The Box Topology
Following an earlier article on products of two topological spaces, we’ll now talk about a product of possibly infinitely many topological spaces. Suppose is a collection of topological spaces indexed by I, and we wish to define a topology on the set What would be a good choice?
If we follow our instincts, the most natural definition would be:
Definition. The box topology on is defined by the basis:
Since our collection Σ is indeed a basis, so the definition makes sense. The only problem is that its behaviour has some anomalies.
Problem 1 : As Terence Tao pointed out in his blog post, if we consider the space of all sequences of real numbers, then there are sequences which one would expect to converge but fail to under the box topology. For example:
Reasonably, one would expect the xi‘s to converge to (0, 0, … ), but if we pick the open subset
then no term of the sequence actually lies in U. The problem is that X has too many open subsets.
Problem 2 : Let I be any index set and X be a topological space. The diagonal map takes to the constant tuple It turns out Δ is not continuous in general if we endow with the box topology! Indeed, the open subset for open gives
which is not open in X in general if I is infinite. The fact that such a natural map is non-continuous makes things rather awkward.
Let’s go back to the drawing board and decide what kind of subsets of can be open. Firstly, we most definitely want the projection maps
to be continuous. This means for any open subset the “open slice” given by is an open subset of X. Let’s define the coarsest topology for which this is true.
Definition. The product topology on X is given by the subbasis of “open slices”:
[ Open slices of X × Y × Z ]
Taking intersections of finitely many elements of this subbasis, we get the basis:
Note that for a finite product the box topology and the product topology are identical.
Taking the complement of an “open slice” gives the “closed slice”
for a closed subset
Now if is a collection of closed subsets, then the product is an intersection of the “closed slices”, so we get:
Proposition 1. The product of closed subsets is closed in the product topology.
Since the box topology is an even finer topology, is also closed in the box topology.
Next, we’ll proceed to give some nice properties of the product topology. Hopefully, these will convince you that we’ve made the “right” choice.
Properties of Product Topology
We’ll give the most important property first.
Universal Property of the Product. Let with projection maps Now, for any topological space Z and function
- f is continuous at if and only if is continuous at z, for each i.
The forward direction is obvious since each is continuous.
For the converse, we need to show that for any open subset V of X containing f(x), contains an open subset U which contains z. If we fix a subbasis of X, we may assume V belongs to this subbasis. Hence let for some open subset and index i. This gives:
which indeed contains an open subset containing z since is continuous. ♦
In particular, we have:
Universal Property of the Product (II) : if Z is a topological space and f:Z → X is a map, then f is continuous if and only if πif is continuous for each index i.
In examining the proof of the universal property, the reader may suspect that this property uniquely defines the product topology, and he’d be right!
In what follows, we’ll list some results which can be proven by invoking the universal property.
Applications of the Universal Property
We’ll use the above universal property of the product topology to prove a variety of results.
Proposition 2. The diagonal map is a homeomorphism onto its image.
Continuity: by the universal property, it suffices to show that composing with each projection map gives a continuous map for each i. But this map is simply the identity, which is clearly continuous.
To show that the inverse map Δ(X) → X is continuous, note that it is simply the restriction of a projection map to Δ(X). ♦
Proposition 3. Let be a collection of continuous maps indexed by i. Then the map which takes:
Let and be the projection maps.
To prove f is continuous, the universal property tells us it suffices to show is continuous for each i. But then so it is indeed continuous. ♦
Proposition 4. Let J be a directed set and be a net in indexed by Then
- the net converges converges to if and only if for each fixed i, the net converges in Xi.
[ Feel free to substitute “net” with “sequence” if you’re not too comfortable with it. ]
As we noted earlier, if and only if the map
is continuous at ∞. By the universal property, this is true if and only if for each i, the composition is continuous. But this map takes which is continuous at ∞ if and only if for each i. ♦
Application: consider the earlier sequence for positive integers i, j. If we fix i, then as j → ∞, Hence,
Proposition 5. Suppose we have a collection of collections of topological spaces where the index set J(i) for j may depend on i. Then:
as topological spaces.
Note that set-theoretically, there’s a natural bijection between the two sides. Let f : LHS → RHS and g : RHS → LHS be this bijection and its inverse. We need to show that they’re both continuous.
There are three collections of projection maps, all of which are continuous.
- To show that f is continuous, we need to show is continuous for any i, j. But this map is simply so it is indeed continuous.
- To show that g is continuous, we need to show is continuous for each i. For that, we need to show that is continuous for each But this is simply so it is indeed continuous. ♦
Exercise (on Universal Properties)
Prove the following universal property for subspace. Suppose gets the subspace topology from X. Let be the inclusion map. Then:
- for any topological space Z, a function f : Z → Y is continuous if and only if is continuous.
Suppose is a collection of subsets of topological spaces. There are two ways to form a topology on
- take the subspace topology on each and take the product topology or
- take the product topology on then take the subspace topology
Prove that both constructions give the same topology, using the universal properties of the product and the subspace topologies.
Interior and Closure
Proposition 6. If is a collection of subsets of topological spaces, then
- in general;
For the first property, suppose each Yi is open in Xi. Then the RHS is which is not open in general unless Yi = Xi for all but finitely many i. Hence it is not equal to the LHS.
For the second property, the RHS is a closed subset containing so it must also contain For the reverse inclusion, if then it does not lie in and is not a point of accumulation for it. Thus, we can pick a basic open subset:
where Ui=Xi except for finitely many i,
such that Thus for some i, and xi is not a point of accumulation of Yi. Furthermore, since we have so
Prove that in the box topology, we do have:
Thus there’s at least one aspect of the box topology which triumphs the product topology.
Product of Metric Spaces
Finally, we ask: if each is a metric space, is the topological product necessarily metrisable also? For finite products, we knew this to be true. It turns out for countably infinite products this is still true.
The first step is to replace each di with a bounded metric. Recall that for any metric space (X, d), replacing d with gives a metric d’ which is topologically equivalent and bounded by 1. Multiplying by a suitable constant, we ensure that each di is bounded by 2–i for i = 1, 2, 3, … and define a metric on by:
It’s not hard to check that d is a metric bounded by 1.
The only question is whether the induced topology is identical to the product topology.
First, let’s consider each open slice: where is open. If is in this slice, then so there’s an open ball Now it’s clear that the open ball in X : So U is indeed open in the metric topology.
Conversely, consider an open ball N(a, ε) in the metric topology, where as before. Pick N such that 2–N < ε/2. Then the set:
is open in the product topology, and the distance between a and any point in this set is bounded by Thus this set is contained in the open ball N(a, ε).