What Are Standard Ways to Define Topology on Function Spaces?

[Tac-Tics]

Given two topological spaces X and Y, there is a standard way to define a topology on X x Y called the product topology.

Given a subset S of X, there is a standard way to define a topology on S called the subset topology.

Given an equivalence relation ~ on X, there is a standard way to define a topology on X/~ called the quotient topology.

Are there any standard ways to define a topology various function spaces X->Y?

 

[Moo Of Doom]

The most common topology on the space of functions from X to Y is probably the product topology, which corresponds to the topology of point-wise convergence. One can think of the space of functions from X to Y, usually written Y^X, as a product of topological spaces: \prod_{x \in X} Y_x where in this case Y_x = Y for all x. The product topology on Y^X is generated by a basis consisting of sets of the form \prod_{x \in X} U_x where U_x is open in Y for each x, and U_x = Y for all but finitely many x. Interestingly, the topology on X does not play a role here at all, so X does not even need to be a topological space, and may simply be any set.

The topology generated by basis elements of the form \prod_{x \in X} U_x, without the requirement that U_x = Y for all but finitely many x, by the way, is called the box topology. The box topology is usually too fine to be of interest, which is why the product topology is usually preferred.

In the case where Y is a metric space, one can define the topology of uniform convergence on Y^X, which is metrizable. The metric is given by d(f, g) = \min(\sup_{x \in X} d(f(x), g(x)), 1).

Hope this helps, and is at least somewhat understandable.

 

[g_edgar]

You can look up "k-space", which is a class of topological spaces where there is a reasonable function-space topology. "Reasonable" can be explained as in category theory. That is, for the function space F(X,Y) of functions from X to Y (whatever it is), there should be a natural identification between F(X \times Y, Z) and F(X,F(Y,Z)) .