Topology Course Lecture Notes

Aisling McCluskey and Brian McMaster

August 1997

Chapter 4
Product Spaces

A common task in topology is to construct new topological spaces from other spaces. One way of doing this is by taking products. All are familiar with identifying the plane or 3-dimensional Euclidean space with ordered pairs or triples of numbers each of which is a member of the real line. Fewer are probably familar with realizing the torus as ordered pairs of complex numbers of modulus one. In this chaper we answer two questions:

How do the above product constructions generalize to topological spaces?
What topological properties are preserved by this construction?

4.1 Constructing Products

The process of constructing a product falls naturally into two stages.

The first stage, which is entirely set-theoretic, consists in describing an element of the underlying set of the product. This task is primarily one of generalizing the notion of ordered pair or triple.
The second stage is describing what open sets look like. This will be done by describing a subbasis for the topology. The guiding goal is to provide just enough opens sets to guarantee the continuity of certain important functions.

4.1.1 Set-Theoretic Construction

Suppose throughout that we are given a family of topological spaces {(X_i, T_i): i Î I } where I is some non-empty `labelling' or index set.

Our first task is to get a clear mental picture of what we mean by the product of the sets X_i. Look again at the finite case where I = {1,2, ¼,n}. Here, the product set

X = X₁ ×X₂ ×X₃ ×¼×X_n =

n
Õ
i = 1

X_i = {(p₁,p₂, ¼,p_n):p_i Î X_i, i Î I }.

i.e. the elements of X are the functions x:I ® È_{i = 1}ⁿX_i such that x(1) Î X₁, x(2) Î X₂, ..., x(n) Î X_n i.e. x(i) Î X_i "i where, for convenience, we usually write x_i instead of x(i). In this form, the definition extends immediately to any I, finite or infinite i.e. if { X_i: i Î I } is any family of sets, then their product is

{x:I ® È_{i Î I}X_i for which x(i) Î X_i "i Î I }

except that we normally write x_i rather than x(i).

Then a typical element of X = ÕX_i will look like: (x_i)_{i Î I} or just (x_i). We will still call x_i the i^th coordinate of (x_i)_{i Î I}. ( Note that the Axiom of Choice assures us that ÕX_i is non-empty provided none of the X_i's are empty.)

4.1.2 Topologizing the Product

Of the many possible topologies that could be imposed on X = ÕX_i, we describe the most useful. This topology is 'just right' in the sense that it is barely fine enough to guarantee the continuity of the coordinate projection functions while being just course enough allow the important result of Theorem .

Definition 1 For each i Î I, the i^th projection is the map p_i:ÕX_i ® X_i which `selects the i^th coordinate' i.e. p_i((x_i)_{i Î I}) = x_i.

An open cylinder means the inverse projection of some non-empty T_i-open set i.e. p_i^-1(G_i) where i Î I, G_i ¹ Æ, G_i Î T_i.

An open box is the intersection of finitely many open cylinders Ç_{j = 1}ⁿ p_{i_j}^-1(G_{i_j}). The only drawable case I = {1,2} may help explain:
(Here will be, eventually, a picture!)

We use these open cylinders and boxes to generate a topology with just enough open sets to guarantee that projection maps will be continuous. Note that the open cylinders form a subbase for a certain topology T on X = ÕX_i and therefore the open boxes form a base for T; T is called the [Tychonoff]product topology and (X, T) is the product of the given family of spaces. We write (X,T) = Õ{ (X_i, T_i): i Î I } = Õ_{i Î I}(X_i, T_i) or even T = Õ_{i Î I}T_i.

Notice that if Ç_{j = 1}ⁿp_{i_j}^-1(G_{i_j}) is any open box, then without loss of generality we can assume i₁, i₂, ... i_n all different because if there were repetitions like

¼Çp_{i_k}^-1(G) Çp_{i_k}^-1(H) ¼

we can replace each by

¼Çp_{i_k}^-1(G ÇH) Ç¼

and thus eliminate all repetitions.

