# Open Sets in the Product Toplogy

• Dec 6th 2007, 07:47 PM
mpetnuch
Open Sets in the Product Toplogy
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 $\displaystyle \mathcal{K}$ be a set of continuous functions from $\displaystyle (X, \tau)$ into $\displaystyle [0,1]$.

Define $\displaystyle K : X \to [0,1]^\mathcal{K}$ by $\displaystyle K(x)(f) = f(x)$.

I then translated it into what I think is right in notation that I am familiar with. Here it is:

Consider $\displaystyle \{ f_\alpha \}_{\alpha \in J}$, an indexed set of continuous functions $\displaystyle f_\alpha : X \to [0,1]$ and define $\displaystyle f : X \to [0,1]^J$ by $\displaystyle f(x) = (f_\alpha(x))_{\alpha \in J}$.

Assuming that is right, my next problem is to show that $\displaystyle K$ or in my notation $\displaystyle f$ is continuous. But what are the open sets in $\displaystyle [0,1]^J$. I think they are $\displaystyle \pi^{-1}_\alpha(U)$ where $\displaystyle U$ is open in $\displaystyle [0,1]$? Are these open sets functions? I understand that the "points" in $\displaystyle [0,1]^J$ are functions (i.e. the $\displaystyle f_\alpha$'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.
• Dec 7th 2007, 06:45 AM
Rebesques
If I understand what you are saying, you just pick open sets$\displaystyle A_j\subset [0,1]$ and show that the set $\displaystyle f^{-1}(\times_j A_j)$ is open in $\displaystyle X.$

To this end, notice that $\displaystyle f^{-1} (\times_j A_j)=\cup_j f_j^{-1} (A_j)$ and that the $\displaystyle f_j$ are continuous.
• Dec 7th 2007, 07:52 AM
Opalg
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 $\displaystyle [0,1]^{\mathcal{K}}$ is a collection of coordinates $\displaystyle z=(z_f)_{f\in\mathcal{K}}$ indexed by the functions in $\displaystyle \mathcal{K}$ (where $\displaystyle z_f\in[0,1]$ for each coordinate z_f).

A typical neighbourhood of z in the product topology is the set N(z) consisting of all elements $\displaystyle w=(w_f)\in [0,1]^{\mathcal{K}}$ which differ from z by less than ε at a finite number of coordinates, say the coordinates $\displaystyle f_1,\:f_2,\,\ldots,f_n$. Thus $\displaystyle N(z) = \{w\in [0,1]^{\mathcal{K}} : |w_{f_j} - z_{f_j}|< \varepsilon\;(1\leqslant j\leqslant n)\}$.

Now we can start to look at the map $\displaystyle K:X\to [0,1]^{\mathcal{K}}$, which is defined by $\displaystyle (K(x))_f = f(x)$. The condition for K(y) to lie in the neighbourhood N(K(x)) is $\displaystyle |(K(y))_{f_j} - (K(x))_{f_j}|<\varepsilon$ for 1≤j≤n, or in other words $\displaystyle |f_j(y)-f_j(x)|<\varepsilon$. 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 $\displaystyle f_j\in\mathcal{K}$, and these functions are all continuous, it should be easy enough to see that this can be done.