Examples of topological spaces

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1 Examples of topological spaces John Terilla Fall 2014 Contents 1 Introduction 1 2 Some simple topologies 2 3 Metric Spaces 2 4 A few other topologies on R 1 and R The Zariski Topology 4 6 Continuous functions 4 7 Limit and boundary points 6 8 Sequences and separation 6 9 Sequences and closure 8 10 Sequences and continuity 8 11 Separation, closure, and continuity 9 12 Sequences and first countable spaces 9 1 Introduction I m assuming that students know linear algebra and have had some abstract algebra, for example, principal ideals appear in these notes without definition. Later, I ll be assuming students know what a group is, how to define the quotient of a group by a normal subgroup, etc... 1

2 I m also assuming that students have some basic working knowledge about sets. Notions such as function and countable appear in these notes without definition. It s assumed that students know these things, as well as concepts such as equivalence relation, partial order, Cartesian product, surjective, etc... There s usually a review of basic set theory in the first chapter of a general topology book, for example Chapter 0 of [2] or Chapters 1, 2, 3 in [3]. 2 Some simple topologies Example 1. Let X be any set. The discrete topology on X is defined to be 2 X. The indiscrete topology on X is defined to be {, X}. Problem 1. Draw a diagram of all the topologies on a three point set indicating which are contained which. Problem 2. The integers Z are given the discrete topology unless specified otherwise. There s another topology on Z for which sets az + b = {a + nb : n N} for a Z\{0} and b Z, together with, are open. Furstenberg [1] used this topology in a delightful proof that there are infinitely many primes. One can check that the sets az + b are also closed in this topology. Since every integer except ±1 has a prime factor Z \ { 1, 1} = S(p, 0). primes p Since the left hand side is not closed (no nonempty finite set can be open) there must be infinitely many closed sets in the union on the right. Therefore, there are infinitely many primes. 3 Metric Spaces A metric space is a pair (X, d) where X is a set and d : X X R satisfies d(x, y) 0 for all x, y X, d(x, y) = d(y, x) for all x, y X, d(x, y) + d(y, z) d(x, z) for all x, y, z X d(x, y) = 0 if and only if x = y for all x, y X. 2

3 The function d is called a metric. If (X, d) is a metric space, x X, and r > 0, the ball centered at x of radius r is defined to be B(x, r) = {y X : d(x, y) < r}. The balls {B(x, r)} form a basis for a topology on X called the metric topology. Example 2. For any x, y R n, d(x, y) = n (x i y i ) 2 i=1 where x = (x 1,..., x n ) and y = (y 1,..., y n ) defines the usual metric on R n and the induced metric topology on R n is called the usual topology on R n. Example 3. Let C([0, 1]) denote the set of continuous functions on [0, 1]. The following define metrics on C([0, 1]): d(f, g) = ρ(f, g) = sup f(x) g(x). x [0,1] 1 0 f g. Example 4. Let l 2 be the set of all the real valued sequences {x n } for which n=1 x2 n converges. Then d : l 2 l 2 R defined by d ({x n }, {y n }) = (x n y n ) 2 defines a metric. Example 5. In any normed vector space, the function d(x, y) = x y defines a metric. Some of the examples above are metrics arising this way. But there are metric spaces that are not linear. For example, if (X, d) is a metric space and Y is any subset of X, (Y, d) will be a metric space, where d is the metric on X restricted to Y Y. Problem 3. Let (X, d) be a metric space. Show that for any x X and any r > 0, the set {y X : d(x, y) r} is closed. Give an example to show that n=1 {y X : d(x, y) r} may not equal B(x, r). Problem 4. Prove that every metric space is first countable. See definition 6 in these notes. 3

