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In topology, the cartesian product of topological spaces is can be topologized in several ways. The canonical way of doing it is using the product topology, because it fits rather nicely with the categorical notion of a product. Another way to do it is with the box topology. The box topology has a somewhat more obvious definition than the product topology, but it satisfies fewer desirable properties. In general, the box topology is finer than the product topology, although the two agree in the case of finite direct products (or when all but finitely many of the factors are trivial). Let I be an arbitrary index set and suppose Xi is a topological space for every i in I. Set X = Π Xi, the cartesian product of the sets Xi. Now consider the topology generated by the collection B = { Π Ui | Ui open in Xi}, the cartesian products of every open set of every Xi. We call this the box topology (so called because in the case of Rn, the basis sets look like boxes or unions thereof). It is easily verified that B is actually a basis for the topology. One often wonders how this topology compares with the product topology. The basis sets in the product topology have almost the same definition as the above, except with the proviso that all but finitely many Ui are equal to the whole space Xi This may seem like a rather irksome detail, but the product topology satisfies a very desirable property for maps fi : Y → Xi into the component spaces: the product map f: Y → X defined by the component functions fi is continuous if and only if all the fi are continuous. This does not always hold in the box topology, because it is in general a much finer topology, so therefore mapping into the range space makes it much harder for functions to be continuous. This actually makes the box topology very useful for providing counterexamples — many qualities such compactness, connectedness, metrizability, etc., if possessed by the factor spaces, are not in general preserved in the product with this topology. (More specific examples here would be useful...) An instructive example (given by Munkres) is the following: Let Rω denote the countable cartesian product of R with itself, i.e. the set of all sequences in R. Let f : R → Rω be the product map whose components are all the identity, i.e. f(x) = (x, x, x, ...). Obviously the component functions are continuous. Consider the open set U = Π (-1/n,1/n). If f were continuous, it would have to contain an interval (-ε, ε) about 0 (since f(0) = (0,0,0,...) is in U). The image of this interval must, in turn, be contained in U. But the image of (ε,ε) is its own countable cartesian product. So (ε,ε) is contained in (-1/n, 1/n) for every n, an impossibility. So f is not continuous though all its components are. In general, a cartesian product of a space X with itself over an indexing set S is precisely the space of functions from S to X; the product topology yields the topology of pointwise convergence, sequences of functions converge if and only if they converge at every point of S. The box topology, once again due to its great profusion of open sets, makes convergence very hard. A cute way of visualizing convergence in this topology is to think of functions from R to R — a sequence of functions converges to a function f in the box topology if, when looking at the graph of f, given a set of "hoops", that is, vertical open intervals surrounding the graph of f above every point on the x_axis, eventually, every function in the sequence "jumps through all the hoops." For functions on R this looks a lot like uniform convergence, in which case all the "hoops", once chosen must be the same size. But in this case one can make the hoops arbitrarily small, so one can see intuitively how "hard" it is for sequences of functions to converge.
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