Topology - Washington State University

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Math 490-01 Notes 5

Topology We will begin our study of topology from a set-theoretic point of view. As the subject expands, we will encounter various notions from analysis such as compactness, continuity, connectedness, and sequential convergence, as well as 'homeomorphism', which is a more geometric notion.

Def. N5.1 A topology on a set X is a subset of the power set 2X with the following properties:

(1) , X ; (2) {Ui i I} = Ui ; (closure under arbitrary unions)

iI

(3) U, V = U V . (closure under finite intersections) The members of are called -open sets, and the pair (X, ) is called a topological space. There are many familiar topological spaces encountered in analysis, the most well-known being (R, u), where u denotes the usual topology on R consisting of unions of open intervals.

For any non-empty set X, let (X) denote the set of all topologies on X. Certain topologies that can be defined on any set X are:

(a) Discrete topology: d = 2X; (b) Indiscrete topology: i = {, X}; (c) Cofinite topology: cof = {} {X - F F is finite and F X}; (d) Cocountable topology: coc = {} {X - C C is countable and C X}; Note that if X is finite, then coc = cof = d; if X is countable, coc = d. The set (X) of all topologies on X is partially ordered by set inclusion, so that 1 2 iff 1 2. The condition 1 2 can be stated as "2 is finer than 1" or "1 is coarser than 2".

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Relative to this ordering, the discrete topology d is the largest (or finest) topology on X, and the indiscrete topology i is the smallest (or coarsest). u For any set X, i cof coc d. But lest you be misled into thinking that (X) is simply ordered, note that u and coc are not comparable when X = R, since R - Q is coc-open but not u-open, whereas (0, 1) is u-open but not coc-open. The ordering of the 5 topologies d, i, cof, coc and u on R is shown in the diagram on the right.

d coc

cof i

A poset (A, ) is defined to be a lattice iff for every pair x, y A, sup{x, y} and inf{x, y}

both exist in A. A poset (A, ) is called a complete lattice iff for every non-empty subset

B A, sup B and inf B both exist in A. If A denotes the set of all proper subsets of some

non-empty set S, ordered by inclusion, then A is a poset but not a lattice. If S is infinite,

then the set of all finite subsets of S, ordered by inclusion, is a lattice but not a complete lattice. The power set 2X, ordered by inclusion, is a complete lattice.

Prop. N5.1 For any set X, (X) is a complete lattice.

Proof : Let T = {i i I} (X), and let = T . Claim 1: is a topology on X. Indeed, and X are in i for all i, and hence in T . If U is any set of subsets of , then U i for all i I, and so U i for all i I, since each i is a topology. Thus U . Similar reasoning works for finite intersections, proving claim 1. Claim 2: = inf T . If i for all i, then . But since itself is a lower bound of T in (X) by claim 1, = inf T . Claim 3: Define U (T ) to be the set of all upper bounds of T in (X), and let ? = U (T ). Then ? = sup T . Since d U (T ), U (T ) = , so ? = U (T ) is a topology on (X) (by claim 1) such that i ? for all i I, so ? is an upper bound for T . Finally, ? is clearly the least of the upper bounds of T .

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One might wonder why, in the preceding proof, we did not try sup T = T . The reason is that unions of topologies are not generally topologies. For example, consider the case where X = {a, b, c}. Several of the members of (X) are shown pictorially in Munkres on page 76. If we label three of those topologies as follows:

1 = 2 = 3 = then we can write 1 2 = 2 3 = Note that 1 2 is not a topology, because it is not closed under unions, and 2 3 is not a topology because it is not closed under finite intersections. In general, a union of topologies does not satisfy either of these conditions. One can, however, describe ? in the preceding proof as the collection of sets consisting of all unions of finite intersections of sets in T .

Basis and Subbasis for a Topology The preceding discussion raises the following important question: Given an arbitary collection B of subsets of X, is there a unique smallest topology B containing B? The answer is yes. The most simple and direct way to construct B from B is to treat B as a subbasis, which is defined as follows. A collection of sets B is said to be a subbasis for a topology B on X iff B covers X, and B consists precisely of all unions of finite intersections of members of B. B is called a basis for topology B on X iff B covers X, and B consists of all unions of members of B. Note that the requirement that B cover X (i.e. X B) is needed to guarantee that X B.

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One way to get a basis for a topology is to start with a subbasis S and let BS be the set of all finite intersections of members of S; then BS will be a basis, and S = BS . However, a basis need not be closed under finite intersections. Almost all topologies used in analysis have a basis consisting of "open balls" relative to some metric or norm, and these are not usually closed under finite intersections.

Most topologies we'll use as examples will be defined by first specifying a basis. However, we shall also encounter some topological properties which are defined by specifying the existence of a certain type of basis for the topology. Thus, when asking which collections B form a basis, two questions naturally arise:

(1) When is B 2X a basis for some topology on X? (2) Given (X), when is B a basis for ? The answer to question (1) is actually given as the definition of a basis in Munkres, on page 78. The answer to question (2) is given in Lemma 13.2.

Theorem N5.1 Let B 2X. Then: (1) B is a basis for some topology B on X iff B covers X and, for all U, V B and for all x U V , there exists W B such that x W U V ; (2) If (X) and B , then B is a basis for iff B covers X and for all U and for all x U , there exists V B such that x V U .

Proof : (1)(=) Assume B is a basis for a topology B on X. Then each member of B, including X, is a union of members of B, so B covers X. If U, V B, then U, V B, so U V B, which means U V is a union of members of B. Therefore for any x U V , there is a set W B such that x W U V .

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(1)(=) Assume B covers X, and let B consist of all unions of members of B. We need to show that B is a topology. "... all unions of members of B" includes the union over the empty set , which is , so B. Also, X B because B covers X, by assumption. If {Ui i I} B is an arbitrary collection of subsets from B, then Ui is a union of unions of members of B, and so by definition of B, Ui B. If U, V B, then by assumption, x U V = Wx B such that x Wx U V . Therefore, we can write U V = {Wx x U V }, which is a union of members of B, so U V B. (2)(=) This follows from (1), since B is assumed to be a basis for a topology. (2)(=) We must show that every set in is a union of members of B. If U and x U , then by assumption there is a Vx B such that x Vx U . Thus U = {Vx x U }.

In general, there are many bases for a given topology. The most familiar basis for the usual topology u on R is B = {(a, b) a, b R, a < b}. However, Theorem 1 part (2) can be used to show that B = {(a, b) a, b Q, a < b} is also a basis for u. The latter basis B is countable, and the existence of a countable basis is an important property of a topology called second countability, which we shall study later.

Suppose A and B are two topological bases on the same set X, and each B B is a union of members of A. Then each B-open set is also A-open, and B A. Of course, if it happens that B A and A B, then B = A. Lemma 13.3 in Munkres (stated here as Prop. 2) addresses these ideas.

Prop N5.2 Let A and B be two topological bases on X. The following are equivalent: (1) B A; (2) x X and B B, (x B B = A A such that x A B).

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