4 4 A few other topologies on R 1 and R 2. Example 6. Check that the sets [a, b) for a < b form a basis for a topology on R. This topology is called the lower limit topology, or the Sorgenfrey topology, or the uphill topology, or the half-open topology, and it probably goes by other names too. Compare this lower limit topology to the ordinary topology. Example 7. In general, the intervals (a, b) = {x X : a < x < b} define a topology on any totally ordered set (include the intervals (a, ) and (, b)). The set R is totally ordered and the order topology on R coincides with the usual topology. The lexicographic order makes R 2 totally ordered. Compare the lexicographic order topology on R 2 to the ordinary one. Example 8. Any set X has a cofinite topology where a set U is open if and only X \ U is finite (or if U = ). The open sets in the cocountable topology are those whose complement is countable. Compare the cofinite and cocountable topologies on R to the usual one. 5 The Zariski Topology Problem 5. Let R be a ring (commutative, with 1) and let spec(r) denote the set of prime ideals of R. The Zariski topology on spec(r) is defining the sets V (E) = {p spec(r) : E p} for any E R to be closed. Check that these closed sets do define a topology on spec(r) and sketch a picture of spec(c[x]) and spec(z). 6 Continuous functions Definition 1. A function f : X Y between metric spaces is called continuous at a point x X if and only if for every ɛ > 0 there exists a δ > 0 so that d(f(x), f(x )) < ɛ whenever d(x, x) < δ. Equivalently, f is continuous at x if and only if for every ɛ > 0 there exists a δ > 0 so that B(x, δ) B(f(x), ɛ). The function f is called continuous if it is continuous at every point x X. The concept that f : X Y is continuous can be defined without reference to the points of X. Here s a slightly better definition. Better Definition. A function f : X Y is continuous if and only if for every open set U Y, f 1 (U) is open in X. It s left as an exercise to check that both definitions are equivalent. Although they are equivalent, the second one is better because it makes sense for any function f : X Y between topological spaces. It s redundant, but for the record: 4

5 Definition 2. A function f : X Y between two topological spaces is continuous if and only if for every open set U Y, f 1 (U) is open in X. The argument that shows the equivalence between the first definition and the better definition of continuous functions between metric spaces generalizes and proves the following: Proposition 1. A function f : X Y between topological spaces is continuous if and only if f 1 (B) is open for every B in a basis for the topology on Y. Here are a few examples Example 9. Let X be a topological space. If S is any set with the discrete topology, then any function f : S X is continuous. For example, a sequence x : N X is continuous. (Unless otherwise specified, the natural numbers N is given the discrete topology = metric topology on N inherited from the metric topology on R). As in other subjects, if x : N X is a sequence, the notation x n is used to denote x(n) and {x n } may be used to denote the sequence x : N R. Example 10. If (X, d) is a metric space and x X, then the function f : X R defined by f(y) = d(x, y) is continuous. One way to study a topological space X is to study the continuous functions from X or the continuous functions to X. For example, the fundamental group of X looks at functions from the circle to X. Also, the set hom(x, R) of continuous functions from X to R is often interesting. Using the definition of addition and multiplication in R, the set hom(x, R) becomes an algebra. It s a good exercise to check: (1) that constant functions are continuous, (2) the sum and product of continuous functions is continuous. Theorem 1. For any topological space X, the idenity id X : X X is continuous. For any topological spaces X, Y, Z and any continuous functions f : X Y and g : Y Z, the composition g f : X Z is continuous. This theorem, though easy to prove, is important since it establishes that topological spaces and continuous functions form a category. The set of continuous functions from X to Y is denoted hom(x, Y ). Since topological spaces forms a category, there is a definition of the equivalence. Definition 3. A continuous function f : X Y is an equivalence, or homeomorphism, if there exists a continuous function g : Y X satisfying f g = id Y and g f = id X. In this case, the spaces X and Y are called equivalent, or homeomorphic. Topology, then, is essentially the study of properties that are preserved under homeomorphisms. Such properties are called topological. 5

6 Example 11. Not every continuous bijection is a homeormorphism. For example, the identity function id : (R, τ discrete ) (R, τ usual ) is a continuous bijection that is not a homeomorphism. Consistent with the themes of category theory, a topological space X is determined by the continuous functions from X. To see this, let S = {0, 1} with the topology {, {1}, S} S is called the Serpinski space. Now, for any open set U X, the function χ U : X S defined by χ U (x) = { 1 if x U, 0 if x / U is a continuous function and every continuous function from X S is a χ U for some open set U. Therefore, a copy of the entire topology of X is encoded in the set hom(x, S). 7 Limit and boundary points For easy reference, a few standard definitions are stated here. Let A be a subset of a topological space X. The closure of A is the smallest closed set containing A and we denote it by A. The interior of A is the largest open set contained in A and we denote it by A. A point x X is a boundary point of A iff every open set containing x contains both a point of A and a point of X \A. We denote the set of boundary points of A by A. A point x X is a limit point of A iff every open set containing x contains a point of A, not equal to x. We denote the set of limit points of A by A. A is called dense if A = X and is called nowhere dense if ( A ) =. Problem 6. Here s a clever problem from [2], due originally to Kuratowski. Let A be a subset of a topological space X. What is the maximum number of distinct sets that can be obtained by iteratively applying complement and closure? Find a subset A of the real numbers that realizes this maximum number. Note: The number is finite, although that is not obvious. 8 Sequences and separation Recall Definition 4. Let X be a topological space. A sequence in X is a function x : N X. We usually write x n for x(n) and may denote the sequence {x n }. A sequence {x n } converges to z X if and only if for every open set U containing z, there exists an N N so that if n N, x n U. If {x n } converges to z X we write {x n } z. 6