It is routine to check that if T_n is the usual topology on Rⁿ, and T the usual topology on R, then

(R, T) ×(R, T) ×¼(R, T) = (Rⁿ, T_n)

as one would hope!

Lemma 1 In a product space (X, T), N is a neighbourhood of p Î X iff there exists some open box B such that p Î B Í N.

Lemma 2 For each i Î I,

(i) p_i is continuous
(ii) p_i is an open mapping.

Proof

(i) Immediate.
(ii) Given open G Í X, then G is a union of basic open sets {B_k:k Î K} in X, whence p_i(G) is a union of open subsets {B_kⁱ:k Î K} of X_i and is therefore open. (The notation here is intended to convey that B_kⁱ is the 'component along the i-th coordinate axis' of the open box B_k.)

Theorem 1 A map into a product space is continuous iff its composite with each projection is continuous.

Proof Since the projections are continuous, so must be their composites with any continuous map. To establish the converse, first show that if S is a subbase for the codomain (target) of a mapping f, then f will be continuous provided that the preimage of every member of S is open; now use the fact that the open cylinders constitute a subbase for the product topology.

Worked example Show that (X, T) ×(Y, S) is homeomorphic to (Y, S) ×(X, T).
Solution
Define

f:X ×Y ® Y ×X

g:Y ×X ® X ×Y
} by

f(x,y) = (y,x)

g(y,x) = (x,y).
Clearly these are one-one, onto and mutually inverse. It will suffice to show that both are continuous.

p₁°f = p₂^¢; p₂°f = p₁^¢. Now p_i^¢ is continuous for i = 1,2 and so f is continuous! Similarly, g is continuous.

Worked example Show that the product of infinitely many copies of (N, D) is not locally compact.
Solution
We claim that no point has a compact neighbourhood. Suppose otherwise; then there exists p Î X, C Í X and G Í X with C compact, G open and p Î G Í C. Pick an open box B such that p Î B Í G Í C. B looks like Ç_{j = 1}ⁿp_{i_j}^-1(G_{i_j}). Choose i_n+1 Î I \{i₁,i₂, ¼,i_n}; then p_{i_n+1}(C) is compact (since compactness is preserved by continuous maps).

Thus, p_{i_n+1} Î p_{i_n+1}(B) = X_{i_n+1} Í p_{i_n+1}(C) Í X_{i_n+1} = (N, D). Thus, p_{i_n+1}(C) = (N, D) ¼ which is not compact!

4.2 Products and Topological Properties

The topological properties possessed by a product depends, of course, on the properties possessed by the individual factors. There are several theorems which assert that certain topological properties are productive i.e. are possessed by the product if enjoyed by each factor. Several of these theorems are given below.

4.2.1 Products and Connectedness

Theorem 2 Any product of connected spaces must be connected.

Proof is left to the reader.

4.2.2 Products and Compactness

Theorem 3 [Tychonoff's theorem] Any product of compact spaces is compact i.e. compactness is productive.

Proof It suffices to prove that any covering of X by open cylinders has a finite subcover. Suppose not and let C be a family of open cylinders which covers X but for which no finite subcover exists. For each i Î I, consider

{G_{i_j}:G_{i_j} Í X_i and p_i^-1(G_{i_j}) Î C }.

This cannot cover X_i (otherwise, X_i, being compact, would be covered by finitely many, say X_i = G_i₁ ÈG_i₂ È¼ÈG_{i_n}, whence

X = p_i^-1(X_i) =

p_i^-1(G_i₁ È¼Èp_i^-1(G_{i_n}).

all in C, contrary to the choice of C

Select, therefore, z_i Î X_i \È{ those G_{i_j}¢s }; consider z = (z_i)_{i Î I} Î X. Since C covered X, z Î some C Î C. Now C = p_k^-1(G_k) for some k Î I and so p_k(z) = z_k Î G_k, contradicting the choice of the z_i's.

To prove the above without Alexander's Subbase Theorem is very difficult in general, but it is fairly simple in the special case where I is finite. Several further results show that various topological properties are 'finitely productive' in this sense.

