Suppose the topology on each space is given by a basis . The collection of all sets of the form , where for each , will serve as a basis for the box topology on .
The collection of all sets of the same form where for finitely many indices of alpha and for all the remaining indices, will serve as a basis for the product topology on .
How am I supposed to prove this?
there are a couple of things to note, here:
first of all, if the index set J is finite, you get the same topology in either case.
for reasons that are often not made clear from the definition itself, the product topology is the largest (finest) topology that makes the projection functions continuous.
so what can go wrong in the box topology? suppose you have .
clearly, the set is open in the box topology. if is the diagonal function: f(x) = (x,x,....,x), each component function is continuous (the component functions are just the identity on ).
but f itself is NOT continuous. suppose it were. the origin is in U, so a pre-image of U must contain some interval containing 0 (since we are supposing the pre-image is open, if f is to be continuous). but any such interval is mapped to its countable cartesian product of itself, and so for SOME n, is mapped outside of U.
so the pre-image of U is {0}, a closed set, and f is not continuous, every though every projection (component) of f is.
with an infinite cartesian product, the box topology has "too many sets". limiting the product topology to just finite products of base elements, makes it just small enough so the projection functions are continuous. this is typical of topology, you want to make the domain set toplogy "just fine enough" and the target (image space) set topology "just coarse enough". with a product space, we are given the target topologies (presumably they are given topologies on the spaces that form the product), so the only topology we have any real freedom to define, is on the product space.
is not each interval (-1/n,1/n) a basis element (open interval) in the standard topology for the real numbers? we're just taking a (countably) infinte product of such open intervals, one for each n. each one is a basis set. such products of basis sets ARE the basis elements of the box topology. basis elements are open (every set in a basis also belongs to the topology, the topology is formed from (arbitrary) unions of these sets).
a continuous function f:X-->Y (where X,Y are topological spaces) is one where the pre-image of an open set U in the topology for Y, , is open in the topology for X.
in my example, X is the real numbers. an open set contains a neighborhood (think: basis element) containing each of its points (if a set S is open, and x is in S, then there is a neighborhood containing x entirely within S). now the point (0,0,...,0) of is clearly contained in U (since 0 is contained in each interval (-1/n,1/n)).
so is contained in . if this pre-image is open (which is has to be if f is to be continuous), there must be SOME interval K containing 0 contained within , say (-ε,ε).
but now we're in a fix: having decided upon our epsilon, we can see that for large enough n, say n > M = 1/ε, (-1/n,1/n) does not contain (-ε,ε).
but surely, since , . however f(K) = (-ε,ε) x (-ε,ε) x.....x (-ε,ε) (countably many times), and this can't possibly lie within U (once we get past the M-th copy...we're in trouble).