Math 535 - General Topology Fall 202 Homework 6 Solutions Problem. Let F be the field R or C of real or complex numbers. Let n and denote by F[x, x 2,..., x n ] the set of all polynomials in n variables with coefficients in F. A subset C F n of n-dimensional space will be called Zariski closed if it is the zero locus of some polynomials: C = V (S) := {x F n f(x) = 0 for all f S} for some S F[x,..., x n ]. Note: The zero locus V (S) is sometimes called the algebraic variety associated to S, hence the letter V. For example, in R 2, the subset V (x 2 + x 2 2 9) R 2 is the circle of radius 3 centered at the origin, which is therefore a Zariski closed subset. By convention, let s say S is not allowed to be empty, though you will show in part (a) that it doesn t matter. a. Show that the notion of Zariski closed subset does define a topology on F n, sometimes called the Zariski topology. The entire space F n is Zariski-closed: F n = V (0). Note: This is why we might as well allow S to be empty: V ( ) = F n = V (0). The empty subset F n is Zariski-closed: = V (). Since Zariski-closed subsets are defined as (arbitrary) intersections V (S) = V (f) f S of basic closed set V (f), it suffices to check that a finite union of basic closed sets is an intersection of basic closed sets. For any polynomials f and g, we have V (f) V (g) = {x F n f(x) = 0 or g(x) = 0} = {x F n f(x)g(x) = 0} = V (fg).
b. Show that the Zariski topology is strictly coarser (i.e. smaller) and the usual metric topology on F n. Any polynomial function f : F n F is metrically continuous, therefore its zero set V (f) = f ({0}) is metrically closed. This prove T Zar T met. To show that the inequality is strict, consider the subset C := {x F n x n 0} (or the real part Re(x n ) 0 in case F = C). Then C is metrically closed. However, C is not Zariski-closed. To prove this, let V (f) be a Zariski-closed subset containing C, so that f vanishes on C. For any fixed a,... a n F, the polynomial f(a,..., a n, x n ) in one variable x n vanishes for infinitely many values of x n, thus is the zero polynomial. Since the a i were arbitrary, this implies f = 0 and thus V (f) = F n. Therefore the Zariski closure of C is C = F n C. 2
c. Show that the Zariski topology on F n is T. For any a F n, the singleton {a} is the Zariski-closed set {a} = V (x a, x 2 a 2,..., x n a n ). d. Show that the Zariski topology on F n is not T 2, i.e. not Hausdorff. It suffices to show that any two basic open subsets have a non-empty intersection. Equivalently, any two basic closed subsets have a union which is not all of F n. For any (non-zero) polynomials f and g, we have V (f) V (g) = V (fg). Since fg is not the zero polynomial, there is a point a F n satisfying (fg)(a) 0, so that a / V (fg). e. In the one-dimensional case n =, show that the Zariski topology on F is the cofinite topology. (T cof T Zar ) Since the Zariski topology is T, every finite subset of F n is Zariski- closed. (T Zar T cof ) Let C F be Zariski-closed. If C is not all of F, then C V (f) for non-zero polynomial f. But a non-zero polynomial f(x ) in one variable has (at most) finitely many zeroes, so that V (f) is finite, as is C. 3
Problem 2. Two points x and y in a topological space X are topologically distinguishable if there is an open subset U X that contains one of the points but not the other. A space X is T 0 if any distinct points are topologically distinguishable. Two points x and y are topologically indistinguisable if they are not topologically distinguishable, which amounts to x and y having exactly the same neighborhoods. One readily checks that topological indistinguishability is an equivalence relation on X, which we will denote by x y. The Kolmogorov quotient of X is the quotient KQ(X) := X/, where topologically indistinguisable points become identified. In particular, X is T 0 if and only if X is homeomorphic to its Kolmogorov quotient. a. Show that the Kolmogorov quotient satisfies the following universal property.. The natural map π : X KQ(X) is continuous. 2. KQ(X) is a T 0 space. 3. For any T 0 space Y and continuous map f : X Y, there is a unique continuous map f : KQ(X) Y satisfying f = f π, i.e. making the diagram X π KQ(X)! f f Y commute. In other words, KQ(X) is the closest T 0 space which X maps into.. π : X KQ(X) is continuous, since it is the canonical projection to the quotient space. 2. Let a, b KQ(X) be distinct points. Pick representatives x, y X of a and b respectively. Since a and b are distinct, x and y are topologically distinguishable. Let U X be an open subset that distinguishes x and y, say WLOG x U and y / U. By definition of, U is a union of equivalence classes (i.e. z U implies that any z z is also in U), which is saying π π(u) = U. Therefore π(u) is open in KQ(X) and contains π(x) = a. However π(u) does not contain b, since every u U is distinguishable from y, so that π(u) π(y) = b. 3. By the universal property of the quotient topology, if suffices to show that such a map f : X Y is constant on equivalence classes, i.e. x x implies f(x) = f(x ). Assuming f(x) f(x ), there is an open U Y that distinguishes f(x) and f(x ) since Y is T 0. Let s say WLOG f(x) U and f(x ) / U. Then f (U) X is an open that distinguishes x and x, as x f (U) and x / f (U). We conclude x x. 4