Theorem 4 If (X₁, T₁), (X₂, T₂), ..., (X_n, T_n) are finitely many sequentially compact spaces, then their product is sequentially compact.

Proof
Take any sequence (x_n) Î X. The sequence ( p₁(x_n))_{n ³ 1} in sequentially compact X₁ has a convergent subsequence p₁(x_{n_k}) ® l₁ Î X₁. The sequence (p₂(x_{n_k}))_{k ³ 1} in sequentially compact X₂ has a convergent subsequence (p₂(x_{n_{k_j}}))_{j ³ 1} ® l₂ Î X₂ and p₁(x_{n_{k_j}}) ® l₁ also.

Do this n times! We get a subsequence (y_p)_{p ³ 1} of the original sequence such that p_i(y_p) ® l_i for i = 1,2, ¼, n. It's easy to check that y_p ® (l₁, l₂, ¼, l_n) so that X is sequentially compact, as required.

Lemma 3 `The product of subspaces is a subspace of the product.'

Proof
Let (X, T) = Õ_{i Î I}(X_i, T_i); let Æ Ì Y_i Í X_i for each i Î I. There appear to be two different ways to topologise ÕY_i:

{\em either} (i) give it the subspace topology induced by ÕT_i
{\em or} (ii) give it the product of all the individual subspace topologies (T_i)_{Y_i}.

The point is that these topologies coincide-if G_i₀^* is open in (T_i₀)_{Y_i₀} where i₀ Î I i.e. G_i₀^* = Y_i₀ ÇG_i₀ for some G_i₀ Î T_i₀, a typical subbasic open set for (ii) is

{(y_i) Î

Y_i : y_i₀ Î G_i₀^* }

which equals

Y_i Ç{ (x_i) Î

X_i: x_i₀ Î G_i₀ Î T_i₀,i₀ Î I }

Y_i Ç{ a typical open cylinder in

X_i }

which is a typical subbasic open set in (i). Hence, (i) = (ii).

Theorem 5 Local compactness is finitely productive.

Proof
Given x = (x₁,x₂,¼,x_n) Î (X, T) = Õ_{i = 1}ⁿ(X_i,T_i), we must show that x has a compact neighbourhood. Now, for all i = 1, ¼, n, x_i has a compact neighbourhood C_i in (X_i,T_i) so we choose T_i-open set G_i such that x_i Î G_i Í C_i. Then

x Î

G₁ ×G₂ ×¼×G_n

Ç₁ⁿp_i^-1(G_i)

C₁ ×C₂ ×¼×C_n

compact subset of ÕX_i

i.e. x has C₁ ×C₂ ×¼×C_n as a compact neighbourhood. (Note that the previous lemma is used here, to allow us to apply Tychonoff's theorem to the product of the compact subspaces C_i, and then to view this object as a subspace of the full product!) Thus, X is locally compact.

Lemma 4 [`(ÕY_i)]^T = Õ[`(Y_i)]^T_i ( in notation of previous lemma).

Proof Do it yourself! (`The closure of a product is a product of the closures.')

4.2.3 Products and Separability

Theorem 6 Separability is finitely productive.

Proof
For 1 £ i £ n, choose countable D_i Í X_i where [`(D_i)]^T_i = X_i. Consider D = D₁ ×D₂ ×¼×D_n = Õ₁ⁿD_i, again countable. Then [`D] = [`(ÕD_i)]^T = Õ[`(D_i)]^T_i = ÕX_i = X.

Notice that the converses of all such theorems are easily true. For example,

Theorem 7 If (X, T) = Õ_{i Î I}(X_i, T_i) is

(i) compact
(ii) sequentially compact
(iii) locally compact
(iv) connected
(v) separable
(vi) completely separable

then so is every `factor space' (X_i, T_i).

Proof For each i Î I, the projection mapping p_i: X ® X_i is continuous, open and onto. Thus, by previous results, the result follows.