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1. Introduction This Week We Will Cover the Topic of Product Spaces. Recall That the Ca CLASS NOTES FOR WEEK 8 (MAY 22-26, 2000) 1. Introduction This week we will cover the topic of product spaces. Recall that the Cartesian product of two sets X × Y is defined as the space of all pairs of elements (x, y) such that x ∈ X, y ∈ Y : X × Y := {(x, y) : x ∈ X,y ∈ Y }. More generally, if X1,... ,Xn are a finite collection of sets, the Cartesian product X1 × Xn can be defined as X1 × . × Xn := {(x1,... ,xn) : xi ∈ Xi for all 1 ≤ i ≤ n}. This product is sometimes abbreviated as n Y Xi := X1 × . × Xn. i=1 For instance, 3 Y{0, 1} = {(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)} i=1 is the space of all binary sequences of length 3. Even more generally, if one has an arbitrary collection {Xα}α∈A of sets, then one can define the product Y Xα := {(xα)α∈A : xα ∈ Xα for all α ∈ A}. α∈A For instance, ∞ Y{0, 1} i=1 is the space of all infinite binary sequences (sequences consisting only of 0s and 1s, such as (0, 1, 0, 1,... ).) We’ve just defined what a product means for sets, but we would also like to define what products mean for topological spaces. For instance, if X and Y are topological spaces, what kind of topology does X × Y get? What are the open sets, convergent sequences, etc? We’ll first discuss the product of two spaces. The product of n spaces is pretty much the same (just replace 2s by ns throughout). The infinite product case is more tricky conceptually and will be left to the Wednesday class. 1 2 CLASS NOTES FOR WEEK 8 (MAY 22-26, 2000) 2. Product of two spaces Let X1 and X2 be two topological spaces. The space X1 × X2 is the set of all pairs (x1, x2), where x1 ∈ X1 and x2 ∈ X2. Traditionally, we draw X1 as a horizontal set, X2 as a vertical set, and X1 × X2 as the rectangle with these co-ordinates. This picture can be a bit mis-leading though, since X1 and X2 are not necessarily one-dimensional. To make X1 × X2 into a topological space, we need to specify what the open sets of X1 ×X2 are. Well, if U1 is open in X1, and U2 is open in X2, then it seems plausible that U1 × U2 is open in X1 × X2. (The sets U1 × U2 look like open rectangles). Exercise 2.1. Show that this is actually the case when X1, X2 are metric spaces. However, these rectangles are not enough to form a topology. For instance, the union of two rectangles is not always a rectangle (e.g. take the union of (0, 1)×(0, 2) and (0, 2) × (0, 1)). However, they do form a base for a topology. Exercise 2.2. Show that the set {U1 ×U2 : U1 open in X1,U2 open in X2} satisfies the axioms for a base (see (4.1), (4.2) on p. 70 of the text). Definition 2.3. The product topology on X1 × X2 is defined to be the topology generated by the base {U1 × U2 : U1 open in X1,U2 open in X2}. In other words, a subset of X1 ×X2 is considered to be open in the product topology if and only if it is the union of open rectangles of the form U1 × U2, where U1 is open in X1 and U2 is open in X2. How is the product topology on X1 × X2 related to the topology of X1 and X2? The most direct relationship comes via the projection maps π1 : X1 × X2 → X1 and π2 : X1 × X2 → X2 defined by π1(x1, x2) := x1 π2(x1, x2) := x2. Theorem 2.4. The projections π1 and π2 are continuous and open. (A map is called open if the image of every open set is open; it’s like continuity, but in reverse). Proof We’ll just prove the theorem for π1, as the proof for π2 is similar. To show that π1 is continuous, we have to show that the inverse image of any open set in X1 is an open set in X1 × X2. So take any open set U1 in X1. The inverse −1 image π1 (U1) is the set of all points in X1 × X2 which project down to X1. If you think about it, that set is just U1 × X2. Since U1 is open in X1 and X2 is open in X2, U1 × X2 is open in X1 × X2. So π1 is continuous. Now to show that π1 is open. Let’s take any open set V in X1 × X2. We have to show that π1(V ) is open. Since V is