Let Ci be a finite collection of component spaces and build P, the product space, as follows. As a set, P is the cross product of all the component sets Ci. For instance, a point in 3 space is given by 3 coordinates, where each coordinate specifies a point in one of the three component spaces. This is what we mean by cross product; a point in x and a point in y and a point in z gives a unique point in the product space.
If S is a set in the product space, the projection onto the ith component is the set of points from Ci that are present in S. The projection of a sphere in 3 space into any of the coordinate axes gives a line segment whose length is the diameter of the sphere. Or, think of 3 space as the topological product of the xy plane and the z axis. Now the projection of the sphere into the xy plane is a disk bounded by the equator of the sphere. In most applications, the projection keeps one or two coordinates and throws the others away.
The topology of P is defined by a base, where a base set in P is given by the cross product of open sets in the component spaces. For instance, cross the open unit intervals in x y and z to get the interior of the unit cube in 3 space, which is an open set.
If the component topologies are defined by bases, we can use those base sets. We don't have to cross every open set with every other open set to build a base for P; it is enough to cross base open sets. To see this, let S be the cross product of arbitrary open sets in the component spaces. Replace each open set with a union of base sets. Take any of the base sets in the first open set, cross with a base set from the second open set, cross with a base set from the third open set, and so on. Call this set Q. But there are many such sets; all the ways you can select a base set from each of the open sets. Each of these sets Qj is open by definition, and their union gives the cross product of the original open sets. Thus S is open.
We still need to show the base is valid. Let z be a point in the intersection of two base open sets in P. Project z onto each component, giving the points z1 z2 z3 … in the component spaces C1 C2 C3 etc. Project the two open sets onto the first component and find two base open sets in C1, both containing z1. Since this is a base, there is a third open set Q1 containing z1, and inside both open sets. Find Q2 Q3 … similarly and take their cross product, giving Q. Now Q is a base open set in P that contains z, and is intirely inside both of the original open sets. Thereforee the topology is valid.
Recall that open sets in the real line are based on open intervals at arbitrary locations and having arbitrary lengths. (You can use rational end points if you want a countable base.) Cross intervals with intervals with intervals to build a base for the product topology. Thus the topology of R3 is based on open boxes at arbitrary locations and with arbitrary dimensions. I know, we usually base our open sets on balls at arbitrary centers and having arbitrary radii, but when you get to metric spaces you will see that these different bases, balls or boxes, actually produce the same topology.
If S is open in P it is the union of base sets, each having open projections in the component spaces. Thus the projection of an open set onto any component is open. However, the converse is not true. The annulus 1 ≤ r < 2 has open projections on the x and y axes, yet it is not an open set in the plane.
The above notwithstanding - projection is a continuous function. Cross an open set in the projected dimensions with the entire space (in the remaining dimensions) to find the preimage, which, being the cross product of open sets, is open in the product space. Combine this with the previous result, and projection is bicontinuous.
If product is a binary operator on topological spaces, verify that product is commutative and associative. No matter how we put the spaces together, the base is still the cross product of base open sets in the component spaces.
That wasn't too painful, but let's assume there are infinitely many component spaces Ci. The product, as a set, is an infinite cross product. This is called a direct product, and it might be empty. Yes, you can cross infinitely many sets, each as big as the universe, and get nothing. But that usually doesn't happen. First, the component spaces are often the same, and when that happens you always get a nonempty product set. Take infinitely many copies of R1, for instance, and build R∞. Each point is defined by an infinite sequence of coordinates.
Even if the spaces are different, we usually assume the axiom of choice, hence the product set is nonempty. This is an assumption of sorts, but it's a safe assumption, so most people make it and move on. I'll do the same. The product space exists and is nonempty.
There are two possible topologies for the product space P, strong and weak. The strong topology is defined as above. A base set in P is the cross product of base sets in the component spaces. Reason as we did above to show this is a valid base, and the topology of P includes all possible cross products of open sets in the component spaces.
In the weak topology, each base set in P is built from component base sets as before, but only finitely many of these base sets are used. The other components are not constrained. Returning to R∞, we might select open intervals from components 1, 3, and 7. All the other components are taken in their entirety. Let Q be the resulting cross product, where three coordinates are restricted to open intervals and the other coordinates are unconstrained. This is an open set in the weak topology. If we try to restrict all coordinates to (0,1), that is an open set in the strong topology. Once again, verify that the base for the weak topology is valid, and the topology of P includes all finite cross products of open sets in the component spaces, where the remaining components are unconstrained.
