Math 190: Fall 2014 Homework 4 Solutions Due 5:00pm on Friday 11/7/2014

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1 Math 90: Fall 04 Homework 4 Solutions Due 5:00pm on Friday /7/04 Problem : Recall that S n denotes the n-dimensional unit sphere: S n = {(x 0, x,..., x n ) R n+ : x 0 + x + + x n = }. Let N S n denote the north pole N = (, 0,..., 0). Look up stereographic projection and prove that S n {N} is homeomorphic to R n. Solution: We define functions f : S n {N} R n and g : R n S n {N} by the following formulas: f(x 0, x,..., x n ) = x 0 (x, x,..., x n ), g(y, y,..., y n ) = + (, y,..., y n ), where = y + + yn. Our first task is to show that f and g are well-defined. Indeed, since x 0 for (x 0, x,..., x n ) S n {N}, f is well defined. Moreover, for (y,..., y n ) R n, we have ( ) + 4y + + 4yn = ( + ), so g is well defined. It is evident that both f and g are continuous functions. We check that they are mutually inverse. To do this, for (x 0, x,..., x n ) S n {N}, we have that g(f(x 0, x,..., x n )) = g( (x,..., x n )) x 0 = + x + +x n ( x 0 ) ( x + + x n ( x 0 ), x x 0,..., = ( x 0) ( x 0 x 0 ( x 0 ), x x n,..., ) x 0 x 0 = ( + x 0 + x 0, x,..., x n ) = (x 0, x,..., x n ). On the other hand, for (y,..., y n ) R n, we have that f(g(y,..., y n )) = f( + (, y,..., y n )) = ( + = (y,..., y n ). y +,..., y n + ) x n x 0 ) Problem : Let B n denote the open unit ball of radius B n = {(x, x,..., x n ) R n : x + x + + x n < }

2 in R n centered at the origin. Prove that B n is homeomorphic to R n. Solution: We define functions f : B n R n and g : R n B n as follows: f(x,..., x n ) = tan(π x /) (x,..., x n ), x arctan /π g(y,..., y n ) = (y,..., y n ), where the right hand sides are interpreted to be 0 when (x,..., x n ) = 0 or (y,..., y n ) = 0. It is clear that both f and g are well defined and continuous. For (x,..., x n ) B n we have g(f(x,..., x n )) = g( tan(π x /) (x,..., x n )) x = arctan(tan(π x /)) (x,..., x n ) π x = (x,..., x n ). Similarly, for (y,..., y n ) R n we have arctan /π f(g(y,..., y n ) = f( (y,..., y n )) = tan(arctan()) (y,..., y n ) = (y,..., y n ). Problem 3: Let I = [0, ]. Prove that the cylinder S I is homeomorphic to the annulus A = {(x, y) R : x + y 4}. Solution: We define functions f : S I A and g : A S I as follows. f : (x, y) t ((t + )x, (t + )y), g : (u, v) u + v (u, v) ( u + v ) It is clear that both f and g are well defined and continuous. Let (x, y) t S I. Then g(f((x, y) t)) = g((t + )x, (t + )y) = (t + ) x + y ((t + )x, (t + )y) (t + x + y ) = (x, y) t.

3 3 Let (u, v) A. Then f(g(u, v)) = f( u + v (u, v) ( u + v )) = (u, v). Problem 4: Let X and Y be topological spaces and let f, g : X Y be continuous functions. Let D X be a dense subset (that is, D = X) and suppose that f(d) = g(d) for all d D. If Y is Hausdorff, prove that f = g. If Y is not Hausdorff, is it necessarily true that f = g? Solution: Suppose Y is Haudsorff and there exists x X such that f(x) g(x). Let U and V be disjoint neighborhoods (in Y ) of f(x) and g(x), respectively. Then f (U) g (V ) is a neighborhood of x; since D is dense in X, there exists d D f (U) g (V ). Then f(d) = g(d) U V, which is impossible because U and V are disjoint. The conclusion does not necessarily hold if Y is not Hausdorff. Indeed, let X and Y equal {a, b} with the indiscrete topology. Then the functions f, g : X Y defined by f(a) = a, f(b) = a, g(a) = a, g(b) = b are both continuous, {a} is a dense subset of X, f(a) = g(a), but f g. Problem 5: (Exercise 9.6 in Munkres) Let (x () α ) α J, (x () α ) α J,... be a sequence of points in the product space α J X α. Prove that this sequence converges to a point (x α ) α J in the product topology if and only if for each α J, the sequence x () α, x () α,... converges to x α. Is the same true if we use the box topology on α J X α? Solution: Suppose that (x (n) α ) α J (x α ) α J in the product topology. Let α 0 J and let U α0 be a neighborhood of x α0. Then U = α J U α is a neighborhood of (x α ), where U α = X α for α α 0. For any n we have that (x (n) α x (n) α 0 ) αinj U if and only if U α0, so U α0. We conclude that there exists N such that n > N implies x (n) α 0 α 0 x α0. that x (n) On the other hand, suppose that x (n) α x α for all α J. A typical basic open set containing (x α ) αinj has the form B = α J U α, where U α is a neighborhood of x α in X α for all α J and U α = X α unless α lies in some finite set {α,..., α k }. There exist N,..., N k > 0 such that n > N i implies x (n) α i U αi for all i k. Let N = max(n,..., N k ). We have that n > N implies that (x α (n) ) α J B. We conclude that (x (n) α ) α J (x α ) αinj. This does not necessarily hold in the box topology. For example, consider the countably infinite product R ω = R R. Let x (n) be the sequence x (n) = (0, 0,..., 0,,,... ), where there are n 0 s. Then for each k we have x (n) k 0. On the other hand, the neighborhood (in the box topology) of (0, 0,... ) given by ( /, /) ( /, /) contains none of the points x (n).

