementary notes on p-norms. Prove that B p 0,ɛ) is not compact. 1

2 a) Suppose ɛ > 0. Since X is totally bounded, we may find a finite collection of centres {x 1,..., x n } such that X = n i=1 Bx i,ɛ/). Some of these balls may not intersect A; without loss of generality suppose that the balls not meeting A are exactly the balls having centres x m+1,..., x n for some m. Then, for each i {1,...,m}, choose a i Bx i,ɛ/). Consider Ba i,ɛ). This ball contains Bx i,ɛ/) by virtue of the triangle inequality. Therefore A m i=1 Ba i,ɛ). b) This is a metric space, so it suffices to show there is a sequence in this set having no convergent subsequence. Define x n ) by setting x n = 0,0,0...,ɛ,0,...), the ɛ being in the n-th place. Then the distance between any two different terms of this sequence is 1/p ɛ, and therefore it is has no Cauchy subsequence. This proves that B p 0,ɛ) is not sequentially) compact. 3. By a discrete set in R, we mean a closed subset D R such that the subspace topology on D is discrete. a) If D is discrete, prove D is countable. b) Let F be a countable set of points in R with the usual topology. connected. Hint: how many lines pass through a point x R \ F? Prove that R \ F is path a) An uncountable discrete space is metrizable by the discrete metric), and all subsets are closed. Therefore there is no countable dense subset, so the space is not separable, and therefore not second countable these concepts coincide for metrizable spaces). On the other hand, any subset of a second countable metric space such as R ) is second countable, by simply restricting a basis. A more elementary rephrasing of this argument: Let B be a countable basis for R. Let D be a discrete subset of R. For each d D, there is an open neighbourhood U d of d such that U d D \ {d} =. There is a basic open neighbourhood V d B such that d V d U d. But now the assignment d V d gives an injective function D B, so D is countable. b) Let x 0, y 0 ) be a point of R. There are uncountably many distinct lines passing through this point, since the lines mx x 0 ) = y y 0 as m R varies forms an uncountable family. Therefore, for any x 0, y 0 ) F there are at least two distinct lines L 1 and L contained in R \ F and containing x 0, y 0 ). For any other point x 1, y 1 ) F, we can also find a line M containing it and contained in R\F. Since M cannot be parallel to both L 1 and L, it must meet one of them, say L 1. Hence x 0, y 0 ) and x 1, y 1 ) lie in the path connected subset L 1 M of R\F, so they are in the same path component. Since the pair of points is arbitrary, R \ F is path connected.

3 4. Let X,d) be a metric space. a) Let K > 0 be a constant. A function f : X X is K -Lipschitz if df x), f y)) K dx, y) for all x, y X. Show a K -Lipschitz function is continuous. b) Suppose X is complete and K < 1. If f is a K -Lipschitz function, show that there exists a unique element x X such that f x) = x. c) Give an example of a function f : R R such that df x), f y)) < dx, y) for all distinct x, y R, but such that f x) > x for all x. a) Since X is a metric space, we can use ɛ δ proofs. Suppose ɛ > 0 is given. Let δ = ɛ/k. Then if dx, y) < δ, it follows that df x), f y) K dx, y) < ɛ, so f is continuous. b) Let x X. Consider the sequence x, f x), f f x)), f 3 x), f 4 x),...). We claim this sequence is Cauchy. Let ɛ > 0 be given. Let C = dx, f x)). Then df n x), f n+1 x)) K n C by an easy induction. Consequently for m n, we have and since df n x), f m x)) C m K i < m K i K i = K n+1 1 K we obtain df n x), f m n+1 CK x)) < 1 K. The right hand side tends to 0 as n, so by choosing N sufficiently large, we can ensure that n,m > N implies that df n x), f m x)) < ɛ. Therefore the sequence is Cauchy. Since X is complete, for any x, the sequence f n x)) n converges to a limit, α x. Moreover, f n+1 )x)) n, a subsequence of that sequence, also converges to α x. But f n+1 x)) n is also the result of applying the continuous function f to the original sequence f n x)) n, so it converges to f α x ). Since X is metric, limits of convergent sequences are unique, and it follows α x = f α x ). Finally, if α x and α are two fixed points of f, then dα x,α) = df α x ), f α)) K dα x,α). Since K < 1, this results in a contradiction unless dα x,α) = 0. c) There are a number of solutions. For instance, suppose h : R 0, ) is a continuously-differentiable function such that h x) 1,0) for all x R. Such functions exist: hx) = π/ arctanx) is an example, since h x) = 1/1 + x ) 1,0). By the mean value theorem, hy) hx) = h z) 1,0) y x so it follows that if y > x, then hy) < hx) and hx) hy) < y x. Then define f x) = x + hx). With this definition f x) > x for all x R. Moreover, if x < y, we have hx) hy)) < y x, so that f x) < f y). Then df x), f y)) = dx, y) + hy) hx) < dx, y), as required. 3

4 5. a) Let X be a non-compact locally compact Hausdorff space. Let X = X { } denote the onepoint compactification of X, and let ι : X X be the canonical inclusion. Let φ : X K be some compactification of X, where K is Hausdorff and φ is an open map. 1. Prove that the function f : K X given by { ιx) if y = φx) f y) = otherwise. is continuous. b) Define S n as the subset {x R n+1 x = 1}. Write x = x 1..., x n ). Consider the function f : R n S n x 1 x x n f x) = 1 + x, 1 + x,..., 1 + x, 1 ) x 1 + x You may assume this function is continuous. i. Construct a continuous function g : S n \ {0,0,...,0, 1)} R n such that g f = id R n. It may help to solve the equation Z = 1 x for x 1+ x. You may assume that addition, subtraction, multiplication and division are continuous functions R R. ii. Prove that f : R n S n is a one-point compactification of R n. a) Identify X with its image ιx ) X. Consider the open sets of X. These come in two flavours: U X X open in X ) and U = X \C where C X is compact. In the first case, f 1 U ) = φu ) is open in K ; in the second f 1 U ) = K \ φc ), and since C is compact in X, it follows φc ) is compact in K and therefore closed in K. The result follows. b) i. Define y 1 g y) = g y 1,..., y n+1 ) =, 1 + y n+1 y y n,..., 1 + y n y n+1 ). We then have x = so that 1+ x 1+ x g f x) = x 1,..., x n ). ii. One can approach this question in two ways. We show that the S n is a Hausdorff compactification of R n given by adjoining one point. This proves that S n is a one-point compactification. The other approach not taken here) is to show that the topology on S n agrees with the topology we gave to the one-point compactification. 1 The hypothesis φ is an open map was missing in this question as posed. It was therefore impossible to answer. If φx ) K is not open, then f 1 { }) = K \ φx ) is not closed, but { } is a closed subset of X, so f cannot be continuous 4

5 We know that S n is compact and Hausdorff. We show that f restricts to a homeomorphism of R n with f R n ) = S n \ {0,...,0, 1)}. One method is to prove that f g is the identity on S n \ {0,...,0, 1)}. We calculate g y) = 1 y n+1 )/1 + y n+1), so that 1/1 + g y) ) = 1 + y n+1)/. Then f g y) = 1 + y n+1 y 1 y y n,,..., 1 + y n y n y n+1 ) = y 1, y,..., y n ). The complement of the image of f is one point. It remains to be shown that S n \{0,...,0, 1)} is dense in S n, but S n \ {0,...,0, 1)} is not compact, and therefore not closed in S n, so its closure must be the whole space S n. 5