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Home , Product topology, Subspace topology

Lecture 6: the subspace and product , their defining properties and an introduction to

Saul Glasman September 19, 2016

The examples of last time illustrate a general principle: Principle 1. Finer topologies are easier to map out of, and coarser topologies are easier to map into. Finding the right is often a matter of balancing fineness and coarseness so that we have a good supply of maps both in and out. We’ll see plenty of examples of this. So far we have a fairly limited range of example of topological spaces - the real numbers and various artificial examples, and that’s about it. We’ll now start introducing constructions that will massively increase our range of available topological spaces. First, we’ll describe how to put topologies on of existing topological spaces. Definition 2. Let X be a and K ⊆ X be a . Then the on K is the topology such that L ⊆ K is open if and only if there’s some open subset U ⊆ X such that U ∩ K = L. Let i : K → X be the inclusion of the subset K - this is a boring that’s basically the identity function, right, we just changed the domain. Let’s notice that the definition of the subspace topology is set up exactly so that i is continuous when K is equipped with the subspace topology. All this continuity requires that whenever U ⊆ X is open, U ∩K is an open subset of K, and that’s what we legislated above. So, Fact 3. The subspace topology on K is the coarsest topology on K which makes i : K → X continuous. Why would we want as coarse a topology as possible? Note that if we chose the finest topology instead, we’d just end up with the discrete topology on K, since all functions out of the discrete topology are continuous. The topology on K has to be fine enough that i is continuous, but coarse enough to retain all the structure of K. And this definition allows us to prove the following theorem, which is the most important property of the subspace topology:

1 Theorem 4. Let Y be a topological space. Let Hom(Y,K) be the set of con- tinuous functions from Y to K, and let HomK (Y,X) be the set of continuous functions f : Y → X such that f(Y ) ⊆ K. Then composition with i gives a bijection i◦ : Hom(Y,K) → HomK (Y,X). Proof. First, suppose g : Y → K is continuous. Then i ◦ g is continuous, and its image is in K, so the function is well defined with the domain and codomain I gave. It’s clear that i◦ is injective. To prove surjectivity, let f ∈ HomK (Y,X). As a function, there is certainly some function g : Y → K such that f = i ◦ g; we just need to prove that g is continuous. Let L ⊆ K be open; then there is some U ⊆ X with U ∩ K = L. It follows that g−1(L) = f −1(U) and so g−1(L) is open, since f is continuous. As an example of the power of the subspace topology, note for example that this allows us to put a topology on the circle for the first time:

1 2 2 2 S = {(x, y) ∈ R | x + y = 1} or indeed, the n-sphere:

n n+1 2 2 2 S = {(x1, x2, ··· , xn+1) ∈ R | x1 + x2 + ··· + xn+1 = 1}. Later in the course we’ll see that the topological space S1 is really essentially different from other spaces we’ve encountered such as R. Example 5. Here’s an example showing that the open sets in the subspace topology might be slightly surprising at first. Consider the interval [0, 1] ⊆ R with its subspace topology. Then, since 1 1 1 (− , ) ∩ [0, 1] = [0, ), 2 2 2 1 the half-open interval [0, 2 ) is an open subset of [0, 1]. In fact, the open subin- tervals {(a, b) | 0 ≤ a < b ≤ 1 together with the half-open intervals {[0, a) | 0 < a ≤ 1} ∪ {(a, 1] | 0 ≤ a < 1} form a basis for the topology on [0, 1]. That’s enough about the subspace topology for now. Let’s move onto our next example of a topology: the . Given two topological spaces X and Y , the task is to put a topology on the X × Y . We’ve already seen an example of this, with the topology we defined on R2 (with two different bases). Let’s see how it works in general.

2 Definition 6. Let X and Y be topological spaces. Then the product topology on X × Y is the topology generated by the basis

BX×Y = {U × V | U ⊆ X open,V ⊆ Y open.}

In the case where X = Y = R, this gives the open rectangle basis for the topology on R2. Remark 7. Note that

(U0 × V0) ∩ (U1 × V1) = (U0 ∩ U1) × (V0 ∩ V1).

So BX×Y is closed under finite intersections, and so it is indeed a basis for a topology. Lemma 8. Let pX : X × Y → X, pY : X × Y → Y be the functions:

pX (x, y) = x, pY (x, y) = y.

Then the product topology is the coarsest topology on X × Y for which pX and pY are continuous. Proof. If U ⊆ X is open, then

−1 pX (U) = U × Y, which is a basis set, so pX is continuous. Obviously the proof that pY is con- tinuous is identical. If T is a topology on X × Y for which pX and pY are continuous, then for −1 −1 each open U ⊆ X and V ⊆ Y , pX (U) = U × Y and pY (V ) = X × V are open in T . But then (U × Y ) ∩ (X × V ) = (U × V ) is open in T . So T contains BX×Y , and thus T contains the product topology.

Again, we see the balance between coarseness and fineness, and again, it buys us a theorem characterizing continuous functions into X × Y : Theorem 9. Let T be a topological space. Define a function

φ : Hom(T,X × Y ) → Hom(T,X) × Hom(T,Y ) by φ(f) = (pX ◦ f, pY ◦ f). Then φ is a bijection. Thus, giving a into X×Y is equivalent to giving a pair consisting of a continuous function into X and a continuous function into Y .

3 Proof. Again, φ is clearly injective. To prove that φ is surjective, suppose gX : T → X and gY : T → Y are functions. We must show that the function g : T → X × Y given by g(t) = (gX (t), gY (t)) is continuous. Remember that because the preimage of a is the union of the preimages, continuity can be checked on a basis. Thus we must show that for each open U ⊆ X, V ⊆ Y , g−1(U × V ) is open in T . But

−1 −1 −1 g (U × V ) = gX (U) ∩ gY (V ) which is open. We’ll take a brief break from describing new kinds of topological spaces to answer the following question. When we had a bijection between two sets, we decided we could consider them essentially the same. When can we consider topological spaces ”the same”? The answer: Definition 10. Let X and Y be topological spaces. A continuous function f : X → Y is called a homemorphism if f is a bijection and its inverse f −1 is also continuous. Clearly if f is a , then the inverse function f −1 is also a homeomorphism. If there is a homeomorphism from X to Y , we say X and Y are homeomorphic and treat them as pretty much the same.

Example 11. R and R>0 are homeomorphic. Indeed, consider the function

exp : R → R>0. Then exp is a homeomorphism, since it’s a bijection, and its inverse ln is also continuous (you’ve probably seen the continuity of exp and ln in calculus).

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