If I understand what you are saying, you just pick open sets and show that the set is open in
To this end, notice that and that the are continuous.
I have a problem that is two fold. First, I am not comfortable with the notation that my textbook uses. I am familiar with another notation of product space (from say Munkres which I read on my own first). Secondly, I am not really comfortable wit the product toplogy. I can get away most of the time, i.e., when the products are finite and hence equal to the box topology but when I must use the product topology I get a headache :-) So I am hoping that maybe somebody here could help me.
First off here, is the begining of the problem I am trying to solve:
Let be a set of continuous functions from into .
Define by .
I then translated it into what I think is right in notation that I am familiar with. Here it is:
Consider , an indexed set of continuous functions and define by .
Assuming that is right, my next problem is to show that or in my notation is continuous. But what are the open sets in . I think they are where is open in ? Are these open sets functions? I understand that the "points" in are functions (i.e. the 's), but what are the open sets. Are they an uncountable number of continuous functions. Let me stop here before I start sounding really dumb.
Thanks for the help.
The way to get your head round the product topology is to remember that a neighbourhood of a point can only control finitely many of the coordinates of that point.
But before you can consider neighbourhoods, you have to be clear about what a point in the product space looks like. A point z in is a collection of coordinates indexed by the functions in (where for each coordinate z_f).
A typical neighbourhood of z in the product topology is the set N(z) consisting of all elements which differ from z by less than ε at a finite number of coordinates, say the coordinates . Thus .
Now we can start to look at the map , which is defined by . The condition for K(y) to lie in the neighbourhood N(K(x)) is for 1≤j≤n, or in other words . To prove that K is continuous, we have to show that this condition will hold whenever y is close enough to x in the space X. But since the condition only involves finitely many functions , and these functions are all continuous, it should be easy enough to see that this can be done.