b. Show that X is regular if and only if its Kolmogorov quotient KQ(X) is T 3. Lemma. T 0 and regular implies T 3. Proof. Let us show that the space X is T 2. Let x, y X be distinct points. Since X is T 0, there is an open U X distinguishing x and y, say WLOG x U and y / U, i.e. y U c. Since U c is closed and disjoint from x, by regularity there are open sets U x and V satisfying x U x, U c V and U x V =. We have y U c V, so that U x and V are neighborhoods that separate x and y. ( ) Note that KQ(X) is automatically T 0. By the lemma, it suffices to show that KQ(X) is regular. Let C KQ(X) be closed, and a / C. Then the preimages π (a) and π (C) are disjoint subsets of X, and π (C) is closed. Pick any representative x π (a). Since X is regular, there are open subsets U, V X satisfying x U, π (C) V, and U V =. By the argument in part (a), π(u) and π(v ) are open in KQ(X) and contain a and C respectively (since π is surjective). Moreover, π(u) and π(v ) are disjoint. Indeed, U and V are disjoint and open, so that any points u U and v V are distinguishable, i.e. π(u) π(v). ( ) Let C X be closed and x / C. Because open subsets of X are unions of equivalence classes, the same holds for closed subsets. Therefore π(c) KQ(X) is closed and disjoint from the point π(x). Since KQ(X) is regular, there are open subsets U, V KQ(X) satisfying π(x) U, π(c) V, and U V =. Therefore π (U) and π (V ) are open in X, contain x and C respectively, and are disjoint. 5
Problem 3. Let (X, d) be a metric space. a. Let S X be a non-empty subset and consider the function f S : X R defined by f S (x) = d(x, S). Show that f S is Lipschitz continuous with Lipschitz constant, i.e. f S (x) f S (y) d(x, y) for all x, y X. In particular, f S is continuous. For every x, y X and s S, we have d(x, s) d(x, y) + d(y, s) and taking the infimum over s S yields d(x, S) d(x, y) + d(y, S) which can be rewritten as d(x, S) d(y, S) d(x, y). Interchanging the role of x and y, we also obtain d(y, S) d(x, S) d(x, y) and therefore f S (x) f S (y) = d(x, S) d(y, S) d(x, y). b. Show that closed subsets of X can be precisely separated by functions, i.e. for any A, B X disjoint closed subsets of X, there is a continuous function f : X [0, ] satisfying f(a) = 0 for all a A f(b) = for all b B f(x) (0, ) for all x / A B. First assume A and B are non-empty. Then treat the case B = separately. Hint: Recall the equivalence d(x, S) = 0 if and only if x S. When A and B are non-empty. Then the functions f A, f B : X R are welldefined and continuous. Consider the function f : X R defined by f(x) = This function satisfies the desired properties. f A (x) f A (x) + f B (x). 6
f is well-defined since the denominator is strictly positive on X: f A (x) + f B (x) = 0 f A (x) = 0 and f B (x) = 0 x A = A and x B = B x A B =. f is continuous, since the sum f A + f B is continuous, so that the quotient f = continuous on X. f A f A +f B is f takes values in [0, ], by the inequalities 0 f A (x) f A (x) + f B (x) for all x X. f satisfies: f satisfies: f(x) = 0 f(x) = f A (x) f A (x) + f B (x) = 0 f A (x) = 0 x A = A. f A (x) f A (x) + f B (x) = f A (x) = f A (x) + f B (x) f B (x) = 0 x B = B. When B = is empty. Then replace f B by the constant function : f(x) = f A(x) + f A (x). As in the previous case, the denominator + f A (x) is strictly positive on X, f is continuous, takes values in [0, ], and vanishes precisely on A. It remains it check: f(x) = f A(x) + f A (x) = f A (x) = + f A (x) x = B. Remark. A space is called perfectly normal if its closed subsets can be precisely separated by functions. A space is called T 6 if it is T and perfectly normal. We have just shown that every metric space is T 6. 7