7 Definition 5. Three fundamental separation axioms: A topological space X is T 0 iff for every pair of points x, y X there exists an open set containing one, but not both of them. A topological space X is T 1 iff for every pair of points x, y X there exists open sets U and V with x U, y V with x / U and y / V. A topological space X is T 2, or Hausdorff, iff for every pair of points x, y X there exists open sets U and V with x U, y V with U V =. Observe that each separation axiom defines a topological property. Now, a few examples: Example 12. Let A = {1, 2, 3} with the topology τ = {, {1}, {1, 2}, A}. Then the constant sequence 1, 1, 1, 1,... converges to 1; it also converges to 2 and to 3. Observe that A is not T 1 : there is no open set around 1 separating it from 2. Example 13. Consider Z with the cofinite topology. For any m Z, the constant sequence m, m, m,... converges to m and only to m. For if l m, the set R \ m is an open set around l contiaining no elements of the sequence. The sequence {n} = 1, 2, 3, 4,... converges to m for every m Z. To see this, let m be any integer and let U be a neighborhood of m. Since Z \ U is finite, there can only be finitely many natural numbers in R \ U. Let N be larger than the greatest natural number in R \ U. Then, if n > N, n U, proving that {n} m. Example 14. Consider R with the usual topology. If {x n } x, then {x n } does not converge to any number y x. To prove it, we can find disjoint open sets U and V with x U and y V (we can be explicit if necessary: U = (x c, x + c) and V = (y c, y + c) where c = 1 2 x y ). Then, there is a number N so that x n U for all n N. Since U V =, V cannot contain any x n for n N and hence {x n } does not converge to y. Now, a few theorems: Theorem 2. If X is T 1, then for any x X, the constant sequence x, x, x,... converges to x and only to x. Proof. Suppose X is T 1 and x X. It s clear that x, x, x,... x. Let y x. Then there exists an open set U with y U and x / U. Therefore, x, x, x,... cannot converge to y. Theorem 3. In a Hausdorff space, limits of sequences are unique. Proof. Let X be Hausdorff, let {x n } be a sequence with {x n } x and {x n } y. If x y, then there are disjoint open sets U and V with x U and y V. Since {x n } x there is a number N so that x n U for all n N. Since {x n } y there 7

8 is a number K so that x n U for all n K. Let M = max N, K. Since M N and M K we have x M U and x M V contradicting the fact that U and V are disjoint. Now, in fact, T 1 spaces are characterized by the property that constant sequences have only one limit. Theorem 4. If X is any topological space for which every constant sequence x, x, x,... converges to x and only to x, then X is T 1. Proof. If X is not T 1, there exist two distinct points x and y for which every open set around y contains x. Then the sequence x, x, x,... y. One might speculate that Hausdorff spaces are characterized by the property that convergent sequences have unique limits, but this isn t true. See Theorem Sequences and closure Let X be a topological space and A X. Theorem 5. If {x n } is a sequence in A that converges to x, then x A. Proof. Suppose {x n } x and each x n A. Recall that A = A A. So it suffice to prove that if x / A, then x A. Let U be a neighborhood of x. Since {x n } x, there exist infinitely many, and in particular one, x n U. This shows that there exists an element of A, not equal to x, in every open set around x. That is, x A. The converse of Theorem 5 is false. Example 15. Consider R with the cocountable topology and let A = [0, 1]. Then A = R but there s no sequence in A that convergest to 10, for example. Theorem 6. A closed set contains the limits of all its convergent sequences Proof. This follows from Theorem 5 since a closed set equals its closure. 10 Sequences and continuity Let X and Y be topological spaces and f : X Y be a function. Theorem 7. If f is continuous and {x n } is a sequence in X with {x n } x, then {f(x n )} f(x). Proof. Suppose f is continuous and {x n } x. Let U be an open set containing f(x). Since f is continuous, there exists an set V containing x with f(v ) U. Since {x n } x, there exists a natural number N so that n n x n V. Then f(x n ) U for n N, proving that {f(x n )} f(x). 8