Note that the weak and strong topologies are the same when finitely many spaces are involved.
Q. So why do we need the weak topology? Why not define P using the strong topology and be done with it?
A. We want topological spaces to form a concrete category, with continuous functions acting as morphisms. If the product space is to be a product in the category, the topology must be weak.
Q. What?!
A. Oh never mind. Here's the idea, without all the category theory.
Let P be R∞, as described earlier. Map R1 into P by setting f(x) = x,x,x,x,x… and so on through all the components. Compose f with any projection and get a continuous function, in fact the identity map. But is f continuous into P? Let Ni be the open neighborhood in the ith component, centered at 0 with radius 1/i. In the strong topology, the direct product of these neighborhoods is an open set containing the origin. Yet the preimage is the origin, which is closed, hence f is not continuous. With the weak topology, the product function is continuous iff the component functions are all continuous. Let's prove this.
Let S be a base open set in P with preimage W. A point z is in W iff z maps to something in S, iff each component function maps z into the projection of S, iff z is in the preimage of each projection. With the weak topology, there are finitely many projections to consider. When the preimages of the projections are open sets, their intersection is open, hence W is open. When the component functions are continuous, so is the product function.
Conversely, assume f is continuous and let Q be a base open set in one of the component spaces. Cross Q with all the other components to get S, an open set in P. By the above, the preimage of S is the intersection of the preimage of Q, which is simply the preimage of Q. Both preimages are the same, hence the preimage of Q is open. Each component function is continuous.
The weak topology also offers advantages when dealing with convergent sequences. If z is the limit of a sequence s in the product space P, let zi be the ith component of z in Ci, and let si be the projection of the sequence s in Ci. Select a base open set in Ci that contains zi and cross with all the other components to get an open set in P that contains z. This contains most of s, i.e. all the terms of s after some index n, and the same holds in the space Ci. Thus the limit of the projected sequence is the projection of the limit.
Conversely, let zi be the limit of the component sequence si, for each component space Ci in the product. Given a base open set about z, find the base open set Qi containing zi in the space Ci. Now Qi contains the entire sequence beyond some index ni. Since there are finitely many components to consider, finitely many base open sets Qi building the open set about z, there are finitely many indexes ni, and we can set n to the largest of these. Now everything in s, beyond n, lies in the open set about z, and s converges to z. A sequence comverges in P iff it converges per component.
We'll use these results again and again in multi-variable calculus, often without thinking about it. That is, f1 approaches 5 and f2 approaches 7 and f3 approaches 9, so naturally the 3 dimensional vector function f approaches the point 5,7,9.
Unless otherwise stated, a product space has the weak topology.
The direct product of j copies of a space S is written Sj, and depends only on the cardinality of j. We can assume j is a cardinal number. That's where we get the notation Rn, for the product of n copies of the real line, with the implied topology. Similarly, R∞ is the product of infinitely many copies of the real line.
Here is a simple application of the product topology. Let S be a space and let P be the product space S cross S. Let D, a subset of P, consist of the points x,x for all x in S. This is the diagonal set.
If the complement of D is open, select any point q,r not in D, pick any base open set in the complement that contains q,r, take the projections in S and S, and find disjoint open sets in S separating q and r. Conversely, if q and r can always be so separated, the complement of D is covered by open sets and is open. The diagonal map in S cross S is closed iff S is Hausdorff.
The complement of an open cross product is the union of C1 cross everything, with everything cross C2. An open set is the union of these open cross products, and thus a closed set is the intersection of corresponding pairwise unions of closed cross products. For example, let S be the diagonal in the xy plane. Imagine x ≤ u cross everything (half plane) union y ≥ u cross everything, together making 3 quarters of the plane. Intersect these for all u and get the half plane on and above the diagonal. Then intersect with the half plane on and below the diagonal, leaving S.
Yes, this is a coproduct in category theory.
Let P be the finite product of discrete spaces. Every point in every component is open, every point in the cross product is open, and P is discrete. This is not the case for an infinite product. Unless the components are trivial, the points of P will not be open.
The coproduct of discrete spaces is discrete.