Problem 4. metrizable. In this problem, we will show that a countable product of metrizable spaces is a. Let (X, d) be a metric space. Consider the function ρ: X X R defined by Show that ρ is a metric on X. ρ(x, y) = d(x, y) + d(x, y). Write ρ = h d where h: R 0 R 0 is the rescaling function defined by We check the three properties of a metric.. Positivity: since h(t) 0 for all t 0. h(t) = t + t. ρ(x, y) = h (d(x, y)) 0 for all x, y X ρ(x, y) = 0 h (d(x, y)) = 0 d(x, y) = 0 x = y where we used the property h(t) = 0 t = 0. 2. Symmetry: ρ(y, x) = h (d(y, x)) = h (d(x, y)) = ρ(x, y). 3. Triangle inequality: First note that h satisfies the sublinearity condition h(s + t) h(s) + h(t) 8
for all s, t 0. Indeed, we have: h(s) + h(t) = Thus the triangle inequality for d = s + s + t + t s( + t) + t( + s) ( + s)( + t) = s + t + 2st + s + t + st s + t + st + s + t + st s+t st +s+t s + t + + s + t + st = = (s + t)( + s + t) + (s + t)st ( + s + t + st)( + s + t) (s + t)( + s + t + st) ( + s + t + st)( + s + t) = s + t + s + t = h(s + t). d(x, y) d(x, z) + d(z, y) along with the fact that h is non-decreasing implies ρ(x, y) = h (d(x, y)) h (d(x, z) + d(z, y)) h (d(x, z)) + h (d(z, y)) = ρ(x, z) + ρ(z, y). 9
b. Show that the metric ρ from part (a) induces the same topology on X as the original metric d. Given h(0) = 0, continuity of h at 0 means that for any ɛ > 0, there is a δ > 0 guaranteeing h(t) < ɛ whenever t < δ. Substituting t = d(x, y), we obtain ρ(x, y) < ɛ whenever d(x, y) < δ, i.e. B d δ (x) B ρ ɛ (x) where the superscript denotes which metric is being used. This proves that every ρ-open is d-open. Since h is continuous and strictly increasing, it is a homeomorphism of a (small) neighborhood of 0 onto a (small) neighborhood of h(0) = 0. The local inverse h satisfies h (0) = 0 and is continuous at 0, therefore the argument above applies again. For any ɛ > 0, there is a δ > 0 guaranteeing h (s) < ɛ whenever s < δ. s = ρ(x, y), we obtain d(x, y) < ɛ whenever ρ(x, y) < δ, i.e. so that every d-open is ρ-open. B ρ δ (x) Bd ɛ (x) Substituting Remark. We could also have used the formula ρ(x, y) = min{d(x, y), }. The goal was just to find a metric ρ which is topologically equivalent to d and is bounded. 0
c. Let {(X i, d i )} i N be a countable family of metric spaces, where each metric d i is bounded by, i.e. d i (x i, y i ) for all x i, y i X i. Write X := i N X i and consider the function d: X X R defined by d(x, y) = i= 2 i d i(x i, y i ). Show that d is a metric on X. (First check that d is a well-defined function.) bounded: For any x, y X, the series defining d(x, y) has only non-negative terms and is i= 2 i d i(x i, y i ) hence the series converges, so that d(x, y) is well-defined. We check the three properties of a metric. i= 2 i =. Positivity: d(x, y) = i= 2 i d i(x i, y i ) i= 2 i (0) = 0 d(x, y) = i= 2 i d i(x i, y i ) = 0 2 i d i(x i, y i ) = 0 for all i N d i (x i, y i ) = 0 for all i N x i = y i x = y. for all i N 2. Symmetry: d(y, x) = = i= i= = d(x, y). 2 i d i(y i, x i ) 2 i d i(x i, y i )
3. Triangle inequality: For any x, y, z X, we have d(x, y) = = i= i= i= 2 i d i(x i, y i ) 2 i (d i(x i, z i ) + d i (z i, y i )) 2 i d i(x i, z i ) + = d(x, z) + d(z, y). i= 2 i d i(z i, y i ) 2
d. Show that the metric d from part (c) induces the product topology on X = i N X i. (T met T prod ) Let ɛ > 0 and consider the open ball B ɛ (x) centered at a point x X. We want to find an open large box U = i N U i inside B ɛ (x). Let N N be large enough to guarantee the inequality i=n+ 2 i < ɛ 2. For the indices i N, take the radii ɛ i := ɛ and consider the open large box U = 2 i N U i defined by { B ɛi (x i ) if i N U i := X i if i > N. Note that U is indeed open in the product topology, since U i X i is open for all i N and U i X i for finitely many indices i. For any point y U, its distance to x is d(x, y) = = < < = = ɛ 2 i= N i= N i= N i= N i= N i= 2 i d i(x i, y i ) 2 i d i(x i, y i ) + 2 i d i(x i, y i ) + 2 i d i(x i, y i ) + ɛ 2 2 i ɛ i + ɛ 2 ɛ 2 i 2 + ɛ 2 N i= < ɛ 2 + ɛ 2 = ɛ which proves the inclusion U B ɛ (x). 2 i + ɛ 2 i=n+ i=n+ 2 i d i(x i, y i ) 2 i () 3
(T prod T met ) Let U = i N U i be an open large box and x U. We want to find an open ball around x satisfying B ɛ (x) U. By definition of large box, there is an index N N satisfying U i = X i for all i > N. For the indices i N, we have x i U i and U i X i is open, so that there is a radius ɛ i > 0 satisfying x i B ɛi (x i ) U i. Take the radius ɛ = 2 N min{ɛ,..., ɛ N } and consider the open ball B ɛ (x). We claim B ɛ (x) U. Let y B ɛ (x). For indices i N, the point y satisfies 2 i d i(x i, y i ) j= = d(x, y) < ɛ 2 N ɛ i 2 i ɛ i 2 j d j(x j, y j ) or equivalently d i (x i, y i ) < ɛ i, which implies y i B ɛi (x i ) U i. For the remaining indices i > N, there is no constraint on y i, namely y i U i = X i automatically. This proves y i N U i = U and therefore the inclusion B ɛ (x) U. 4