9 11 Separation, closure, and continuity Here s a good problem. It doesn t use sequences, but it does nicely involve separation, closure, continuity. Problem 7. Let X and Y be topological spaces and suppose f : X Y is continuous, surjective, and open. Prove that Y is Hausdorff if and only if is closed in X X. {(x, x ) X X : f(x) = f(x )} 12 Sequences and first countable spaces Sequences are important tools in analysis. Indeed much of analysis can be characterized using sequences. For example, Theorem 8. f : R n R m is continuous if and only if for every sequence {x n } in R n with {x n } x, the sequence {f(x n )} f(x). A common way to show that a given function is not continuous is to find a sequence {x n } converging to a point x R n for which the sequence {f(x n )} doesn t converge to f(x). Comparing Theorems 7 and 8, one sees that for the topological spaces X = R n and Y = R m, the converse to Theorem 7 is true, but the converse of Theorem 7 is not true in general. There are spaces X and Y and discontinuous functions f : X Y for which {f(x n )} f(x) for every convergent sequence {x n } x. Before getting into examples, let us first isolate the important topological property possessed by R n that makes the converse of Theorem 7 true. The feature is that R n has a countable neighborhood base: Definition 6. We say a topological space X is first countable if for every x X, there exists a countable collection {U x,n } n=1 of neighborhoods of x with the property that if U is any neighborhood of x, there exists a U x,n with U x,n U. The collection {U x,n } is called a countable neighborhood base. Also, see definition 3.4 in [4]. Observe that being first countable is a topological property. Example 16. Every metric space is first countable. To see this, let X be a metric space and recall that a set U is open in the metric topology if and only if for every x U, there exists an ɛ > 0 so that B(x, ɛ) U. Since for any ɛ > 0, there exists a natural number n with 0 < 1 n < ɛ, we see that the collection ( U x,i = B x, 1 ), for x X and n N n defines a countable neighborhood base for X. 9

10 Theorem 9. Suppose X and Y are first countable. Then f : X Y is continuous if and only if for every sequence {x n } in X with {x n } x, the sequence {f(x n )} f(x). Proof. It was already proved that if f is continuous and {x n } is a sequence in X with {x n } x then {f(x n )} f(x). The only thing to prove is the converse. So, suppose f is not continuous. So, there is a point x X and a neighborhood U of f(x) so that for every neighborhood V of x, f(v ) U. Let {V x,i } be a neighborhood base of x. For each n N, we have f(v x,1 V x,n ) U, so there exists a point x n V x,1 V x,n with f(x n ) / U. By this construction {x n } x, and {f(x n )} f(x). Since R n is a metric space, it s first countable so Theorem 8 is simply a special case of Theorem 9. One might summarize the above discussion and the previous theorem as saying that sequences suffice to describe continuous functions between first countable spaces. More is true: sequences essentially characterize the topology of first countable spaces. Theorem 10. Let X be a first countable space. Then X is Hausdorff if and only if the limits of convergent sequences are unique. Proof. It was already proved that if X is Hausdorff, the limits of convergent sequences are unique. So, suppose that X is not Hausdorff. Then, there exist two points x, y X for which every neighborhood of x intersects every neighborhood of y. Let {U x,i } and {V x,i } be countable neighborhood bases of x and y guaranteed by the first countability of X. We construct a sequence {z n } as follows: for each n N, U x,1 U x,n is a neighborhood of x and V x,1 V x,n is a neighborhood of y and therefore they must intersect. Choose a point z n U x,1 U x,n V x,1 V x,n. Exercise: verify that the sequence {z n } converges to both x and y. Theorem 11. Let X be a first countable space and let A X. A point x A if and only if there exists a sequence {x n } in A with {x n } x. Proof. It was already proved that if {x n } is a sequence in A with {x n } x, then x A. So suppose that x A. If x A, then the constant sequence x, x, x,... is a sequence in A converging to x. If x A then we find a sequence {x n } converging to x using a countable neighborhood base {U x,i }. For each n N, the set U x,1 U x,n is an open set containing x. Because x A, there exists a point of A, call it x n, with x n U x,1 U x,n. The remaining check that the sequence thus defined converges to x is left as an exercise. Theorem 12. A set in a first countable space is closed if and only if it contains the limits of all its convergent sequences. Proof. This is an immediate corollary of the previous theorem. 10

11 References [1] H. Furstenberg. On the infinitude of primes. Amer. Math. Monthly, 62(5):353, May [2] J. L. Kelley. General Topology. Van Nostrand, [3] Seymour Lipschutz. Schaum s Outline of Theory and Problems of General Topology. McGraw-Hill, [4] J.P. May. An outline summary of basic point set topology. may/misc/topology.pdf. [5] S. Willard. General Topology. Addison-Wesley,