open, it is the union of open rectangles. We CLASS NOTES FOR WEEK 8 (MAY 22-26, 2000) 3 can write this as α α V = [ U1 × U2 α∈A where A is an index set (finite or infinite, it doesn’t matter) and for each α ∈ A, α α U1 is an open set in X1 and U2 is an open set in X2. Of course we can assume that α α the sets U1 and U2 are non-empty (if any of these sets were empty, they wouldn’t contribute anything to the union and we could just throw them out). α α α It is clear that π1(U1 × U2 )= U1 for each α ∈ A, so α π1(V )= [ U1 α∈A (Exercise: prove this!). The right-hand side is the union of open sets, and is therefore open. Thus π1 is open. So the product topology has the nice property that the projections π1, π2 are continuous. In fact, it is the smallest topology with this property; we threw in the barest minumum of open sets in X1 × X2 which were required in order to make these maps continuous. If we made the topology any smaller, at least one of π1 and π2 would fail to be continuous. (Exercise: prove this!) The above theorem can be used to prove many theorems of the form If X1 × X2 have [insert property here], then X1 and X2 must individually have [insert property here]. For instance, if X1×X2 is connected, then π1(X1×X2) is connected, since the image of a connected set under a continuous map remains connected. But π1(X1 × X2)= X1, so X1 is connected. Similarly X2 is connected. Conversely, if X1 and X2 both have some topological property, then one can usually prove the same property for the product X1 × X2. For instance, the product of two connected sets is connected, two Hausdorff spaces is Hausdorff; the product of two compact sets is compact, and so forth. These are a little trickier to prove, though, and I’ll skip over them. Note that X1 × X2 doesn’t actually contain X1 or X2 directly. (X1 × X2 consists of pairs of elements, whereas X1 and X2 consist of individual elements). However, for every x2 ∈ X2, X1 × X2 contains the set X1 ×{x2}, which is homoeomorphic to X1. Exercise 2.5. For each x2 ∈ X2, show that the map fx2 : X1 → X1 ×{x2} defined by fx2 (x1) = (x1, x2) is a homeomorphism from X1 to X1 ×{x2}. (Of course, X1 ×{x2} is given the relative topology induced by X1 × X2). Thus X1 × X2 consists of many “horizontal slices”, each of which is homeomorphic to X1. One can similarly divide X1 × X2 into vertical slices, each of which is homeomorphic to X2. 4 CLASS NOTES FOR WEEK 8 (MAY 22-26, 2000) Suppose f : Y → X1 × X2 is a continuous map from some topological space Y to a product space X1 ×X2. This map has two components, π1 ◦f : Y → X1 and π2 ◦f : Y → X2. For instance, if f : [0, 2π] → R × R is the curve f(t) = (cos(t), sin(t)), then we can break it into the components π1 ◦ f(t) = cos(t) and π2 ◦ f(t) = sin(t). Since f, π1, and π2 are all continuous, we see that π1 ◦ f and π2 ◦ f are continuous. In other words, the components of a continuous function are also continuous. The converse is also true: Theorem 2.6. Let f : Y → X1 × X2 be such that π1 ◦ f and π2 ◦ f are both continuous. Then f is also continuous. −1 Proof Let V be any open set in X1 × X2. Our job is to show that f (V ) is an open set in Y . Let y be any point in f −1(V ). We have to show that y is an interior point of f −1(V ). Well, since y ∈ f −1(V ), we know that f(y) ∈ V . Since V is open, we therefore know that f(y) is an interior point of V . Since the product topology uses open rectangles as a base, there must exist an open rectangle U1 × U2 inside V which contains f(y). −1 −1 Since U1 × U2 is inside V and contains f(y), the set f (U1 × U2) is inside f (V ) and contains y. This will allow us to show that y is an interior point of f −1(V ) −1 provided that f (U1 × U2) is open. If we knew that f was continuous, then we’d be done, since U1 × U2 is continuous. But this is exactly what we’re trying to prove! So we can’t use that. However, we do know that π1 ◦ f and π2 ◦ f are continuous, so we should try to use that.
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