4 4 Problem 6: (Exercise 9.7 in Munkres) Let R ω = R R R be the set of all real sequences (x, x, x 3,... ) and let R R ω be the subset of real sequences (x, x, x 3,... ) such that x i 0 for only finitely many values of i. Compute the closure of R inside R ω in both the product and box topologies. Solution: We claim that R is dense in the product topology. Indeed, let x = (x, x,... ) R ω R. Let U = U U... be a basic open neighborhood of x, that is, U i is an open neighborhood of x i for all i and there exists N such that n N implies U n = R. Then U contains (x, x,..., x N, 0, 0,... ) R and x lies in the closure of R. We conclude that R = R ω in the product topology. We claim that R is closed in the box topology. Indeed, let x = (x, x,... ) R ω R. Then x i 0 for infinitely many values of i. Consider the neighborhood of x given by U = U U..., where U i = R if x i = 0, U i = (0, x i ) if x i > 0, and U i = (x i, 0) if x i < 0. Then U contains none of the points in R. We conclude that R = R in the box topology. Problem 7: Let (X, d) be a metric space and let x 0 X be a point. Prove that the function d : X X R 0 is continuous and prove that the function f : X R 0 given by f(x) = d(x, x 0 ) is continuous. Solution: We have that f = d i, where i : X X X is given by i(x) = (x, x 0 ). By your first midterm, i is continuous, so it is enough to show that d is continuous. To see that d is continuous, let (x, y) X X and let ɛ > 0. It is enough to show that there exist δ, δ > 0 such that for all x B d (x, δ ) and for all y B d (y, δ ), we have that d(x, y ) d(x, y) < ɛ. Indeed, let δ = δ = ɛ/. Then for all x B d (x, δ ) and for all y B d (x, δ ), we have that d(x, y ) d(x, x) + d(x, y) + d(y, y ) < ɛ/ + d(x, y) + ɛ/ = d(x, y) + ɛ. Similarly, we have that d(x, y) < d(x, y ) + ɛ. We conclude that for all x B d (x, δ ) and for all y B d (y, δ ), we have that d(x, y ) d(x, y) < ɛ. Problem 8: Let M(n, R) denote the space of n n real matrices (regarded as R n in the Euclidean topology). Explain why the determinant function det : M(n, R) R is continuous. Use this to prove that the collection of invertible matrices GL(n, R) = {A M(n, R) : det(a) 0} (the notation here means general linear group ) is an open set and the special linear group SL(n, R) = {A M(n, R) : det(a) = } is closed. Solution: We claim that for any n n matrix A, the determinant det(a) is a polynomial function in the entries of A. One way to see this is using induction on n and the expansion of a determinant along a row or column. Alternatively, one could use the definition of a determinant as an alternating sum of permutation monomials. Either way, polynomial functions are continuous, so det is a continuous function. Since det

5 is continuous, we know that GL(n, R) = det (R {0}) is open (because R {0} is open in R) and SL(n, R) = det ({}) is closed (because {} is closed in R). Problem 9: (Exercise. in Munkres) Let X and Y be metric spaces with metrics d X and d Y, respectively. Let f : X Y be a function such that d X (x, x ) = d Y (f(x ), f(x )) for all x, x X. Prove that f is an imbedding. Solution: We first claim that f is injective. Indeed, if x, x X and f(x ) = f(x ), we necessarily have 0 = d Y (f(x ), f(x )) = d X (x, x ), so that x = x. We now claim that f is continuous. Indeed, let x X and ɛ > 0. If x X and d X (x, x ) < ɛ, then d Y (f(x), f(x )) = d X (x, x ) < ɛ. We conclude that f is continuous. Let g : f(x) X denote the inverse function to f. For any x in X and ɛ > 0, we have that for y = f(x ) f(x) such that d Y (f(x), y ) < ɛ: d X (g(f(x)), g(y )) = d Y (f(x), y ) < ɛ. We conclude that g is also continuous. Therefore, f is an imbedding. 5